solution-for-wb-jee-2012 Maths (Page 2 - 20)

8/2/2019 solution-for-wb-jee-2012 Maths (Page 2 - 20)

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Q. 1 Q. 60 carry one mark each

1. The maximum value of |z| when the complex number z satisfies the condition2

z 2z

is

(A) 3 (B) 3 2 (C) 3 1 (D) 3 1

Ans : (C)

Hints :2 2 2

z | z | | z | 2z | z | | z |

2| z | 2 | z | 2 0 | z | 3 1

2. If

50

253 3i 3 x iy2 2

, where x and y are real, then the ordered pair (x, y) is

(A) (3, 0) (B) (0, 3) (C) (0, 3) (D)1 3

,2 2

Ans : (D)

Hints :

50 50503 3 3 i

i 32 2 2 2

50253 Cos iSin

6 625 50 503 Cos iSin

6 6

25 1 33 i2 2

3. Ifz 1

z 1is purely imaginary, then

(A)1

| z |2

(B) | z | 1 (C) |z| =2 (D) |z|=3

Ans : (B)

Hints : Letz 1

iz 1

, such that R .

On taking Componendo and Dividendo

1 iz

1 i;

1 iz 1

1 i

Code-

WBJEE - 2012 (Answers & Hints) Mathematics

Regd. Office:Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-110075 Ph.: 011-47623456 Fax : 011-47623472

(1)

ANSWERS & HINTSfor

WBJEE - 2012

SUB : MATHEMATICS


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4. There are 100 students in a class. In an examination, 50 of them failed in Mathematics, 45 failed in Physics, 40 failedin Biology and 32 failed in exactly two of the three subjects. Only one student passed in all the subjects. Then thenumber of students failing in all the three subjects

(A) is 12 (B) is 4

(C) is 2 (D) cannot be determined from given information

Ans : (C)Hints : n(M P B) n(M) n(P) n(B) n(M P) n(PMB) n(B P) n(M P B)

given n(M) = 50 , n (P) = 45, n(B) = 40; n(M P) n(P B) n(B M) 3(M P B) 32

99 = 50 + 45 + 40 (32 + 3 n (M P B) )+ n(M P B) ; 2n(M P B) = 36 32; n(M P B) = 2

5. A vehicle registration number consists of 2 letters of English alphabets followed by 4 digits, where the first digit is notzero. Then the total number of vehicles with distinct registration numbers is

(A) 2 426 10 (B)26 10

2 4P P (C)26 10

2 3P 9 P (D)2 326 9 10

Ans : (D)

Hints :26 26 9 1010 10

. Total No. of ways = 262 9 103

6. The number of words that can be written using all the letters of the word IRRATIONAL is

(A) 310!

(2!)(B) 2

10!

(2!)(C)

10!

2!(D) 10!

Ans : (A)

Hints : IRRATIONAL. There are 2 I, 2 R, 2 A, one T, one O, One N, One L. Total Number of words = 310!

(2!)

7. Four speakers will address a meeting where speaker Q will always speak after speaker P. Then the number of waysin which the order of speakers can be prepared is

(A) 256 (B) 128 (C) 24 (D) 12

Ans : (D)

Hints : The number of ways in which speaker P and Q can deliver their speach is 4C2.

Total number of ways = 4C2 2! = 12

8. The number of diagonals in a regular polygon of 100 sides is

(A) 4950 (B) 4850 (C) 4750 (D) 4650

Ans : (B)

Hints : Number of Diagonals in n sided Polygon is equal to nC2 n. Total number of Diagonals = 100C

2 100=4850

9. Let the coefficients of powers of x in the 2nd, 3rd and 4th terms in the expansion of (1+x)n, where n is a positive integer,be in arithmetic progression. Then the sum of the coefficients of odd powers of x in the expansion is

(A) 32 (B) 64 (C) 128 (D) 256

Ans : (B)

Hints : Co-efficient of 2nd, 3rd and 4th terms are nC1, nC

2, nC

3. It is given 2.nC

2= nC

2+ nC

3n = 7.

Sum of Co-efficient of odd power of x = 26


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10. Let f(x) = ax2 + bx + c, g(x) = px2 + qx + r, such that f(1) = g(1), f(2) = g(2) and f(3) g(3) = 2. Then f(4) g(4) is

(A) 4 (B) 5 (C) 6 (D) 7

Ans : (C)

Hints : Let a p = w, b q = y and c r = z

f(1) = g(1) w + y + z = 0; f(2) = g(2) 4w + 2y + z = 0; f(3) = g(3) + 2 9w + 2y + z = 2

w = 1, y = 3, z = 2

f(4) g(4) = 16 w + 4y + z = 6

11. The sum 1 1! + 2 2! + .... + 50 50! equals

(A) 51! (B) 51! 1 (C) 51! + 1 (D) 2

Ans : (B)

Hints : Let

50

r 1

S r.r ! ;

50

r 1

(r 1 1)r! ;

50

r 1

(r 1)! r ! ; 51! 1

12. Six numbers are in A.P. such that their sum is 3. The first term is 4 times the third term. Then the fifth term is

(A) 15 (B) 3 (C) 9 (D) 4

Ans : (D)Hints : Let 5d, 3d, d, + d, + 3d, + 5d are six numbers in A.P.

6 = 3 =1

2; ( 5d) = 4 ( d) d =

3

2. Hence

1 33d 3 4

2 2

13. The sum of the infinite series1 1.3 1.3.5 1.3.5.7

1 ....3 3.6 3.6.9 3.6.9.12

is equal to

(A) 2 (B) 3 (C)3

2(D)

1

3

Ans : (B)

Hints : Letn 1 1.3 1.3.5 1.3.5.71 x 1 ....

3 3.6 3.6.9 3.6.9.12

21 n(n 1) 1.3 2 1nx , x x ,n3 2 3.6 3 2

14. The equations x2 +x +a=0 and x2 +ax +1=0 have a common real root

(A) for no value of a (B) for exactly one value of a

(C) for exactly two values of a (D) for exactly three values of a

Ans : (B)

Hints : x2 + x + a = 0 ..............(A); x2 + ax +1 = 0 ..............(B) (A) (B);

(1a)x + (a1) = 0; (1a) (x1) = 0 if a 1, x = 1, so, only solution is a = 2

15. If 64, 27, 36 are the Pth , Qth and Rth terms of a G.P., then P + 2Q is equal to

(A) R (B) 2R (C) 3 R (D) 4 R

Ans : (C)

Hints : tP

= 64, a . rP1 = 64.............(1); tQ

= 27, a . rQ1 = 27.............(2)

tR

= 36, a . rR1 = 36.............(3)

(2)2 (1)(3)3 ; 2 Q + P = 3R


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16. If and are the roots of the equation x2 + px + q = 0 where , , p and q are real, then the roots of

the equation (p2 4q) (p2x2 + 4 px) 16 q = 0 are

(A)1 1 1 1

and (B)1 1 1 1

and

(C)1 1 1 1

and (D) and

Ans : (A)

Hints : 2 = P, = q; so the given equation is 4 (4 2x2 8 x) 16 + 16 =0

2 2 2x (2 )x ( ) 0 ;1 1

x

17. The number of solutions of the equation log2(x2 + 2x 1) = 1 is

(A) 0 (B) 1 (C) 2 (D) 3

Ans : (C)

Hints : x2 + 2x 1 = 2 x2 + 2x 3=0 (x + 3) (x 1) = 0; x = 1, 3

18. The sum of the seriesn n n

1 2 n

1 1 11 C C ..... C

2 3 n 1is equal to

(A)n 12 1

n 1(B)

n3 2 1

2n(C)

n2 1

n 1(D)

n2 1

2n

Ans : (A)

Hints : (1+x)n = C0

+ C1x + C

2x2 + ...............+ C

nxn; Integrating

1 1n 2 n0 1 2 n0 0

(1 x) dx (C C x C x ..... C x )dx ;

n 131 2 n

0

CC C C2 1C ..........

n 1 n 1 2 2 3 n 1

19. The value ofr 2

1 2 .... (r 1)

r !is equal to

(A) e (B) 2e (C)e

2(D)

3e

2

Ans : (C)

Hints :r 2 r 2

r(r 1) 1 1 1 1 1........

2 r 2 r 2 2 0 1 2e

2

20. If

1 2 1P

1 3 1 , Q = PP

T

, then the value of the determinant of Q is equal to

(A) 2 (B) 2 (C) 1 (D) 0

Ans : (A)

Hints :T

1 11 2 1 6 8

PP 2 31 3 1 8 11

1 1. So det (PPT) = 6664=2.


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21. The remainder obtained when 1!+2!+...+95! is divided by 15 is

(A) 14 (B) 3

(C) 1 (D) 0

Ans : (B)

Hints : Starting from 5! all the other numbers are divisible by 15 since 15 will be a factor

So 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33

So the remainder on dividing 33 by 15 is 3.

22. If P,Q,R are angles of triangle PQR, then the value of

1 cosR cosQ

cosR 1 cosP

cosQ cosP 1is equal to

(A) -1 (B) 0 (C)1

2(D) 1

Ans : (B)

Hints : On expanding the determinant,

= 1 + cos2p + cos2Q + cos2R + 2cosP cosQ cosR

= 1 cos2 P + cos2Q + cos2Q + cos2R + (cos(P+Q) + cos(P Q)) cosR

now P + Q = R

So = 1+ cos2P + cos2Q + cos2R cos2R cos(P Q) cos(P + Q)

= 1 + cos2P + cos2Q cos2P + sin2Q

= 1 + 1 = 0

23. The number of real values of for which the system of equations x + 3y+5z = x

5x+y+3z = y

3x+5y+z = z

has infinite number of solutions is

(A) 1 (B) 2 (C) 4 (D) 6

Ans : (A)

Hints :

x(1 ) 3y 5z 0

5x (1 )y 3z 0

3x 5y (1 )z 0

So for infinite solution,

1 3 5

5 1 3 0

3 5 1

1 1 2 3

9 3 5

9 1 3 0 C C C C

9 5 1

making two zeroes,

0 2 2

0 4 2 0

9 5 1

(9 ) ( 2 + 4 + 4 + 2 + 8) = 0

(9 ) ( 2 + 6 + 12) = 0

= 9

24. The total number of injections (one-one into mappings) from 1 2 3 4a ,a ,a ,a to 1 2 3 4 5 6 7b ,b ,b ,b ,b ,b ,b is

(A) 400 (B) 420 (C) 800 (D) 840

Ans : (D)

Hints : So total = 7 6 5 4 = 840 one-one into functions


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25. Let

1010 r

r

r 0

(1 x) c x and

77 r

r

r 0

(1 x) d x If5 3

2r 2r 1

r 0 r 0

P c and Q d , thenP

Qis equal to

(A) 4 (B) 8 (C) 16 (D) 32

Ans : (B)

Hints :

10 10 10 10 10 10

0 2 4 6 8 10P C C C C C C= 2101 = 29

Q = 7 7 7 7 7 1 61 3 5 7C C C C 2 2

So9

6

2P 8Q 2

26. Two decks of playing cards are well schuffled and 26 cards are randomly distributed to a player. Then the probability

that the player gets all distinct cards is

(A) 52C26/104C

26(B) 2 X 52C

26/104C

26

(C) 213 x 52C26

/104C26

(D) 226 x 52C26

/104C26

Ans : (B)

Hints : Total no. of way of distribution = 104 26C

Favourable no. of ways of ditribution = (selecting any one decki) (selecting 26 cards out of it)

= 52262 C

So probability =

52

26

10426

2 C

C

27. An urn contains 8 red and 5 white balls. Three balls are drawn at random. Then the probability that balls of bothcolours are drawn is

(A)40

143(B)

70

143(C)

3

13(D)

10

13

Ans : (D)

Hints : Total no. of selection = 133C

So probability =

8 5 8 5

1 2 2 1

133

C C C C 10

13C

28. Two coins are avilable, one fair and the other two-headed. Choose a coin and toss it once; assume that the unbiased

coin is chosen with probalility3

4. Given that the outcome is head, the probability that the two-headed coin was

chosen is

(A)3

5(B)

2

5(C)

1

5(D)

2

7

Ans : (B)

Hints : B is the event of selecting a biased coin

UB is the event of selecting an unbiased coin

3 1P(UB) , P(B)

4 4

B P(B H)P

H P(H)

HP(B).PB

H HP(B).P P(UB).PB UB

=

11

241 3 1 5

14 4 2


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29. Let be the set of real numbers and the funtions f: and g: be defined by f(x)=x2 + 2x-3 and g(x) =

x+1. Then the value of x for which f(g(x)) = g(f(x)) is

(A) -1 (B) 0 (C) 1 (D) 2

Ans : (A)

Hints : fog(x) = (x + 1)2 + 2(x + 1) 3 = x2 + 4x

= x2 + 4x

gof(x) = x2 + 2x 3 + 1 = x2 + 2x 2

fog(x) = gof(x)

x2 + 4x = x2 + 2x 2

x = 1

30. If a,b,c are in arithmetic progression, then the roots of the equation ax2 - 2bx + c = 0 are

(A) 1 andc

a(B) -

1

aand -c (C) -1 and -

c

a(D) -2 and -

c

2a

Ans : (A)

Hints : a 2b + c = 0

So x = 1 is a root

Now product of roots = ca

So the other root is = c a

31. If sin-1x + sin-1y + sin-1z =3

2, then the value of x9 + y9 + z9 - 9 9 9

1

x y zis equal to

(A) 0 (B) 1 (C) 2 (D) 3

Ans : (C)

Hints :1 1 1sin x sin y sin z

2

x = y = z = 1

32. Let p,q,r be the sides opposite to the angles P,Q,R respectively in a triangle PQR, if r2sin P sin Q = pq, then thetriangle is

(A) equilateral (B) acute angled but not equilateral

(C) obtuse angled (D) right angled

Ans : (D)

Hints : 1p q r

2RsinP sinQ sinR

(R1: Circum radius)

2

r sin PsinQ pq

2

1 1 1(2R sinR) sinPsinQ (2R sinP)(2R sinQ)

2sin R 1 1 1[ P Q 0, R 0]

sin R 1 [sinR 1]

0R 90


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33. Let p,q,r be the sides opposite to the angles P,Q,R respectively in a triangle PQR. Then 2pr sinP Q R

2equals

(A) p2+q2+r2 (B) p2+r2-q2 (C) q2+r2-p2 (D) p2+q2-r2

Ans : (B)

Hints :

P Q R

2prsin 2 = 2pr sin Q2 P R Q

= 2pr cosQ = p2 + r2 q2

34. Let P 2, 3 , Q 2,1 be the vertices of the triangle PQR. If the centroid of PQR lies on the line 2x+3y = 1, then the

locus of R is

(A) 2x + 3y=9 (B) 2x - 3y =7 (C) 3x + 2y=5 (D) 3x - 2y=5

Ans : (A)

Hints : Let R(h, k)

Co-ordinate of centroid G2 2 h 3 1 k h k 2

, ,3 3 3 3

G lies on the line

h k 22 3 13 3

2h + 3k 6 = 3 2h + 3k = 9

35.

x

x 0

1lim

1 x 1

(A) does not exist (B) equals loge( 2) (C) equals 1 (D) Lies between 10 and 11

Ans : (B)

Hints :

x

x 0

1Lt

1 x 1

=x

x 0 x 0

1Lt Lt 1 x 1

x

= elog 2 = 2loge = loge(2)

36. If f is a real-valued differentiable function such that f(x) f(x) < 0 for all real x, then

(A) f(x) must be an increasing function (B) f(x) must be a decreasing function

(C) f(x) must be an increasing function (D) f(x) must be a drecreasing function

Ans : (D)

Hints : f(x) f (x) < 0

f(x) < 0 ; f (x) > 0 or f(x) > 0 ; f (x) < 0 ............... (ii)

So possible graphs of f(x) are

|f(x)| is decreasing function in both cases.

37. Rolles theorem is applicable in the interval [-2,2] for the function.

(A) (A) f(x) = x3 (B) f(x) = 4x4 (C) f(x) = 2x3 +3 (D) f(x) = x

Ans : (B)

Hints : For f(x) = 4x4 f(2) = f(2) & f (0) exist


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38. The solution of

2

2

d y dy25 10 y 0,

dxdxy(0) = 1, y(1) = 2e1/5 is

(A) y = e5x + e5x (B) y = (1 + x)e5x (C)x

5y (1 x)e (D)x

5y (1 x)e

Ans : (C)

Hints : Let y = emx is a trial solution

dym

dx

225m 10m 1 0

1m (equal roots)

5

The differential solution

x

5y (A Bx)e (A & B are two arbitrary constant)

Initially y (0) = 1 0(A 0) e 1 A 1

y (1) = 2 e1/51 1

5 5(A B)e 2e

B = 2 A = 2 1 = 1

Required solution y = (1 + x)ex/5

39. Let P be the midpoint of a chord joining the vertex of the parabola y2 = 8x to another point on it. Then the locus of Pis

(A) y2 = 2x (B) y2 = 4x (C)2

2x y 14

(D)2

2 yx 14

Ans : (B)

Hints : Let P(h, k) be the midpoint of chord AB through vertex (0, 0) of parabola y2

= 8x

Co-ordinate of B (2h 0, 2k 0) (2h, 2k)

B (2h, 2k) lies on y2 = 8x

4k2 = 8 2h

k2 = 4h

40. The line x = 2y intersects the ellipse2

2x y 14

at the points P and Q. The equation of the circle with PQ as

diameter is

(A)2 2 1x y

2(B) x2 + y2 = 1 (C) x2 + y2 = 2 (D)

2 2 5x y2

Ans : (D)

Hints : Solving the given equations then common points are P1

2,2

and Q1

2,2

Eqn. of Circle PQ as

diameter1 1

(x 2)(x 2) (y )(y )2 2

= 0 x2 - 2 + y2 -1

2= 0 x2 + y2 =

5

2


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41. The eccentric angle in the first quadrant of a point on the ellipse

2 2x y1

10 8at a distance 3 units from the centre of

the ellipse is

(A)6

(B)4

(C)3

(D)2

Ans : (B)

Hints : P 10 cos , 2 2 sin , centre (0, 0)

OP 3 (given)

10cos2 + 8sin2 = 9 [apply distance formula]

4as angle in 1st quadrant

42. The transverse axis of a hyperbola is along the x-axis and its length is 2a. The vertex of the hyperbola bisects the linesegment joining the centre and the focus. The equation of the hyperbola is

(A) 6x2 y2 = 3a2 (B) x2 3y2 = 3a2 (C) x2 6y2 = 3a2 (D) 3x2 y2 = 3a2

Ans : (D)

Hints :ae

Vertex at , 02

aea e 2

2

2 22 2

2 2

b be 1 2 1

a a

b2 = 3a2

Equation of hyperbola 3x2 y2 = 3a2

43. A point moves in such a way that the difference of its distance from two points (8, 0) and (8, 0) always remains 4.

Then the locus of the point is

(A) a circle (B) a parabola (C) an ellipse (D) a hyperbola

Ans : (D)

Hints : Let P(h, k) be the moving point

2 2 2 2h 8 k (h 8) k 4 ........................ (i)

Conjugate equation of (i)

2 2 2 2 32h(h 8) k (h 8) k 8h4

................ (ii)

[ using the form A2 B2 = 32h32h 32h

A B 8hA B 4

]

(i) + (ii) and squaring (h + 8)2 + k2 = (2 + 4h)2

or h2 + 16h + 64 + k2 = 4 + 16h2 + 16h

15h2 k2 = 60

44. The number of integer values of m, for which the x-coordinate of the point of intersection of the lines 3x + 4y = 9 andy = mx + 1 is also an integer, is

(A) 0 (B) 2 (C) 4 (D) 1

Ans : (B)

Hints : From given equation5

x3 4m

x is integer

3 + 4m = 5 or 1

m = 2 or m = 1


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45. If a straight line passes through the point ( ) and the portion of the line intercepted between the axes is divided

equally at that point, thenx y

is

(A) 0 (B) 1 (C) 2 (D) 4

Ans : (C)

Hints : Let, the equation of line bex y 1a b

, passes through P( )

P( ) is the mid point of the portion intercepted between the axes

a 0

2;

b 0

2

a = 2 ; b = 2x y

12 2

x y

2

46. The equation y2 + 4x + 4y + k = 0 represents a parabola whose latus rectum is

(A) 1 (B) 2 (C) 3 (D) 4

Ans : (D)

Hints : y2 + 4x + 4y + k = 0(y + 2)2 = 4 (x + 1 k/4)

Latus rectum = 4

47. If the circles x2 + y2 + 2x + 2ky + 6 = 0 and x 2 + y2 + 2ky + k = 0 intersect orthogonally, then kis equal to

(A)3

2 or2

(B)3

2 or2

(C)3

2 or2

(D)3

2 or2

Ans : (A)

Hints :

2g1g

2+ 2f

1f2

= c1

+ c2

2k2 k 6 = 0

k = 2, 3/2

48. If four distinct points (2k, 3k), (2, 0), (0, 3) (0, 0) lie on a circle, then

(A) k < 0 (B) o < k < 1 (C) k = 1 (D) k > 1

Ans : (C)

Hints : Equation of circle x(x 2) + y (y 3) = 0

(2k, 3k) will satisfy

So, K = 1

49. The line joining A (b cos , b sin ) and B (a cos , a sin ), where a b , is produced to the point M(x, y) so that

AM : MB = b : a. Then xcos y sin2 2

is equal to

(A) 0 (B) 1 (C) 1 (D) a2 + b2

Ans : (A)

Hints :abcos abcos basin absin

x , yb a b a

ab abx 2sin .sin , y 2cos .sin

b a 2 2 b a 2 2

xcos ysin 02 2


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50. Let the foci of the ellipse

22x y 1

9subtend a right angle at a point P. Then the locus of P is

(A) x2 + y2 = 1 (B) x2 + y2 = 2 (C) x2 + y2 = 4 (D) x2 + y2 = 8

Ans : (D)

Hints : P(h, k) be a point on the ellipse,

2 2e

3 .

S1

2 2, 0 S2

2 2, 0 , m1m

2= 1

k k1

h 2 2 h 2 2

h2 + k2 = 8

51. The general solution of the differential equationdy x y 1

dx 2x 2y 1is

(A) loge|3x + 3y + 2| + 3x + 6y = c (B) log

e|3x + 3y + 2| 3x + 6y = c

(C) loge|3x + 3y + 2| 3x 6y = c (D) log

e|3x + 3y + 2| + 3x 6y = c

Ans : (D)

Hints : Put x + y =

52. The value of the integral /2/6

1 sin2x cos2x dxsin x cos x

is equal to

(A) 16 (B) 8 (C) 4 (D) 1

Ans : (D)

Hints :

/2

/6

1 sin2x cos 2xdx

sin x cos x

/ 2 2 2 2

/6

(sinx cos x) cos x sin xdx

sin x cos x=

/2

/ 6

(sin x cos x cos x sin x)dx1

2 1 12

53. the value of the integral2

1010

1dx

1 (tan x)is equal to

(A) 1 (B)6

(C)8

(D)4

Ans : (D)

Hints :

101/2

101 101

0

cos / 2 xI dx

sin( / 2 x) cos ( / 2 x)

I4

54. The integrating factor of the differential equation e edy

3x log x y 2log xdx

is given by

(A) 3e

(log x) (B) e elog (log x) (C) elog x (D)1

3e(log x)

Ans : (D)

Hints :dy

3x logx y 2logxdx

dy 1 2y

dx 3x logx 3x

I.F. =1

dx3xlogxe =

1

31

loglogx3e (logx)


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55. Number of solutions of the equation tan x + sec x = 2 cos x, x [0, ] is

(A) 0 (B) 1 (C) 2 (D) 3

Ans : (C)

Hints : tanx + secx = 2cosx

2sin2x + sinx 1 = 0

sinx =12

56. The value of the integral4

0

sin x cos xdx

3 sin2xis equal to

(A) loge2 (B) log

e3 (C)

1

4log

e2 (D)

1

4log

e3

Ans : (D)

Hints :

/4

0

sin x cos xdx

3 sin2x

=

/4

2

0

sin x cos x

(sinx cos x) 4dx

Let sinx cosx = t

(cox + sinx)dx = dt

0

2

1

dt 1log3

44 t

57. Let

x

ex

3 1y sinx log (1 x), x 1

3 1. Then at x = 0,

dy

dxequals

(A) 1 (B) 0 (C) 1 (D) 2

Ans : (A)

Hints :

x

ex

3 1y sinx log (1 x) x 1

3 1

x 0

dy1

dx

58. Maximum value of functionx 2

f(x)8 x

on the interval [1, 6] is

(A) 1 (B)9

8(C)

13

12(D)

17

8

Ans : (D)

Hints : 2dy 1 2

[1, 6]dx 8 x

f(x) is decreasing

Max. value will be at x = 1

max

1 17f 2

8 8


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(14)

59. For3

x ,2 2

the value of1d costan

dx 1 sinxis equal to

(A)1

2(B)

1

2(C) 1 (D) 2

sin

(1 sin x)

Ans : (B)

Hints :1d cos xtan

dx 1 sinx

1

sin xd 2

tandx

1 cos x2

d x 1

dx 4 2 2

60. The value of the integral

2

|x|

2

(1 2 sin x)e dx is equal to

(A) 0 (B) e2 1 (C) 2(e2 1) (D) 1

Ans : (C)

Hints :

2 2 2|x| |x| |x|

2 2 2

(1 2sinx)e dx e dx 2sinx e dx

=2x

02.e 0 = 2(e2 1)

Q. 61 Q. 80 carry two marks each

61. The points representing the complex number z for whichz 2

argz 2 3

lie on

(A) a circle (B) a straight line (C) an ellipse (D) a parabola

Ans : (A)

Hints :z 2

argz 2 3

locus of z is circle

62. Let a, b, c, p,q, r be positive real numbers such that a, b, c are in G.P. and ap = bq = cr. Then

(A) p, q, r are in G.P. (B) p, q, r are in A.P. (C) p, q, r are in H.P. (D) p2, q2, r2 are in A.P.

Ans : (C)

Hints : b2 = ac ; 2logb = loga + log c

ap = bq ;loga q

logb p. . . . . . (ii)


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and bq = cr ;logc q

logb r. . . . . . (iii)

Hence

2 1 1

q r p; p, q, r are in H.P.

63. Let Skbe the sum of an infinite G.P. series whose first term is k and common ratio is

k(k 0)

k 1. Then the value of

k

kk 1

( 1)

Sis equal to

(A) loge4 (B) log

e2 1 (C) 1 log

e2 (D) 1 log

e4

Ans : (D)

Hints :k

Sk k(k 1)k

1k 1

;

k

kk 1

( 1)

S

=1 1 1 1

.......2 2.3 3.4 4.5

=1 1 1

1 2[1 .....]2 3 4

= e e1 2log 2 1 log 4

64. The quadratic equation 2x2 (a3+ 8a 1)x + a2 4a = 0 possesses roots of opposite sign. Then

(A) a 0 (B) 0 < a < 4 (C) 4 a < 8 (D) a 8

Ans : (B)

Hints : 0

a(a 4) 0

0 a 4

65. If loge(x2 16) log

e(4x 11), then

(A) 4 < x 5 (B) x < 4 or x > 4 (C) 1 x 5 (D) x < 1 or x > 5

Ans : (A)

Hints : x2 16 > 0

x ( , 4) (4 , ) .......(i)

4x 11 > 0 ;11

x ,4

.......(ii)

loge(x2 16) log

e(4x 11)

(x 5) (x + 1) 0 ; x [ 1, 5] ....... (iii)

x (4, 5] 4 < x 5

66. The coefficient of x10 in the expansion of 1 + (1 + x) + ......... + (1 + x)20 is

(A) 19C9

(B) 20C10

(C) 21C11

(D) 22C12

Ans : (C)

Hints : 1 + (1 + x) + ......... + (1 + x) 20

=21(1 x) 1

x

coefficient of x10 in

21(1 x) 1

x x =21C

11


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67. The system of linear equations

x + y + z = 3

x y 2z = 6

x + y + z =

has

(A) infinite number of solutions for 1 and all (B) infinite number of solutions of = 1 and = 3

(C) no solution for 1 (D) unique solution for = 1 and = 3

Ans : (B)

Hints :

1 1

1 1 2

1 1 1

+ 1

if = 1 then = 0

so equation has infinite number of solution for = 1 and = 3

68. Let A and B be two events with P(AC) = 0.3, P(B) = 0.4 and P(A BC) = 0.5. Then P(B/A BC) is equal to

(A)1

4(B)

1

3(C)

1

2(D)

2

3

Ans : (A)

Hints : P(A BC) = P(A) P(A B) ; P (A B) = 0.2

C

C

C

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

P(A B )= C C

P(B A)

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

0.2 0.2 1

0.7 0.6 0.5 0.8 4

69. Let p, q, r be the altitudes of a triangle with area S and perimeter 2t. Then the value of1 1 1

p q ris

(A)s

t(B)

t

s(C)

s

2t(D)

2s

t

Ans : (B)

Hints : Area of1

ABC base height2

1S a p

2;

1 a

p 2s........ (i)

similarly1 b

q 2s

...... (ii)

1 c

r 2s........ (iii)

(i) + (ii) + (iii)

1 1 1 b c a 2t t

p q r 2s 2s s


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70. Let C1and C

2denote the centres of the circles x2 + y2 = 4 and (x 2)2 + y2 = 1 respectively and let P and Q be their

points of intersection. Then the areas of triangles C1PQ and C

2PQ are in the ratio

(A) 3 : 1 (B) 5 : 1 (C) 7 : 1 (D) 9 : 1

Ans : (C)

Hints : Equation of chord S2

S1

= 0 ; x2 + y2 4x + 3 = 0

7x

4

C1T = Perpendicular distance of point (0, 0) from

7x

4is

4

1 2

C2T = Perpendicular distance of point (2, 0) from

7x

4is

1

4

area of C1

PQ = 11

C T PQ2

=1 7

PQ2 4

area of C2

PQ = 21

C T PQ2

=1 1

PQ2 4

1

2

area of C PQ

area of C PQ = 71= 7 : 1

71. A straight line through the point of intersection of the lines x + 2y = 4 and 2x + y = 4 meets the coordinate axes at Aand B. The locus of the midpoint of AB is

(A) 3(x + y) = 2xy (B) 2(x + y) = 3xy (C) 2(x + y) = xy (D) x + y = 3xy

Ans : (B)

Hints : Point of intersection of given lines4 4

x , y3 3

Equation of lines AB isx y

1a b

It is passes through

4 4

,3 3 , so,

4 4

13a 3b .... (1)

(h1, k) be the mid point of AB

a = 2h , b = 2k

from (1)2 2

13h 3k

locus of (h, k) is 2(x y) 3xy

72. Let P and Q be the points on the parabola y2 = 4x so that the line segment PQ subtends right angle at the vertex.If PQ intersects the axis of the parabola at R, then the distance of the vertex from R is

(A) 1 (B) 2 (C) 4 (D) 6

Ans : (C)

Hints : PQ subtants right angle at vertex.

22 2

21 1

so, t1

t2

= 4

equation of PQ is

211 12

1

2ty 2t x t

t 4

it intersect the x axis so, put y = 0

we get x = 4


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73. The incentre of an equilateral triangle is (1, 1) and the equation of one side is 3x + 4y + 3 = 0. Then the equation of thecircumcircle of the triangle is

(A) x2 + y2 2x 2y 2 = 0 (B) x2 + y2 2x 2y 14 = 0

(C) x2 + y2 2x 2y + 2 = 0 (D) x2 + y2 2x 2y + 14 = 0

Ans : (B)

Hints :2 2

3 1 4 1 3r

3 4= 2

For equalateral triangle R = 2r = 2 2 = 4

equation of circumcircle

(x 1)2 + (y 1)2 = 42 ; x2 + y2 2x 2y 14 = 0

74. The value of

1

n

n

(n!)lim

nis

(A) 1 (B) 21

e(C)

1

2e(D)

1

e

Ans : (D)

Hints :

1n

nn

nA lim

n

log A = nn

1 nlim log

n n=

n

1 1.2.........nlim log

n n...........n=

1

0logxdx = 1

A = e1 ;1

Ae

75. The area of the region bounded by the curves y = x3, y =1

x, x = 2 is

(A) 4 loge2 (B) e

1log 2

4

(C) 3 loge2 (D) e

15log 2

4Ans : (D)

Hints : Area of region2

3

1

1x dx

x= e

15log 2

4

76. Let y be the solution of the differential equation

2dy yx

dx 1 y logxsatisfying y(1) = 1. Then y satisfies

(A) y = xy 1 (B) y = xy (C) y = xy + 1 (D) y = xy + 2

Ans : (B)

Hints : . 21 dx 1 1

logxx dy y y

; logx=

2

d 1 1.dy y y

;1

dy logyyIF e e y

Ans. 21

y y.dyy

y logy c

y.logx = logy + c x = 1, y = 1. c = 0

log (x)y = logy ; y = (x)y


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77. The area of the region, bounded by the curves y = sin1x +x (1 x) and y = sin 1x x(1 x) in the first quadrant, is

(A) 1 (B)1

2(C)

1

3(D)

1

4

Ans : (C)

Hints :

1 12

0 0A 2x(1 x) (2x 2x )dx

=

1

3

78. The value of the integral

5

1

x 3 1 x dx is equal to

(A) 4 (B) 8 (C) 12 (D) 16

Ans : (C)

Hints :5 5

1 1| x 3 | dx |1 x | dx =

3 5 5

1 3 1(3 x)dx (x 3)dx (x 1)dx = 12

79. If f(x) and g(x) are twice differentiable functions on (0, 3) satisfying f (x) = g (x), f (1) = 4, g (1) = 6, f(2) = 3, g(2) =9,then f(1) g(1) is

(A) 4 (B) 4 (C) 0 (D) 2

Ans : (B)

Hints : f (x) = g (x) f(x) = g(x) 2x + D

f (x) = g (x) + c f(2) = g(2) 4 + D

f (1) = g (1) +c 3 = 9 4 + D

4 = 6 + c 2 = D

c = 2

f (x) = g (x) 2 f(x) g(x) = 2x 2

f(1) g(1) = 2 2 = 4

80. Let [x] denote the greatest integer less than or equal to x, then the value of the integral

1

1

(| x | 2[x]) dx is equal to

(A) 3 (B) 2 (C) 2 (D) 3Ans : (A)

Hints :1 1

0 1I 2 | x | dx 2 [x]dx

=1 0 1

0 1 02 x dx 2 ( 1)dx 0dx = 2 1 00 1(x ) 2(x) 0

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

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