AIEEE Solved Model Exam Test Physics, Mathematics and Chemistry

www.iitscholars.com XII Pass(8-9 Pass) Batches MATHEMATICS CHEMISTRY PHYSICS Q. No. Answer Q. No. Answer Q. No. Answers 1 C 36 A 71 C 2 C 37 A 72 C 3 B 38 D 73 D 4 A 39 B 74 C 5 B 40 B 75 C 6 D 41 A 76 C 7 A 42 B 77 C 8 B 43 C 78 B 9 D 44 C 79 D 10 D 45 A 80 A 11 A 46 A 81 B 12 A 47 B 82 A 13 A 48 A 83 D 14 A 49 A 84 D 15 A 50 A 85 D 16 A 51 C 86 C 17 B 52 D 87 C 18 C 53 D 88 B 19 C 54 B 89 D 20 D 55 B 90 B 21 A 56 B 91 A 22 A 57 A 92 D 23 C 58 D 93 B 24 C 59 B 94 D 25 D 60 B 95 C 26 C 61 B 96 C 27 C 62 B 97 D 28 D 63 A 98 D 29 C 64 B 99 A 30 A 65 C 100 B 31 B 66 A 101 A 32 C 67 B 102 B 33 A 68 A 103 B 34 A 69 B 104 C 35 C 70 A 105 A www.iitscholars.com HINTS & SOLUTIONS Mathematics, Physics, Chemistry 1. x2(y – 1) + x(y – 1) + (y – 2) = 0 For real x, D 0 2. 2y = 2 – 2x > 0 3. h(x) = f[g(x)] = x 4. 2 2 a b − + + c £ a sin x + b cos x + c + 2 2 a b c + + 5. LCM of (2p, p) = 2p 6. 7. ex ® 0 (sin x. cot x) 8. 1/3 1/3 2 2 1/5 1/5 4 4 1 1 1 x 1 x x x 1 1 1 1 x 1 x x −− + − − ®¥ = + − + 9. ax2 + bx + c = a (x – a) (x – b) So required limit 2 2 2 2 a(x )(x ) 2a sin (x ) 2 x a(x )(x ) 4 2 −a −b −b ®a −a −b 10. It is a continuous function. 11. f(0+) = f(0) 12. x 4 1 tan x lim 4x p ® − − p 13. x 1 sin x cos x lim 1 sin x cos x ®p − + + + www.iitscholars.com 15. 1 2 3 y cos cos x sin x 13 13 − = + = cos–1(cos a cos x + sin a sin x) y = cos–1cos(a − x) or cos–1cos(x – a) y = a − x or x − a 1 2 y cos x 13 − = − or 1 2 x cos 13 − − dy 1 or 1 dx = − + 21. For Monotonic increasing dy 0 dx ³ 22. f(x) = x2 + ax + 1 f ¢(x) = 2x + a for increasing function f ¢(x) > 0 2x + a > 0 …..(i) 1 x 2 2 2x 4 2 + a 2x + a 4 + a …..(ii) By 2x + a a + 2 …..(iii) By (i) & (iii) a + 2 > 0 a > − 2 23. For max. value tanA = tanB A = B 25. Min. value of 2 D 1 f (x) m(b) 4a 1 b = − = = + So 0 < m(b) 1 26. Y X O 27. For horizontal tangent dy 0 dx = 28. dx y dy 31. For differentiable 1 + |x| ¹ 0 Valid for x Î R 32. x y | f (x) f (y) lim | x y | x y ® − £ − − www.iitscholars.com |f ¢(x)| 0 f ¢(x) = 0 So f(x) = constant k f(x) = k f(0) = 0 (so, k = 0) \ f(x) = 0 \ f(1) = 0 34. 35. f ¢(1−) ¹ f ¢(1+) f ¢(0−) = f ¢(0) = f ¢(0+) 71. By ( ) n a S ut 2n 1 2 = + − 72. ( ) 2 u sin 2 80 T 16sec g 10 q × = = = Again 2u sin 2usin30 T 16 u 160m/s g 10 q = = = At T 3 t ,v u cos30 160 80 3m/s 2 2 = = = = 73. By conservation of energy at top position v = 0 74. v a t D = D 75. By smax 1 T f & T m g = = where 1 m is mass of hanging part. 76. Displacement = 0 work = 0 77. ag W k U = D + D f 0 U U U = + D = − i 81. 3 2 t v t ,at t 0 v 0 3 = = = = x at t = 2 v = 4 89. 3 v 8t = www.iitscholars.com at t = 0, v = 0 ( ) 2 2 f 1 k m v v 2 D = − i at t = 1, v = 8 = 64 J 91. net W K = D [ ] [ ] ( ) 2 2 1 m 20 10 2 = − 300J = + 93. F ma = F a m = 2 v 0 2as = + 2 F v 2 d m = 2 1 1 F K.E. mv m 2 d 2 2 m = = i.e., independent of ‘m’. 96. N + F = Mg N Mg F = − N = 50 − F 20 N = µ 20 [50 F) = µ − 20 0.5[50 F] = − F = 10 Newton 99. 2 1 mu E 2 = at highest point K.E. = ( )2 1 m u cos30 2 2 1 3 mu 2 4 = 3 E 4 = 105. W = DU F q ( ) FLcos 45 Mg L Lcos 45 = − ( ) F mg 2 1 = − F N 20 Mg fsmax