Aieee 2012 Solution (19th may)

Detailed solution given below

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Q10

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Q11

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Q12

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Ans 4 (actual ans 24mm)

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Q29

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Q61

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Q62

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Q63

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Q64

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Q65

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Q66

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Q67

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Q68

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Q69

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Q70

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Q72

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Q73

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Q74

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Q75

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Q76

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Q77

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Q78

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Q79

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Q84

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Q86

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Q87

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Q88

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Q89

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Ans 2

By

Saurav gupta

Electronics amp telecomm Engg(2nd year)

Jadavpur university

Q90

1)

2)

3)

4)

Ans 4

AIEEE PHYSICS SOLUTION (EXAM-19TH MAY 2012)

BY

SAURAV GUPTA

ELECTRONICS amp TELECOMM ENGG(2ND YEAR)

JADAVPUR UNIVERSITY

Q1 mv + mvrsquo = 2m x 200

v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0

S = vrsquot + frac12 gtsup2

490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx

10ˉsup2 g = k 05 x 10ˉsup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil

= 10ˉsup2 g + B x 10ˉsup1

K xrsquo = Frsquo

2g x 8 x 10ˉsup3 = 10ˉsup2 g + B x 10ˉsup1

B = 06 T directed out of the page

Ans 3

Q3 No interference as slit width is comparable to the

wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2

Q9 Mgh = frac12 mvsup2 + frac12 I ωsup2

Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q10 For matter waves λ =hp

For photons λ = hcE

Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1

Q13 A V = ArsquoVrsquo

VrsquoV = AArsquo

= 5

P ρ + vsup22 = constant (Bernoulli rsquos theorem)

P ρ + vsup22 = Prsquo ρ + vrsquosup22

Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3

Q14 T Vʸˉsup1= constant

T V ⁰⁵ = Trsquo (v4)⁰⁵

Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm

˳int v dv = -k m int⁰ r dr

Vsup2 = kxsup2m

X prop vradicm

X1x2 prop radic(m1m2)

Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q17 Self explanatory

Ans 1

Q18 Fκ= mg sinα

13 mg cosα = mg sinα

Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

No of significant figures should be same

Ans 4

Q20 V = 7 Vʳ

V[1- exp(-tRC)]= 7V exp(-tRC)

Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25

1f = 1f1 + 1f2 ndash df1f2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k

frac12 kx˳sup2 = frac12 μv˳sup2

Where μ=m2m(m+2m)reducedmass

= 2m3

frac12 k x˳sup2 = 13 m v˳sup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2 mv˳sup23x˳sup2

Ans 4

Q23 KE = frac12 I ωsup2

= frac12 Lsup2I as L=Iω

Now L = constant

KE prop 1I

Ans 3

Q24 1λ = R(1nᵢsup2- 1nrsquosup2)

1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4

λsup3sup2 λsup3sup1= 64

λsup2sup1λsup3sup1 = 12

Ans 4

Q25 Y = a cos(kx-ωt)

Yrsquo = -a cos(-kx-ωt)

= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1

E1 = frac12 n1 x C1 x V1sup2

= 8C1 Vsup2n1

Case 2

E2 = frac12 C1 x n2 x Vsup2

E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2

Q30 q = area of i-t diagram

q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1

AIEEE MATHS SOLUTION (EXAM-19TH MAY 2012)

BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k

AD = icirc + ĵ + (z-1) k

Area of tetrahedron= 16 [AB AC AD]

0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1

Q62 Both statements are true


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

But statement 2 is not correct explanation of 1

Ans 4

Q63 (asup2+bsup2+csup2)psup2-2p(ab+bc+cd)+(bsup2+csup2+dsup2)le 0

Coeff of psup2 lt 0 amp Dle0

But

Coeff of psup2 is gt 0 hence

(asup2+bsup2+csup2)psup2-2p(ab+bc+cd)+(bsup2+csup2+dsup2) = 0

Also p is real

Hence D = 0

(ab+bc+cd)sup2 = (asup2+bsup2+csup2)(bsup2+csup2+dsup2)

By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q64 F(x) = x expx(1-x)

Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx

ln(xsup2+1)⁰⁵ +c

Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)

Tr = radicr+1 -radicr

T1 = radic2 -radic1

T2 = radic3 -radic2

T15 = radic16 - radic15

Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Σ Tr = radic16 -radic1

= 3

Ans 4

Q67 A = 12hellip1000

Σ isup2 1sup2+2sup2+3sup2+hellip+nsup2

Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now

For (2n+1)3 to be an integer 2n+1 has to be a multiple of 3

ie possible values of n are 147hellip

=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Let total no of possible values be x

1000 = 1 + (x-1) 3

X =334

Probability = 3341000

= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)

cosA = (bsup2+csup2-asup2)2bc

=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689

Symmetric order pairs =

(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20

Can also be obtained in this way

No of Symmetric order pairs = no of total order pairs ndash no of reflexive order pairs

= 5sup2 - 5

= 20

Total no of symmetric relation = 2sup2⁰

Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q72 [Ysup1 ⁵ + xsup1 sup1⁰]⁵⁵

Tr = ᶯCr (ysup1 ⁵)ʳ (xsup1 sup1⁰)⁵⁵ˉʳ

For the powers of x and y to be free from radicals

r = 5q amp 55-r = 10p pqϵ z

5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73

Aˢ = π (2)sup24 ndash frac12 x 2 x 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2

Aᴮ= π(2)sup2 ndash (π-2)

= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4

Q74 P q equiv (pq) Ʌ (qp) formula

Can also be checked by truth table

Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0

|AB ndashBA| =γδ(α-β)sup2

If α =β

Then matrix is not invertible

Statement 1 ---false

Also matrix can never be identity matrix

Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y

slope of chord m = (1-0)(0-2) = - frac12

-x4y =- frac12

X =2y

Substituting this value in eqᶯ (1) we get

X = plusmnradic2

Y = plusmn 1radic2

|P1P2| = radic (2radic2)sup2 + (2radic2)sup2

= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ

max value = radic3sup2 + 4sup2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4

Q79 Area = ˳intsup1 (xsup2-xsup3) dx +intᴾ (xsup3 - xsup2) dx

16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

dAdt = ldbdt + bdldt

-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q83 Xsup2 -(sinα -2)x ndash( 1+sinx) = 0

Let αβ be roots of the eqᶯ

F(α)= αsup2 + βsup2 =(α+β)sup2 -2αβ

= (sinα-2)sup2 + 2(1+sinα)

= sinsup2α -2 sinα +6

Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1

Q84 Frsquo(x) =sin(logx)

Y = f[(2x+3)(3-2x)]

Yrsquo = f lsquo[(2x+3)(3-2x)] 12(3-2x)sup2

= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2

d= (3icirc + 2ĵ - 2k)radic17

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)

Shortcut ndash by seeing the fig 3rd point must lie in 3rd quadrant

By option checking

Ans 3

Q88 ddx G(x) = exptanxx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now

intsup1 sup2 2 exptanπxsup2x dx

Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Area = 2 x frac12 4 x 2

= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Distance between lines = | (1 ndash 32)radic2 |

= radic2 frac12

Now

|(1-a-1)radic2|lt12radic2

-frac12 lt a lt frac12 -----(1)

Also | (1+a-32)radic2 | lt frac12radic2

0 lt a lt 1-----(2)

Combining (1) amp (2)

0 lt a lt frac12

Ans 4

Q2

1)

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4)

Ans 3

Q3

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Ans 2

Q4

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Q5

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Q10

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Q11

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Q12

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Q13

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Q14

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Q27

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Q28

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Q29

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Q30

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Q61

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Q62

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Q63

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Q64

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Q65

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Q66

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Q67

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Ans 1

Q68

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Q69

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Ans 2

Q70

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Q71

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Ans 3

Q72

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Q73

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Ans 4

Q74

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Ans 1

Q75

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Q76

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Ans 3

Q77

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Ans 3

Q78

1)

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Ans 4

Q79

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Ans 4

Q80

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Ans 2

Q81

1)

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Ans 2

Q82

1)

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Ans 4

Q83

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Ans 1

Q84

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Q85

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Ans 1

Q86

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Ans 4

Q87

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Ans 3

Q88

1)

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Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q4

1)

2)

3)

4)

Ans 2

Q5

1)

2)

3)

4)

Ans 3

Q6

1)

2)

3)

4)

Ans 3

Q7

1)

2)

3)

4)

Ans 2

Q8

1)

2)

3)

4)

Ans 2

Q9

1)

2)

3)

4)

Ans 3

Q10

1)

2)

3)

4)

Ans 4

Q11

1)

2)

3)

4)

Ans 4

Q12

1)

2)

3)

4)

Ans 1

Q13

1)

2)

3)

4)

Ans 3

Q14

1)

2)

3)

4)

Ans 3

Q15

1)

2)

3)

4)

Ans 2

Q16

1)

2)

3)

4)

Ans 4

Q17

1)

2)

3)

4)

Ans 1

Q18

1)

2)

3)

4)

Ans 2

Q19

1)

2)

3)

4)

Ans 4

Q20

1)

2)

3)

4)

Ans 4

Q21

1)

2)

3)

4)


Q22

1)

2)

3)

4)

Ans 4

Q23

1)

2)

3)

4)

Ans 3

Q24

1)

2)

3)

4)

Ans 4

Q25

1)

2)

3)

4)

Ans 2

Q26

1)

2)

3)

4)

Ans 1

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 4

Q29

1)

2)

3)

4)

Ans 2

Q30

1)

2)

3)

4)

Ans 1

Q61

1)

2)

3)

4)

Ans 1

Q62

1)

2)

3)

4)

Ans 4

Q63

1)

2)

3)

4)

Ans 3

Q64

1)

2)

3)

4)

Ans 3

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 4

Q67

1)

2)

3)

4)

Ans 1

Q68

1)

2)

3)

4)

Ans 4

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 3

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 4

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 1

Q75

1)

2)

3)

4)

Ans 2

Q76

1)

2)

3)

4)

Ans 3

Q77

1)

2)

3)

4)

Ans 3

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q6

1)

2)

3)

4)

Ans 3

Q7

1)

2)

3)

4)

Ans 2

Q8

1)

2)

3)

4)

Ans 2

Q9

1)

2)

3)

4)

Ans 3

Q10

1)

2)

3)

4)

Ans 4

Q11

1)

2)

3)

4)

Ans 4

Q12

1)

2)

3)

4)

Ans 1

Q13

1)

2)

3)

4)

Ans 3

Q14

1)

2)

3)

4)

Ans 3

Q15

1)

2)

3)

4)

Ans 2

Q16

1)

2)

3)

4)

Ans 4

Q17

1)

2)

3)

4)

Ans 1

Q18

1)

2)

3)

4)

Ans 2

Q19

1)

2)

3)

4)

Ans 4

Q20

1)

2)

3)

4)

Ans 4

Q21

1)

2)

3)

4)


Q22

1)

2)

3)

4)

Ans 4

Q23

1)

2)

3)

4)

Ans 3

Q24

1)

2)

3)

4)

Ans 4

Q25

1)

2)

3)

4)

Ans 2

Q26

1)

2)

3)

4)

Ans 1

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 4

Q29

1)

2)

3)

4)

Ans 2

Q30

1)

2)

3)

4)

Ans 1

Q61

1)

2)

3)

4)

Ans 1

Q62

1)

2)

3)

4)

Ans 4

Q63

1)

2)

3)

4)

Ans 3

Q64

1)

2)

3)

4)

Ans 3

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 4

Q67

1)

2)

3)

4)

Ans 1

Q68

1)

2)

3)

4)

Ans 4

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 3

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 4

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 1

Q75

1)

2)

3)

4)

Ans 2

Q76

1)

2)

3)

4)

Ans 3

Q77

1)

2)

3)

4)

Ans 3

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q8

1)

2)

3)

4)

Ans 2

Q9

1)

2)

3)

4)

Ans 3

Q10

1)

2)

3)

4)

Ans 4

Q11

1)

2)

3)

4)

Ans 4

Q12

1)

2)

3)

4)

Ans 1

Q13

1)

2)

3)

4)

Ans 3

Q14

1)

2)

3)

4)

Ans 3

Q15

1)

2)

3)

4)

Ans 2

Q16

1)

2)

3)

4)

Ans 4

Q17

1)

2)

3)

4)

Ans 1

Q18

1)

2)

3)

4)

Ans 2

Q19

1)

2)

3)

4)

Ans 4

Q20

1)

2)

3)

4)

Ans 4

Q21

1)

2)

3)

4)


Q22

1)

2)

3)

4)

Ans 4

Q23

1)

2)

3)

4)

Ans 3

Q24

1)

2)

3)

4)

Ans 4

Q25

1)

2)

3)

4)

Ans 2

Q26

1)

2)

3)

4)

Ans 1

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 4

Q29

1)

2)

3)

4)

Ans 2

Q30

1)

2)

3)

4)

Ans 1

Q61

1)

2)

3)

4)

Ans 1

Q62

1)

2)

3)

4)

Ans 4

Q63

1)

2)

3)

4)

Ans 3

Q64

1)

2)

3)

4)

Ans 3

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 4

Q67

1)

2)

3)

4)

Ans 1

Q68

1)

2)

3)

4)

Ans 4

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 3

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 4

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 1

Q75

1)

2)

3)

4)

Ans 2

Q76

1)

2)

3)

4)

Ans 3

Q77

1)

2)

3)

4)

Ans 3

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q10

1)

2)

3)

4)

Ans 4

Q11

1)

2)

3)

4)

Ans 4

Q12

1)

2)

3)

4)

Ans 1

Q13

1)

2)

3)

4)

Ans 3

Q14

1)

2)

3)

4)

Ans 3

Q15

1)

2)

3)

4)

Ans 2

Q16

1)

2)

3)

4)

Ans 4

Q17

1)

2)

3)

4)

Ans 1

Q18

1)

2)

3)

4)

Ans 2

Q19

1)

2)

3)

4)

Ans 4

Q20

1)

2)

3)

4)

Ans 4

Q21

1)

2)

3)

4)


Q22

1)

2)

3)

4)

Ans 4

Q23

1)

2)

3)

4)

Ans 3

Q24

1)

2)

3)

4)

Ans 4

Q25

1)

2)

3)

4)

Ans 2

Q26

1)

2)

3)

4)

Ans 1

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 4

Q29

1)

2)

3)

4)

Ans 2

Q30

1)

2)

3)

4)

Ans 1

Q61

1)

2)

3)

4)

Ans 1

Q62

1)

2)

3)

4)

Ans 4

Q63

1)

2)

3)

4)

Ans 3

Q64

1)

2)

3)

4)

Ans 3

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 4

Q67

1)

2)

3)

4)

Ans 1

Q68

1)

2)

3)

4)

Ans 4

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 3

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 4

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 1

Q75

1)

2)

3)

4)

Ans 2

Q76

1)

2)

3)

4)

Ans 3

Q77

1)

2)

3)

4)

Ans 3

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q13

1)

2)

3)

4)

Ans 3

Q14

1)

2)

3)

4)

Ans 3

Q15

1)

2)

3)

4)

Ans 2

Q16

1)

2)

3)

4)

Ans 4

Q17

1)

2)

3)

4)

Ans 1

Q18

1)

2)

3)

4)

Ans 2

Q19

1)

2)

3)

4)

Ans 4

Q20

1)

2)

3)

4)

Ans 4

Q21

1)

2)

3)

4)


Q22

1)

2)

3)

4)

Ans 4

Q23

1)

2)

3)

4)

Ans 3

Q24

1)

2)

3)

4)

Ans 4

Q25

1)

2)

3)

4)

Ans 2

Q26

1)

2)

3)

4)

Ans 1

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 4

Q29

1)

2)

3)

4)

Ans 2

Q30

1)

2)

3)

4)

Ans 1

Q61

1)

2)

3)

4)

Ans 1

Q62

1)

2)

3)

4)

Ans 4

Q63

1)

2)

3)

4)

Ans 3

Q64

1)

2)

3)

4)

Ans 3

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 4

Q67

1)

2)

3)

4)

Ans 1

Q68

1)

2)

3)

4)

Ans 4

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 3

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 4

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 1

Q75

1)

2)

3)

4)

Ans 2

Q76

1)

2)

3)

4)

Ans 3

Q77

1)

2)

3)

4)

Ans 3

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q16

1)

2)

3)

4)

Ans 4

Q17

1)

2)

3)

4)

Ans 1

Q18

1)

2)

3)

4)

Ans 2

Q19

1)

2)

3)

4)

Ans 4

Q20

1)

2)

3)

4)

Ans 4

Q21

1)

2)

3)

4)


Q22

1)

2)

3)

4)

Ans 4

Q23

1)

2)

3)

4)

Ans 3

Q24

1)

2)

3)

4)

Ans 4

Q25

1)

2)

3)

4)

Ans 2

Q26

1)

2)

3)

4)

Ans 1

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 4

Q29

1)

2)

3)

4)

Ans 2

Q30

1)

2)

3)

4)

Ans 1

Q61

1)

2)

3)

4)

Ans 1

Q62

1)

2)

3)

4)

Ans 4

Q63

1)

2)

3)

4)

Ans 3

Q64

1)

2)

3)

4)

Ans 3

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 4

Q67

1)

2)

3)

4)

Ans 1

Q68

1)

2)

3)

4)

Ans 4

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 3

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 4

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 1

Q75

1)

2)

3)

4)

Ans 2

Q76

1)

2)

3)

4)

Ans 3

Q77

1)

2)

3)

4)

Ans 3

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q18

1)

2)

3)

4)

Ans 2

Q19

1)

2)

3)

4)

Ans 4

Q20

1)

2)

3)

4)

Ans 4

Q21

1)

2)

3)

4)


Q22

1)

2)

3)

4)

Ans 4

Q23

1)

2)

3)

4)

Ans 3

Q24

1)

2)

3)

4)

Ans 4

Q25

1)

2)

3)

4)

Ans 2

Q26

1)

2)

3)

4)

Ans 1

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 4

Q29

1)

2)

3)

4)

Ans 2

Q30

1)

2)

3)

4)

Ans 1

Q61

1)

2)

3)

4)

Ans 1

Q62

1)

2)

3)

4)

Ans 4

Q63

1)

2)

3)

4)

Ans 3

Q64

1)

2)

3)

4)

Ans 3

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 4

Q67

1)

2)

3)

4)

Ans 1

Q68

1)

2)

3)

4)

Ans 4

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 3

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 4

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 1

Q75

1)

2)

3)

4)

Ans 2

Q76

1)

2)

3)

4)

Ans 3

Q77

1)

2)

3)

4)

Ans 3

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q20

1)

2)

3)

4)

Ans 4

Q21

1)

2)

3)

4)


Q22

1)

2)

3)

4)

Ans 4

Q23

1)

2)

3)

4)

Ans 3

Q24

1)

2)

3)

4)

Ans 4

Q25

1)

2)

3)

4)

Ans 2

Q26

1)

2)

3)

4)

Ans 1

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 4

Q29

1)

2)

3)

4)

Ans 2

Q30

1)

2)

3)

4)

Ans 1

Q61

1)

2)

3)

4)

Ans 1

Q62

1)

2)

3)

4)

Ans 4

Q63

1)

2)

3)

4)

Ans 3

Q64

1)

2)

3)

4)

Ans 3

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 4

Q67

1)

2)

3)

4)

Ans 1

Q68

1)

2)

3)

4)

Ans 4

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 3

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 4

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 1

Q75

1)

2)

3)

4)

Ans 2

Q76

1)

2)

3)

4)

Ans 3

Q77

1)

2)

3)

4)

Ans 3

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q22

1)

2)

3)

4)

Ans 4

Q23

1)

2)

3)

4)

Ans 3

Q24

1)

2)

3)

4)

Ans 4

Q25

1)

2)

3)

4)

Ans 2

Q26

1)

2)

3)

4)

Ans 1

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 4

Q29

1)

2)

3)

4)

Ans 2

Q30

1)

2)

3)

4)

Ans 1

Q61

1)

2)

3)

4)

Ans 1

Q62

1)

2)

3)

4)

Ans 4

Q63

1)

2)

3)

4)

Ans 3

Q64

1)

2)

3)

4)

Ans 3

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 4

Q67

1)

2)

3)

4)

Ans 1

Q68

1)

2)

3)

4)

Ans 4

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 3

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 4

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 1

Q75

1)

2)

3)

4)

Ans 2

Q76

1)

2)

3)

4)

Ans 3

Q77

1)

2)

3)

4)

Ans 3

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q24

1)

2)

3)

4)

Ans 4

Q25

1)

2)

3)

4)

Ans 2

Q26

1)

2)

3)

4)

Ans 1

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 4

Q29

1)

2)

3)

4)

Ans 2

Q30

1)

2)

3)

4)

Ans 1

Q61

1)

2)

3)

4)

Ans 1

Q62

1)

2)

3)

4)

Ans 4

Q63

1)

2)

3)

4)

Ans 3

Q64

1)

2)

3)

4)

Ans 3

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 4

Q67

1)

2)

3)

4)

Ans 1

Q68

1)

2)

3)

4)

Ans 4

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 3

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 4

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 1

Q75

1)

2)

3)

4)

Ans 2

Q76

1)

2)

3)

4)

Ans 3

Q77

1)

2)

3)

4)

Ans 3

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q27

1)

2)

3)

4)

Ans 4

Q28

1)

2)

3)

4)

Ans 4

Q29

1)

2)

3)

4)

Ans 2

Q30

1)

2)

3)

4)

Ans 1

Q61

1)

2)

3)

4)

Ans 1

Q62

1)

2)

3)

4)

Ans 4

Q63

1)

2)

3)

4)

Ans 3

Q64

1)

2)

3)

4)

Ans 3

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 4

Q67

1)

2)

3)

4)

Ans 1

Q68

1)

2)

3)

4)

Ans 4

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 3

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 4

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 1

Q75

1)

2)

3)

4)

Ans 2

Q76

1)

2)

3)

4)

Ans 3

Q77

1)

2)

3)

4)

Ans 3

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q29

1)

2)

3)

4)

Ans 2

Q30

1)

2)

3)

4)

Ans 1

Q61

1)

2)

3)

4)

Ans 1

Q62

1)

2)

3)

4)

Ans 4

Q63

1)

2)

3)

4)

Ans 3

Q64

1)

2)

3)

4)

Ans 3

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 4

Q67

1)

2)

3)

4)

Ans 1

Q68

1)

2)

3)

4)

Ans 4

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 3

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 4

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 1

Q75

1)

2)

3)

4)

Ans 2

Q76

1)

2)

3)

4)

Ans 3

Q77

1)

2)

3)

4)

Ans 3

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q61

1)

2)

3)

4)

Ans 1

Q62

1)

2)

3)

4)

Ans 4

Q63

1)

2)

3)

4)

Ans 3

Q64

1)

2)

3)

4)

Ans 3

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 4

Q67

1)

2)

3)

4)

Ans 1

Q68

1)

2)

3)

4)

Ans 4

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 3

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 4

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 1

Q75

1)

2)

3)

4)

Ans 2

Q76

1)

2)

3)

4)

Ans 3

Q77

1)

2)

3)

4)

Ans 3

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q63

1)

2)

3)

4)

Ans 3

Q64

1)

2)

3)

4)

Ans 3

Q65

1)

2)

3)

4)

Ans 2

Q66

1)

2)

3)

4)

Ans 4

Q67

1)

2)

3)

4)

Ans 1

Q68

1)

2)

3)

4)

Ans 4

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 3

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 4

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 1

Q75

1)

2)

3)

4)

Ans 2

Q76

1)

2)

3)

4)

Ans 3

Q77

1)

2)

3)

4)

Ans 3

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q66

1)

2)

3)

4)

Ans 4

Q67

1)

2)

3)

4)

Ans 1

Q68

1)

2)

3)

4)

Ans 4

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 3

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 4

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 1

Q75

1)

2)

3)

4)

Ans 2

Q76

1)

2)

3)

4)

Ans 3

Q77

1)

2)

3)

4)

Ans 3

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q69

1)

2)

3)

4)

Ans 2

Q70

1)

2)

3)

4)

Ans 3

Q71

1)

2)

3)

4)

Ans 3

Q72

1)

2)

3)

4)

Ans 4

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 1

Q75

1)

2)

3)

4)

Ans 2

Q76

1)

2)

3)

4)

Ans 3

Q77

1)

2)

3)

4)

Ans 3

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q72

1)

2)

3)

4)

Ans 4

Q73

1)

2)

3)

4)

Ans 4

Q74

1)

2)

3)

4)

Ans 1

Q75

1)

2)

3)

4)

Ans 2

Q76

1)

2)

3)

4)

Ans 3

Q77

1)

2)

3)

4)

Ans 3

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q75

1)

2)

3)

4)

Ans 2

Q76

1)

2)

3)

4)

Ans 3

Q77

1)

2)

3)

4)

Ans 3

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q78

1)

2)

3)

4)

Ans 4

Q79

1)

2)

3)

4)

Ans 4

Q80

1)

2)

3)

4)

Ans 2

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q81

1)

2)

3)

4)

Ans 2

Q82

1)

2)

3)

4)

Ans 4

Q83

1)

2)

3)

4)

Ans 1

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q84

1)

2)

3)

4)

Ans 2

Q85

1)

2)

3)

4)

Ans 1

Q86

1)

2)

3)

4)

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Q87

1)

2)

3)

4)

Ans 3

Q88

1)

2)

3)

4)

Ans 4

Q89

1)

2)

3)

4)

Ans 2

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

By

Saurav gupta


Jadavpur university

Q90

1)

2)

3)

4)

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


v + vrsquo = 400

400 + vrsquo = 400

Vrsquo = 0


490 = 0 + frac12 98 x tsup2

T = 10 sec

Ans 1

Q2 Mg = kx



BY

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JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

K = 2

Frsquo = mg + Bil


K xrsquo = Frsquo



Ans 3


wavelength of light

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q4 P = [3P˳- P˳VV˳]

PV = [3P˳- P˳VV˳]V

for

PV=max

d(PV)dV = 0

V = 32 V˳

Hence

P =32P˳

PV=nRT

T= (32P˳) (32 V˳)nR

= 94 P˳V˳nR

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q5 E= hcλ

P = E x N

N = PE

= 10⁴λhc

≃ 25 x 10sup3sup1

Ans 3

Q6

15V

6Ω

4Ω

12Ω


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1Ω

R eq = 3Ω

I = 153

= 05 A

V|4Ω = 15 ndash 05

= 1 volt

P= Vsup2R

= frac14

Ans 3

Q7 Self explanatory

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

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JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

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JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

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JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q8

X prop tsup3

V proptsup2

a propt

Ans 2


Now I = 25 mRsup2

Solving we get

Vsup2 = 10gh7

Ans 3


BY

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JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

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JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

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JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

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JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

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JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY



Ans 4

Q11 B= 2ĵ q= 2μC

V = (2icirc + 3ĵ)x 10⁶

F = q(v x B)

= 8k

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q12 fact

Ans 1


VrsquoV = AArsquo

= 5



Where ρ= 10sup3

Solving we get

V= 5


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 3



Trsquo= 2T

Ans 3

Q15 g = 43πGrρ

gpropr

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q16

r v

r= extension

a= - krm

v dvdr = -k rm


Vsup2 = kxsup2m

X prop vradicm


Ans 4

x

m


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 1

Q18 Fκ= mg sinα


Cotα = 3

Ans 2

Q19


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q20 V = 7 Vʳ


Solving we get

t = 3RC ln2

Ans 4

Q21 F = 25



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

25 = frac12 + frac12 - d4

d = 24cm

= 24mm

Ans 4

Q22 V˳

23m k



= 2m3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



Now L = constant

KE prop 1I

Ans 3


1λsup3sup1 = 8R9


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

1λsup3sup2 = 5R36

1λsup2sup1 = 3R4



Ans 4



= a cos(kx+ωt+π)

Ans 2

Q26


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

d sinθ = nλ

d x frac12 = 5 x 10ˉ⁷

d =10ˉ⁶

Ans 1

Q27 λ = 1240E nm

λ1 = 12402 = 620nm

λ2 = 12404 = 310nm

λ3 =12406 ≃ 207nm

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4

Q29 Case 1

V1=4Vn1


= 8C1 Vsup2n1

Case 2


E1 =E2

C2 = 16C1n1 n2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 2


q= frac12 x 4 x 01

= 02

iR = dфdt

intdф = intidt R

Δф = qR

= 02 x 10

=2

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q61 A =(111) B=(122) C=(212) D=(22z)

AB = ĵ +k

AC = icirc + k



0 1 1

1 0 1 = 0

1 1 z-1

z = 3

Ans 1



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4



But



Also p is real

Hence D = 0


By option cheking

abcd are in GP

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Frsquo(x)ge0

[X(1-2x) +1]ge0

(2x+1)(x-1)ge 0

X ϵ [- frac12 1]

Ans 3

Q65 xsup2-x

Xsup3 -xsup2 +x -1

X(xsup2-1)

X⁴-1

int dx

int dx


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

int x(xsup2+1) dx


Ans 2

Q66 1

radicr + radicr+1

radicr -radicr+1

r-(r+1)


T1 = radic2 -radic1

T2 = radic3 -radic2


Tr =

Tr =


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 3

Ans 4



Σ i 1 + 2 + 3 + hellip+n

= (2n+1)3

Now



=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


1000 = 1 + (x-1) 3

X =334


= 0334

Ans 1

Q68 (b+c)11 = (c+a)12 = (a+b)13

Solving we get

a168 = b144 = c120

a7 = b6 = c5 = λ(say)


=15


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 4

Q69 Z z 1

Δ =frac14 Iz -iz 1

Z+iz z-iz 1

= frac12 |z|sup2

Ans 2

Q70 A = 34689


(34)(43)(36)(63)(38)(83)(39)(93)


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

(46)(64)(48)(84)(49)(94)(68)(86)(69)

(96)(89)(98) = 20



= 5sup2 - 5

= 20


Ans 3


Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY





5q + 10p = 55

(Pq) = (15)(34)(53)(72)(91)(110) = 6

Ans 4

Q73



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

= π -2


= 3π +2

A ˢAᴮ = (π-2)(3π+2)

Ans 4



Ans 1

Q75


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

0 αγ

Βδ 0

0 βγ

αδ 0

0 γ(α-β)

δ(β-α) 0


If α =β




Ans 2

AB=

BA=

AB-BA=


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q76 Ysup2 = 1- xsup24 ---(1)

dydx = -x4y


-x4y =- frac12

X =2y


X = plusmnradic2

Y = plusmn 1radic2


= radic10

Ans 3


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q77 X = 7

(4+7+2+8+6+a)6 = 7

a = 15

MD = (3 + 0+5+1+1+8)6

= 3

Ans 3

Q78 0 1 4

1 0 -3

Cosθ sinθ 0

= -3cosθ + 4sinθ



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

=5

Ans 4


16 = 112 + (p⁴-1)4 ndash (psup3 -1)3

P = 43

Ans 4

Q80 dldt =2 ms

dbdt = -3 ms

1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


-5 = -3l + 4

3l = 9

l =3

Ans 2

Q81

-2a a+b a+c

b+a -2b b+c

c+a b+c -2c

by solving

= 4(a+b)(a+c)(b+c)

Ans 2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


Ans 4






Frsquo(α) = 0

Α =π2


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Ans 1


Y = f[(2x+3)(3-2x)]


= 12(3-2x)sup2 sin[log(2x+3)(3-2x)]

Ans 2

Q85


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

P(100)

Q(2λ+1 -3λ-18λ-10)

dr of PQ = (2λ-3λ-18λ-10)

PQ | line

(2λ)2+(-3λ-1)(-3)+(8λ-10)8 = 0

λ= 1

Q=(3-4-2)

Ans 1


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q86 Let d = (lmn)

d | (2icirc - ĵ + 2k)

2l ndash m + 2n = 0 -----(1)

a = icirc + ĵ - k

b = 2icirc + ĵ - k

a x b = -ĵ -k

d | a x b

-m ndashn= 0

M +n= 0-----(2)

Solving (1) amp (2)

l3 = m2 = n-2


Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

Q87

(-23)

(00) (5-1)


By option checking

Ans 3



BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY

now


Put πxsup2 = z

2πx dx = dz

int exp(tanp)p dp

= int ddp G(p) dp

G(π4) -G(π16)

Ans 4

Q89 (-42) (42)

frac14

π4

π16

π4

π16


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= 8

Ans 2

Q90

(1a)

X+y = 1

X + y = 32


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4


BY

SAURAV GUPTA


JADAVPUR UNIVERSITY


= radic2 frac12

Now




0 lt a lt 1-----(2)


0 lt a lt frac12

Ans 4

Aieee 2012 Solution (19th may)

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Transcript of Aieee 2012 Solution (19th may)