"Maths BITSAT Question Bank PDF"
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This document is a question bank for the BITSAT exam, focusing on mathematics. It includes a variety of math problems, potentially covering different topics within the subject.
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M AT H E M AT I CS QUESTION BITSAT 2023...
M AT H E M AT I CS QUESTION BITSAT 2023 BANK ST INFORMATI O 1. If Re z 2 z 2 , then the locus of z is (A) parabola (B) circle (C) ellipse (D) hyperbola 2. If then the equation CODEX X2–abx – a2– 0 has (A)one positive root and one negative root (B) Both positive roots (C) Both negative roots (D) Non-real roots 3. If a +2b + 3c = 12, (a, b, R+), then the maximum value of ab2 c3 is (A) 23 (B) 24 (C) 2 6 (D) 25 4. Sum of n terms of the infinite series 1.32 252 3.72 .... is n n (A) 6 n 1 6n2 14n 7 (B) 6 n 1 6n2 14n 5 n (C) n 1 2n 1 3n 1 (D) 4n3 4n2 n 6 5. If log75 = , log5 3 = b and log32 = c, then the logarithm of the number 70 to the base 225 is 1 a abc 1 a abc (A) (B) 2a 1 b 2a 1 b 1 a abc 1 a abc (C) (D) 2a 1 b 2a 1 b 6. The maximum number of points of intersection of 10 circles is (A) 80 (B) 90 (C) 85 (D) 95 : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-1 C1 C C C C 7. 2 2 3 3 4 4 ....20 20 C0 C1 C2 C3 C19 (A) 120 (B) 260 (C) 210 (D)180 0 x p x q 8. If p # q # rand x p 0 x r 0, then the value of x which satisfy the x q x r 0 equationis (A) x = p (B) x = q (C) x = r (D) x = 0 x 3 2 9. Matrix A 1 y 4 , if xyz = 60 and 2 2 z 8x + 4y] + 3z = 20, then A(adj A) is equal to CODEX 64 0 0 88 0 0 68 0 0 34 0 0 0 88 0 (A) 0 64 0 (B) (C) 0 68 0 (D) 0 34 0 0 0 64 0 0 88 0 0 68 0 0 34 10. If f(x) = 4x – x2, x R, and f(a + 1) – f(a – 1) = 0, then is equal to (A) 0 (B) 2 (C) 1 (D) 3 11. Which of the following is not an equivalence relation in z? (A)aRb a + b is an even integer (B) aRb a – b is an even integer (C) aRb a < b (D) aRb a = b 12. Which of the following is always true? (A) ~ p ~ q p q (B) p q ~ q ~ p (C) ~ p ~ q p ~ q (D) ~ p q p q q p 13. The solution of the inequation 4 x 0.05 7.2 x 4, x R is 7 (A) 2 , (B) 2, (C) 2, (D) None of these 2 n 14. If cos3 x.sin 2x am sin mx is identity in x, then m 1 3 1 (A) a3 = ,a2 = 0 (B) n = 6, a1 = 8 2 3 1 (C) n = 5, a1 = (D) Σam = 4 4 : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-2 15. Total number of solutions of 1 cot x cot x , x 0.3 is equal to sin x (A) 1 (B) 2 (C) 3 (D) 0 The minimum value of sin1 x cos 1 x is equal to 3 3 16. 3 53 93 113 (A) (B) (C) (D) 32 32 32 32 17. The origin is shifted to (1, 2). The equation y2– 8x – 4y + 12 = O changes to y2 = 4ax,then a is equal to (A) 1 (B) 2 (C) –2 (D) –1 18. The equations of the bisector of the angles between the straight lines 3x + 4y + 7 = 0 and 12x +5y – 8 = are CODEX (A) 7x+ 9y + 17=0, 99x + 77y +51 = 0 (B) 7x – 9y – 17 = 0, 99x + 77y – 51 = 0 (C) 7x – 9y – 17 = 0, 99x +77y + 51 = 0 (D) None of the above 19. Equation of circle which passes through the points (1, –2) and (3, – 4) and touch the X-axis is (A)x2 + y2 + 6x + 2y + 9 = 0 (B)x2 + y2 + 10x + 20y + 25 = 0 (C) x2 + y2 + 6x + 4y + 9 = 0 (D) None of the above 20. If x = 9 is the chord of contact of the hyperbola x2– y2 = 9, then the equation of the corresponding pair of tangent is (A) 9x2– 8y2 + 18x – 9 = 0 (B)9x2– 8y2–18x + 9 = 0 (C)9x2– 8y2–18x – 9 = 0 (D)9x2– 8y2 + 18x + 9 = 0 21. The points with position vectors 10iˆ 3ˆj,12iˆ 15ˆj and aiˆ 11jˆ are collinear, if a is 82 (A) – 8 (B) 4 (C) 2 (D) 9 22. Let a, b,cbe vectors of lengths 3, 4, 5 respectively and a be perpendicular to (b + c), bto (c + a) and c to (a + b), then the value of (a + b + c) is (A) 2 5 (B) 2 2 (C) 10 5 (D) 5 2 23. For non-zero vectors a, b, c; a b . c a b c holds if and only if (A) a. b = 0, b. c = 0 (B)b. c = 0, c. a = 0 (C)c. a = 0, a. b = 0 (D)a. b = b. c = c. a = 0 : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-3 24. Angle between the diagonals of a cube is (A)/3 (B)/2 (C) cos1 1/ 3 (D) cos1 1/ 3 25. Consider the two lines x 1 y 2 z 1 L1 : 3 1 2 x2 y 2 z3 and L 2 : 1 2 3 The unit vector perpendicular to both the lines L1 and L2 is ˆi 7ˆj 7kˆ ˆi 7ˆj 5kˆ ˆi 7ˆj 5kˆ 7iˆ 7ˆj kˆ (A) (B) (C) (D) 99 5 3 5 3 99 26. The distance between the line r 2iˆ 2ˆj 3kˆ ˆi ˆj 4kˆ and the planea. ˆi 5ˆj kˆ 5 is (A) 10 9 CODEX (B) 10 3 3 (C) 10 3 (D) None of these 27. Two cards are drawn from a pack of 52 cards. What is the probability that either both are red or both are kings? (A) 7/13 (B) 63/221 (C) 55/221 (D) 3/26 1 1 28. If A and B are two independent events such that P A and P B , then which 2 5 of the following is correct? A 1 A 5 A B (A) P (B) P (C) P 0 (D) All of these B 2 A B 6 A ' B' 29. Box I contain 5 red and 2 blue balls, while box II contains 2 red and 6 blue balls. A fair coin is tossed. If it turns up head, a ball is drawn from box I, else a ball is drawn from box II. The probability ball drawn is from box I, if it is blue, is (A) 27/56 (B) 8/29 (C) 21/29 (D) 29/56 2 30. For a random variable X, E(X) = 3 and E(x ) = 11. The variable of x is (A) 8 (B) 5 (C) 2 (D) 1 31. The sum of 10 items is 12 and the sum of their squares is 18, then the standard deviation will be (A) – 3/5 (B) 6/5 (C) 4/5 (D) 3/5 32. The height of the chimney when it is found that on walking towards it 50 m in the horizontal line through its base, the angle of elevation of its top changes from 30° to 60° is (A) 25m (B) 25 2 m (C) 25 3 m (D) None of these : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-4 1 cos x 2 33. The value of lim is x 0 1 cos x (A)1/2 (B) 2 (C) 2 (D) None of these ax 2 1, x 1 34. If f x 2 is differentiable at x = 1, then x ax b, x 1 (A)a = 1, b = 1 (B) a = 1, b = 0 (C) a = 2, b = 0 (D) a = 2, b = 1 35. The slope of the tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point t = 2 is (A) 7/6 (B) 5/6 (C) 6/7 (D) 1 1 36. 1 2 sin x dx is equal to x x 2 3 tan tan 2 3 (A) 2 3 log 1 2 x tan 2 3 2 CODEX c (B) 2 3 log 2 x tan 2 3 2 c x 2 3 tan 1 2 (C) log c (D) None of these 3 x tan 2 3 2 1 log 1 x 37. 0 1 x2 dx is equal to 1 (A) log 2a (B) log (C) log 2 (D) None of these 8 8 2 4 38. The area of one curvilinear triangle formed by curves y = sin x, y = cos x and x-axis, is (A)2 sq units (B) 2 2 sq units (C) 2 2 sq units (D) None of the above x y 1 dy x y 1 39. Solution of , given that y = 1 when x = 1 is x y 2 dx x y 2 x y x y 2 2 2 2 (A) ln 2x y (B) ln 2x y 2 2 x y x y 2 2 2 2 (C) ln 2x y (D) ln 2 x y 2 2 : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-5 1 x dy 40. If y sin 2 tan1 , then is 1 x dx x x (A) 1 (B) –1 (C) (D) x 12 1 x2 41. The maximum value of the function Y = x(x – 1)2, is (A) 0 (B) 4/27 (C) –4 (D) None of these dy 42. The solution of x 3 4x 2 tan y e x secy satisfying y(1) = 0, is dx (A) tan y = (x – 2)ex logx (B) sin y = ex (x – 1)x–4 (C)tan y = (x – 1)ex x–3 (D) sin y = ex (x – 1)x–3 43. The runs of two players for 10 innings each are as follows A 58 59 60 54 65 66 52 75 69 52 B 94 CODEX 26 92 The more consistent player is 65 96 78 14 34 98 13 (A) player A (B) player B (C)both player A and B (D) None of the above 44. The linear programming problem minimise z = 3x + 2y subject to constraints x + y 8,3x + 5y 15, x 0 and y 0, has (A)one solution (B) no feasible solution (C)two solution (D) infinitely many solutions 45. Find the area enclosed by the loop in the curve 4y2 = 4x2 – x3. (A) 128/5 (B) 15/128 (C) 130/17 (D) 17/130 46. If a be a root of the equation 4x2 + 2x – 1 =0, then the other root of the equation is (A) 4a3 + 2a (B) 4a2 – 2a (C) 4a3 – 3a (D) 4a3 + 3a 47. If A = {x : x is a multiple of 4} and, B = {x : x is a multiple of 6}, then A B consists of multiples of (A) 16 (B) 12 (C) 8 (D) 4 1 48. If |w| = 2, then the set of points z = w is contained in or equal to the set of w points z satisfying (A) Im (z) = 0 (B) Im z 1 (C) Re z 2 (D) z 3 1 cos 1 cos x 49. The value of lim is x0 x4 1 1 1 1 (A) (B) (C) (D) 6 8 10 12 : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-6 50. Let a1,a2 ,.......a40 , be in AP and h1,h2 ,.......h10 be in HP. If a1 = h1 = 2 and a10 = h10 = 3, then a4h7 is (A) 2 (B) 3 (C) 5 (D) 6 1 5 2x , is 9 9 51. The number of terms in the expansion of 1 5 2x (A) 5 (B) 7 (C) 9 (D) 10 52. The number of different seven-digit numbers that can be written using only the three digit 1, 2 and 3 with the condition that the digit 2 occurs twice in each number is (A) 7 C2 25 (B) 7 p z 25 (C) 7 C2 5 2 (D) None of these 53. Given 2x – y + 2z = 2, x – 2y + z = –4 x + y + z = 4, then the value of such that the given system of equation has no solution is (A) –3 (B) 1 (C) 0 (D) 3 CODEX 1 1 1 4 2 2 54. Let A 2 1 3 and10B 5 0 If B is the inverse of A, then the value of 1 1 1 1 2 3 is (A) 4 (B) –4 (C) 3 (D) 5 7 55. If x 0, , then the value of cos 1 1 cos 2x 2 2 sin 2 x 48 cos2 x sin x is equal to (A) x cos 1 7 cos x (B) x sin 1 7 cos x (C) x cos 1 6 cos x (D) x cos 1 7 cos x 56. A running track of 440 ft is to be laid out enclosing a football field, the shape of which is a rectangle with a semi-circle at each end. If the area of the rectangular portion is to be maximum, then the lengths of its side are (A) 70 ft and 110 ft (B)80 ft and 120 ft (C) 35 ft and 110 ft (D) 35 ft and 120 ft dy dx tan x y sec x sin x, find general solution 2 57. (A) y tan x log cos ecx cot x cos x c (B) y sec 2 x tan x c (C) y log sec x tan x cos ec x c (D) y tan2 x sin x c 58. If the straight line y = mx + c touches the parabola y2 – ax + 4a3 = 0, the c is a a a a (A) am (B) am (C) a 2m (D) a 2m m m m m : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-7 59. A normal is drawn at the point P to the parabola y2 = 8x,which is inclined at 60° with the straight line y = 8. The the point P lies on the straight line (A) 2x y 12 4 3 0 (B) 2x y 12 4 3 0 (C) 2x y 12 4 3 0 (D) None of these 1 60. The value 1 dx, is x 1 x 2 3 5 4 1 1 1 1 4 x 1 4 3 x 1 4 4 x 1 4 1 2x 1 4 (A) C (B) C (C) C (D) C 3 x 2 4 x 2 3 x 2 3 4x 3 61. The area of the region bounded by the parabola (y – 2)2 = (x – 1), the tangent to the parabola at the point (2, 3) and the X-axis is (A) 3 (B) 6 (C) 9 (D) 12 62. ˆ uand CODEX vˆ are two non-collinear unit vectors such that uˆ vˆ 2 uˆ vˆ 1. Then the value of uˆ vˆ is equal to uˆ vˆ uˆ vˆ (A) (B) uˆ vˆ (C) uˆ vˆ (D) 2 2 63. A six faced die is a biased one. It is thrice more likely to show an odd numbers than show an even number. It is thrown twice. The probability that the sum of the numbers in two throws is even, is 5 5 1 (A) (B) (C) (D) None of these 9 8 2 1 64. The sum of all the solution of the equation cos cos cos , 0,6 3 3 4 100 (A) 15 (B) 30 (C) (D) None of these 3 2 2 65. Let be the solution of 16sin 16cos 10 in 0, . If the shadow of a vertical pole 4 1 is of its height, then the altitude of the sum is 3 (A) (B) (C) 2 (D) 2 3 66. For each parabola y = x2 + px + q, meeting coordinate axes of 3-distinct points, If circles are drawn through these points, then the family of circles must pass through (A) (1, 0) (B) (0, 1) (C) (1, 1) (D) (p, q) : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-8 67. The number of ways of arranging letters of the word HAVANA so that V and N do not appear together is (A) 40 (B) 60 (C) 80 (D) 100 68. Let a1, a2, a3 …. Be a harmonic progression with a1 = 5 and a20 = 25. The least positive integer n for which an< 0, is (A) 22 (B) 23 (C) 24 (D) 25 3x 1 z 1 69. If the plane 3x + y + 2z + 6 = 0 is parallel to the line 3y , then the 2b a value of 3a + 3b is 1 3 (A) (B) (C) 3 (D) 4 2 2 70. Let a, b be the solutions of x2 + px + 1 = 0 and c, d be the solution of x2 + qx + I = 0 If (a – c) (b – c) and (a + d) (b + d) are the solution of x2 + x + = 0, then is equal to (A) p + q 1 CODEX tan 1 (B) p – q tan a b 1 (C) p2 + q2 (D) q2 – p2 71. If then tan 1 tan 1 b a (A) a = 1 , b = 1 (B) a = sin 2, b = cos 2 (C) a = cos 2, b = sin 2 (D) None of these 1 1 1 x x e 2 ex 72. The value of lim is x 0 x2 11 11 e (A) e (B) e (C) (D) None of these 24 24 24 73. The locus of the mid-point of the chord if contact of tangents drawn from points lying on the straight line 4x – 5y = 20 to the circle x2 + y2 = 9 is (A) 20 x 2 y2 36x 45y 0 (B) 20 x 2 y2 36x 45y 0 (C) 36 x 2 y 20x 45y 0 2 (D) 36 x 2 y 20x 45y 0 2 x 2 dx 74. Let f x and f 0 0, then the value of f(1) be 1 x2 1 1 x2 (A) log 1 2 (B) log 1 2 4 (C) log 1 2 4 (D) None of these 75. The mean of five observations is 4 and their variance is 5.2. If three of these observations are 1.2 and 6, then the other two are (A) 2 and 9 (B) 3 and 8 (C) 4 and 7 (D) 5 and 6 : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-9 76. In a sequence of 21 terms, the first 11 terms are in AP with common difference 2 and the last 11 terms are in GP with common ratio 2. If the middle term of AP be equal to the middle term of the GP. Then the middle term of the entire sequence is 10 10 32 32 (A) (B) (C) (D) 31 31 31 31 77. If p a,p b,r c and the system of equations px + ay + az = 0 bx + qy + bz = 0 cx + cy + rz = 0 p p p has a non-trivial solution, then the value of is pa qb r c 1 (A) 1 (B) 2 (C) (D) 0 2 78. CODEX 1 If g x x 2 x 2and go f x 2x 2 5x 2, the f(x) is equal to 2 (A) 2x – 3 (B) 2x + 3 (C) 2x2 + 3x + 1 (D) 2x2 – 3x + 1 n 1 sin 8 icos 8 79. The smallest positive integral value of n such that is purely 1 sin icos 8 8 imaginary, is equal to (A) 4 (B) 3 (C) 2 (D) 8 80. A house subtends a right angle at the window of a opposite house and the angle of elevation of the window from the bottom of the first house is 60. If distance between two houses be 6 m, then the height of the first house is (A) 8 3 m (B) 6 3 m (C) 4 3 m (D) None of these 81. A spherical balloon is filled with 4500 cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 72 cubic meters per minute then the rate (in meters per minute) at which the radius of the balloon decreases 49 min after the leakage began is 9 7 2 (A) (B) (C) (D) 9 7 9 9 sin 3B 82. If in a ABC, 2b2 = a2 + c2, then is equal to sinB c 2 2 c 2 a2 c 2 a2 2 a2 c 2 a2 (A) (B) (C) (D) 2ca ca ca2 2ca : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-10 83. If the sum of the coefficients in the expansion of (x + y)n is 1024, then the value of the greatest coefficient in the expansion is (A) 356 (B) 252 (C) 210 (D) 120 84. The area enclosed by the curves y = sin x + cos x and y = |cos x – sin x| over the interval 0, is 2 (A) 4 2 1 (B) 2 2 2 1 (C) 2 2 1 (D) 2 2 2 1 sin sin 85. If , , 0, and If , , are in AP, then is equal to cos cos (A) sin (B) cos (C) cos (D) 2 cos CODEX : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-11 Answer Key 1. (A) Let z = x + iy Re(z + 2) z 2 x 2 2 Re(x + iy + 2) x iy 2 x 2 y2 Squaring on both sides, (x + 2)2 = (x – 2)2 + y2 X2 + 4 + 4x = x2 + 4 – 4x + y2y2 = 8x The locus of z is a parabola. 2. (A) Let be the roots of x2 – abx – a2 = 0 CODEX Where, + = = – 2 ,which shows product of roots < 0, i.e. one root must be negative and the other must be positive. Hence, equation has one positive root and one negative root. 3. (C) Given, a + 2b + 3c = 12 a, b, c, R+ As, AM GM abbc c c 6 2 3 ab c 6 12 6 2 3 ab c ab2c 3 26 6 Hence, the maximum value of ab2c3 is 26. 4. (A) an of the series = n(2n + 1)2 n N an n 4n2 4n 1 an 4n3 4n2 n Sn an 4n3 4n2 n n n 1 4n n 1 2n 1 n n 1 2 4 2 6 2 n n 1 2 4 2n 1 2n 2n 1 2 3 n n 1 6n2 6n 8n 4 3 2 3 : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-12 n n 1 6n2 14n 7 2 3 n 6 n 1 6n2 14n 7 5. (D) loge 70 loge 7 loge 5 loge 2 log225 70 loge 225 loge 25 loge 9 loge 5 loge 2 1 loge 7 loge 7 loge 5 loge 3 2 2 loge 7 loge 7 1 log7 5 log3 2 log5 3 log7 5 2 log7 5 log5 3 log7 5 2 a ab CODEX 1 a abc 1 a abc 2a 1 b 6. (B) Two circles intersect maximum at two distinct points. Now, two circles can be selected in 10C2 ways. The total number of points of intersection are 10C2 × 2 10 9 2 90 1 2 7. (C) C1 C n, 2 n 1 , C3 n 2 and so on. C0 C1 2 C2 3 C1 C C C n n 1 2 2 3 3 .....n n n C0 C1 C2 Cn1 2 On putting n = 20, C1 C C C n n 1 2 2 3 3 .....n n n C0 C1 C2 Cn1 2 8. (C) 0 x p x q Given, x p 0 x r 0 xq x r 0 x p x r xp 0 x p x q 0 xq 0 x q x r : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-13 x p x q x r x q x p x p x r 0 x r x p x q x q x p 0 x r x 2 px qx pq x 2 qx px pq 0 x r 2x 2 2pq 0 x r 0 or x 2 pq 0 Either x = r or pq 9. (C) A.adj A A I A xyz 8x 3 z 8 2 2 2y A xyz 8x 3z 4y 28 CODEX 60 20 28 68 adj A always exists whenever (A)–1 exists. 1 A. adjA A I 1 0 0 68 0 0 68 0 1 0 0 68 0 0 0 1 0 0 68 10. (B) Given, f(x) = 4x – x2 F(a + 1) – f(a – 1) 4 a 1 a 1 4 a 1 a 1 2 2 4a 4 a2 1 2a 4a 4 a2 1 2a 8 4a 0 a2 11. (C) Let R = {(a, b) : a + b is an even integer a, b Z} For a Z, a + a = 2a is an even integer. (a, a) R a Z R is reflexive. Let(a, b) R a + b is an even integer. (b + a) is an even integer (b, a) R : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-14 R is symmetric. Let (a, b), (b, c) R (a + b) and (b + c) are even integers. (a + b) + (b + c) = (a + c + 2b) is an even integer. (a + c + 2b) – 2b = (a + c) is an even integer. (a, c) R R is transitive. R is an equivalence relation. Let R = {(a, b) : (a – b) is an even integer, a, b Z}. For a Z, a – a = 0 is an even integer. (a, a) R a Z R is reflexive. Let (a, b) R (a – b) is an even integer. CODEX –(a – b) is an even integer. (b, a) R (b – a) is an even integer. (b, a) R R is symmetric. Let (a, b), (b, c) R (a – b) and (b – c) are even integers. (a – b) + (b – c) is an even integer. (a – c) is an even integer. (a, c) R R is transitive, R is an equivalence relation. Let R a,b : a b, a , b Z Let a,b : a b, a , b Z Let a Z. a a is false. R is not reflexive. R is not an equivalence relation. Let R a,b : a b, a , b Z. It is quite easy to check that R is an equivalence relation. 12. (B) Since, ~ p q ~ p ~ q and ~ p q ~ p ~ q So, options (a) and (d) are not true. : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-15 ~ p q q ~ q, so option (c) is not true. Now, p q ~ p q ~ q ~ p ~ ~ p ~ q q ~p ~pq p q ~ q ~ p 13. (A) 4–x+ 0.5 – 7.2–x< 4 Let 2–x = t The equation becomes 2t 2 7t 4 0 t 4 2t 1 0 1 1 2 t 4 t 0 t 4 2 2 CODEX Since, t 2 x 0 x R 0 t 4 0 2 x 22 As, 2x is an increasing function, – x < 2 or x > – 2 Solution is (– 2, ). 14. (A) cos 3x 3 cos x cos3 x. sin 2x sin 2x 4 1 3 sin 5x sin x sin 3x sin x 8 8 1 3 1 sin x sin 3x sin 5x 4 8 8 1 3 Here, n 5, a1 , a2 0, a3 , 4 8 a4 0, a5 1/ 8 15. (B) 1 cot x cot x sin x 1 Let cot x 0 cot x cot x 0 sin x 1 0 which is not possible. sin x : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-16 1 Let cot x 0 cot x cot x sin x 1 2 cot x sin x 1 2 8 cos x x , 2 3 3 The number of solutions is 2. 16. (A) Let I = (sin–1 x)3 + (cos–1x)3 sin1 x cos1 x sin1 x cos x sin x cos x 2 2 1 1 1 2 sin1 x cos1 x 3 sin1 x sin1 x 2 2 2 3 2 4 2 CODEX 2 sin1 x 3 sin1 x 2 2 2 2 1 3 sin x sin1 x 2 2 16 16 4 2 2 1 3 sin x 2 16 4 2 1 sin x 4 0 2 For minimum value put sin1 x 0 4 2 3 Minimum values 0 2 16 32 17. (B) Let P(x, y) be the original position of the points w.r.t the original axes. Let us move the origin at new position to (h, k). Hence, the position of the same point P in the new system is x’ = x – h y’ = y – k Here, (h, k) = (1, 2) x’ = (x – 1), y’ = (y – 2) According to the question, Y2 – 8x – 4y + 12 = (y – 2)2 – 4a(x – 1) : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-17 y2 – 8x – 4y + 12 = y2 – 4y + 4 – 4ax + 4a On comparing respective coefficients, we get 4a = 8 a = 2 18. (D) Equations of the bisectors of the angles between the given straight lines are given 3x 4y 7 12x 5y 8 by 9 16 144 25 13 3x 4y 7 5 12x 5y 8 39x 52y 91 60x 25y 40 Taking positive signs, 39x 52y 91 60x 25y 40 21x 27y 131 0 21x 27y 131 0 CODEX Taking negative signs, 39x 52y 91 60x 25y 40 99x 77y 51 0 19. (B) Since, the circle touches X-axis, (x h)2 y k k 2 2 …..(i) Also, it passes through the points (1, –2) and (3, –4). (1 – h)2 + (– 2 – k)2 = k2 ….. (ii) 2 2 2 and (3 – h) + (– 4 – k) = k ….. (iii) Subtracting Eq. (iii) from Eq. (ii), we get h=k+5 On solving these equations, we get k = – 10, – 2 and h = – 5, 3 By putting the values of (h, k) = (– 5, –10) or (3, – 2) In Eq. (i), we get x2 + y2 + 10x + 20y + 25 = 0 or x2 + y2 – 6x + 4y + 9 = 0 20. (B) x = 9 meets the hyperbola at 9,6 2 and 9, 6 2. Then, the equations of tangent at these points are 3x 2 2y 3 0 and 3x 2 2y 3 0. The combined equation of these two tangent is : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-18 9x2 – 8y2 – 18x + 9 = 0. 21. (D) Let the points the A 10iˆ 3ˆj , B 12iˆ 15ˆj and C aiˆ 11ˆj AB 2iˆ 18 ˆj and AC a 10 ˆi 8 ˆj Since , A, B and C are collinear, then 2 18 82 a a 10 8 9 22. (D) We have , a 3, b 4 and c 5. It is given that a b c ,b c a and c a b CODEX a. b c 0, b. c a 0 and c. a b 0 a. b a. c b.c b. a c. a c. b 0 or a. b b. c c.a 0 (adding all the above equations) 2 2 2 2 Now, a b c a b c 2 a. b b. c c.a 32 42 52 50 abc 5 2 23. (D) We have, a b . c a b c a b c sin cos a b c sin cos 1 and 0 2 a b and c || nˆ n b and c both a and b. a, b c are mutually perpendicular. a. b = b. c = c. a = 0 24. (C) Let each edge of cube be a then coordinates of the vertices of cube are O(0, 0, 0), A(a, 0, 0), B(0, a, 0), C(0, 0, a), N(a, a, 0), p (a, a, a), L(0, a, a), M(a, 0, a) : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-19 Z C L M P O Y B A N X Direction ratios of the diagonals OP, AL, BM and CN are (a, a, a), (a, – a, a) and (a, a, – a). Let be the acute angle between diagonals OP and AL. a1a2 b1b2 c1c 2 cos a b12 c12 2 1 a22 b22 c 22 a2 a2 a2 CODEX a a a a a a a 2 a2 a2 a 2 a 2 a 2 a2 3a2 3a2 a 3 a 3 1 1 cos cos1 3 3 25. (B) Given lines are x 1 y 2 z 1 x2 y2 z3 L1 : and L 2 : 3 1 2 1 2 3 Now, convert into vector form L1 : ˆi 2ˆj kˆ 3iˆ ˆj 2kˆ L2 : 2iˆ 2ˆj 3kˆ ˆi 2ˆj 3kˆ Line L1 comparing with a1 + b1, and L2 comparing with a2 + b2, then we have b1 3iˆ ˆj 2kˆ and b2 ˆi 2ˆj 3kˆ Perpendicular to both b1 and b2 = b1 × b2 ˆi ˆj kˆ 3 1 2 1 2 3 ˆi 3 4 ˆj 9 2 kˆ 6 1 ˆi 7ˆj 5 kˆ : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-20 ˆi 7ˆj 5 kˆ Re quired unit vector 1 7 5 2 2 2 ˆi 7ˆj 5kˆ ˆi 7ˆj 5kˆ 75 5 3 26. (B) The given line is r = a + tb, where a 2iˆ 2ˆj 3k, ˆ b ˆi ˆj 4kˆ and given plane is r. n d, where n ˆi 5ˆj k, ˆ d 5. Since, b.n = 1 – 5 + 4 = 0 given line is parallel to given plane. The distance between the line and the plane is equal to length of the perpendicular CODEX from the point a 2iˆ 2ˆj 3kˆ on the line to the given plane. Required distance 2iˆ 2ˆj 3kˆ . ˆi 5ˆj kˆ 5 1 25 1 2 10 3 5 10 27 3 3 27. (C) LetE1 = event of getting both red cards E2 = event of getting both kings and E1E2 = event of getting 2 kings of red cards 26 4 C2 325 C2 6 P E1 52 , P E 2 52 C2 1326 C2 1326 2 C2 1 and P E1 E2 52 C2 1326 P(both red or both kings) = P(E1E2) P E1 P E2 P E1 E2 325 6 1 330 55 1326 1326 1326 1326 221 28. (D) Since, A and B are independent events. 1 P A / B P A 2 : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-21 A P A A B P A B P A B PA 1/ 2 1/ 2 1/ 2 5 P A B 1 1 1 1 1 1 6 6 2 5 10 2 5 10 10 AB AB Similarly, P P 0 A ' B' A B ' 29. (B) Let E1 = coin shows head, E2 = coin shows tail, A = drawn ball is blue 1 P E1 P E2 2 CODEX P(A/E1) = Probability of drawing a blue ball from bag I = 2/7 P(A/E2) = Probability of drawing a blue ball from bag II = 6/8 By Baye’s theorem, we have 30. (C) Given, E(X) = 3 and E(X2) = 11 Variable of X = E(X2) – [E(X)]2 = 11 – 32 = 11 – 9 = 2 31. (D) Given, x1 + x2 + x3 +…+ x10 = 12 and x12 x 22 .... x10 2 18 2 1 1 2 n x2 x n 2 1 1 9 36 9 18 12 10 10 5 25 25 Standard deviation = 3/5 32. (C) Let PQ be the chimney whose height is h metres. Q h 30° 60° A 50 m B X P : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-22 PQ h In BPQ, tan 60 3 BP x h 3x.... i PQ and in APQ, tan 30° AP 1 h 3 50 x 50 x h 3 50 x 3x x 25 [using Eq. (i)] Height of the chimney 25 3 m 33. (C) The required limit lim CODEX 1 cos x 1 cos x 2 2 x0 1 cos x 1 cos x 2 sin x 2 2 lim x . 1 x0 sin x / 2 2 1 cos x 2 2 x 2 1 2 1 2 2 4 34. (C) f 1 h f 1 f ' 1 0 lim h 0 h a 1 h 1 1 a 2 lim h0 h lim a h2 2h 2a h 0 h f 1 h f 1 f ' 1 0 lim h0 h 1 h a 1 h b 1 a 2 lim h 0 h h2 2h ah b lim h 0 h : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-23 2 a, if b 0 Thus, 2a 2 a, b 0 a2 b0 35. (C) dx dy We have, 2t 3 and 4t 2 dt dt dy dy dt 4t 2 dx dx 2t 3 dt Thus, slope of the tangent to the curve at the point t = 2 is dy 4 2 2 6 dx 2 2 3 7 t 2 36. (C) 1 CODEX Let I = dx 1 2 sin x x 2 tan On putting sin x 2 , we get 2 x 1 tan 2 1 I dx x 4 tan 1 2 x 1 tan2 2 x 1 tan2 2 dx x x 1 tan 2 4 tan 2 2 x sec 2 2 dx x x 1 tan 2 4 tan 2 2 1 x 2 2 d tan 2 x tan 2 2 3 dt x 2 , Where t tan 3 2 t 2 2 2 : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-24 1 t2 3 2. log c 2 3 t2 3 x 2 3 tan 1 2 log c 3 x tan 2 3 2 37. (A) Let 1 log 1 x / 4 log 1 tan I dx sec 2 d 0 1 x 2 0 1 tan2 [Let x = tan dx = sec2d] / 4 log 1 tan d 0 / 4 0 CODEX log 1 tan d 4 / 4 1 tan log 1 d 0 1 tan / 4 2 log d 0 1 tan / 4 log2 1d I 0 2I log 2 I log 2 4 8 38. (C) Required area Y y = cos x y = sin x X’ X O 𝛑 𝛑 Y’ 𝟒 𝟐 / 4 / 2 sin x dx cos x dx 0 / 4 cos x 0 sin x / 4 /4 /2 : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-25 2 1 1 2 2 2 sq units 39. (B) x y 1 dy x y 1 x y 2 dx x y 2 dy d 1 d 1 1 or 1 dx dx 2 dx 2 d 1 1 2 2 2 dx 1 2 2 2 d 2 2 4 or 2 dx 2 2 d 2 dx 2 2 2 CODEX 1 2 d 2dx 2 On integrating, we get 1 log 2 2 2x c 2 1 y x log x y 2 c 2 or 2 Given, y = 1 when x = 1 x y 2 0 2 1 1 0 log 2 c or y x log 2 2 2 x y 2 2 or log 2 x y . 2 40. (D) 1 x Given, y = sin 2 tan1 1 x 1 cos 2 Let x cos 2 y sin 2 tan1 1 cos 2 2 sin2 y sin 2 tan1 2 cos2 : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-26 y sin 2 tan1 tan y sin 2 y 1 cos2 2 y 1 x2 dy 1 x 2x dx 2 1 x 2 1 x2 41. (B) We have, y = x(x – 1)2 dy x 1.1 2x x 1 2 dx 3x 2 4x 1 x 1 3x 1 dy For maximum or minimum, 0 CODEX dx x 1 3x 1 0 1 x ,1 3 d2 y d2 y Now, 6x 4 2 2 0 dx 2 dx x 1 3 1 y is maximum when x and maximum value is 3 4 y x 1 27 3 d2 y Also, 2 20 dx x 1 y is minimum when x = 1. 42. (B) dy 4 ex We have, cos y sin y 3 dx x x Let sin y = t dy dt cos y dx dx dt 4 ex t 3 dx x x 4 x dx IF e e 4 log x x 4 : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-27 The solution is ex tx 4 x 4. dx xex e x x3 sin y x 4 xex ex c x 1, y 0 sin y e x x 1 x 4 c 0 43. (A) Let us make the table from the given data Player A Player B xi xi – 61 (xi – 61)2 yi yi – 61 (yi – 61)2 58 –3 9 94 33 1089 59 –2 4 26 –35 1225 60 CODEX –1 1 92 31 961 54 –7 49 65 4 16 65 4 16 96 35 1225 66 5 25 78 17 289 52 –9 81 14 –47 2209 75 14 196 34 –27 729 69 8 64 98 37 1369 52 –9 81 13 –48 2304 x i 610 x i 61 526 2 y i 610 y i 61 11416 2 For player A, Mean xi 610 61 n 10 x 61 2 i SD N 526 7.25 10 y i 610 61 n 10 y 61 2 i SD N 11416 10 : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-28 = 33.79 Since, SD for player A is 7.25 < SD for player B is 33.79. Hence, player A is more consistent player. 44. (B) Table for equation x + y = 8 is x 0 8 y=8–x 8 0 Table for equation 3x + 5y = 15 is x 0 5 15 3x 3 0 y 5 y CODEX 8 7 6 5 4 3 2 1 x' x 1 2 3 4 5 6 7 8 y' It can be concluded from the graph, that there is no point which can satisfy all the constraints simultaneously. Therefore, the problem has no feasible solution. 45. (A) Substitute 0 for y in the equation 4y2 = 4x2 – x3, 0 = x2(4 – x) x = 0, x = 4 It means curve makes the loop symmetric about X-axis between 0 and 4. 4 2 4 4 Area 2 y dx 4x 2 x 3 dx x 4 x dx 0 2 0 0 Let 4 x t dx dt 4 Area 4 t t dt 0 4 2 2 4 4 t t t dt 4 t 3/2 t 5/ 2 0 3 5 0 : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-29 8 2 64 64 2 128 8 32 64 3 5 3 5 15 15 46. (C) 4x2 + 2x – 1 =0 1 1 , .... i 2 4 Also, 4a3 + 2a – 1 = 0 as is a root and we have to prove that b = 43 – 3. 1 2a 3a 2a2 2a 1 1 2 4a 2 4a 32 1 2a 4a 1 1 1 2a 2 2 1 fromEq. i CODEX Now, 2 Hence, = 43 – 3. 47. (B) Let x A B. Then, x A and x B x is a multiple of 4 and x is a multiple of 6 is a multiple of 4 and 6 both x is a multiple of 12. 48. (D) Wehave, 2 2 cos isin 1 z 1 2 cos i sin cos i sin 2 3 5 z cos i sin 2 2 9 25 17 z 3 4 4 2 Hence, option (d) is correct. 49. (B) 1 cos x 1 cos 1 cos x 2 sin2 2 : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-30 x 2 sin2 sin2 2 202 400 Similarly, 21: ? 21 3 24 242 576 50. (D) Let d be the common difference of the AP. Then, a10 3 a1 9d 3 1 2 9d 3 d 9 1 7 a 4 a1 3d 2 3 3 CODEX 1 1 1 Let D be the common difference of , ,... h1 h2 h10 Then h10 = 3 1 1 1 1 9D h10 3 2 3 1 1 9D D 6 54 1 1 1 1 7 6D h7 h1 2 9 18 18 h7 7 7 18 a 4h7 6 3 7 51. (A) We have, 1 5 2x 1 5 2x 9 9 C C 5 2x .... C 5 2 2 9 9 0 0 8 2x 8 Clearly, it has 5 terms. 52. (A) Other than 2, remaining five places can be filled by 1 and 3 for each place. The number of ways for five places is 2 × 2 × 2 × 2 × 2 = 25. For 2 selecting 2 places out of 7 is 7C2. Hence, the required number of ways is 7C2 × 25. : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-31 53. (D) Since, the system has no solution. 2 1 2 1 2 1 0 1 1 2 2 1 1 1 2 3 0 4 2 1 6 0 3 9 3 54. (D) 1 1 1 A 2 1 3 1 1 1 1 1 1 CODEX A 2 1 3 1 1 1 11 3 1 2 3 1 2 1 4 5 1 10 T 4 5 1 4 2 2 Adj A 2 0 2 5 0 5 2 5 3 1 2 3 4 2 2 B A 1 1 5 0 5 A 1 1 Adj A 10 A 1 2 3 4 2 2 10B 5 0 5 1 2 3 Hence, = 5 55. (A) 7 y cos1 1 cos 2x 2 sin 2 x 48 cos2 x sin x cos1 7cos x cos x 1 49 cos2 x 1 cos2 x cos1 cos x cos1 7 cos x cos x 7 cos x x cos1 7 cos x : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-32 56. (C) D x C r r r r r r A x B Perimeter = 440 ft 2x r r 440 2x 2r 400 A = Area of the rectangular portion = x2r 440 2x 1 Ax 400x 2x 2 Let dA 1 400 4x 0 dx CODEX d2 A x 110 for which 0 dx 2 A is maximum where x = 110 440 2x 440 220 2r 70 22 / 7 r 35 ft and x 110 ft 57. (A) Given that dy dx tan x y sec x sin x, 2 dy tan x y sec 2 x sin x dx dy sec 2 x sin x y dx tan x tan x sec 2 x Here,P andQ cos x tan x sec 2 x IF e P dx dx e tan x e log tan x logcot x IF e cot x Now, y IF Q IF dx C : 353, Rajeev Gandhi Nagar, Instrumentation Limited Colony, Kota, Rajasthan 324005 DOWNLOAD OUR APP & GET UNLIMITED PRACTICE FOR FREE : www.atpstar.com PAGE NO.-33 y.cot x cos x.cot x C 1 sin2 x y.cot x C sin x y.cot x log |cosec x – cot x| + cos x + C 58. (D) Solving the given equations, (mx + c)2 = 4ax – 4a3 m2 x 2 2mc.x c 2 4ax 4a3 m2 x2 2mc 4a x c 2 4a3 0 Since the straight line touches the parabola at a point, so the discriminant = 0 2mc 4a 4m2 c 2 4a3 0 2 CODEX m2c 2 16amc 16a2 4m2c 2 16a3m2 0 mc a a2m2 0 a mc a a2m2 c a 2m m 59. (C) For the parabola y2 = 4 ax, the equation of normal at P(am2, – 2am) is y = mx – 2am – am3. Here, m tan 60 3 P a 3 3 6, 4 3 2
