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West Bengal Institute of Technology (WBIT) 2012-2nd Year B.Tech Wbjee Mathematics year- - Question Paper

Wednesday, 17 July 2013 11:55Web

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See attachment.

68010

Duration : Two Hours

Maximum Marks : 100

M 2012

MULTIPLE CHOICE QUESTIONS SUBJECT : MATHEMATICS

[ Q. 1 to 60 carry one mark each ]

. -1 . -1 . -1 3 n 9 + 9 + 9__1

1. If sin x + sin y + sin z = , then the value of^{x y} 999 is equal to

2 x y z

A. 0 B. 1 C. 2 D. 3

2. Let p, q, r be the sides opposite to the angles P, Q, R respectively in a triangle PQR. If _r² _si_n P _si_n Q = pq, then the triangle is

A. equilateral B. acute angled but not equilateral

C. obtuse angled D. right angled

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

F P - Q + RI ²p^rsin|--- equals

2

A. p² +q² + r² B. p² + r² - q² C. q² + r² - p² D. p² + q² - r²

4. Let P (2, -3), Q (-2, 1) be the vertices of the triangle PQR. If the centroid of APQR lies on the line 2x + 3y = 1, then the locus of R is

A. 2x + 3y = 9 B. 2x - 3y = 9 C. 3x + 2y = 5 D. 3x - 2y = 5

5. lim

A. does not exist B. equals log_e (n² j C. equals 1

D. lies between 10 and 11

6. 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 decreasing function

7. Rolles theorem is applicable in the interval [-2, 2] for the function A. f (x) = x³ B. f (x) = 4x⁴ C. f (x) = 2x³ + 3

D. f (x) = n |x|

2 1

8. The solution of 25y - 10 + y = 0, y(0) = 1, y(1) = 2e⁵ is

² dx

dx

x

A. y = e^5x + e

C. y = (1 + x)e

-5x

5x

B. y = (1 + x)e

D. y = (1 + x)e ⁵

9. 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 P is

2

^D' ^{x2 +} T = ¹

x² 2 ,

^C. _T + y² = ¹

A. y² = 2x

B. y² = 4x

2

_x

10 .The line x = 2y intersects the ellipse + y = 1 at the points P and Q. The equation of the circle with

PQ as diameter is

2 + 2 ¹ A. ^x +^y = -

2 , 2 ⁵ ^D* ^x +^y = 2

B. x² + y² = 1

C. x² + y² = 2

2 2 x y

11. The eccentric angle in the first quadrant of a point on the ellipse -+-= 1at a distance 3 units from

the centre of the ellipse is

%

^a7

%

^C7

12. The transverse axis of a hyperbola is along the x-axis and its length is 2a. The vertex of the hyperbola bisects the line segment 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

13. 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

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

A. 0 B. 2 C. 4 D. 1

15. If a straight line passes through the point (a,P) and the portion of the line intercepted between the axes is

divided equally at that point, then

is

A. 0 B. 1 C. 2 D. 4

16. The maximum value of |z| when the complex number z satisfies the condition

is

A. 3 B. 3 +/2 C. 3 +1 D. 3 _ 1

25

C. (0, -3)

B. (0, 3)

D.

A. (-3, 0)

z 1

18. If-is purely imaginary, then

z +1

A. |z| = 1 B. |z| = 1 C. |z| = 2 D. |z| = 3

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

A. is 12 B. is 4

C. is 2 D. cannot be determined from the given information

20. A vehicle registration number consists of 2 letters of English alphabet followed by 4 digits, where the first digit is not zero. Then the total number of vehicles with distinct registration numbers is

A. 26² x 10⁴ B. ²⁶ p₂ x¹⁰ P₄ C. ²⁶ P₂ x 9 x¹⁰ P₃ D. 26² x 9 x 10³

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

10! 10! 10!

^A. (2!)³ B. (2!)² C2l" ^D. ^10!

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

A. 256 B. 128 C. 24 D. 12

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

A. 4950 B. 4850 C. 4750 D. 4650

24. Let the coefficients of powers of x in the 2^nd, 3^rd and 4^th 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

25. 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

26. The sum 1 *1! + 2 * 2! + ... + 50 x 50! equals

A. 51! B. 51! - 1 C. 51! + 1 D. 2 * 51!

27. 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

₁ 1 1.3 1.3.5 1.3.5.7

28. The sum of the infinite series ¹ is equal to

3 3.6 3.6.9 3.6.9.12

A. V2 B. V3 C. D.

29. 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

30. 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. 3R D. 4R

31. The equation y2 + 4x + 4y + k = 0 represents a parabola whose latus rectum is A. 1 B. 2 C. 3 D. 4

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

3 3 3 3

A. 2 or - B. -2 or - C. 2 or D. -2 or

2 2 2 2

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

A. k< 0 B. 0 < k < 1 C. k = 1 D. k> 1

34. The line joining A(b cos a, b sin a) and B(a cos P, a sin P), where ab, is produced to the point M(x, y)

a+P . a+P

so that AM : MB = b : a. Then xcos + ysm

2 2

A. 0 B. 1 C. -1 D. a² + b²

_x2

35. Let the foci of the ellipse + y² = 1 subtend 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

_ dy x + y + 1

36. The general solution of the differential equation = 2_x + 2y + 1 is

A. log_e |3x + 3y + 2|+3x + 6y = c B. log_e |3x + 3y + 2|-3x + 6y = c

C. log_e |3x + 3y + 2|-3x - 6y = c D. log_e |3x + 3y + 2|+3x - 6y = c

p^/2 F1 + sin2x + cos2x J _A

37. The value of the integral : ^dx is equal to

Jn/6 V sinx + cosx J

A. 16 B. 8 C. 4 D. 1

ft 1 4

38. The value of the integral J 1 + 01 is equal to

ft ft ft

A. ¹ B. - C. ^D 4

dy

39. The integrating factor of the differential equation 3xlog_ex + y = 21og_ex is given by

dx

^B. ^loge^(loge^x)

C. log_ex

^D. (^loge^x)

40. Number of solutions of the equation tanx + secx = 2cosx,x e[0, n]is A. 0 B. 1 C. 2

D. 3

41. The value of the integral

n

4

|* sin x + cos x ,

J 3 + sin2x ^{dx is e}q^{ual to}

C. 4^loge²

A. log_e 2

B. log_e 3

^D. ₄^loge³

^/3-1" _v3^x + 1,

^sinx + ^loge (² + ^x)^{, x} > -¹. Then at x = 0, equals

42. Let y =

A. 1 B. 0 C. -1

x 2

43. Maximum value of the function f(x) = + x on the interval [1, 6] is A. 1

D. -2

D ^D. 8

_C 13 ^C. 12

the value of {tan ¹ , ^cosx } is equal to dx L 1 + sinxj

2

sinx

C. 1

D.

(1 + sin x)

45. The value of the integral J(1 + 2sinx)el^xl dx is equal to

-2

A. 0 B. _e2 _-1 C. 2(e² - j

46. If (a + /p j and ( a - Jfi j are the roots of the equation _x² + p_x + q = 0 where a, p, p and q are real, then the roots of the equation (p² - 4qj(p²x² + 4pxj - 16q = 0 are

D. 1

1 J_ ^a + #.

A.

C.

and

\

/ \ 1 1

G __+1 j

and

J

H ^-# K

B.

_vVa P y

and

_vVa

p

\

f \ 11

d. (ya+ypj and

47. The number of solutions of the equation log2(x² + 2x -lj = 1 is

A. 0

C. 2

B. 1

D. 3

1 n 1n 1 n

48. The sum of the series 1 + - Ci + - C-> + .... +--_T C is equal to

2¹3² n + 1ⁿ

3(2ⁿ - j 2n

2ⁿ+¹ -1 n +1

2ⁿ +1 n +1

2ⁿ +1 2n

A.

C.

B.

D.

1 + 2 +.....+ (r - 1) .

. The value of

is equal to

49

r!

r=2

_C e ^C. 2

A. e

B. 2e

^D' T

1 2 1 1 3 1

T

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

50. IfP =

A. 2 B. -2 C. 1

51. The remainder obtained when 1! + 2! + ... + 95! is divided by 15 is A. 14 B. 3 C. 1

D. 0

-1 cosR cosQ cosR -1 cosP cos Q cos P -1

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

is equal to

A. -1 B. 0 C. 2

53. The number of real values of a for which the system of equations

D. 1

x + 3y + 5z = ax 5x + y + 3z = ay 3x + 5y + z = az

has infinite number of solutions is

A. 1 B. 2 C. 4 D. 6

54. The total number of injections (one-one into mappings) from ja1, a2, a3, a4} to {b1, b2 , b3 , b4 , b5 , b6 , b7 } is

A. 400 B. 420 C. 800 D. 840

7 5 3

10 . , 7 r p

55. Let (1 + x)¹⁰ = c_rx^r and ^{+ x} = 2~i^dr^x If ^P = 2~i^2r and ^Q = 2~i^d2r+1, then q is equal to

r=0 r=0 r=0 Q

r=0

A. 4 B. 8 C. 16 D. 32

56. Two decks of playing cards are well shuffled and 26 cards are randomly distributed to a player. Then the probability that the player gets all distinct cards is

^A. ⁵²C26 / ¹⁰⁴C26 ^B. 2 x ⁵²C26 / ¹⁰⁴C26

^C. 2¹³ x ⁵²C26 / ¹⁰⁴C26 ^D. 2²⁶ x ⁵²C26 / ¹⁰⁴C26

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

_A _40 _B .70 _C 3 _D 10

^A. 143 ^B. 143 ^C. 13 13

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

3

the unbiased coin is chosen with probability 4. Given that the outcome is head, the probability that the twoheaded coin was chosen is

3 2 12

^{A. 5 B. 5 C. 5 D. 7}

59. Let R be the set of real numbers and the functions f : R R and g : R R be defined 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

60. If a, b, c are in arithmetic progression, then the roots of the equation _ax2 _ 2b_x + _c = 0 are

A. 1 and B. _ and -c C. -1 and _^c D. -2 and -

a a a 2a

[ Q. 61 to 80 carry two marks each ]

dy y²

61. Let y be the solution of the differential equation x =- satisfying y(1) = 1. Then y satisfies

dx 1 - y log x

A y = x^y1 B. y = x^y C. y = x^y+¹ D. y = x^y+²

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

1 1 1 A. 1 B. - C. - D. -

2 3 4

63. The value of the integral J [|x - ³ +1^{1 - x}l]^dx is equal to

A. 4 B. 8 C. 12 D. 16

64. 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

65. Let [x] denote the greatest integer less than or equal to x, then the value of the integral J (|x| - 2[x])dx

is equal to A. 3

C. -2

B. 2

D. -3

z - 2

66. The points representing the complex number z for which arg

= lie on 3

v ^z + ² J

A. a circle B. a straight line C. an ellipse D. a parabola

67. Let a, b, c, p, q, r be positive real numbers such that a, b, c are in G.P. and a^p = b^q = c^r. 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.

k

68. Let Sk be the sum of an infinite G.P. series whose first term is k and common ratio is

rc (_-1) k

the value of - is equal to

(k > 0). Then

k +1

Sk

k=1

^A. loge 4

C. 1 - loge ²

B. loge ² - 1

D. 1 - loge 4

2/3 \ 2

69. The quadratic equation 2x -1 a + 8a -11x + a - 4a = 0 possesses roots of opposite sign. Then

B. 0 < a < 4

C. 4 < a <

A. a < 0

Da > 8

70. If log_e ex² - 16j < log_e (4x - 11), then

A. 4 < x < 5

C. -1 < x < 5

B. x < -4 or x > 4

D. x < -1 or x > 5

20

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

A. 19C

B. ²⁰C

C. 21C

D. ²²C

,9 j. v.1₀

72. The system of linear equations

Xx + y + z = 3 x - y - 2z = 6 -x + y + z = has

A. Infinite number of solutions for X -1 and all

B. Infinite number of solutions for X = -1 and = 3

C. No solution for X -1

D. Unique solution for X = -1 and = 3

11

12

73. Let A and B be two events with P(A^C) = 0.3, P(B) = 0.4 and P(A n B^C) = 0.5. Then P(B\A u B^C) is

equal to 1

^A' 4

1

^B- 3

2

^D-i

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

P q ^r

2s

t

B. -

s

D.

t

I ..2 A 4- _

s

A. -t

75. Let C1 and C2 denote 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 C₂PQ are in the ratio A. 3 : 1 B. 5 : 1 C. 7 : 1 D. 9 : 1

76. A straight line through the point of intersection of the lines x + 2y = 4 and 2x + y = 4 meets the coordinates axes at A and 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

77. 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

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

A. x² + y² - 2x - 2y - 2 = 0

C. x² + y² - 2x - 2y + 2 = 0

B. x² + y² - 2x - 2y - 14 = 0

D. x² + y² - 2x - 2y + 14 = 0

1

(_n!)ⁿ

79. The value of li_m -- is

n

A. 1

B.

2

e

1

D. -

e

Q 1

80. The area of the region bounded by the curves y = x³, y = , x = 2 is

x

B. + log_e 2 4 ^e

C. 3 - loge 2

A. 4 - loge 2

^D. ¹⁵ - loge ²

68010

M 2012

SUBJECT : MATHEMATICS

KEY ANSWERS WITH EXPLANATIONS

1. Given sin ¹ x + sin ¹ y + sin ¹ z = 3. 2

-1 -1 -1 2

: sin ¹ x = sin ¹ y = sin ¹ z =

: x = y = z = 1

9 9 9 1

x + y + z

999 x .y .z

= 1 +1 +1 - j = 3 -1 = 2 Ans. C.

2. r².sinP.sinQ = pq

₂ p q ^{r2 X} 2R2R = ^pq

r² = 4R² r = 2R

side = 2 x circumradius = diameter So, triangle is right angled triangle. Ans. D.

3. Given P + Q + R = 180

: P + R = 180-Q

(180-2Q'

= sin(90-Q) = cosQ

sin

2

Now 2pr cos Q

p^z + r^z - q^z ₂ 2 2

= 2pr -2pr = p + - q

Ans. B.

4. Let R = (h, k)

h k - 2 3^, 3

'2 - 2 + h 1 - 3 + k

3

. ²3+lij² =¹

f+^{k - 2} =¹ 2h + 3k = 9

Required locus 2x + 3y = 9 Ans. A.

% ^x -1

5. lim

+ x - 1

(%^x - i)(vr+x+ij

= lim

1 + x - 1

x0

%^x - 1

= lim

x Q ^x

= (log2 %) x 2 = log₂ %² Ans. B.

6. Given f(x).f '(x) < Q Vx e R and f(x) is differentiable

f(x) is continuous function and f(x) and f'(x) are opposite of sign either f(x) >0 or f(x) < 0 but It can not cut the x-axis

When f(x) > 0 then f '(x) < 0 . f(x) is decresing function When f(x) < 0 then f '(x) > 0 . f(x) is increasing function

We can say that |f(x)| is decreasing function.

Ans. D.

7. Check the options individually. Take option (b) f(x) = 4x⁴

Now, Rolles theorem is applicable. Ans. B.

8. Let y = e^mx

dy

mx

= me

dx

d²y dx²

= m²e^mx

25m²e^mx - 10me^mx + e^mx = 0 25m² - 10m +1 = 0

(5m -1)² = 0

1 1

^{m =} 55

.'. General solution is y = (A + Bx)e ⁵ y (0) =1

1 = A .1 A = 1

1

y(1) = 2e ⁵

2e 5 = (A + B)e 5

2 = (1 + B)

B = 1

x

.. y = (1 + x)e⁵

Ans. C.

9. y² = 8x y² = 4ax 4a = 8 a = 2

, 2t² + 0,2 ₊2 u

h = 2 = t² or t² = h

, 4t + 0 _{0+ +} k

k = 2 = 2t or t =

. k² _h .. = ^h

. k² = 4h y² = 4x Ans. B.

10. x² + y² = 1

2

4y

+ y² = 1

4

2y² = 1 2 = ¹

y =

2

y = -

x =

P

- 42 -

^, ~ 42,

equation of circle PQ as diameter is (x - 42 j(x + 42 j +

Q

1

= 0

^y+ 42,

42,

x² - 2 + y² - 2 = 0 2 2 5 ^x2 + y² =

Ans. D.

11. Let the pt on the first quadrant T0cos 0, 48 sin 0 j distance from the centre = 3 10cos² 0 + 8sin² 0 = 9 8 + 2cos² 0 = 9

₂ 1 1 cos 0 = cos 0 = + [_v pt lie on 1st quadrant]

2 v2

0 = *

4

Ans. B.

2 2

12. Let the hyperbola = 1

a² b²

Length of the transverse axis = 2a vertex = (a, 0)

(f, ⁰) = (a, 0)

focus (ae, 0) centre = (0, 0)

ae

= a

e² = 4

2

b

1 + -^b2 = 4 b² = 3a a²

2

_x2 _y²

Equation : ---= 1

a² 3a²

Ans. D.

13. The distance between (8, 0) and (-8, 0) = 16 > 4.

According to the defination of hyperbola the locus is a hyperbola. Ans. D.

f=+m: Hh1⁹} ^3x+^4(mx+^1f=⁹

14. m e I

(3 + 4m)x = 5

5

e I

3 + 4m

.. 3 + 4m = -5, - 1, 1, 5 when _{m e} I . m can take only two values m = -1, - 2.

Ans. B.

x y

15. Let the equation of the line + t- = 1

a b

Its passes through (a, p)

(0,b)
(a, 0)'

a+p = 1

ab

(a, p) is the mid points of (a, 0) and (0, b)

a . n b

a = and p =

a = 2a, b = 2p

Equation of the line = 1 ^y = 2

ⁿ 2a 2p a p

Ans. C.

16.

= 2

<^|z| +R

Now,

2

^|z|+|z| - ²

|z|² -2|z| + 2 - 0

|z|² -2|z| + 1 < 3

(|z|-1)² < 3

-S < (|z|-1) <V3

1 -V3 < |z|< 1 + V3 1 -V3

v |z|- 0 Ans. C.

x + iy

2 2

Ans. D.

z 1

--r = ik, k e R, k

z + 1

2z ik +1 . ..

2 = ik1^{; b}y comp.-div.

1 + ik

z = i-

1 - ik

¹ - ikl V1 + k²

Ans. B.

19. n(M) = 50 = No. of failed in maths. n(P) = 45 n (B) = 40

n(M0P) + n(M0B) + n(P 0B) - 3n(M0 P 0B) = 32

We have to find n(M 0 P 0 B)

Total no of student =100 n(M U P U B) = 99

n(M) + n(P) + n(B) - |n(M 0 P) + n(M 0 B) + n(P 0 B)} + n(M 0 P 0 B) = 99 50 + 45 + 40 - {32 + 3n(M 0 P 0 B)} + n(M 0 P 0 B) = 99 135 - 32 - 2n(M 0 P 0 B) = 99 2n(M 0 P 0 B) = 4

n(M 0 P 0 B) = 2 Ans. C.

20. 1 2 3 4 5 6

two alphabet we can choose 26² ways. and 1st number we can choose 9 ways. next 3 numbers we can choose 10³ ways.

Ans. D.

21. I - 2 R - 2 A - 2 T - 1 N - 1

O - 1 L - 1

10!

Number of words -

(2!)³

Ans. A.

22. Required number of ways is which the order of speakers can be prepared

=n

= ~!

= 4

= 2

= 12 [Taking speakers P & Q as identical]

Ans. D.

23. No. of diagonals in a regular polygon = ¹⁰⁰C₂ -100

= 100x9 -100 2

= 50 x 99 - 100 = 4950 - 100 = 4850 Ans. B.

24. ⁿC₁, ⁿC₂, ⁿC₃ are in A.P.

_ n - 2) +

-2?-=--+ n

-2 6

n² - 3n + 2

or, n - 1 =--+ 1

or, 6n - 6 = n¹ - 3n + 2 + 6 or, n² - 9n +14 = 0 n = 7, n = 2 not acceptable.

2ⁿ 2⁷ sum = ²- = = 2⁶ = 64

Ans. B.

25. f(x) = ax² + bx + c, g(x) = px² + qx + r

(a - b) + (b - q) + (c - r) = 0

Now f(1) = g(1) a + b + c = p + q + r

(1)

4 (a - b) + 2 (b - q) + (c - r) = 0

f(2) = g(2) 4a + 2b + c = 4p + 2q + r

3(a - b) + (b - q) = 0

(2)

[using (1)] f(3) - g(3) = 2

9(a - p) + 3(b - q) + (c - r) = 2 8(a - p) + 2(b - q) = 2 [using (1)]

4(a - p) + (b - q) =1 (a - p) =1 (using (2))

Now, f(4) - g(4) = 16(a - p) + 4(b - q) + (c - r)

= 15 (a - p) + 3(b - q) (using (1))

{v a - p = 1 } ^{ b - q = -3j

= 15.(1) + 3(-3)

= 15 - 9 = 6

Ans. C.

26. 1x1! + 2 x 2! + 3 x 3! + .... + n x n!

= (n + 1)! - 1

1 x 1! + 2 x 2! + ... + 50 x 50!

= 51! - 1

Ans. B.

27. Six numbers are in A.P.

a - 5d, a - 3d, a - d a + d, a + 3d, a + 5d 6a = 3

1

a =

2

- 5d = 4H - d

2 H 2

--5d = 2 - 4d 2

d = - -

2

Fifth term = a + 3d

2

1 9

2 2 =-8 = 2 = -4 Ans. D.

28. (1 + x)ⁿ = 1 + nx +

ⁿ (^{n - 1}) 2

x +.

2!

1

comparing, nx =

3

nx (nx - x) 1 3

3 6

2

1

--x

'3 H 3

29. Let a be the common root

2

a + a + a = 0

₂

a + aa + 1 = 0

(~)_

a(l a) + a 1 = 0

(l a)(a l) = 0

Either a = 1 or, a =1

Put a =1 in the 1st equation

1 + 1 + a = 0 a = - 2 Put a = 1

. ² + x + 1 = 0 x² + x + 1 = 0

They have no real common root.

Put a = -2

x² + x - 2 = 0 &x² -2x+ 1 = 0 they have a one real common root. Ans. B.

or, x = 1

30. Let A be the 1st term & r be the c.r.

A . r^{p -}1 = 64 = 2⁶ A . r^{q -}1 =27 = 3³ A . r^{R -}1 = 36 = 2² . 3²

1 ^p~¹ Now, 2 = A ⁶ r ⁶

1 ^q~¹

3 = A ³ r ³

1 R~1

2 3 = A ² r ²

1 p~¹ 1 q~¹ 1 R ~1

A⁶ r⁶ A³ r³ = A² r²

^p~¹+2q~2 3R ~3

r 6 6 = r 6

p 1 + 2q 2 3R 3 6 = 6 p + 2q = 3R Ans. C.

31. y + 4y = -4x - k

y + 4y + 4 =-4x + 4 - k

k - 4

(y + 2)² =-4x - (k - 4)

= -4

x + -

Y² = -4 AX L.R. = 4A = 4 unit.

Ans. D.

³². ^Apply, ²(gig2 + ^f1^f2 ) = ^C1 + C2

2 (1X 0 + k k) = 6 + k

2k² = 6 + k

or, 2k² - k - 6 = 0

or, 2k² - 4k + 3k - 6 = 0 or, 2k(k - 2) + 3(k - 2) = 0

or, (k - 2)(2k + 3) = 0

3

^k = ², ^- j Ans. A.

33. Equation of circle is (x - 2)(x - 0) + (y - 0)(y - 3) = 0

x² - 2x + y² - 3y = 0 x² + y² - 2x - 3y = 0 4k² + 9k² - 4k - 9k = 0

13k² = 13k .. k = 0, 1 k = 1 Ans. C.

ab cos P ab cos a b a

= _u^ab = (cos p- cos af b - a ^{v r} '

ab

34. x =

a + P a P

2 sin--sin-

b a

2

ab sin P ab sin a

y

b a ab

a + P P a

2 cos--sin-

b a

2

. a+P sin¹¹

x

y

a+P

2

cos

a+P . a+P

xcos:--+ y sin-= 0

2

Ans. A.

2 2

35. x_+z_ = 1

9 1

a = 3, b = 1

2

e² = 1

2

a

242

e =

3

= 22

ae

^mps ^{x m}ps' = ¹

y 0 y 0

-X-

x + 2%/2 x 2/2

2

y

= 1

2

x 8

y² = - x² + 8

x² + y² = 8 Ans. D.

dy _ x + y + 1

36.

dx 2 (x + y) + 1

Let x + y = z

dy dz

1 + = dx dx

dy = dz 1 dx dx

dz z + 1 --1 =

dx 2z + 1

dz z +1

+1

dx 2z +1

_ z +1 + 2z +1

= 2z+1

3z + 2

2z +1

or, -dz = dx

3z +1

2 f ^3z + 2 or, -dz = I dx + c

J 17+t^dz = J' J-

2 j* ^3z + ² - \

or, -- dz = x + c

3J 3z + 2

2 1 1fd (3z + 2)

f ³

or, z----^:-¹ = x + c

3 3 3 J 3z + 2

21

z--log 3z + 2 _ x + c

or,

3 9 ¹

2(x + y) - -1log|3x + 3y + 2| = x + c

6x + 6y - log|3x + 3y + 2| = 9x + c'

or, 3x - 6y + log|3x + 3y + 2| = c Ans. D.

or,

%

² (sinx + cosx)² + (cosx + sinx)(cosx - sinx)

³⁷ z

%

6

%

2

j[(cosx + sinx) + (cosx - sinx)]dx

%

6

%

= 2sinx ²

6

. % . %

= 2

sin--sin

2 6

= 2 1 -1

2,

= 2.1 = 1

2

Ans. D.

%

2

1

-dx

/ \101 sinx

0 1 +

cosx

101

%

2

j

( cos x)

dx

%

2

Zz

dx

I x101 / n101

sin x + cos x

( sin x)

101

a a

Apply j f (x)dx = j f (a - x)dx 0 0

%

2

= j

2I = I dx = _x|2 = %

0

I= *

4

dy y 2

39. +---=

dx 3xlog_e x 3x

, dx I = _e ^{3x lo}ge ^x

1 j ^d (lge ^x )

= _e 3 loge x

¹l^og (l^oge ^x )

= e ³

= _e^loge (^loge ^x) 3

1

= (^loge ^x) 3 Ans. D.

sin x 1

40.--1--= 2cosx

cosx cosx

sinx + 1 = 2cos² x = 21 - sin² xj

sinx + 1 = 2(1 + sinx)(1 - sinx)

(1 + sinx)[ 1 - 2(1 - sinx)] = 0 (1 + sin x)(1 - 2 + 2 sin x) = 0 (1 + sinx)(2 sinx - 1) = 0 sinx = -1

1

sin x =

2

= % 5k

" ^x = 6 ^, 6

Ans. C.

n

4

41 f sinx + cosx_d ' J 3 + sin2x ^dx 0

n/4

J

(sin x + cos x)dx 4 - (sin x - cos x)

x

0

%

4

z

1

0

dz

2² - z²

Putting sinx - cosx = z (cosx + sinx)dx = dz,

2 + z 2 - z

1

log

2 x 2

0

= [^log^{(3) - lo}g⁽¹⁾]

= ^log⁽³⁾ Ans. D.

'3* -1"

sinx + log_e (1 + x)

42. y =

3^x +1 v V

3^x +1 - 2 / .

(sinx) + log_e (1 + x)

3^x +1

^F1 - ²

sinx + log_e (1 + x)

3^x +17

2sinx , s

= sinx---+ log 1 + x

3^x +1

(3^x +1 j cos x - (sin x)3^x log_e 3

dy ->

-ir = cosx - 2-dx

e³ 1

+

1+x

(^3x+¹)

dy

dx

1 . 2 - 0 + 1

= 1 - 2-1--

2² 1 + 0

x=0

= 1 - 1 + 1 = 1 Ans. A.

43. ^f(x) . x + x

^{f ,(x)}=8^- 48 x

For max & min.

f'(x) = 0

1 - a=0

8 x²

1 = A

8 = x² x² = 16

x = +4 [x e [1, 6]] f '(4^-j > 0 f '(4 +) < 0

at x = 4 f(x) is max.

f(4) = 4+2 = 1+1 = 1

^w 8 4 2 2

Ans. A.

-1

tan

2

44. Exp. = 4~ dx

xx cos 2 + sin 2

d_

dx

tan

xx cos 2 + sin 2

-1¹ - tan2

tan

dx

x

1 + tan

2)

= 3T _T^tan-l M ^{f -} 2

f 3f

^- 2 < ^x < ~2

n x 3n

^- 4 < 2 ^< ~4~

n x 3n

>-->--

4 2 4

n n n n

>--->--

2 4 2 2

= _d_ F n_-x

= ^{0 -} 2

dxU 2

Ans. B.

2

45. J (1 + 2sinx)edx

2 2 = J edx + 2 J sinxe^|x|dx

2 -2

2

= 2 x Je^|x|dx + 0

x

0 |2

= 2e

10

= 2(e² - 1) Ans. C.

46. x² + px + q = 0 roots are a + yfp and a - #

2

^Fp² - 4q¹

v ⁴ y

(p² - 4q) = ^4p

a -p = q p = a -q =--q =

Now, 4p(4a²x² - 8ax) - 16(a² - p) = 0 16a²px² - 32apx - 16a² + 16p = 0

2 2 1 1 _n

x²--x--+ _r = 0

a pa²

1 1

+ -

11

Sum of the roots = =

Product of the roots =

j__1

a² p

47. x² + 2x - 1 = 2

x² + 2x = 3 x² + 2x + 1 = 4 (x +1)² = 2²

x +1 = +2 x =1, -3 Ans. C.

C C C

48. C₀ + -¹ + ++-

⁰ 2 3 n +1

/1 Z*1 Z*1 Z*1 Z*1

I (1 + x)ⁿdx = I C₀dx + I Cjxdx + I C₂x²dx+--+ I C_nxⁿdx J0 J0 J0 J0 J0

(1 + x)ⁿ⁺¹

n +1

¹ |1 X²

= Cri + ^C1²r

¹ 3

+ _C ^x3 + ^C2 3

1

+

_xn+1

+C_n ^x . ⁿ n +1

0

C_r

⁰ ++- ~ⁿ

n +1 12 Ans. A.

(r - 1)r

2 r!

r=2

T¹ -

² (r - ²)^!

1 Z

² t? (r - ²)^!

11

1

= e 2

2ⁿ⁺¹ -1 = C₀ C

n +1

49.

Ans. C

1 2 1 1 3 1

50. P =

Q = PP^T

			"1	1"
"1 2	1
		X	2	3
1 3	1_
			1	1_

1 + 4 + 1 1 + 6 + 1 1 + 6 + 1 1 + 9 + 1

6 8

8 11

|Q = 66 - 64 = 2 Ans. A.

51. 1! +2! +3! +4! = 1 +2 + 6 + 24

= 33

33

15

2

30

3

Required remainder = 3 Ans. B.

K

52. Putting P = Q = R = y

1

2

-1

1

2

1

2

1

2

-1

1

2

1

2

-1 cosR cosQ cosR -1 cosP cos Q cos P -1

Ans. B.

53. (1 -afx + 3y + 5z = 0 5x + (1 - afy + 3z = 0 3x + 5y + (1 - afz = 0

1 -a 3 5

5 1 -a 3

3 5 1 -a

(1 -a)³ + 27 +125 - 45(1 -a)

1 -a³ - 3a(1 -a) + 152 - 45 + 45a = 0 3a² - 3a - a³ + 108 + 45a = 0 - a³ + 3a² + 42a + 108 = 0 a³ - 3a² - 42a - 108 = 0

a	b	c
c	a	b
b	c	a

= (a³ + b³ + c³ - 3abcj

It has one real solution and two imaginary solution.

Ans. A.

54. n(A) _ 4 n(B) _ 7

₇ 7!

no. of mappings _ ⁷p4 = f

= 3! x 4 x 5 x 6 x 7

= 20 x 6 x 7 _

_ 840

Ans. D.

55. 2¹⁰ = C₀ + C₁ + ... + C

10

= ^C0 + ^C2 + ^C4 + ^C6 + ^C8 + ^C10 = ²

2⁷ = d0 + d1 + ... + dy

= d1 + d + d5 + d7 = 2⁶

P 2⁹ ₃ ^p = 2³ = 8

" Q 2⁶

Ans. B.

56. Total no. of possible outcomes _ ¹⁰⁴C26, which are equally likely.

Number of casses that the player gets all distinct cards _ (²C²⁶ x ⁵²C26

= 2²⁶ x ⁵²C,

26

2²⁶ x ⁵²C-

26

Required probability = 104

26

57. caseI 2R, 1W caseII 1R, 2W

⁸C₂ x ⁵C₁ + ⁵C₂ x ⁸C₁

8R

5W

13

C

3

8.7 - 5.4 _a

~Y x 5 +~y x 8

13 x 12 x 11

6

(140 + 80) x 6

220 13 x 22

10

13

13 x 12₂ x11 Ans. D.

58. Let X = the event that outcame is head Given that

A B H/T H/H

^p(^a)=3

^p=(^b)=4

¹/4 x 1 _

¹/4

³ +1 5 4

3 11 i

4 ^x 2 + 4 ^x1

= ¹/4

1

_4

5

2

5

3 + 2 _

^{8 8} 2

Ans. B.

59. f : R R g : R R

f(x) = x² + 2x - 3 = x² + 2x + 1 - 4 = (x +1)² - 4 g(x) = x + 1 fg(x) = f (x + 1)

= (x + 2)² - 4

= x² + 4x + 4 - 4 = x² + 4x g[^f(^x)] = g (^x + ¹)^{2 - 4}

= (x +1)² - 4 +1 = (x +1)² - 3

= x² + 2x + 1 - 3 = x² + 2x - 2

x² + 4x = x² + 2x - 2 2x = -2 x = -1 Ans. A.

60. 2b = a + c

ax² - 2bx + c = 0

ax² - (a + c)x + c = 0

ax - ax - cx + c = 0 ax(x -1) - c(x -1) = 0 (ax - c) (x -1) = 0

c i

x = , x = 1 a

Ans. A.

2

dy

^y

61. x=

logx = t

1 dx = x' dy dy

dx 1 - ylogx 1 dx 1 1,

or,

dy	V	y
dt	1	t
dy	V	y
dt	t =	1
dy	y =	y²

IF. = e^Z y^dy = e^logy = y j d(t.y) = j ydy + k

t.y = logy + k y. logx = logy + k

1 x 0 = log 1 + k k = 0

y = xy

y. ^loge ^x = ^loge y Ans. B.

62. Solving, sin^-1 x + x(1 - x) = sin^-1 x - x(1 - x) or, 2x(1 - x) = 0 x = 0, 1

Required area = j_Q2x(1 - x)dx

1

= 2

0

2 3 x x

~2 3

²¹2^-3) =^{2 x} 6=3 ^sq. ^unit.

Ans. C.

63.

Area = 2 x 2 + y(2+6).2

= 4 + 1x 8 x 2

= 4 + 8 = 12 sq. unit Ans. C.

64. h(x) = f (x) - g(x)

h' (x) = f' (x) - g' (x)

h"(x) = f "(x) - g"(x)

h " (x) = 0

.. h' (x) = c

f' (x) - g' (x) = c

^f' (1) - g' (1) = c

4 - 6 = c c = -2

f' (x) - g' (x) = -2 f (x) - g(x) = -2x + c'

f(2) - g(2) = -2.2 + c'

3 - 9 = -4 + c'

-6 = -4 + c c' = -2

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

f(1) - g(1) = -2.1 - 2

= -4 Ans. B.

1

65. J(|x| - 2[x])dx

-1

1 1 J |x|dx - 2 J[x]dx -1 -1

0 1

. -x )dx + | xdx - 2

-1 0

J(-x)dx + J xdx - 2 J(-1)dx + 0

-1

r 2 o	0	r 2 o	1
x	+	x	+
2		2
-	-1	.	0

+ 2[x]-1

= 3

Ans. A.

^f z - 2 I K

66. arg

v z + 2 ) 3

z lies on a circle Ans. A.

67. a, b, c are in G.P.

b² = ac .. 2 log b = log a + log c

Now, a^p = b^q = c^r

: p log a = q log b = r log c log a q log c q

log b p log b r

From (1)

1 1 2 + = p ^r q

: p, q, r in H.P. Ans. C.

k

Z

k=1

68.

^sk

(-1)¹

Z-

k=1-

k

1 -

k + 1

(-1)" k (k + 1)

Z

k=1

11

k k +1

Z (-1)¹

k=1

1111111111

-1 + +-----+ +-----+ + .

2233445566

2 2 2 2

--+---+ .

3 4 5 6

1111

+---+ .

3 4 5 6

= 2

11111

----1-----1--.

2 3 4 5 6

= 2

-1

11111

--+---+---.

= -2

- 1

2 3 4 5 6

11111

1--+---+---.

= -2

+1

23456

= -2 jn2 + 1 = 1 - jn4 Ans. D.

69. .. O lies between the roots f(0) < 0

a - 4a < 0

a (a - 4) < 0

0 < a < 4 Ans. B.

.2

. loge (x² - 16) < loge (4x - 11)

70.

x² - 16 < 4x - 11

x² - 4x - 5 < 0

x² - 5x + x - 5 < 0 (x - 5)(x + 1) < 0

-1

-1 < x < 5

U

4

4x - 11 > 0 x >

x² - 16 > 0 x > 16

x > 4, x < -4

Req. Answer 4 < x < 5 Ans. A.

71. 1 + (1 + x) + (1 + x) +.... + (1 + x)

₁ (1 + x)²¹ - 1 . (1 + x) - 1

(1 + x)²¹ - 1

x

21

. Required coefficient _ C₁₁ [From N^r found coefficien of x¹¹] Ans. C.

72. To get infinite no of solution.

u _ 3

		X	1	1
Now		1	-1	-2	= 0
		-1	1	1
	-1	-2		-2	1		1 -1
X			+1			+
	1	1		1	-1	+	-1 1

= 0

or,

, X (-1 + 2) + 1( +2 - 1) + 1(1 - 1) = 0

or, X + 1 = 0 .-. X = -1 Ans. B.

73. p(A^C j = 0.3 P (A) = 0.7

P (B) = 0.4 P(B^C j = 0.6

P( A o B^C j = 0.5 P(B/ AU B^C j

p(b n(A U B^C jj

p(a u B^c j P((B o A) u ) _ P(A u B^c j

P (B n A)

P(A u B^C j

P(A) - P(A n B^C j P(A u B^C j

_0.7 - 0.5_

P (A) + P (B^C j - P (A n B^C j

- 0.2 _ j_

= 0.7 + 0.6 - 0.5 = 4 Ans. A.

74. A

S = ap = bq = cr

2 2 2

where a, b, c are the sides of the triangle

a + b + c = 2t

,_T 111abc2tt

Now i---+ =--1---1--= = -

p q r 2s 2s 2s 2s s

Ans. B.

75. x² + y² = 4

(x - 2)² + y² = 1 equation of common chord.

x² + y² - 4 - (x - 2)² + y² - 1 = 0

x² + y² - 4 - x² - 4x + 4 + y² - 1j = 0

x² + y² - 4 - x² - y² + 4x - 3 = 0

4x = 7

7

^x = 4 Here R = C1Q = 2

0

^RQ = J⁴-19 = T⁵

PQ =

V15

2

_acpq=nH =

^AC1^PQ 2 4 2 16

_{ac pq}=1 1 V15 = yi5

^AC2PQ 242 16

. ACjPQ 7

AC₂PQ 1 Ans. C.

76.

4

Any line through this point is y - = m

x -

3

4 (1 - m)

4(m - 1) ^A

-- 1

3m

and B

2 (m - 1)

Then coordinates of A & B are A

0

3

2 (1 - m)

and k

Let P (h, k) be the mid-point of AB. Then h =

3m 3

Eliminating m from the above two repations, we get 2(h + k ) = 3hk . The required locus is 2(x+y) = 3xy.

Ans. B.

77. Any chord of a parabola y² = 4ax, which subtends right angle at the vertex, always pass through a fixed point (4a, 0) on the axis of the parabola.

.. In this case R is (4, 0)

The distance of R from the vertex is 4.

Ans. C.

78.

The incentre, circumcentre, centroid of an equlateral triangle are same. |3(1) + 4 b¹) + 3|

= 2

Inradius =

I.

'3* + 4 Circumradius = 4

Circumcircle is (x 1)² + (y -1)² = 4

ⁱ.^e.^, x² + y² - 2x - 2y - 14 = 0

Ans. B.

1

79. Let L = lim

n!

H nⁿ K

n

1

logL = lim log

n-ro n

F1	2	n 1
	n	n ^K
2 +		n
		+ log
n		a

= lim

nro n

= logx dx J0

= [^(xlog^{x - x}0 = -1

.-. L = e^-1 = ¹

- ]Clog F -

n < \ n

e

= lim ¹

nro n

1

r_

n

r=1

Ans. D.

1

x =--

2

/

n

1

n =--

2

1

2

3

x101 / N101

(cosx) + (sinx)

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