Vidyamandir Classes · Vidyamandir Classes VMC | JEE Main-2020 1 Solutions |7th January Morning...

18
Vidyamandir Classes VMC | JEE Main-2020 1 Solutions |7 th January Morning SOLUTIONS JEE Main – 2020 | 7 th January 2020 (Morning) PHYSICS SECTION – 1 1.(2) A logic gate is reversible if we can recover input data from the output. 2.(1) 2000kg m = 4000 r f N = ? = v 4000 20000 T = + , 24000 T N = . P Tv = ; 60 746 24000 u = , 1.9 m/s u = 3.(1) 2 0 0 I It It = 0 IA = = BA n 0 0 (1 2) =− = − R d V nAI t dt 1 0 at sec 2 R V t = = 2 0 0 0 0 2 (1 2) 2 R R nI R V I t R R = = 4.(4) 0 0 0 ˆ ˆ cos 60 sin 60 2 2 2 E x y = + 0 3 1 ˆ ˆ 1 2 2 2 y x = 5.(4) 8 6000 10 cm = For 2 nd minimum 2 sin 2 d = 3 4 d = So, for 1 st minimum, 1 sin d = 1 3 sin 4 d = = 1 25.65 (from sin table) = , 1 25 6.(1) Energy conservation : 2 2 1 1 2 2 2 GMm GMm mu mv R R + = + 2 GM V u R = (i) Momentum conservation: 9 10 10 2 2 T orbital m m GM GM V V R R = =

Transcript of Vidyamandir Classes · Vidyamandir Classes VMC | JEE Main-2020 1 Solutions |7th January Morning...

Page 1: Vidyamandir Classes · Vidyamandir Classes VMC | JEE Main-2020 1 Solutions |7th January Morning SOLUTIONS JEE Main th– 2020 | 7 January 2020 (Morning) PHYSICS SECTION – 1 1.(2)

Vidyamandir Classes

VMC | JEE Main-2020 1 Solutions |7th January Morning

SOLUTIONS

JEE Main – 2020 | 7th January 2020 (Morning)

PHYSICS

SECTION – 1

1.(2) A logic gate is reversible if we can recover input data from the output.

2.(1) 2000kgm=

4000rf N=

?=v

4000 20000T = + , 24000T N=

.P T v= ;

60 746 24000 u = , 1.9 m/su =

3.(1) 20 0I I t I t= −

0 IA = = BA n

0 0 (1 2 )

= − = − −R

dV nAI t

dt

1

0 at sec2

RV t= =

2

0 0

0 0

2(1 2 )

2

RR

nI RVI t

R R

− = = −

4.(4) 0 0 0

ˆ ˆcos60 sin 602 2 2

E x y −

= + −

0

3 1ˆ ˆ1

2 2 2y x

= − −

5.(4) 86000 10 cm− =

For 2nd minimum 2sin 2d = 3

4d

=

So, for 1st minimum, 1sind = 1

3sin

4d

= =

1 25.65 (from sin table) = , 1 25

6.(1) Energy conservation :

2 21 1

2 2 2

GMm GMmmu mv

R R

−− + = +

2 GM

V uR

= − …(i)

Momentum conservation:

9

10 10 2 2T orbital

m m GM GMV V

R R

= =

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2 [By (1)]10

r

m GMV m u

R= −

Kinetic energy ( )2 2 21 81100 100 (1)

2 10 20 2T r

m m GM GMV V u

R R

= + = + − =

7.(4) 1 2

1 2

1 2mixmix

mix 1 2

P Pp

V V V

n C n CC

C n C n C

+ = =

+

1 21 2

1 2mix

1 1 2

1 2

1 1

1 1

R Rn n

n R n R

+

− − =

+ − −

On rearranging, we get, 1 2 1 2

mix 1 21 1 1

n n n n+= +

− − −

mix

5 3 2

1 1/ 3 2 / 3= +

− mix mix

5 171 1.42

12 12 − = = =

8.(1) Time-period 2 r

Tv

= =

2

0

1

2 a n nT

V Z Z

= ;

30

21

2 a nT

V Z

=

3T n

3

1 12 13

2 2

18

8

T nT T

T n= = = ; 16

2 12.8 10 secT −=

142 16

17.8 10

12.8 10f

−=

9.(2) Loss in P.E. = Gain in K.E.

2 21 1

2 2mgh mv Iw= +

Put, (no slipping)v wr=

2

2 2 21 1

2 2 2

mrmgh mw r w= +

2 23 1 4

4 3= =

ghmgh mw r w

r

10.(3) Equal number of magnetic field lines enters the circular region & comes out of infinite plane excluding

circular area. So, magnetic flux are equal in magnitude but opposite in direction.

0i = −

11.(1) For damped oscillation : 0ma bv kx+ + =

2

20

d x dxm b kx

dtdt+ + = …(i)

For LCR series circuit 0di q

iR Ldt C

− − − =

2

2

10

d q dqL R q

dt Cdt+ + = …(ii)

Comparing equation (i) and (ii), 1

, ,L m C R bk

12.(4) 0

1e

L Dm

f f

= +

, if final image is least distance of distinct vision

150 25

375 15 ef

= +

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750

2.17 cm 21.7 mm 22 mm345

ef = = =

Also, 0 e

L Dm

f f

=

if final image is at infinity.

150 25

3755 ef

=

, 22 mmef =

13.(4) 0 0 ,E B C C= = speed of light in vacuum

8 80 3 10 3 10 V/mE −=

0 9 V/mE = 3 10 ˆ9sin(1.6 10 48 10 ) V/mE x t k= +

14.(1) 2AB CDI I md= +

2 2

12 16

ml ml= +

27

48

ml=

Radius of gyration 27 7

48 48

ABI ll

m= = =

15.(4) 1

0.42.5

I A A= =

1 0.22

II A= =

16.(4) Take origin at 1 kg mass y = vertical, x = horizontal

1 0 1.5 3 2.5 0

cm 0.9 cm5

X +

= = ; 1 0 1.5 0 2.5 4

2cm5

Ycm + +

= =

17.(1) 20 cosI I= ,

200 cos

10

II=

1

cos 0.31 0.70710

= =

45 &90 45 − ; 71.6 = Angle rotated 90 71.6 18.4= − =

18.(4) All 'dc s are in series

1 1 1 1

....c dc dc dc= + + +

1 1

c dc=

0k Adc

dx

=

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0 00

1 1ln |1 |

(1 )

ddx

d dc x AK K A= = +

+

1d

2

ln|1 |2

dd d

+ = −

0

11

2

d d

c K A

= −

; 0 1

2

= +

K A dc

d

19.(1) 6.0m g= , 60 cml = , 21.0 mmA = , 190 msv −= , 11 216 10 NmY −=

2Tv T v= =

;

2v lY

A l

=

2 2 3

6 11

6 10 8100

10 16 10

v l mvl m

AY AY

= = =

5 33 10 m = 0.03 10 m = 0.03 mm− −=

20.(1) 11 1 2 2 2 1

2

VPV PV P P

V

= =

1.41

13

atm

=

Work

3 5

1 1 2 2

310 1.01 10 1

4.65

1 0.4

PV PVJ

− − −

= = −

Work 90J

Closest answer is 90.5 J

SECTION – 2

21.(600) 300 2

1900 3

= = − =in

W

Q ;

31800

2inQ W J= = ; 600low inQ Q Q J= − =

22.(60) 6 5 62 12 2 5 10 5 10 60 10− − − = + = + = 60C =

23.(10) = −A PKE PE PE A Bmgh mgh= − (2 1) 1 10 1 10mg J= − = =

24.(175) x yz z xyB A B A = + 3 25 4 25 175Wb= + =

25.(11) 1240

4310

photonE ev= =

No. of electrons emitted . .

photon

I An

E=

53 11

19

6.4 10 110 1 10

4 1.6 10

−−

= =

11x =

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CHEMISTRY

SECTION – 1

1.(2) It is Gay Lussac law of gaseous volume

2.(4) 2 3 3CS CH COCH+

A B

A……..A (non polar –non polar)

B……..B (polar –polar)

A……B (non polar…..polar)

So

A......AA......B

B......B

So, it is non ideal solution showing positive deviation.

So, volume should be greater than 200ml

3.(1) E Cu / Cu 0.34V++ = E Cu / Cu 0.522V+ =

Cu 2e Cu++ + → E 0.34V = -(1) oCuE / Cu ?++ + =

Cu e Cu+ → E 0.522V = -(2)

Cu e Cu++ ++ → oE ?=

(1)-(2)

Cu 2e Cu++ + → E 0.34V = -(3)

Cu Cu e++→ + oE 0.522V= − -(4)

Cu e Cu++ ++ → E ? = -(5)

3 4 5G G G + =

G nfE = −

3G 2 F .34 = − ( )4G 1 F 522 = − −

4G 1 f E = −

( ) ( ) ( )2f .34 1 .522 1 f E − + − − = −

E 0.158V =

4.(2) Vitamins Deficiency

Vitamin 2B -(Riboflavin) Cheilosis

Vitamin 1B - (Thiamine) Beriberi

Vitamin 6B - (Pyridoxine) Convulsions

Vitamin C - (Ascorbic acid) scurvy)

5.(2) Diammine chloride (methanamine) (Platinum II) chloride

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6.(2)

7.(1)

8.(4) n 5= Ss,Sp,Sd,Sf ,Sg

2 2n 5 25= =

No. of orbitals =1 3 5 7 9 25+ + + + =

Each orbital has one electron with 5

1m

2= +

9.(4) Potassium is alkali metal

It always show +1 oxidation state

22

22 2 2

2 2

K O 2K O

K O 2K O

KO K O

+ −

+ −

+ −

⎯⎯→ +

⎯⎯→ +

⎯⎯→ +

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10.(3) (A)

3 3

H3 3 3 3

33 2

CH CH| |

CH C C CH CH C C C CH| | | |

CHCH OHon

+

+

− − − ⎯⎯⎯→ − − − −

3 3

33

H|

CH C C CH| |

CHCH

+

− − − 1,2methyl

shift⎯⎯⎯⎯⎯

3

3 3

3

CH|

CH C — C CH| |

CH H

+

− −

(B) ( )

alcohol3 3 3 3E2

3 3

CH CH CH CH CH C C CH| | | |

BrCH CH H

⎯⎯⎯⎯→− − − − = −

(C) ( )CH O K3 3

3 3 3

3 3

CH CH CH CH CH CH CH CH| | |

BrCH CH

− +

− − − ⎯⎯⎯⎯⎯⎯→ − − =

( )3 3CH O K− + ⎯⎯→Sterically hindered base (Hoffmann elimination)

(give Hoffmann major product)

(D)

11.(1) Due to lanthenoid contraction, size of 4d size of 5d series except La

12.(2) Fact

13.(2) Fact

14.(4) Dipole moment of 4CH & CC 4 0=

Dipole moment of 3CHc 0

15.(4) ( )c f Br I

16.(2)

17.(2) Fact

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18.(2)

19.(4) B→ Guanidine Type strongest organic base (conjugate acid is stablised by equivalent Resonance)

(three resonating structure)

A→ 2NH CH NH− = (also guanidine type but two equivalent resonating structure of conjugate

acid) (so less base then (B)

C→ 3 3CH NHCH (Aliphatic amine)

Order of basic nature B A C

Order of bkP B A C

20.(4) MOT can explain the nature of bonding due to synergic bonding 4[Ni(CO) ] .

SECTION – 2

21.(2)

22.(10.60)

2 4NaOH H SO

(A) (B)

+

Molarity of 34

NaOH 1040 100

−= =

Molarity of 3

2 4

9.8H SO 10

98 100

−= =

2NaOH + 2 4 2 4 2H SO Na SO H O→ +

310 M−

310 M−

40 10

Moles 340 10−= moles 310 10−=

Limiting Reagent is 2 4H SO

So, moles of NaOH left ( ) ( )3 340 10 10 10 2− −= −

320 10−

Molarity of 3

420 10OH 4 10

50

−− −

= =

OHP 4 log4 3.398= − =

HP 14 3.398 10.602= − =

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23.(23.03)

2yt 6.93years=

2.303 100t log

K 10=

0.693

K 0.16.93

= =

2.303

t 1 23.03.1

= = years

24.(–2.70) ( ) ( )gA 2B→

U 2.1kcal, s 20cal / k = = , T 300K=

G H T S = −

H U ngRT = + ng 2 =

( )( )g 3300 300 20 = −

3300 6000= −

2700cal= −

2.7kcal= −

25.(1.67)( ) ( ) ( )

2 3 2Hot &conc x y

3Cl 6NaOH 5NaCl NaClO 3H O+ → + +

3 3NaCl AgNO AgCl NaNO+ →

Y is N 3NaClO 3Na ClO+ −

O Cl 0

O

− − =

Bond order =

(in Resonance)

noof band

1no.of bond

= +

2

13

= +

5

1.673

= =

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MATHEMATICS

SECTION – 1

1.(1) 1 x ydye e

dx

−− =

y y xdy

e e edx

− =

y x ydy

e e edx = +

ye t=

y dy dt

edx dt

=

xdt

e tdx

= +

xdt

t edx

− =

I.F. 1dx xe e

− −=

x x xt e e e dx− − =

xt e x c− = +

x xt xe ce= +

y x xe xe ce= +

0, 0x y= =

1 0 c= + 1c =

At 1x =

( 1)y xe e x= +

(2)ye e=

log log 2ye ee e=

log 2 1ey = +

2.(4) Given that plane passes through the point 1 1 1 2 2 2

(2, 1, 0) (4, 1, 1),

, , , ,x y z x y z and

3 3 3

(5, 0, 1).

, ,x y z

Equation of plane passes through 3 non-collinear point

1 1 1

2 1 2 1 2 1

3 1 3 1 3 1

0

x x y y z z

x x y y z z

x x y y z z

− − −

− − − =

− − −

2 1 0

4 2 1 1 1 0 0

5 2 0 1 1 0

x y z− − −

− − − =

− − −

2 1

2 0 1 0

3 1 1

x y z− −

=

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( 2)(0 1) ( 1)( 1) ( 2) 0x y z− + − − − + − = 2 1 2 0x y z− + − − =

2 3x y z− − =

Now image of (2, 1, 6)R w.r.t 2 3x y z+ − = is

2 2 22 1 6 2(2 1 12 3)

1 1 6

− − − − + − −= = =

x y z

x

2 2 22 1 64

1 1 2

x y z− − −= = =

2 6x = 2 5y = 2 2z = −

3.(4) Let x be a random variable and k be the value of assigned x for 3, 4, 5k =

k 0 1 2 3 4 5

( )P k 1

32

12

32

11

32

5

32

2

32

1

32

Now expected value ( )= xP k

1 12 11 5 2 1

1 ( 1) ( 1) 3 4 532 32 32 32 32 32

= − + − + − + + + 24 28 4 1

32 32 8

− += =

4.(4) 4y mx= +

Let 1

y mxm

= + is tangent to 2 4y x=

Now solving with 2 2x by=

2 1

2x b mxm

= +

2 2

2 0b

x bmxm

− − =

0x = (because line touches the curve)

2 2

( 2 ) 4 1 0b

bmm

− − − =

2 2 8

4 0b

b mm

+ = 2 2 8

4b

b mm

= −

3 2bm = − 3 2

mb

= −

1/ 32

mb

= −

Now

1/ 3

42

b − =

642

b − =

128b = −

2nd method

Given 4y mx= + ........ (i)

Let the equation of common tangent of 2 4y x= & 2 2x by= is

1

y mxm

= + ........ (ii)

(i) & (ii) are identical

1

4m =

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So the line 1

44

y x= + is also common tangent of 2 2x by=

Solving 2 162

4

xx b

+ =

22 16x bx b= +

22 16 0x bx b− − =

0D =

2 4 2 ( 16 ) 0b b− − =

2 128 0b b+ =

128b = − & 0b = (which is not possible)

So 128b = −

5.(2) 125 124 249 49 ..... 49 49 1+ + + + +

Clearly given series in G.P.

126 126 63 631(49 1) 49 1 (49 1)(49 1)

49 1 48 49

− − + −= =

Greatest value of 63k =

6.(4) Given digits are 1, 3, 5, 7, 9

For digits to repeat we have 5 choices, hence total members are, 51

6! 5 6!

2! 2C

=

7.(3) Given equation, k k kx y a+ =

Differentiating w.r.t. x,

1 1 0k k dy

kx kydx

− −+ =

1kdy x

dx y

= −

........ (i)

& given that

1/ 3dy y

dx x

= −

........ (ii)

(i) & (ii) are tangent

1 1/ 3kx x

y y

− −

=

1

13

k − = − 2

3k =

8.(3) Given equation 2( 1) tan 2 tan 1k x x k+ − = −

2( 1) tan 2 tan ( 1) 0+ − + − =k x x k

Sum of roots 2

tan tan1

b

a k

+ = − =

+

Product of roots 1

tan tan1

c k

a k

− = =

+

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2 2tan ( ) (tan( )) 50+ = + =

22

tan tan 150 5011 tan tan

11

kk

k

+ + = = = = −− − +

22

1 501 1

1

kk k

k

+ =+ − +

+

2

502

=

2 100 = 10 =

9.(2) Given equation,

2 1 0x x+ + =

2, =

Taking =

2

2 4

1 1 11

13

1

A

=

2 2

2 2 2 4 4 2

2 3 3 4 3

1 1 1 1 1

1 1 1

1 1 1

A

+ + ++ + + = ++ + + + + + + + + + +

3 0 01

0 0 33

0 3 0

=

2A I= 4A I=

Now, 31 28 3 4 7 3( )A A A A A= = 3 3I A A =

10.(3) The system of linear equations

2 2 0x ay az+ + =

2 3 0x by bz+ + =

2 4 0x cy cz+ + = has non zero solution

Then 0,D =

2 2

2 3 0

2 4

a a

b b

c c

=

1 2

1 3 0

1 4

a a

b b

c c

=

1( 4 ) 2 ( ) (4 3 ) 0bc bc a c b a c b− − − + − =

3 4 2 2 4 3 0bc bc ac ab ac ab− − + + − =

2 2 3 0bc ac ab ab− + + − =

2bc ab ac− − = −

2bc ab ac+ =

2 1 1

b a c= +

1 1 1, ,

a b c are in AP

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11.(2) Given statement

( ) ( )p q q p

p q p q p q p ( )( )p q p q −

T T T F F F

T F F F T F

F T T T T T

F F T T T T

( ) ( )p q p q p =

12.(3) 2

tan cot 1( ) 2

1 tan siny x

x

+ = +

+

2 2

2

sin cos

1cos sin2sin sin

1cos

x +

+

+

2 2

2 2 2

2

cos sin

sin cos 12

cos sin sin

cos

+

+ +

2

2

cos 12

sin cos sin x

+

22cot cosec +

21 cot 2cot+ + 1 cot+ (1 cot )− +

for 3

,4

2cosec

dy

d=

dy

d at

25 5cosec 4

6 6

= = =

13.(3) z x iy= +

1 ( 1)

2 2 (2 1)

z x iy

z i x y i

− − +=

+ + +( 1) (2 (2 1) )x iy x y i= − + − +

(2 (2 1) ) (2 (2 1) )x y i x y i= + + − +

2 2

2 2

2 2 2Re( ) 1

4 4 4 1

x x y yz

x y y

+ + += =

+ + +

2 2 2 22 2 2 4 4 4 1x x y y x y y− + + = + + +

2 22 2 2 3 1 0x y x y+ + + + =

2 2 3 1

02 2

+ + + + =x y x y

2 2 1 9 1 5

4 16 2 4r g f c= + − = + − =

Circle whose diameter is 5

.2

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14.(1) Using LMVT for [ 7, 1]x − −

( 1) ( 7)

2( 1 7)

f f− − −

− +

( 1) 3

26

f − +

( 1) 9f −

Using LMVT for [ 7, 0]x −

(0) ( 7)

2(0 7)

f f− −

+

(0) 3

27

f + (0) 11f

(0) ( 1) 20f f+ −

15.(2)

Required area = Total area − shaded area

11 23/ 2

0 0

22 ( ) 2

3 2

xx x dx x

= − − = − −

2 1 12 (12 1)

3 2 6

= − − = −

16.(4) Distance between foci 2 6ae= =

3ae = ........ (i)

Distance between directrixes 2

a a

e

= − −

2

12a

e= 6

a

e= ........ (ii)

Multiply (i) and (ii)

2 18a =

Dividing (i) by (ii)

2 1

2e =

Now, 2

2

21

be

a= −

21

12 18

b= − 2 9b =

Length of 22 2 9

3 23 2

bLR

a

= = =

Page 16: Vidyamandir Classes · Vidyamandir Classes VMC | JEE Main-2020 1 Solutions |7th January Morning SOLUTIONS JEE Main th– 2020 | 7 January 2020 (Morning) PHYSICS SECTION – 1 1.(2)

Vidyamandir Classes

VMC | JEE Main-2020 16 Solutions |7th January Morning

17.(4) Let the numbers 2 , , , , 2a d a d a a d a d− − + +

Sum 5 25a= = 5a =

Product 2 2 2 2( 4 )( ) 2520a d a d a= − − =

2 2(25 4 )(25 )5 2520d d− − =

2 2(4 25)( 25) 504d d− − =

Solving we get 1d = or 11

2d =

Now 11

,2

d = since 11 1

52 2

a d− = − = −

Therefore, largest number 11

2 5 2 162

a d

= + = + =

18.(1) 2( ) 1g x x x= + −

2( ( )) ( ) ( ) 1g f x f x f x= + −

24 5 51

5 4 4g f f f

= + −

Also, 2( ( )) 4 10 5g f x x x= − +

25 5 5 25 50 5

4 10 5 54 4 4 4 4 4

g f

= − + = − + = −

2 5 51

4 4f

= − = −

25 1 1 5

14 2 4 4

f

+ − − = −

25 1

04 2

f

+ =

5 1

4 2f

= −

19.(1 & 3) ( 1 ) ( )f a b x f x x R+ + − =

1( ( ) ( 1)

b

a

I x f x f x dxa b

= + ++ . . . (i)

1( ) ( ( ) ( 1 ))

b

a

I a b x f a b x f a b x dxa b

= + − + − + + + −+ . . . (ii)

1( ) ( ( 1) ( ))

b

aI a b x f x f x dx

a b= + − + +

+ . . . (iii)

(i) + (iii)

2 ( ( 1) ( )

b

a

a bI f x f x dx

a b

+= + ++

2 ( 1) ( )

b b

a a

I f x dx f x dx= + +

Page 17: Vidyamandir Classes · Vidyamandir Classes VMC | JEE Main-2020 1 Solutions |7th January Morning SOLUTIONS JEE Main th– 2020 | 7 January 2020 (Morning) PHYSICS SECTION – 1 1.(2)

Vidyamandir Classes

VMC | JEE Main-2020 17 Solutions |7th January Morning

2 ( 1 ) ( )

b b

a a

I f a b x f x dx= + + − +

( ) ( )

b b

a a

I f x dx f a b x dx= = + −

( ) ( 1)

b b

a a

f x dx f x dx= +

1 1

1 1

( 1) ( )

b b

a a

I f x dt f x dx

− +

− +

= + =

20.(2) Since a bisects the angle between and ,b c

ˆˆ ˆ ˆ ˆ 4ˆ ˆ( )

2 3 2

i j i j ka b c

+ − += =

ˆˆ ˆ4 2 4

3 2

i j ka

+ +=

or

ˆˆ ˆ2 4 4

3 2

i j ka

+ −=

On comparing, we get

4, 4 = = or 1, 2 = = −

ˆˆ ˆ4 2 4a i j k= + + or ˆˆ ˆ2 2a i j k= + −

2 2 2 0a k + = − + =

SECTION – 2

21.(36) 3

/ 2 12

3 3 12lim

3 3

x x

x xx

− −→

+ −

2

2

273 12

3lim1 3

3 3

+ −

x

x

x

x x

2

/ 22

3 12 3 27lim

3 3

x x

xx→

− +

/ 22

(3 3)(3 9)lim

(3 3)

x x

xx→

− −

/ 2 2 2

/ 22

(3 3)((3 ) (3) )lim

3 3

x x

xx→

− −

/ 2 / 2

2

/ 22

(3 3)(3 3)(3 3)lim (3 3)(3 3) 36

(3 3)

x x x

xx→

− − += − + =

Page 18: Vidyamandir Classes · Vidyamandir Classes VMC | JEE Main-2020 1 Solutions |7th January Morning SOLUTIONS JEE Main th– 2020 | 7 January 2020 (Morning) PHYSICS SECTION – 1 1.(2)

Vidyamandir Classes

VMC | JEE Main-2020 18 Solutions |7th January Morning

22.(30) Let 2 2 20 1 2(1 ......)(1 .......) .......x x x x a a x a x− + + + + = + + +

Put 1x =

0 1 2 21(2 1) ......+ = + + + + nn a a a a

Put 1x = −

0 1 2 3(2 1)(1) .......n a a a a+ = − + −

0 2 24 2 2( ...... )nn a a a+ = + + +

0 2 2...... 2 1na a a n+ + + = +

2 1 61n+ = 30n =

23.(18) (1, 2,...... ) 10Var n =

22 2 2 21 2 3 ...... 1 2 ......10

n n

n n

+ + + + + + + − =

2( 1)(2 1) ( 1)

106 2

n n n n n

n n

+ + + − =

2 1 120n − = 11n =

Now, (2, 4, 6......2 ) 16Var m =

(1, 2, 3...... ) 4Var m =

18m n+ = 2 1 48m − = 7m =

24.(5)

Since ,APC APB and BPC have equal area, P is a centroid of .ABC

1 2 3 1 2 3 17 8, ,

3 3 6 3

x x x y y yP

+ + + + =

Length of line segment

217 7 8 1

5units6 6 3 3

PQ

= − − − − =

25.(3) ( ) 2 | 3|f x x= − −

Clearly, ( )f x is non differentiable at points 1, 3x x= = and 5x =

( ( )) 2 || 2 | 3|| 3|f f x = − − − −

( ( )) ( ( 1)) ( (3)) ( (5)) 1 1 1 3

x S

f f x f f f f f f

= − + + = + + =