Rt Solutions-Practice Test Papers 1 to 14 Sol

7/30/2019 Rt Solutions-Practice Test Papers 1 to 14 Sol

1/61PAGE # 1

Practice Test Paper-1Q.1

[Sol. Given (2 + sin ) (3 + sin ) (4 + sin ) = 6.As, L.H.S 6. Equality can hold only if sin = 1

=

2

3,

2

7 sum =

2

3+

2

7= 5 = k (Given)

So, k = 5 Ans.]

Q.2

[Sol. M1

M2

M3

M4

M5

M6

| G1

G2

G3

| H1

H2

H3

H4

H5

H6

H7

Number of ways of selecting 4 books = 6C4

+ 6C3 10C

1+ 6C

2 10C

2+ 6C

1 10C

3= 1610. Ans.

Alternatively: Total number of ways of selection of four books

total number of ways of selecting four non maths books.

= 16C410C4 = 1610 Ans.]

Q.3

[Sol. Let f '(x) = (x 1) (x + 1) = (x21)

Integrating on both sides, we get f(x) =

x

3

x3

+ c

Now, f(1) = 1 1 =3

2+ c ......(i)

1

1

1

3

1

x

y

O

Similarly, f (1) = 3 3 =3

2+ c ......(ii)

From (1) and (2), we get 2c = 2 c = 1 and = 3

f(x) =

x3

x

3

3

+ 1 = x

3

3x + 1So, f(2) = 8 6 + 1 = 3 Ans.]

Paragraph for Question 4 to 6

[Sol. Let P ' (x) = A(x 1)(x 2)(x 3)

or P ' (x) = A(x36x2 + 11x 6)

Integrating both sides with respect to x, we get

P(x) =4

A(x48x3 + 22x224x) + B .....(1)

Given, P(0) = 0 B = 0 ......(2)

and P(

1) = 55 A = 4 .....(3)So, from (1), we getP(x) = x48x3 + 22x224x

= x )24x22x8x( 23

= x(x 4)rootsrealnon

26x4x

(i) Clearly, area of triangle formed by extremum points of P(x) = 1122

1

[HINTS & SOLUTIONS]


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PAGE # 2

(ii) As, P(x) + P(x) = functioneven

24 x44x2

So,

1

1

dx)x(P)x(P = 1

0

24dxx44x22 =

1

0

24 dxx22x4 =

1

0

35

3

x22

5

x4

=

3

22

5

14 =

15

11034 =

15

1134=

15

452Ans.

(iii) From above graph, P(x) has two relative minimum points and one relative maximum point

(A) is incorrectClearly, from above graph, range of P(x) = [9, ) (B) is correctAs, real roots of equation P(x) = 0 are 0 and 4. So sum of real roots = 4 (C) is incorrectAs, P'(x) = 0 has 3 distinct real roots, so P''(x) = 0 has two distinct real roots.

P(x) has two inflection points (D) is incorrect.]

Q.7

[Sol. Statement-1: Let det.(B) 0, then B1 exists.Now, AB = O ABB1 = O A = Oso det.(B) = 0.

Similarly, suppose det.(A) 0, then A

1exists.Now, AB = O A1AB = O B = OBut B is not a null matrix, so det.(A) = O

Statement-1 is true.Statement-2: Obviously Statement-2 is true. For example :

Let M =

20

10and N =

00

55then, MN = O, but neither M = O nor N = O. Ans.]

Q.8

[Sol. Statement-1: The equation of plane through A (1, 1, 1), B (1, 1, 1) and C (1, 3, 5) is

3x z 2 = 0, which clearly, passes through (2, k, 4) k R.Obviously, Statement-2 is true and not explaining Statement-1. ]

Q.9

[Sol. We have

g(x) =

1,0xif,0

1,0x,1x

sin1xx

sinx22

Clearly, g(x) is differentiable x R.(As sum and product of differentiable functions is also differentiable function.)

Now, g '(x) =

1,0x,0

1,0x,1xsin)1x(21xcosxsinx2xcos

Clearly, g'(x) is discontinuous at both x = 0 and x = 1.

Also, g(0) = 0 = g(1) (Given)

So, Rolle's theorem is applicable for g(x) in [0, 1].


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PAGE # 3

Q.10

[Sol. A vector coplanar with kji2a

, kjib

and perpendicular to k6j2i5c

will be

along cba

= acbbca

= kji29kji18 = kj39k9j27

Vector will be along kj3 .

This vector will lie in the plane which will be parallel to it.

Normal of plane will be perpendicular to vector. Ans.]

PART-B

Q.1

[Sol.

(A) Obviously each digit must be 0 or 1 1

Hence number o five digit numbers are = 1 2 2 2 2 = 16 Ans.

(B) Let u = x2 du = 2x dx

2b

12b

du1u

1

2

1Limit = 1tanbtan

2

1Lim 121

b

=

422

1=

8

= L

L120=

8

120

= 15 Ans.

(C) ax + 3y z = a .....(1)

2ax y + z = 2 .....(2)

bx 2y + z = 1 a .....(3)

For no solution, D = 0

D =

12b

11a2

13a

= 0 [Note: D1 = a]

D = 2b a = 0

Also Dx

=12a1

112

13a

= a(1) 3(2 1 + a) 1(3 a)

Dx

= a 3 3a + 3 + a = a

Dx 0 (Given a [1, 8] )

Hence for no solution

b =2

a, a I

sum =2

1(1 + 2 + ......... + 8) =

2

36= 18 Ans.

Note: when a = 0, b = 0 system will have infinite solution.]


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PAGE # 4

PART-C

Q.1

[Sol. We have, f(x) = x33ax2 + 3(a21) x + 1

f '(x) = 3 1aax2x 22 = 3 )1a(x)1a(x = a + 1

= a 1

M m

a 1 a + 1

a 1 > 2 and a + 1 < 4 a > 1 and a < 3So, a ( 1, 3) a1 = 1 and a2 = 3.Hence, (a12 + a22) = (1)2 + (3)2 = 10. Ans.]

Q.2

[Sol. Given, 2xy + 6 4x 3y = 0

2x (y 2) 3 (y 2) = 0 (2x 3) (y 2) = 0

Clearly, image of R (3, 4) in y = 2 is P(3, 0) and in x =2

3is Q (0, 4). y = 2

(3, 4)R(0, 4)

(3, 0)2

3x O

Y

X

4,

2

3

(3, 2)

Inradius of the PQR is, r = s =

2

543

34

2

1

= 1. Ans.]

Q.3

[Sol. Given, f(x) = ex x

0

x2ye)1xx(dy)y('fe ..........(1)

differentiating both the sides

f '(x) = ex ex f '(x) + x

0

yxdy)y('fee 1x2ee1xx xx2

f '(x) = f '(x) + x

0

2xyx xxedy)y('fe

0 = f(x) + (x2x + 1) exex (x2 + x) [Substituting x

0

yxdy)y('fe = f (x) + (x2x + 1) ex from (1)]

f(x) = ex (x2 + x x2 + x 1)

f(x) = ex (2x 1) f(1) = e.Also, f '(x) = ex 2 + (2x 1) ex f '(1) = 2e + e = 3eand f '' (x) = 2ex + (2x 1)ex + 2ex

f '' (1) = 2e + e + 2e = 5e.

Hence f(1) + f '(1) + f '' (1) = 9e k = 9. Ans.]


5/61PAGE # 5


[Sol. Given, h(x) =)x(g

)x(f

h'(x) =)x(g

)x('g)x(f)x('f)x(g2

Clearly, h'(c) = 0 (As, f ' (c) = 0 and g ' (c) = 0)

So, h'(c) =)c(g

)c('g)c(f)c('f)c(g2

[f '(c) > 0; g'(c) < 0 ; f (c) > 0 ; g(c) > 0]

h'(c) > 0|||ly h'(c+) < 0

So, h(x) has a local maximum at x = c. Ans.

Aliter-1: Nr i.e. f (x) is maximum at x = c

and Dr i.e. g (x) is minimum at same x i.e. x = c

Hence)x(g

)x(fis maximum at x = c

Aliter-2: For example, let f (x) = 2x1

1

and g (x) = x2 and c = 0.

So, h (x) =)x1(x

122 = )x(g

)x(f h ' (x) = 222

2

)x1(x

)x21(x2

Clearly, for x < 0, h ' (x) > 0

and for x > 0, h ' (x) < 0 h(x) has a local maximum at x = c]

Q.2

[Sol. Given that 0ckbka21

Now, OQRArea

PQRArea

=

4

cb2

1

acab2

1

P(a)

Q(b)

b a c a

O(0)

cb

ck1bkckbk12121

= 4 (1 + k

1) (1 + k

2) k1k2 = 4

Hence, k1

+ k2

= 3. Ans. ]

Q.3

[Sol. | A | = x + y + z

det.(adj.(adj. A)) = 2

)1n(A.det

= (det. A)4 = 28 34 = 124

| A | = 12

x + y + z = 12Using beggar method x + y + z = 9.

Number of matrix = 11C2 = 55. Ans.]


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PAGE # 6


[Sol.

(i) The given lines (L1

and L2) are parallel and distance

between them (BC or AD) is =5

515 = 2 units.

Let BAC = AB = BC cosec = 2 cosec

B

B1

C

A

D

A1

L2

L1

L

P(4, 3)

and AA1

= AD sec = 2 sec

Clearly, area (||gmAA1BB1)= (AB)(AA

1) = 4 sec cosec

=2sin

8, which is least for =

4

Let slope of line L be m.

So, 1 =

m4

31

4

3m

(4m + 3) = (4 3m) m =

7

1or 7

But m > 0 (Given)

The equation of line L, is (y 3) =7

1(x 4) x 7y + 17 = 0 Ans.(i)

(ii) If line L : x 7y + 17 = 0 is orthogonal to circle x2 + y2 6x + 4y 9 = 0, then line Lmust be normal to the given circle. So, centre of circle (3, 2) must satisfy the line L.Hence, 3 7( 2) + 17 = 0 17 + 17 = 0 = 1 Ans.(ii)

(iii) Equation of circle S having the ends of diameter at (0, 1) and (2, 0) is

x(x + 2) + y(y + 1) = 0 i.e. x2 + y2 + 2x + y = 0

So, equation of circle S', is (x2 + y2 + 2x + y) + (x 7y + 17) = 0

As it passes through (1, 2), so 171412)1(2)2()1(22

= 0

9 + 4 = 0 =4

9

Hence, the equation of circle S', is

(x2 + y2 + 2x + y) 4

9(x 7y + 17) = 0 4x2 + 4y2x + 67y 153 = 0 Ans.(iii)]

Q.7

[Sol. Equating the coefficient of x20, we get

220 a20 = 1 20 20 12a (C)

put, x = 2

1

, we get

0 1020

q2

p

4

1b

2

a

0q2

p

4

1b

2

a1020

b2

a and 0q

2

p

4

1


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PAGE # 7

a + 2b = 0 (B)

b =2

1220 20

.

put, x = 0 we get

1 b20 = q10

1

2020 20

2

12

= q10; 10

20

20

q2

121

;

10

20q

2

1 q =

4

1

Using, 0q2

p

4

1

04

1

2

p

4

1 p = 1

Hence B, C, D. Ans.]

Q.8

[Sol. x

y

3/2

O 44

x= 4

y=3

. Now, verify alternatives. ]

PART-B

Q.1

[Sol.

(A) IN =

0

x9 dxex2

=

0 II

x

I

9

dxex

2

I.B.P.

IN

=

0

x8

2

ex2

+

0

x7 dxex2

8 2

IN

= 0 + 4ID

IN

= 4ID

D

N

I

I= 4 Ans.

(B)MB

x4 13x2 + 36 0 (x29) (x2 4) 0 x [3, 2] [2, 3]Now, let

f(x) = x33x f '(x) = 3(x21) > 0 x [3, 2] [2, 3] f

max.(x = 3) = (3)33(3) = 27 9 = 18. Ans.

(C)MB

Any circle through (2, 2) and (9, 9) is

(x 2) (x 9) + (y 2) (y 9) + (y x) = 0 .........(1)For the point of intersection with x-axis, we put y = 0 in (1), we get

(x 2) (x 9) + 18 x = 0


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PAGE # 8

Put disc. = 0 (11 + )2 4 + 36 = 0 = 23, 1

x =2

1 = 6

So, the absolute value of the difference of x-coordinate of the point of contact = | 6 (6) | = 12 Ans.

(D) y = cos1(3x 4x3) =2

sin1(3x 4x3) =

2

xsin3 1

because sin1(3x 4x3) = 3 sin1x if x

1,

2

1

Hencedx

dy=

2x1

3

2

3xdx

dy

=

4

31

3

= 6 Ans.

Alternatively (D): Clearly,dx

dy= 23

2

)x4x3(1

)x123(1

)x43(x1

)1x4(3

dx

dy

2

2

2

3x

= 6 Ans.]

PART-C

Q.1

[Sol. x

0

x

0 KingUsing

dtxttandtttan)t(f)x(f = 0

x

0

x

0

dtttandtttan)t(f)x(f = 0

Differentiate w.r.t. x

f '(x) + 1)x(f tan x = 0

1)x(f

)x('f

+ tan x = 0

integrate w.r.t. x

ln 1)x(f + ln (sec x) = Cf(0) = 0 C = 0.

Hence

1xcos

1)x(f

f(x) = cos x

1 f(x) = 0 cos x = 1

Hence only solution in

2

,2

is x = 0.

Hence number of solution is 1.

Note that domain of f(x) is

2

,2

Ans.]


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PAGE # 9

Q.2

[Sol. We have, f '' (x) = 12x 4

f ' (x) = 6x24x + cAs, f ' (1) = 0 c = 2 f ' (x) = 6x24x 2 f (x) = 2x32x22x + As, f (1) = 0 = 2Hence, f (x) = 2(x3x2x + 1)

or f (x) = 2(x 1)2 (x + 1).

Now, M (x = 2, y = 6) and f ' (2) = 14.

So, the equation of normal at M is (y 6) =14

1(x 2)

For x-intercept, put y = 0

we get x = 86 Ans.]

Q.3

[Sol. Let y = f 2(x) g2(x)

Now,dx

dy= 2 )x('g)x(g)x('f)x(f = 2 )x(f)x(g)x(g)x(f = 0

f2(x) g2(x) = k (constant)So, f2(3) g2(3) = k (5)2 (4)2 = k k = 9, x RHence, f2() g2() = 9. Ans.]


[Sol. We have

(1 + x)10 = 10C0

+ 10C1

x + 10C2

x2 + 10C3

x3 +........ + 10C9

x9 + 10C10

x10 .......(1)

Also (x 1)10 = 10C0

x1010C1

x9 + 10C2

x8 +........ 10C9

x + 10C10

.......(2)

Multiplying (1) and (2), we get

102 1x = 1010109910110010 xCxC.......xCC

10

109

1091

10100

10 CxC.......xCxC .......(3)

Comparing the coefficients of x10 in (3) , we get

10C5

(1)5 = 210

102

9102

2102

1102

010 CC.......CCC Ans.]

Q.2

[Sol. f '(x) = f(x) = 1)x(f

)x('f

ln (f(x)) = x + c

x = 0, f(0) = 1 c = 0 f(x) = ex

g(x) = ex (x + 1)2ex = ex (x2 + 2x)

1

0

xdx)x(g)x(fe =

1

0

2xdxx2xe =

1

0

2x xe = e Ans.]


10/61

PAGE # 10

Q.3

[Sol. Let, I =

dx

Add&King

xsin1xsin1

x

2

2

2I =

dx

xsinxsin11

xsin122x

22

2

22

So, I =

dx

xsin112

xsin11x

2

22

=

dxx2

1 2=

3

3. Ans.]

Paragraph for question nos. 4 to 6

[Sol.(i) angle between banda is obtuse

hence ba

0 (log3x)2 + 2log

3x 1 0 x (0, )

42 + 4 0 [1, 0]

Volume =6

1

111

11

12

=6

1|23 |

Since, [1, 0]

Volume

2

1

,3

1

(ii) ca

= cb

0c)ba(

cba

kj2i)xlog(3

= kj)xlog(i)x(log33

+ kji

xlogxlog33

; 2 = log3x + and 1 = 1

= 2 log

3x = 0 and log

3x = 2

= 0 = 2, log

3x = 2, = 0

kj2a

ki2b

kjic

7|ji|

|k2ji3|

|ca|

|cb|


11/61

PAGE # 11

(iii) c)ba(

= b2a

b2ab)ca(a)cb(

1cb

, 2ca

log3x + log

3x 1 = 1 ... (1)

log3

x + 2 + 1 = 2

log3x =1 ... (2)

log3x = 1, = 1

kj2ia

kjib

kjic

accbba

= 2cba

= 4. Ans.]

Q.7

[Sol. f(x) =

x xe

1

e

1

22t1

dt

t1

dt xe1

1 ]ttan xe

11 ]ttan =

4)e(tan

4)e(tan x1x1

f(x) = )e(tanx1

+ )e(cotx1

= 2

Now verify alternatives. Ans.]

Q.8

[Sol. cos C =2

1=

64

c6416 2 c2 = 48 c = 34

Also,Asin

a=

Csin

c

Asin

4=

23

34 A = 30

As C = 60, A = 30 B = 90

Now, r =s

=

2

cba

ac2

1

=

cba

ac

=

8344

344

=

3412

316

=

33

34

Now verify alternatives.

(C) r = 33

34

(D) the length of internal angle bisector of angle C is3

8

Hence A, B are correct.]


12/61PAGE # 12

PART-B

Q.1

[Sol.

(A) Given, C : y2 = px3 + q .......(1)

2ydx

dy= 3px2

dx

dy=

y2

px3 2

.......(2)

Put x = 2 and y = 3 in (1), we get 9 = 8p + q .......(3)

Also, from (2), we get)3,2(M

dx

dy

=6

)4(p3= 2p = 4 .......(4)

From (2) and (3), we get p = 2, q = 7.Hence, (p q) = 2 ( 7) = 2 + 7 = 9. Ans.

(B) We have

cosec210

1cosec21

011

V = 4 cosec2 1 2cosec

= 1eccos2

eccos42

=

4

5

4

1eccos4

2

2

V eastl = 14

4

4

5

16

94 ]

(C) We have11

xxsin

1x

xxcos= 0

(sin x x) ( cos x x2) = 0 Either sin x = x or cos x = x2

Hence, number of real roots are 3. Ans.

(D) Given, tan A = 4

1

, tan B = 3

2

, tan C = 5

1

and tan D = d

Now, A + B + C + D = 2

tan (A + B) = tan (C + D) ab1

ba

=1cd

dc

3

2

4

1

1d5

1=

6

11

d5

1

Solving 39 d = 65

d =3

5 3 | tan D | = 3

3

5= 5. Ans.]


13/61

PAGE # 13

PART-C

Q.1

[Sol. Let O be the centre of polygon

Area of rectangle = 4 OA1A

2= 6 .........(1)

and Area OA1A

2=

n

1area of polygon .........(2)

A1

A2

Ak + 1

AkO

(1) and (2) 4

6

= 60n

1

n = 40. Ans.]

Q.2

[Sol. Using tan1 + tan1 + tan1 = tan1

1In L.H.S. we get

tan1

2)c)(bx()bx(ax)ax(c1

bxaxcbxaxc

where c =

8

x

x

1.

Now, 1 = ax

8

x

x

1+ abx2 + bx

8

x

x

1 1 = 222 x

8

bbabxx

8

aa x R

1 = (a + b) + x2

8

baab

On comparing, we geta + b = 1 ..........(1)

and ab =8

ba =

8

1.......(2) (Using (1))

Now, a2 + b2 + 2ab = 1 a2 + b2 +4

1= 1

Hence, 4(a2 + b2) = 3. Ans.]

Q.3

[Sol. Using cosine law in BKC,

2

Bcos =

22

232

14

184

=26

2

93

=212

15=

24

5

B

A K C

c= 2m

m 1

a = 22

23x


14/61

PAGE # 14

Now,

x =2

Bcos

ca

ac2

=

24

5

2m2

)m222(

2

23=

1m

m4

24

5 3 =

1m

m5

3m + 3 = 5m

m =

2

3

AB = 2m c = 3 . Ans.]


[Sol. S =

80tan1

1.......

10tan1

1

0tan1

1333

Note that

90tan1

1

tan1

133 = 1

=

33 cot1

1

tan1

1=

3

3

3 tan1

tan

tan1

1=

3

3

tan1

tan1.

Hence, S = 1 + (1 + 1 + 1 + 1) = 5 A. Ans.]

Q.2

[Sol. Given, f '(x) = 5)x(f2 or5y

dy

= 2dx. On integrating, we get

ln )5y( = 2x + c, if x = 0 ; y = 0 c = ln 5

ln

5

5y= 2x y + 5 = 5e2x y = 5e2x5 f(x) = 5(e2x1)

Now, f(x) + 5 sec2x = 5(e2x + tan2x) = 0

So, no solution exist (0, 2). Ans.]Alternative : f '(x) 2 f(x) = 10

dx

dy2y = 10 ; Integrating factor = e2x

y e2x = 5e2x 5

f(0) = 0 ; c = 5

y = 5e2x 5 f(x) = 5 (e2x1)Hence, f(x) + 5 sec2x = 0 e2x + tan2x = 0 No solution. Ans.]


15/61PAGE # 15

Q.3

[Sol. I = 2

2

12

1

dx1xx

xtan.......(1) put x =

t

1

I =

2

21

2

1

dt

1tt

t

1tan

.........(2)

Now, (1) + (2) gives

2I =

2

2

122

2

3

2

1x

dx

2

I =

2

2

1

1

3

1x2tan

3

2

4

=

0332

=36

2 A. Ans. ]

Paragraph for question nos. 4 and 5

[Sol. 0nar 1

nanr1

1

A

(1,1,1)

L

r = (1, 1, 1)+ (1, 1, 1)

n = i + j1^ ^

jikzjyix = 2x + y = 2 ............ plane

1

Now, kiAB ; A(1, 1, 1); B(0, 1, 0)

Now,

111

101

kji

nVAB2

n = AB V2

V = i j k^ ^ ^

B( j )^

2

1k11j10in2

= kj2in2

Hence, equation of plane 2 is kj2ijr = 0 or x + 2y z = 2

(i) Vector along the line of intersection of planes 1

and 2

is21

nn

which is also perpendicular

to the normal vector of plane 1.

21

nn

= kji

So, required vector =

3

kji6 = 2 kji or 2 1,1,1


16/61PAGE # 16

(ii) If is the acute angle between 1 and 2 then

cos =21

21

nn

nn

=

62

k2j2iji =

62

3=

2

3 = cos1

2

3=

6

. Ans.]

Q.6

[Sol. A is true by the Extreme Value Theorem, B is true by the Intermediate Value Theorem and C is not

always true, D is true because g is continuous.]

Q.7

[Sol.mb

As, cos1x 0 x [ 1, 1]and tan1x > 0, x in (0, )So, a solution must be a positive number only.

Now, cos1x = tan1x

cos1x = cos1

2x1

1 x2 = 2

x1

1

x4 + x21 = 0 x2 =

2

51

x =2

51 (0, 1)

Now verify alternatives.]

Q.8

[Sol. Suppose , be two distinct roots in (0, 1) of the equation x33x + p = 0Let f(x) = x33x + p.

As f() = 0 = f()

f(x) satisfies hypothesis of Rolle's theorem in [, ] (0, 1). So f '(c) = 0 3c2 = 3 c = 1But c must lies between and .Hence p .So, S

1is false.

Also, S2

is obviously true. Ans.]

Q.9

[Sol. We have

f(x) = tr.(A ) =

3

1iii

a = x +x9 + 2

As,

x

9x

2x

9x

x +x

9 6 (as x > 0 is given)

so, f(x) 8 f

Min.(x = 3) = 8

Also, S2

is true and explaining S1also. Ans.]


17/61

PAGE # 17

PART-B

Q.1

[Sol.

(A) Let the point of intersection of curves y2 = 2ax (a > 0) and xy = 24 be P(x1, y1).

Now, y2 = 2ax )y,x(P 11

dx

dy

=

1y

a= m

1... (1)

Also, xy = 24

1

1

)y,x(P x

y

dx

dy

11

= m

2... (2)

As, m1 m

2= 1 x

1= a ...(3)

As, x1y

1= 24 y1 = a

24

So, P

a

24y,ax

11

Also, Point P must satisfy

y12 = 2ax

1

2a

32= 2a2 a4 = 16 a = 2. Ans.

(B) f (x) is differentiable in R

Hence )x(fLimx

must exist and is finite

y = f (x) must have a horizontal asymptoteO

y=1x

y=f(x)(0,4)

as x then only )x(fLimx

will exist

)x('fLimx

= 0

)x('f)x(fLimx

= 3 )x(fLimx

= 3 Ans.

(C) y = mx + (n + 2 x) or y + x (n + 2) mx = 0 )2n(y + (1 m) (x 0) = 0 m R Fixed point is the point of intersection of lines y = n + 2 and x = 0.Family of lines which passes through (0, n + 2)

Hence n + 2 = 3 n = 1. Ans.]


18/61

PAGE # 18

PART-C

Q.1

[Sol. Let the 3 consecutive terms are

a d, a, a + d ; d > 0

hence a2 2ad + d2 = 36 + K ....(1)

a2 = 300 + K ....(2)

a2 + 2ad + d2 = 596 + K ....(3)

now (2) (1) gives

d(2a d) = 264 ....(4)(3) (2) gives

d(2a + d) = 296 ....(5)

(5) (4) gives

2d2 = 32 d2 = 16 d = 4 or d = 4 (rejected as increasing A.P.)Hence from (4)

4(2a 4) = 264 2a 4 = 66 2a = 70 a = 35 K = 352 300 = 1225 300 = 925 Ans.]

Q.2

[Sol. Given, f (y) = f (x) + y

y2 + ay + b = x2 + ax + b + y y2 + y(a 1) x2ax = 0As, y R, so D 0 x R (a 1)2 + 4(x2 + ax) 0 x R 4x2 + 4ax + a22a + 1 0 x RNow, D 0

16a2 16(a2 2a + 1) 0 2a 1 0 a 2

1

So, amax.

=2

1

Hence, maximum value of 100a = 50 Ans.]

Q.3

[Sol. For number of solutions of given equation to be 9, so n must be 5.

Now (3 + cosec2x) + ysin2

2 = 5

possible only if cosec2x = 1 and sin2y = 0

possible values of x =2

,

2

3,

2

5,

2

7

possible values of y = 0, , 2, 3, 4

Hence number of ordered pairs (x, y) = 4 5 = 20 Ans.]

Q.4

[Sol. Put f (x) = t f (t) = 4 t = 2, 0, 4Hence f (x) = 2 2 values of x.

or f (x) = 0 2 values of x.or f (x) = 4 3 values of x.

Hence, the number of real solutions of the equation f(f(x)) = 4 are 7. Ans.]


19/61

PAGE # 19

Practice Test Paper-5

Q.1

[Sol. )x(fLim

2x

=h

(sinh)sinLim

0h = 1 and )x(fLim

2x

= 1h

)1(sinLim

0h

.

So, )x(fLim

2x

does not exist. Ans.]

Q.2

[Sol. 1, x, y G.P. Let x = k, y = k 2

and x, y, 3 A.P. 2y = x + 3 2k2 = k + 3

2k2 k 3 = 0 k = 1,2

3.

x + y = k2 + k =4

1

2

1k

2

.

x + y|max

=4

15

4

14 at k =

2

3. Ans.]

Q.3

[Sol. Possible cases :

(i) 0 0 0 0 0 3 !5

!6

(when digit 0 comes at first place then number of arrangement =!4

!5)

Number of six digit numbers with 0 0 0 0 0 3 =!5

!6

!4

!5= 1

(ii) Similarly for 0 0 0 2 2 1

!2!2

!5

!2!3

!6= 30

(iii) 1 1 1 1 1 2 !5

!6= 6

Total = 1 + 30 + 6 = 37. Ans.]


[Sol. We have, M(0) =

3

200

02

10

004

3

= diag

3

2,

2

1,

4

3


20/61PAGE # 20

222

2

3

2,

2

1,

4

3diag)0(M

333

3

3

2,

2

1,

4

3diag)0(M and so on

Now, n32 )0(M.........)0(M)0(M)0(M

=

n2n2n2

3

2.....

3

2

3

2,

2

1......

2

1

2

1,

4

3......

4

3

4

3diag

So, P =

.....3

2

3

2,........

2

1

2

1,.....

4

3

4

3diagLim

222

n

=

32

1

3

2

,

2

1

1

2

1

,

4

3

1

4

3

diag = diag (3, 1, 2)

Also, Q = diag

2

1,1,

3

1

PQ =

100010001

= I3

(PQ)m = I3 m N

Rm

= M(x) R100

= M(x)

Sk= M(x) k2 )PQ(.....)PQ()PQ( S

360= M(x) 360 I

3= 360 M(x)

(i) adj (adj P) = PP2n

= 6 P = 6P

Padjadj.Tr = Tr (6P) = 36.

(ii) Tr . (R100

) = f(x) = )x(M.Tr

f(x) = x23

2

2

1x

4

1xx3 2

= 3x2 + 2x +12

23.

2

0

dx6

41)x(cosf)x(sinf

= dx6

41

12

23xcos2xcos3

12

23xsin2xsin3

2

0

22

= dxxcosxsin22

0

= 4


21/61PAGE # 21

(iii) Tr. (S360

) = g(x) = 360 )x(M.Tr

=

12

23x2x3360

2=

36

23x

3

2x3360

2

=

9

1

36

23

3

1x3360

2

=

36

19

3

1x3360

2

g(x)|min = 360 3 3619 = 570. Ans.]

Q.7

[Sol. We have

22x2

x1

e.1x)x(f

f ' (x) =

x

32

23

e.1x

1x5x3x1x

1

01

graph of g(x) = x 3x + 5x 13 2

Let g (x) = x33x2 + 5x + 1,

g ' (x) = 3x26x + 5, discriment < 0

so g (x) has only one real root.

Also g (1) g (0) < 0 so one root of g(x) = 0

belong to the interval (1, 0) say

Now sign scheme of f '(x) = x

32

)x(g

23

e.)1x(

1x5x3x1x

At x = 11

ve +ve

At x =

+ve ve

Clearly f (x) has two points of extremum, maxima at x (1, 0) and minima at x = 1.]Q.8

[Sol.

(A) ba

= ca

either cb

or cba

.

(B) ba

= ca

either cb

or cb||a

(C) True

(D) True. Ans.]

PART-B

Q.1

[Sol.

(A) Given cos = 12 sinsin = 12

cos

= cos12

12 cos = sin cos2 = sin + cos cos + sin = 2 cos but given that cos + sin = k cos


22/61

PAGE # 22

Hence, K = 2

L = 11 2 (As, K + L = 11 (given)) 9 < L < 10 2 < log

3L < 3

Sum of integers = 2 + 3 = 5. Ans.

(B) OC = 2 OC2 = 4 h2 + k2 = 4 x2 + y2 = 4.

O 22

C(h,k)2

Y

X Ans.

(C) sin x21sin 1 = sin

3

where = sin1x

1 2x = sin21

cos2

3= 2

xx12

32

2 4x = xx13 2 x3x332 2

(2 3x)2 = 3(1 x2) 4 12x + 9x2 = 3 3x2

Hence, (12x 12x2) = 1.

(D) g (10) = 10

0

f (t) dt = 5

2

0

f (t) dt = 5

1

0

1

0

td)t2(ftd)t(f

[As, f is periodic with period 2.]

a2

0

a

0

a

0dx)xa2(fdx)x(fdx)x(f

=

1

0

1

0

td)t(ftd)t(f (Since f(x) is periodic with period 2 hence f (2 t) = f ( t)) .

=

1

0

1

0

td)t(ftd)t(f (f ( t) = f(t), as f is odd.)

= 0. Ans.]

PART-CQ.1

[Sol. x2 + y2 + y 1 + k (x + y 1) = 0 .......(1)

Hence C passes through the intersection of x2 + y2 + y1= 0 and y = 1 x

Solving

x2 + (x 1)2 + (1 x) 1 = 0

2x23x + 1 = 0

2x22x x + 1 = 0

2x (x 1) (x 1) = 0


23/61

PAGE # 23

x = 1 or x =2

1

y = 0 or y =2

1

Hence, points are (1, 0) and

2

1,

2

1

Now minimum radius will be when the two points are the extremities of the diameter of the circle C.

diametermin

=4

1

4

1 =

2

1

Hence, r |min

=22

1=

8

1 R = 8. Ans.]

Q.2

[Sol. = b82

1 =

2

1 8 a =

2

1 10 c

a = b and c =5

a4.......(1)

cos A =bc2

acb222

=bc2

c2

=

5

a4a225

a16 2

cos A =5

2 sin A =

5

21

From (1)

sin C =5

4sin A =

25

214

=2

1ab sin C =

2

1 h

c c

=2

1a2

25

214=

2

1 10 c

=25

214

16

c252

= 10 c

c2 =21

40 c c =

21

40=

q

p

p + q = 40 + 21 = 61 Ans.


24/61

PAGE # 24

Alternatively: = aha21

a

2= h

a= 8

a

= 4 .....(1)

|||lyb

= 4 .....(2) and

c

= 5 ......(3)

(1) and (2) a = b

andc

a=

4

5=

c

b

a = b =4

c5

s =2

cba =

4

c7

s a =4

c5

4

c7 =

4

c2=

2

c= s b

s c = c4

c7 =4

c3

2 = s(s a)(s b)(s c) =4

c3

2

c

2

c

4

c7.....(4)

Also =2

1 c 10 = 5c

2 = 25c2 .....(5)

(4) and (5) 25c2 =4

c3

2

c

2

c

4

c7 25 =

64

c21 2 c2 =

21

6425

c = AB =21

85=

21

40=

q

p

p + q = 40 + 21 = 61 Ans.]Q.3

[Sol. Let I = 100

1

dxx

)x(f=

100

10

10

1

dxx

)x(fdx

x

)x(f=

100

10

10

1

dxx

x

100f

dxx

)x(f

Put,x

100 = t dxx

1002

= dt tx

dx = dt tdt

xdx

I =

1

10

10

1

dtt

)t(fdx

x

)x(f= 2

10

1

dxx

)x(f= 2 5 = 10. Ans.]


25/61

PAGE # 25


Q.1

[Sol. As, ar.(ABC) = ar. (ACD) + ar. (BCD)

2

1(3) (4) =

2

1(4) (CD) sin 60 + 30sinCD3

2

1

A

C B

D

60

30

b=4

a=3

c=5

CD =334

24

=

39

)334(24 = )334(

13

8 . Ans.]

Q.2

Sol.

100

0k !3k!2k!1k

3k=

100

0k 2k3k2k1!1k

3k

=

100

0k2 9k6k!1k

3k=

2

3k!1k

3k=

100

0k !1k3k

1

=

100

0k !3k13k2k =

100

0k !3k1

!2k1 =

!1031

!21

S =2

1 B. Ans.]

Q.3

[Sol. Clearly, = 0

111 11k

1k1

= 0 1k Ans.]


[Sol.

(i) Given 1, b, c are in A.P. 2b = 1 + c ......(1)Also, 2, 5b, 10c are in G.P.

25b2 = 20c or 5b2 = 4c .......(2)Now, putting c = (2b 1) from (1) in (2), we get

5b2 = 4(2b 1) 5b2 + 8b 4 = 0 b = 2 (As b I)

So, c =

5 f (x) = x22x 5 = (x 1)26Clearly in interval x [0, 4],m = minimum value of f (x) = f

min.(x = 1) = 6

and M = maximum value of f (x) = fmax.

(x = 4) = (4 1)26 = 9 6 = 3.

Hence, (M + m) = 3 + (6) = 3 Ans.(i)


26/61PAGE # 26

(ii) Given g (x) =

upwardopeningparabolais)x(gofGraph

222 3x4ax1a

Also, g (0) = 3

g (x) < 0 is true, for atleast one real x provided a2 + 4 0 2 a 2So, the number of integral values of a are 5 (i.e., a = 2, 1, 0, 1, 2). Ans.(ii)

(iii) We have y = f (x) + h(x) = (x22x 5) + x2 (p 3)x + p = x2 + (1 p)x + (p 5)

As, range of y is [0, ), so put D = 0

(1 p)2 = 4(p 5) 1 2p + p2 = 4p 20 021p6p

RppositiveAlways

2

So, p Ans.(iii)]

Q.7

[Sol. Statement-1 is true and Statement-2 is false. Statement-2 could be true only if f(x) is continuous.

Rolle's Theorem can also be used for validity of Statement-1. Ans.]

Q.8

[Sol. Statement 1: B1

+ B2

+ B3

+ B4

= 10

B1

+ B2

+ B3

+ B4

= 6

Using beggar total number of ways 9C3

Statemen-2: Number of ways of choosing any 3 places from 9 different places is 9C3

Hence S-1 is true, S-2 is true and statement-2 is the correct explanation for statement-1. Ans.]

Q.9

[Sol. S-1 : Let r be the common ratio of geometric progression. So,

S = 8r1

r

x

(Given)

x = 8(r r2) ........(1)

As, r ( 1, 1) {0} so r r2

4

1,2 {0} .......(2)

From equations (1) and (2), we getx = 8 (r r2) (16, 2] {0}

Hence, S - 1 is false and obviously S-2 is true. Ans.]

Q.10

[Sol.(A)O x=a

y

x

point of non-differentiability.

(B) y = x3

f '(0) = 0

O

y

xBut at x = 0, f(x) is increasing.

(C) f "(c) can be zero also. e.g. f(x) =x4.

(D) false e.g. f(x) = x3 at x = 0 is increasing.]


27/61

PAGE # 27

Q.11

[Sol. As, a1

+ a2

+ a3

= 126

and a2

=2

aa31

=

3

126= 42.

The numbers in A.P. are 42 d, 42, 42 + d.

Let g1

= A, g2

=AR; g3

= AR2

AR = 34 and A + AR2 = 85

R34 + 34R = 85 34R2 85R + 34 = 0

34R268R17R + 34 = 0 (R 2)(2R1) = 0For R = 2, A = 17

g1

= 43 + d = 17 d = 26 (Rejected)

for R =2

1, A = 68

g1

= 43 + d = 68 d = 25

Common difference of A.P. = 25 and common ratio of G.P. is2

1Ans.]

PART-B

Q.1

[Sol.

(A) We know that | adj A | = | A |2 for a 3 3 matrix

Given adj A = KAT |adj A| = |KAT| = K3 | A | (|AT| = | A | ) K3 | A | = | A |2

K3 = | A | ; Now det A = 1 (1 4) 2(2 4) + 2 (4 + 2) = 27 k3 = 27K = 3.

(B) Given, f (x) = (2x + 1)50 (3x 4)60

f ' (x) = 220(2x + 1)48 (3x 4)58 (2x + 1)(3x 4)(3x 1)

12

13

43

+ +

Sign scheme of f '(x)

Least positive integer is k = 2(C) By applying continuity and differentiability at x = 1, we get a = 3, b =1.

Hence, (2a + b) = 2 (3) 1 = 5. Ans.]

PART-C

Q.1

[Sol.mb

Sn

= 1 2 + 2 3 + 3 4 + ......... upto n terms

=

n

1n

)1n(n =

n

1n

n

1n

2 nn =2

)1n(n

6

)1n2()1n(n

Sn

=3

)2n()1n(n ......(1)


28/61

PAGE # 28

Also, n1

= .......6543

1

5432

1

4321

1 upto (n 1) terms

=

1n

1n )3n()2n()1n(n

1=

1n

1n )3n()2n()1n(n

n)3n(

3

1

n1

=

1n

1n )3n()2n()1n(

1

)2n()1n(n

1

3

1

n1

=)2n()1n(n3

1

18

1

n1=

nS9

1

18

1 [using (1)]

n1

=n

n

S18

2S 18S

n

n1 Sn + 2 = 0

Hence, 118S1nn

= | 2 | = 2. Ans.

For objective: Put n = 2 ]

Q.2

[Sol. We have f '(x) =h

)x(f)hx(fLim

0h

f '(x) =h

h

02x2f

2

h2x2f

Lim0h

=

h

2

)0(f)x2(f

2

)h2(f)x2(f

Lim0h

=h2

)0(f)h2(fLim

0h

= f '(0) = k (say)

f '(x) = kf(x) = kx + c ; f(0) = 0 c = 0 f(x) = kx

Now, I =

2

0

2dxxsinkx =

2

0

222dxxsinxsinkx2xk

=

2

0

32

3

xk

2

0

2

0

2dxxsindxxsinxk2 =

2k2

3

8k

32

=

k43

k8 32.

Hence I(k) =

k)4(k

3

8 23 .

This is a quadratic in k and its minimum value occurs when k = 316

34

= 2

4

3

f(x) = x4

32

; f( 42) = 3. Ans.]


29/61

PAGE # 29


Q.1

[Sol. m = 5! 2 5! = 10 4! 5!

n = 4! 5!

Hence m = 10n k = 10 Ans.]

Q.2

[Sol. Let k

zj

yi

xr

; x + y + z 12, y 3 and x, z 1(Give y = 3 and 1 to each x and z, add one extra beggar)

x' + y' + z' + u' = 7 Number of possible r

= 10C

3(Using beggar Method) Ans.]

Q.3

[Sol. Let f() = b f1(b) = f ( + h) = b k f1(b k) = + hf (h) = b + k f1(b + k) = h

x

y

O

y = f(x)

( , f( ))

Now L.H.D. of f1(x) at x = f () = b

(f1) ' (b) =b)kb(

)b(f)kb(fLim

11

0k

=

)(f)h(f

hLim

0h

=

h

)(f)h(f

1Lim

0h =

)('f

1 = r

1

||ly Now R.H.D. of f1(x) at x = f () = b

(f1) ' (b+) =b)kb(

)b(f)kb(fLim

11

0k

=

)(f)h(f

)h(Lim

0h

=

h

)(f)h(f

1Lim

0h

=)('f

1

=l

1

Aliter: Verification by taking an example

y = f (x) =

0xforx2

0xforx3

f ' (0) = 3 = l

f ' (0+) = 2 = r

g (y) = x = f1(y) =

0yfor2

y

0yfor3

y

x

y

O

y = f(x)

y = 3x

y = 2x

g ' (0+) = f ' (0+) =3

1=

l

1

g ' (0) = f ' (0+) =21 =

r1

Note: Ifl and r are positive, then L.H.D and R.H.D of f1 arer

1&

1

land if l and r are negative

then L.H.D. and R.H.D arel

1&

r

1. (Think) Ans.]


30/61PAGE # 30

Q.4

[Sol.

x= 1(0,0)

2( ,0)

X

Y

y = sinx

( /2, 1)y = 1 + cos x

y=x +x 10x3 2

Graph of f(x)

1

Clearly, from above graph, f(x) has local maximum at x =2

and absolute maximum at x = 1. Ans.]

Q.5

[Sol. f(x) =

xsinsin

2xcoscos

211

=

2x3,)2x(4

x2

5

2

3x)2x(

2)x2(

2

2

3x,

4x

2x

2

3xx

2x2

2

x2

,x2

x2

x2

2x0,x

2

222

22

2

2

f(x) =

xsinsin

2xcoscos

2

11

O

y

/2

3 /2 2x

2/4

2/4

f(x) =

2,0xx

2

2

,

f(x) =

x2

x2

f(x) =

2

x2

; x

,2

f(x) = ( x)24

2, < x

2

3,

f(x) =4

2 (2x)2

2

3< x 2

At x = , f(x) is not differentiable


31/61

PAGE # 31

At x = , it is local as well as global minimum.

Range :

4,

4

22

,

I =

2

0

dx)x(f =

2

0

2

dxx2

Using king

I =

2

0

2 dxx =2

0

3

3

x

=

24

3Ans.]

Q.6

[Sol. Given,7

b

13

a =

15

k= k (say)

a = 13k, b = 7k, c = 15 k

Now, cos A = bc2

acb222

= )15()7(2

16922549

= 1052

105

= 2

1

A =A = 3

cos B =ac2

bca 222 =

)15()13(2

49225169 =

26

23

cos C =ab2

cba 222 =

)7()13(2

22549169 =

26

1.

As, cos C < 0,

C is obtuse. So ABC is obtuse.

Also,k35

k26k35s2

a2s2s

as

as

srr

1

=k35k9 =

359

Also, 2s

=

)k35()k35(

4Asinbc2

1

=2k3535

2

3k15k74

2

1

=3535

3105= 35:33

Also, tan

2

CB=

cb

cbcot

2

A= 3

k15k7

k15k7

= 322

8= 3

11

4.

Now, verify alternatives. Ans.]


32/61

PAGE # 32

PART-C

Q.1

[Sol. Equation of S is (x 1)2 + (y 1)2 + (x + y 2) = 0x2 + y2 + x(2) + y(2) + 2(1 ) = 0 ........(1)

S=0

P(1,1) y=2xx2 + y2 + 2x + 2y 2 = 0 ........(2)

Given (1) and (2) are orthogonal, so

122

2)1(2

)2(2

= 2 (1 ) 2 2 + 2 = 2

4 = 4 = 1Hence equation of S is x2 + y2x y = 0.

Now, length of tangent from (2, 2) is 2244 = 2. Ans.]

Q.2

[Sol. Given a1

> 0.

Let d = common difference of A.P.

Now, 3a8

= 513

3 (a1 + 7d) = 5(a1 + 12d) 2a1 + 39d = 0 .......(1)As, a1

> 0 so d < 0.

Now, Sn

= d1na22

n1

= d1nd392

n = n40n

2

d 2 = 40020n2

d 2

Clearly, Sn

will maximum when n = 20. (As d < 0). Ans.]

Q.3

[Sol. Given, y = x + 2 .......(1)

Equation of CD

y 2 = 1 (x 0)

y = 2 x .......(2)

Area of triangle ABC =

168

13

14

3

8

120

2

1

=168

3148

120

6

1

P (0,2)1

(0,2)

y=2x

45 y = 2x + 10

P2

3

14,

3

8

x

y

O

C

DP (8, 6)3

y= 2x+10

m= 1

135

=168

2200120

6

1

= 20486

1 =

3

64

6

168 = .

213

64

. Ans.]


33/61

PAGE # 33

Q.4

[Sol. Equating the components,

3x + 2

y + 4

z = x ; 2

x + 2

z = y & 4

x + 2

y + 3

z = z

hence (3 )x + 2y + 4z = 02x y + 2z = 04x + 2y + (3 )z = 0

for non trivial solution

324

22423

= 0

Use : C1 C

1 C3 and C3 C3 2C2

)1(21

)1(20

02)1(

= 0 ( + 1)2

121

20

021

( + 1)2[1(4) 2(2)] = 0 ( + 1)2 ( 8) = 0Hence = 1 or 8 sum = 7. Ans. ]

Q.5

[Sol. Given, ac2ycoscbxsinbar and cbar = 0

cba

(sin x + cos y + 2) = 0

But cba

0, so sin x + cos y = 2, which is possible when sin x = 1 and cos y = 1

For (x2 + y2) to be minimum, x2 =4

2and y2 = 2.

Hence, the minimum value of 220

(x2 + y2) =

2

2

2 4

20=

4520

2

2= 25. Ans.]

Q.6

[Sol. F(x) =

1x,2

)1(g)1(f

1x,2

)1(g)1(f

,11,x),x(g

1x1),x(f

continuous at x = 1 F(1

) = F(1

+

) = F(1)

f(1) = g(1+) =2

)1(g)1(f

4 + a = 1 + b =2

)b1()a4( =

2

ba5 .

b a = 3. .......(1)continuous at x = 1 F(1) = F(1+) = F(1)

b 1 = 4 a =2

)1b()a4(


34/61

PAGE # 34

b 1 = 4 a =2

ab3

a + b = 5 .......(2)(1) and (2)

a = 1, b = 4 a2 + b2 = 17. Ans.]


Q.1

[Sol. Given, z3 + iz2 = (1 + i) z1 z3z1 = i (z1z2)

12

13

zz

zz

= i

12

13

zz

zz

= 6

i

e

A(z )1

C(z )3B(z )2

/6

| z3z

1| = | z

2z

1| ABC is isosceles and A =

6

. ]

Q.2

[Sol. Urn-IBnWm

1

1 ball1

Urn-IIBnWm

2

2 ball1

Mixed lot item1

P(item belongs to Lot-I) =2

1= P(item belong to Lot-II)

P(item from the mixed lot is W) =

22

2

11

1

nm

m

nm

m

2

1

Ans.]

Q.3

Sol. F(x) = dx)x(P = x2013x20128x2 + 8x + C, where C is constant of integration.F(x) = x(x 1) (x20118) + C

F (0) = F(1) = F (81/2011) = C

F'(x) = 0 has atleast two real roots. (Using Rolle's Theorem)

Note that P(x) = 0 has exactly two real roots in

2011

1

8,0x ]

Q.4

Sol. Equation of normal at P is

(y 1) = 2(x 1) 2x + y = 3

At x =4

1, y =

2

7

0,

4

1

y,

4

1

4

1x

P(1, 1)

C

y

radius =22

2

71

4

11

=4

25

16

25 =

4

55Ans.


35/61

PAGE # 35

Paragraph for Question no. 5 to 7

[Sol. Let f(x) = ax3 + bx2 + cx + d

f(0) = 5 d = 5So, f(x) = ax3 + bx2 + cx + 5

f '(x) = 3ax2 + 2bx + cNow, f '(2) = 0 12a 4b + c = 0 .....(1)

and y = f(x) passes through P (

2, 0), so 0 =

8a + 4b

2c + 5 ......(2)Also, f '(0) = 3 c = 3 .....(3)

On solving, we get a =2

1, b =

4

3

f(x) =23 x

4

3x

2

1

+ 3x + 5

(i)x

y

5

4

3

2

1

1O1

2

52

5

y = 3

Clearly, from above graph, we get, number of solutions of equation |)x(|f = 3 are 4. Ans.

(ii) Equation of normal at Q(0, 5) is

(y

5) = 3

1(x

0) x + 3y = 15. Ans.

(iii) Required area =

1

1

dx)x(f = dx5x3x4

3x

2

11

1

23

= dx5x4

32

1

0

2

=2

19. Ans.]

Q.8

Sol. Put x = tan

f(x) = sin1 2sin

=

20;

2sinsin

02

;2

sinsin

1

1

f(x) =

x0;xtan2

1

0x;xtan2

1

1

1


36/61

PAGE # 36

f '(x) =

x0;)x1(2

1

0x;)x1(2

1

2

2

f '(1) =4

1; f ' (1) =

4

1 (A) is correct

Range of f (x) =

4

,0 . Hence (B) is incorrect.

Also f '(x) is an odd function. (C) is correct

x

)x(fLim

0x

2

1

2

1)0(fand

2

1)0(f,As (D) is incorrect]

Q.9

[Sol. The system will always have a solution (x, y) unless the two lines are parallel.

i.e. m = 2m

1 m = 1 ]

PART-C

Q.1

[Sol. A : outcome of the disease is positive

B1

: He has disease d1

B2

: He has disease d2

3

1)B(P

i

B3

: He has disease d3

P(A / B1

) =

10

8

P(A / B2) =

10

6

P(A / B3) =

10

4

P(B1/A) =

10

4

10

6

10

8

3

110

8

3

1

=18

8=

9

4 (p + q) = 4 + 9 = 13. Ans.

P(B2/A) =

18

6

P(B3/A) =

18

4]


37/61

PAGE # 37

Q.2

[Sol. As, 32 rrn

=212

431

kji

= k5j10i10

3

kj2i2n

d = .nonOPofojectionPr

= 3

kj2i2kj2i3

=

3

146 =

3

9= 3. Ans.

Aliter: Equation of plane OQR is 2x 2y + z = 0

So, distance of P from plane OQR =144

1)2(2)3(2

=

3

9= 3. Ans.]

Q.3

[Sol. Equation of normal at P1

(4 cos 1, 3 sin

1) is

7sin

y3

cos

x4

11

... (1)

Also, equation of CQ1

is

y =

1

1

cos

sinx ... (2)

Solving (1) and (2), we get

1cos

x4

1sin

3

1

1

cos

sinx = 7

y

xC

(0, 0)

P (4 cos , 3sin )1 1 1 x +y =162 2

K1

1

Q1 1 1(4 cos , 4sin )

1

cos

x

= 7 x = 7 cos 1, y = 7 sin 1

So, K1 = (7 cos 1, 7 sin 1) CK1 = 7Similarly, CK2

= CK3

= ..... = CKn

= 7

n

1ii

175CK 7n = 175 n =7

175= 25. Ans.]


Q.1

[Sol. Obviously for c (0, 1), f (x) lies obove the g (x)

also x2 = cx3 x = 0 or x =c

1

hence c1

0

32 dxcxx = 3c12

1

x

y

x2x3

O (0,0) x=1c

or 3c12

1=

3

2 c3 =

8

1 c =

2

1

Hence,

2

c

1

c

1= 2 + 4 = 6 Ans.]


38/61

PAGE # 38

Q.2

Sol. Equation of chord of contact with respect to point (4, 2) is

2a

x4 2

b

y2= 1 and with respect to point (2, 1) is 1

b

y

a

x222

.

Now, according to given condition,

2

2

2

2

b1

a

2

b2

a

4

= 1 4

4

a

b=

4

1

2

2

a

b=

2

1

Now, e =2

3

2

11

a

b1

2

2

Ans.]

Q.3

[Sol. P(E) = P( R R W W B or R R R W W or W W W R R)

=

!2!2

!5 5

3

1+

!2!3

!5 5

3

1+ 5

3!2!3

!5

Simplifying P(E) = 53

50 k = 50 Ans.]

Q.4

Sol. 2

= 12

3

= 22 =

14

Hence

2

321

]

Paragraph for Question no. 5 to 7

[Sol. C1

: z + z = 2 | z 1 |

2x = 2 | x 1 + iy |

x2 = (x 1)2 + y2

y2 = 2x 1 y2 = 2

2

1x

C2

: arg )i1(z =

Curve C2 is a ray emanating from (1, 1) and making an angle from the positive real axis. C

1and C

2has exactly one common point

C2

must be a tangent to C1.

C2

: y + 1 = m(x + 1)

Solving, C1

and C2

y2 = 2

1m

1y 1

0,

2

1

(1,1)

my2 = 2(y + 1 m) m

my22y + 3m 2 = 0


39/61

PAGE # 39

Put D = 0 4 4m (3m 2) = 0

3m22m 1 = 0 (3m + 1) (m 1) = 0 m =3

1, 1.

m =3

1rejected m = 1 (As, a (0, ) )

C2

: y + 1 = x + 1 y = xPutting y = x in the curve C

1

x2 = 2x 1 (x 1)2 = 0 x = 1 P (1, 1)Complex number corresponding to P is z

0= 1 + i

(i) | z0 | = 2

(ii)

From above graph , (x1

+ y1) = 3 + 4 = 7 Ans.

(iii) Area of the shaded region

= 1

2

1

dx1x22 =

1

2

1

2

3

22

3 )1x2(2

(1,1)P(z )0

Q(z ')0

4

4

=3

2[1 0] =

3

2sq. units.]

PART-B

Q.1

[Sol.

(A) The equation of circle passing through A(1, 2), B(2, 3) and having least posssible perimeter, is(x 1) (x 2) + (y 2) (y 3) = 0 (Circle described on AB as diameter)

x2 + y2 3x 5y + 8 = 0 ..........(5)If above equation of circle intersects orthogonally the circle x2 + y2 + 2x + 2ky 26 = 0

then, using condition of orthogonality,

we get

k

2

51

2

32 = 8 26 3 5k = 18 5k = 15

k = 3 Ans.

(B) Using the limit y 0 2y

ycos1 =21

We have,

a

2

20x x

)xcos1cos(1

)xcos1cos(1

xcos1cos(1cos1Lim

=2

1

a

42

20x x

)]xcos1[(

)xcos1(

)xcos1cos(1Lim

= a

84

20x x

x

x

xcos1Lim

8

1


40/61

PAGE # 40

l =a8

0xxLim

128

1

For finite limit 8 a 0 a 8 Ans.

(C) If matrix A is non-singular, so det. (A) 0

1053

842

2p31 0 2(p + 2) + 12 0 p 4

So, p R {4}. Ans.

(D)1t

1y

1t

t

dt

dy

I.F. = (t + 1) et

(t + 1) et y = e t + CPut t = 0 and y = 1

C = 0 At t = 1

2e1 y = e1 y =2

1]

PART-C

Q.1

[Sol. Since it is isosceles. So AB = AC

Now, tan2

=

a

r1

r1

= a tan

2= 2 tan 15 = 2 32

Now r2

= 4r1

2Csin

2Bsin

2Asin 111

I

B C

A

2a = 4

B1 C1

A1

r2

r1

90

90

r2

= 4r1

sin 15 sin 15 sin 60

r2

= 2r1

(2sin2 15) sin 60 = 2r1

(1 cos 30) sin 60 = 2r1 2

3

2

31

= 2 2 32

2

32

2

3= 2323 = 1237

Now, (2r2 + 7r1) = 2 1237 + 7 322 =

24 + 28 = 4. Ans.]

Q.2

Sol. Given,

0

22 6dx)x(gxsin)x(g8

6dx.xsin4)x(gdxxsin16

0

22

0

4


41/61

PAGE # 41

6dxxsin4)x(g16

332

0

22

6dxxsin4)x(g6

0

22

g (x) =4 sin2x Max. of g (x) = 4 sin2 x is 4. Ans.]


Q.1

[Sol. Given,

n

0rr

nC1r

2r=

n

0rr

nC1r

11r

= 2n +

n

0r

rn

1r

C= 2n + S (where S =

n

0r

rn

1r

C)

S =

1n

C........

3

C

2

C

1

Cn

n

2

n

1

n0

n

= 1

0

ndx)x1( =

1n

121n

Now, 2n + S = 2n +1n

1

(2n + 1 1) =

1n

1

13n2n

Now,

1n

123nn

=6

128

n = 5 Ans.]

Q.2

[Sol. As,

sin (cos1x) = sin

xsin2

1= cos (sin1x)

Hence,

1

1

11 dcottan , where = cos (sin1x) =2

(2) = . Ans.]

Q.3

[Sol. Putting (, ) in y2 = 4x, we get = 0, 4

but 0 (Given)So, = 4

P = (4, 4) t1 = 2

P( , )P(4,4)

y2

=4x

S(1,0)

Q(9,6)

(0,0)V

m2

m1

y

x

L(1,2)

Hence t2

= 3

112 t

2ttUsing

Q = (9, 6)

Now m1

=3

4; m

2=

4

3

= 90Hence equation of line passing through (1, 2) and inclined at an angle = 90 is x = 1. ]


42/61

PAGE # 42


Sol.

(i) No American together

A1A

2A

3A

4B, C, D, E

For A1

we have four (B, C, D, E) favourable cases out of total cases 7.

Hence, probability =7

4(say) A

1and B are paired and the remaining

are, A2A

3A

4C D E

for A2, favourable cases 3 out of total cases 5

Hence, probability =5

3say AA

2and C are paired and the remaining are

Now, A3

A4

D, E

for A3, favourable cases 2 out of total 3.

Hence probability =3

2.

P (No two Americans together) =35

8

3

2

5

3

7

4 (B)

(ii) P (delegates of the same country form both pairs)A

1A

2A

3A

4E, B, C, D

= 115

1

7

3

[For A1

we have 3 favourable and of 7 and two A2

only 1 favourable out of 5 are for the remaining, no

constraints.]

EBCD

=35

3(B)

(iii) P (delegates of the same country not forming any pair+ forming both pair + forming exactly one pair) = 1

P (forming exactly one pair) = 1

35

3

35

8=

35

24 (C). Ans.

Alternatively: A1, A2, A3, A4, B, C, D, E

n(S) =!4)!2(

!84 = 105

(i) The probability that no two delegates of the same country are paired =105

CCC1

2

1

3

1

4

=

105

234=

35

8

(ii) The probability that delegates of the same country form two pairs =105

33=

35

3

(As A1, A

2, A

3, A

4can be paired in 3 ways and B, C, D, E can be paired in 3 ways.)

(iii) The probability that exactly two delegates of the same country are paired together

=105

34C2

4 =

35

24Ans.]


43/61

PAGE # 43

PART-B

Q.1

[Sol.

(A) (z + i)2 = (z i)2 or (z i)2 ; if (z + i)2 = (z i)2 4zi = 0 z = 0if (z + i)2 = (z i)2 = i2 (z i)2

z + i = i (z i) or i (z i)

(1 i) z = i2 i = 1 iz = 1.

if z + i = iz + i2

(1 + i) z = 1 i = (1 + i)

z = 1

Hence, z = 0 or 1 or 1 3 solutions Ans.

Aliter: Given, 1iz

iz4

iz

iz= (1)1/4

we get, z = 1, 0,

1 ]

(B) We have,21

t

2

t

2= 1 t1 t2 = 4 ; also t2 = t1

1t

2....(1)

t1t2

= 21

t 2 4 + 2 = 21

t 21

t = 2

Also, 22

t = 21

t + 21

t

4+ 4 squaring (1)

90S(a,0)

P(t )1

Q(t )2

x

y

x+a=0

O

22

t = 2 + 2 + 4 = 8

Now, SQ = a(1 +22t ) = a(1 + 8) = 9a and SP = a(1 +

21t ) = a(1 + 2) = 3a

SP

SQ= 3 SQ = 3SP = 3 Ans.]

(C) On differentiating both sides with respect to x, we get

0 sin2x f(sin x) cos x = cos x

f(sin x) =xsin

12

2)2(2

1f

2

Ans.

(D) Given, 4y2 + 2 cos2x = 4y sin2x 4y24y + 1 + cos2x = 0 (2y 1)2 + cos2x = 0

y =2

1and cos x = 0 x =

2

,

2

3

So, two ordered pairs are possible i.e.,

2

1,

2and

2

1,

2

3Ans.]


44/61

PAGE # 44

PART-C

Q.1

[Sol. Given, Re(z) 2 = | z 7 + 2i |

(x 2)2 = (x 7)2 + (y + 2)2

(y + 2)2 = 10

2

9x

So, the given locus is that of a parabola with directrix

x = 2 and focus (7, 2).

x=2

x

2

P(z1)P(7, 3)

Q(z2)

Q(7,7)

(2, )O S(7,2)92

, 2

y

V

Clearly, minimum PQ = l(L.R.) = 10 Ans.]

Q.2

Sol. An =

2/1

0

n

dxx AAn = )1n(2

11n 2

n

An = )1n(2

1

n

A2n

n

=n)1n(2

1

O

2

1

x

y

Hence,

n

1n

nn

n

A2=

2

1

n

1n 1n

1

n

1=

2

1

1n

11

Given,2

1

1n

11 =

3

1 1

3

2=

1n

1

3

1=

1n

1

n = 2. Ans.]

Q.3

[Sol. We have, (3p2pq + 2q2) wvu

= 0

But, wvu

0 3p2 pq + 2q2 = 0Now divide by q2 (assuming q 0)

3t2 t + 2 = 0, where t =q

pand t R

but D < 0, hence no real values of t

only solution is p = 0 and q = 0Hence exacly one ordered pair of (p , q) i.e. (0, 0).

Aliter: We have, (3p2pq + 2q2) wvu

= 0

But, wvu

0 3p2 pq + 2q2 = 0 2p2 + p2 pq +2

2

q

+

4

q7 2= 0

2p2 +2

2

qp

+4

q72

= 0 p = 0, q = 0, p =2

q

This is possible only when p = 0, q = 0

i.e. exacly one ordered pair of (p , q)]


45/61

PAGE # 45


Q.1

[Sol. Given, f(x) = x3ax2 + 2x, x [0, 2]Now, f ' (x) = 3x22ax + 2

f(2) = (2)34a + 4 = 12 4a and f(0) = 0

Using LMVT, we have

02

)0(f)2(f

2

1'f

2

0)a412(2a

4

3

a4

11 = 6 2a a = 6

4

11 a =

4

13. Ans.]

Q.2

[Sol. Equation of line through O(0, 0, 0) and perpendicular to the plane 2x y z = 4, is

1

0z

1

0y

2

0x

= t (let)

Any point on it is (2t, t, t)

As above point lies on the plane 3x

5y + 2z = 6, so

6t + 5t 2t = 6 9t = 6 t =3

2.

Co-ordinates of point of intersection are

3

2,

3

2,

3

4 (x

0, y

0, z

0) [Given]

Hence, (2x03y0 + z0) = 4 Ans.]

Q.3

[Sol. Given, z + w = 0 ..........(1)

and z

2

+ w

2

= 1 .........(2) Putting w = z from (1) in (2), we get

2z2 = 1 z = 2

1.

For z =2

1, w =

2

1and for z =

2

1, w =

2

1.

So, from both possibility, we get

2wz . Ans.]

Q.4[Sol.

We must have |2 4| < 5 5 < 2 4 < 51 < ( 2)2 < 9 0 ( 2)2 < 9 3 < 2 < 3

D(z)

2 F1F2

1 < < 5 0 < < 5 = 1, 2, 3, 4 4 values. ][Note : = 0, 4 is not possible because, | z 2 | + | z 4| = 5 will represent a circle.]


46/61

PAGE # 46


[Sol.mb

Given, f(x) = x2 ex f '(x) = x ex (2 x)

+

0 2

Sign scheme of f '(x) Ox

y

4

1y

2e

4y,2x

x = 2

Graph of f(x) = x e2 x

As, g(x) =

xe2

02

dtt1

)t('f g'(x) =

)e41(

e2)e2('fx2

xx

=

)e41(

)e1(ee8x2

xe2x2 x

.

Now, verify alternatives. ]

Q.8

[Sol.

(A) P(A/B) = P(B/A) P(A) = P(B)

Now, P(A B) = P(A) + P(B) P(A B) P(A B) = P(A) + P(B) P(A B)

P(A B) = 2P(A) 1 > 0; P(A) >2

1 True

(B) P(B) =4

3; P(A/ B) =

2

1

)B(P

)BA(P =

2

1 P(A B) =

4

3

2

1=

8

3

Now, P(A) + P(B) P(A B)= P(A B) 1

P(A) +43

83 1

P(A) 8

5

P(A)]max.

=8

5 True

(C) P(AC BC)C = C4443)aa()aa( = (a4)

C = a1

+ a2

+ a3

= P(A + B)

P(AC BC) = (a1

+ a3

+ a4)C = a

2= P(AB)

Hence P(AC BC) + P(AC BC) = P(A) + P(B)] = 6 233121

= 65 .

(D) Given, A is subset of B.

Now, P(B/A) =)A(P

)AB(P =

)A(P

)A(P= 1.

S

AB

]


47/61

PAGE # 47

Q.9

[Sol. As, 2213 rrrr

= 122221 rrrrrr

= )k3ji(6)kji2(6 = k12i6

Row 3 vector 556

)k2i(6 = k2i or k2i

A =

201112

311

or

201112

311

Tr.(A) = 0, 4.

Also 3232 rrrr = |rr||r||r| 3232 = 5656 = 30.

Since 321 randr,r

are coplanar they are linearly dependent.

133221rrrrrr

= 2321rrr

= 0]

PART-C

Q.1

[Sol. Clearly, the given curves intersect at x = 0,

1k

k2

Required area = dxk

xkxx

1k

k

0

22

2

= 2

k

1k6

1

Now, for maximum area,2

k

1k

will be minimum k +k

1= 2 k = 1. Ans.]

Q.2[Sol. Let = 3h 2, = 3k

3

2= h and

3

= k

As, (h, k) lies on given circle, so

h2 + k22h 4k4 = 0

43

4)2(

3

2

99

)2(22

2 + 2 2 12 = 44 ( 1)2 + ( 6)2 = 44 + 37 Locus of (, ) is (x 1)2 + (y 6)2 = 81,which represents circle whose radius = 9.]


48/61

PAGE # 48

Q.3

[Sol. Let f n(x) =

x2

x

t dten

)x('fn =

nn x)x2(ee2

For maxima and minima, )x(f'n

= 0 nn x)x2( ee2

nnn x)x2( ee2

Taking log on both sides, we get

ln 2 2nxn = xn ln 2 = xn (2n 1) xn =12

2nn

l x =

n

1

n12

2n

l= a

n

Also, f n''

n

1

n 12

2nx

l< 0 f

n(x) is maximum at x =

n

1

n12

2n

l.

Now, ln an

=n

12

2nn

n

ll

=n

)12(n)2n(nn

lll

Hence L = )a(nLim nn

l

=n

)12(n)2n(nLim

n

n

lll=

n

2

112n

n

)2n(nLim

n

n

n

lll

L =

n2

11n2nn

0Lim

n

n

ll

=

ln 2

Hence, eL = 2 Ans.]

Q.4

[Sol.mb

We know that | adj. A1 | = | A1 |2 = 2|A|

1

11A.adj.det =|A.adj|

11 = | A ||

2 = 22 = 4 Ans.]


49/61

PAGE # 49


[Sol. The equation of the tangent is )i..(..........12

3

b

y

2

1

a

x

Auxiliary circle is x2 + y2 = a2.................(ii)

C is the centre.

Combined equation of CL, CM is obtained by homgenising (ii) with (i), i.e.,

x2 + y2a2 0b2y3

a2x

2

Since LCM = 90

1 4

1+1 0

b4

a32

2

4

7

b4

a32

2

90

M

P( =60)

(0, 0)C

y

x

L

7b2 = 3a2 7 a2 (1e2) = 3a2

Hence e =7

2Ans. ]

Q.2

[Sol. Originally the number of white balls in the bag vary from 0 to 100.

P(B0) = P(B

1) = P(B

100) =

101

1

ballW100hasBag:B

ballW2hasBag:BballW1hasBag:B

ballWnohasBag:Binitially

100

2

1

0

A

B0 B1

B2 B100

After the white has been dropped in the bag

A = A B0

+ A B1

+ ........+ A B100

P(A) = P(B0) P(A/B0) + P(B1) + P(A/B2) + ...... + P(B100) P(A/B100)

=101

1

101

101......

101

2

101

1=

101

51

2

102101

101

1

101

1

. Ans.]

Q.3

[Sol. From above figure,

P(C (A B)') = 1

5

1

15

1

10

1 BA

S

C=

30

62330 =

30

19Ans.

Q.4

[Sol. (a + b + c2) (a + b2 + c) = | a + b + c2 |2 = (a2 + b2 + c2ab bc ca) (a b)2 + (b c)2 + (c a)2 = 2 (a = b, | b c | = 1) or (b = c, | a c | = 1) or (c = a, | a b | = 1)| b c | = 1

(b, c) = {(2, 1), (1, 2), (3, 2), (2, 3), (4, 3), (3, 4), (5, 4), (4, 5), (6, 5), (5, 6)}

Required probability =666

103

=36

5Ans.]


50/61

PAGE # 50


[Sol. Let dr's of line L be , so equation of line L is

kcjbiatr

......(1)

If L intersects L1

at P, so shortest distance between them is zero.

0 =

kj2ikcjbiakj2ikcjbiakji2

3

4,

3

10,

3

10

Q

(5, 5, 2)P

0O

L : r =2

L : r =1

8i3

3j + k

+ (2i j+ k)

(2i + j k)

+ (i 2j + k)

121

cba

kji

kji2

= 0 a + 3b + 5c = 0 .......(2)

Similarly, L intersects L2

at Q, so shortest distance between them is zero.

0 =

kji2kcjbia

kji2kcjbiakj33

i8

112cba

kji

k

j

33

i8

= 0

3a + b 5c = 0 .......(3)

On solving (2) and (3), we get2

c

5

b

5

a

.

So, the equation of line L is k2j5i5tr

.

Any point on line L is A' (5t, 5t, 2t). If A' is the point of intersection of L and L1, so A' will also satisfy

L1, we get 1

1t2

2

1t5

1

2t5

2 (5t 2) = 5t 1 10t + 4 = 5t 1 t = 1So, co-ordinates of P are (5, 5, 2).

Similarly, if A' (5t, 5t, 2t) is the point of intersection of L and L2, so A' will also satisfy L2, we get

1

1t2

1

3t5

23

8t5

5t + 3 = 1 2t t =

3

2

So, co-ordinates of Q are

3

4,

3

10,

3

10

(i) Given M (1, 2, 3) and N (2 + , 1 2, 1)

MN = P.v of N P.v ot M = k)4(j)21(i)1(

Also, n

= normal vector of plane k3j4ir

= k3j4i .

Now, nMN

= 0 1 ( + 1) + 4 (1 + 2) + 3 ( 4) = 0

12 = 7 =12

7 Option (B) is correct.


51/61

PAGE # 51

(ii) The normal vector of the plane through P (5, 5, 2) and Q

3

4,

3

10,

3

10and perpendicular to the

plane kjir

+ 1 = 0 is parallel to the vector = kjiPQn

=

111

3

2

3

5

3

5kji

=

111

255

kji

3

1

= k0j3i3

3

1 = ji

The required equation of plane, is 1 (x 5) + 1 (y + 5) + 0 (z 2) = 0

x + y = 0 or jir

= 0 Option (D) is correct

(iii) Volume of tetrahedron OPAB (where O is origin) = OBOAOP6

= |

502

310

255

|6

1

=6

1[5(5) + 5(6) + 2(2)] =

6

51=

2

17. Option (C) is correct.]

Q.8

[Sol. S = {HTH, THH, TTH, HHH, HTT, THT, TTT, HHT}

A = {HTH, HHH, HTT, HHT}

B = {HTH, TTH, HTT, TTT}

C = {HTH, THH, TTH, HHH}

D = {TTH, HTT, THT}

E = {HHH, TTT}

A B C = {HTH}

As, P(C) =2

1, P(D) =

8

3, P(E) =

4

1

2 P(D) =4

3= P(C) + P(E) (A) is correct

Also, A B C = {HTH, THH, TTH, HHH, HTT, TTT, HHT} A B C S, as THT is not included in A B C. A, B, C are not exhaustive (B) is incorrect

Also, P(A B C) =8

1P(A) P(B) P(C)

A, B, C are independent events. (C) is correct

Note that, P(A) = 2

1= P(B) = P(C)

A,B,C are equally likely. (D) is correct Ans.]

Q.9

[Sol. Given, + = p, = q ;1

+ = p1 ,

= q1

; and +1

= p2 ,

= q2

Now, q1

q2

=

= 1. and p1

+ p2

=

1

=

)1()(=

q

p(q + 1)


52/61

PAGE # 52

Also, qq2

=

= 2, as is the root of x2 + px + q = 0.

So, 2 + p + q = 0 qqpqq 22 = 0

Note that (p1 p2)

2 = ()2 2

2

)(

)1(

= (p2 4q)

q

)1q( 2

(qp1 qp

2)2 = (p24q) (q + 1)2. Ans.]

PART-C

Q.1

[Sol. Let c = 14 x, a = 14, b = 14 + x

Now, cos B =13

5 (14 + x)2 = (14 x)2 + 1422(14 x) 14

13

5 x = 1.

So, a = 14, b = 15, c = 13.

Hence, r =s

=

21

84= 4. Ans.]

Q.2Sol. Sn

= C0C

1+ C

1C

2+ ... + C

n1C

n

Sn

= nC0

.nCn1

+ nC1

.nCn2

+ ... + nCn1

.nC0

= 2nCn1

Sn

= 2nCn1

Now,n

1n

S

S =

4

15

1nn2

n2n2

C

C

=4

15

4

15

!n2

)!1n()!1n(

)!2n(!n

)!2n2(

n)2n(

)1n2)(1n(

=8

158(2n2 + 3n + 1) = 15n2 + 30n

16n2 + 24n + 8 = 15n2 + 30n n26n + 8 = 0 (n 4) (n 2) = 0 n = 2 or 4

Sum of all values of n = 6 ]

Q.3

Sol. Given equation of planes are

P1

: x + y + 1 = 1 ... (1)

P2

: x + 2ay + z = 2 ... (2)

P3

: ax + a2y + z = 3 ... (3)

If 3 planes intersect in a line, then

01aa1a21

111

2

Applying C1C

1C

2and C

2C

2C

3

011aaa

11a2a21

100

22


53/61

PAGE # 53

On expanding along C1, we get

(1 2a) (a21) (a a2)(2a 1) = 0 1 + 3a 2a2 = 0 a =2

1, 1.

But, for a = 1, planes P1

and P3

are parallel and for a =2

1, planes P

1and P

2are parallel.

Hence, we conclude that 3 planes will never intersect in a line for any real value of a.


Q.1

Sol. Given, tan1 x + tan1 y + tan1(xy) =12

11...(1)

At x = 1;4

+ 2 tan1 y =

12

11 2 tan1 y =

3

2 y = 3

Differentiate both sides of equation (1) with respect to x, we get

2x1

1

+ 2y1

1

y ' + 2)xy(1

1

(xy' + y) = 0

2

1+

4

1y' +

4

1(y' + 3 ) = 0

2

1+

4

3+

2

1y' = 0 2 + 3 + 2y' = 0

So, y' = 1 2

3. Ans.]

Q.2

[Sol. As, f(0+) = f(0) = f(0) = 0, so f(x) is continuous at x = 0.

Further, f(0 + h) > f(0) and f(0 h) > f(0) where h is sufficiently small positive quantity.

Hence, f(x) has local minimum at x = 0. ]

Q.3

[Sol. We have, dyy

dxydyx2

y

xd = dy

y

x= y + C

As, y(1) = 1 C = 2 y

x= y + 2

Now, y() = 3 3

= 5 = 15. Ans.]

Q.4

[Sol. Given, f (x) =xtan1

9xtan4xtan2

2

=

xtan1

)xtan2(22

+

xtan1

xtan14

2

2

+ 5 = 2 sin 2x + 4 cos 2x + 5

Rf= 205,205

Hence, (M + m) = 10 Ans.


54/61

PAGE # 54

Aliter : f(x) =xtan1

xtan2

2

+

xtan1

xtan42

2

+

xtan1

92

= sin2x + 2 sin 2x + 9 cos2x

= 1 + 4(1 + cos 2x) + 2 sin 2x = 5 + 2 sin 2x + 4 cos 2x. ]


Sol. For the given ellipse, 116

y

25

x22

, 5

3

25

16

1e . So, eccentricity of hyperbola = 3

5

.

Let the hyperbola be, 1B

y

A

x2

2

2

2

... (1)

Then, B2 = A2

19

25=

9

16A2. Also, foci of ellipse are (3, 0).

As, hyperbola passes through (3, 0). So, 2A

9= 1 A2 = 9, B2 = 16

Equation of hyperbola is 116y

9x

22

(i) Vertices of hyperbola are (3, 0) (A) is correct.Focal length of hyperbola = 10 (B) is incorrect.

Equation of directrices of hyperbola are x = 5

9. (C) is incorrect.

(ii) Any point of hyperbola is P(3sec, 4tan).Equation of auxiliary circle of ellipse is x2 + y2 = 25.

Equation of chord of contact to the circle x2 + y2 = 25, with respect to P(3 sec, 4 tan), is3x sec + 4y tan = 25 ... (1)

If (h, k) is the mid point of chord of contact, then its equation is

hx + ky 25 = h2 + k225 hx + ky = h2 + k2 ... (2)As, equations (1) and (2) represent the same straight line, so on comparing, we get

22kh

25

k

tan4

h

sec3

sec =

22 kh

25.

3

h, tan =

4

k

kh

2522

Eliminating , we get,2

22kh

25

16

k

9

h22

= 1. (As, sec2tan2 = 1)

Locus of (h, k) is

22222

25

yx

16

y

9

x

(iii) Required area of quadrilateral = 2(A2 + B2) = 2(9 + 16) = 50 Ans.]


55/61

PAGE # 55

Q.8

Sol. S = {1, 2, 3, ... n}

E1

= No. is divisible by 2.

E2

= No. is divisible by 3.

If n = 6k say n = 6

S = {1, 2, 3, 4, 5, 6}

P(E1) =

6

3=

2

1; P(E

2) =

6

2=

3

1

P(E1 E

2) =

6

1= P(E

1).P(E

2) (B) is correct.

If n = 6k + 2 say n = 8

S = {1, 2, 3, 4, 5, 6, 7, 8}

P(E1) =

8

4=

2

1; P(E

2) =

8

2=

4

1

Here, P(E1E

2) =

8

1= P(E

1) P(E

2) (C) is correct.

Note that : P(E1 E2) = 101

P(E1) P(E2)

Not independent dependent. ]

Q.9

[Sol. 2xydx

dy= x2 + y2 + 1

Put y2 = t

dx

dtx = x2 + t + 1

dx

dt

x

t=

x

1x2

I.F. =x

1

Hencex

t=

dx

x

1x2

2

= x x

1+ C

y2 = x21, C = 0 as y(1) = 0

Now, y (x0) = 3 3 = 20x 1

20

x = 4

x0

= 2 Ans.

Alternatively: We have, 2xydy = (x2 + 1)dx + y2dx

2

2

x

dxydyxy2 = dx

x

11

2

x

1xd

x

yd

2


56/61

PAGE # 56

We get, cx

1x

x

y2

As, y (1) = 0 c = 0 y2 = x21

Now, y (x0) = 3 3 = 20x 1

20

x = 4

x0 = 2 Ans.]

PART-C

Q.1

[Sol.

P(HHH) =

33

6

1

3

11

18

5

2

1

18

13

=

8

1

18

5

8

1

18

13 =

8

1 p + q = 9 Ans.]

Q.2

[Sol. Clearly, d =22

4

3

2

1

2

1

4

3

=16

1

16

1 =

8

1=

22

1

So, | sin | = d22 | sin | = 1

y

x

0,

2

1

2

1,0

4

3,

2

1

2

1,

4

3

y=

x

O

=2

,

2

,

2

3.

Hence, the number of values of are 3.]

Q.3

[Sol. Given,dx

dyx2y = x4 y2 or y

x

2

dx

dy = x3y2. Now dividing by y2, we get

x

1

y

2

dx

dy

y

12

= x3 ........(i)

This is a Bernouli's differential equation, substituting t

y

2

, we get

dx

dt

dx

dy

y

22

. So, equation (i) becomes

x

t

dx

dt

2

1 = x3 t

x

2

dx

dt = 2x3

IF =2xnxn2

dxx

2

xeee2

ll


57/61

PAGE # 57

So, general solution is given by

x2t =6

x26

+ C y

2x2=

3

x6

+ C y

2=

3

x4

+ 2x

C

If x = 1, y = 6 C = 0

3

x

y

2 4

y = 4x

6i.e. f(x) = 4x

6

Now,dxdy

= 24x5 = 5x

24. Hence,

5

1

3xdxdy

=

324

= 8. Ans.]

Q.4

[Sol. Given, y = x2

Now,axdx

dy

=

axx2 = 2a

Equation of tangent is (y a2) = 2a (x a) BOx

y

A (a, a )2

0,2

a

Put y = 0, we get

x = a 2

a=

2

a.

Also, equation of OA is y = ax

Area = a

0

2dx)xax( = k

2

a

2

a2

a

0

32

3

x

2

ax

=

4

ka3

3

a

2

a33

=4

ka3

6

a3

=4

ka3

k =3

2

q

p(Given) (p + q)

least= 5. Ans.]


Q.1

[Sol. Tangent to the parabola

y2 = 4x is y = mx +m

1....(i)

m2x my + 1 = 0,

As, it touches the circle x2 + y2 = 1, soS(1,0)O

y

x

1

mm

1

24

m4 + m21 = 0

m2 = tan2 =2

411 =

4

152

2

15= 2 sin 18 Ans.]


58/61

PAGE # 58

Q.2

Sol. Given, 3

2 III

7dx)x(''f)x3(

32)x('f)x3( +

3

2

7dx)x('f

0 (f '(2)) + f(3) f(2) + 7 f(3) = f '(2) + f(2) + 7 = 4 + (1) + 7 = 10. Ans.

Q.3

[Sol. Let A =

333231

232221

131211

aaa

aaa

aaa

, where aij {0, 1}

As, trace A = 1

So, any two elements in main diagonal are 0 and one element is 1. Also, non-diagonal elements an be 0

or 1.

So, number of matrices = 3 23 = 3 8 = 24. Ans.]

Q.4

Sol. As, z lies on the curve arg(z + i) =4

, which is a ray originating from (i) and lying right side of

imaginary axis making an angle4

with the real axis in anticlockwise sense.

O45

Re(z)

Im(z)

(4+3i)

(43i)

The value of | z(4 + 3i) | + | z (4 3i) | will be minimum when z, 4 + 3i, 4 3i are collinear. Minimum value = distance between (4 + 3i) and (4 3i)

= 22 )33()44( = 3664 = 100 = 10. Ans.]

Q.5

[Sol. Statement-1: We have baxa

0bxa

atbx

, for some scalar t.

atbx

= k)t31(j)t21(i)t2(

Now, xa

= 0 t = 21

2

k

2

i3x

x

=4

1

4

9 =

2

5 Statement-1 is false.

Obviously, Statement-2 is true. ]


59/61

PAGE # 59

Q.6

[Sol. Option (B) is true.

S-1: As, f ' (1+) = 0 = f '(11) f is differentiable at x = 1.So, f is differentaible x R.

S-2: As, f(1 + h) < f(1) < f (1 h).

Where h is sufficienty small positive quantity, so f(x) is decreasing at x = 1.

f(x) has neither local maximum nor local minimum at x = 1.But, S-2 is not explaining S-1. Ans.]

Q.7[Sol. Given, = 2 tan = tan 2

tan =

2tan1

tan2

2x

y

0

0

=

202

0

00

y1x

1xy2

20

20

0

0 y)1x(

)1x(2

2x

1

3x02 y

02 = 3

(1,0) (2,0)x

y

M(x , y )0 0

(0,0)A B

Locus of M is hyperbola 3x2y2 = 3Now, verify alternatives. Ans.]

Q.8

[Sol. Given, f(x) =

0x,0

1x0,)x2(cos1xx1

Note: f (x) is not defined at x = 1 for 0 ......(1)f(x) is continuous at x = 0 but for continuous at x = 1.

f (1) = f(1)

2 hsinhLim 20h

= 0 20h

hLim

= 0

+ 2 > 0 > 2 ......(2)Hence, (1) (2) 0

Note: f (x) is differentiable in (0, 1). Ans.]

Q.9

[Sol. Now, slope of LM =0t

2

= t (given)

2 = t2 = 2 + t2

Let mid point of LM is (h, k)

(h, k)

y

x

slope = m = t(given)

(0, 0)

M(0, ) L (t, 2)

Now, 2h = t and 2k = + 2 = t2 + 4 On eliminating t, we get

k = 2(h2 + 1)

Locus of (h, k) is y = 2(x2 + 1) x2 =2

1(y 2)

Now, verify alternative. Ans.]


60/61

PAGE # 60

PART-B

Q.1

[Sol.

(A) Clearly, ABC is equilateral.

Now, ar. (ABC) =2

zz4

3

=2

z4

33= 348 (Given)

B( z)

A(z)

Re(z)

C( z) 2

2 /3

2 /3

Im(z)

O(0,0)

64z2

8z . Ans.]

(B) Equation of chord of hyperbola 11

y

2

x22

, whose mid-point is (h, k) is

2

hxky =

2

h2

1

k2

(using T = S1)

As, it is tangent to the circle x2 + y2 = 4, so

22

22

k4

h

k2

h

= 2

2

22

22

k4

h4k

2

h

Locus of (h, k) is (x22y2)2 = 4(x2 + 4y2) = 4.

(C) We know that ac = (semi-minor axis)2 = 4

Now,

4

4

dx}x2{ = 2/1

0

dx}x2{16 =

2

18

2

18

dx}x2{ =

2

1

0

dx}x2{16 =

2

1

0

dxx216 =

2

1

0

dxx32

= 322

1

0

2

2

x

= 32

8

1= 4 Ans. ]

PART-C

Q.1[Sol Given, A = (I + B) (I B)1

Now, AT

= T1

)BI(BI

= T1

)BI(

(I + B)T

= 1T)BI( (I + BT) = (I + B)1 (I B)Also, AAT = (I + B) (I B)1 (I + B)1 (I B)

= (I + B) ((I + B) (I B))1(I B) = (I + B) ((I B) (I + B))1 (I B)= (I + B) (I + B)1 (I B)1 (I B) = I I = I2 = I.

| AAT | = | I | = 1 | A |2 = 1As, | A | > 0 | A | = 1.Hence, det.(2A) det. (adj A) = 8 det.( A) (det.( A))2 = 8(1) (1)2 = 7 Ans.]


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Q.2

[Sol. p = sinv

=|c|

cv

=

|c|

cv|c|v222

=3

4)3)(6( =

3

14

B(3,5,2) M Q

A(2,3,1)v=(1, 2, 1)

c=(1, 1, 1)

r

r r

r

= v ^ cp

30p2 = (30) 314

= 140 Ans.]

Q.3

[Sol. 2

4

5

2

0

4

1

0

4

5

4

1

dxxcosdxxsindxxcosdx)x(F

=

2

4

54

5

4

14

1

0

xsinxcosxsin

y

xO 1/4

5/4 2

=

2

1

2

2

2

1=

22

10

10

2

024

10dx)x(F

24

10dx)x(F

24

22

= 5. Ans.]

Rt Solutions-Practice Test Papers 1 to 14 Sol

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Transcript of Rt Solutions-Practice Test Papers 1 to 14 Sol