MATHEMATICS – A4 1. If sin cos y a x b x = + , then 2 2 dy y dx + is (A) Constant (B) Function of x (C) function of y (D) Function of x and y Ans: A) 1 cos sin y a x b x = − ( ) 2 1 2 2 2 2 cos sin 2 sin cos y a x b x ab x x = + − ( ) 2 2 1 2 2 2 2 2 2 2 2 sin cos 2 sin cos cos sin 2 sin cos y y a x b x ab x x a x b x ab x x + = + + + + − 2 2 a b = + 2. If 2 3 ( 1) ( 1)( 2) ( ) 1 ... 2 6 n n n n n n f x nx x x x − − − = + + + + + then ''(1) f = (A) 1 2 n − (B) 1 ( 1)2 n n − − (C) 2 ( 1)2 n n n − − (D) ( 1)2 n n n − Ans: C) ( ) ( ) ( ) ( ) 1 1 ' 1 n n f x x f x n x − = + = + ( ) ( )( ) 2 '' 1 1 n f x n n x − = − + ( ) 2 '' 1 ( 1)2 n f n n − = − 3. If 1 tan / 2 tan / 2 1 A = − and AB I = then B = (A) ( ) 2 sin / 2 . A (B) ( ) 2 cos / 2 . T A (C) ( ) 2 cos / 2 . A (D) ( ) 2 cos / 2 . I An s: B) 1 2 1 tan 1 1 2 1 tan tan 1 2 2 AB I B A adj A A − − = = = = + 2 cos 2 T B A = 4. If 1 2 2 sin 1 x u x − = + and 1 2 2 tan 1 x v x − = − then du dv is (A) 1 (B) 1 2 (C) 2 (D) 2 2 1 1 x x − + Ans: A) ( ) 1 1 1 2 2 2 2 tan sin sin sin sin 2 2 1 1 tan x u x − − − = = = = + + 1 2tan x − = 2 2 1 du dx x = + ( ) 1 1 1 1 2 2 2 2 tan tan tan tan tan 2 2 2 tan 1 1 tan x v x x − − − − = = = = = − − 2 2 1 dv dx x = + 2 2 21 1 21 du du dx x dv dv dx x + = = = + 5. The function f(x) = cot x is discontinuous on every point of the set (A) , 2 n x n Z = (B) ; x n n Z = (C) 2 ; x n n Z = (D) (2 1) ; 2 x n n Z = + Ans: B) f (x) = cot x Is not defined x n = 6. If the function is 1 ( ) 2 f x x = + , then the point of discontinuity of the composite function y = f(f(x)) is (A) 1 2 (B) 5 2 − (C) 5 2 (D) 2 5 Ans: B) ( ) 1 2 2 ( ) 1 1 2 4 2 5 2 2 x x f f x x x x + + = = = + + + + + Discontinuous for 5 2 5 0 2 x x − + = = 7. An enemy fighter jet is flying along the curve given by 2 2 y x = + . A soldier is placed at (3, 2) wants to shoot down the jet when it is nearest to him. Then the nearest distance is (A) 5 units (B) 3 units (C) 6 units (D) 2 units Ans: A) Let (x, y) be point on curve then 2 2 ( 3) ( 2) d x y = − + − 2 2 4 ( 3) d x x = − + 2 4 ( 3) let t x x = − + 3 2( 3) 4 t x x = − + 3 4 2 6 0 x x + − = 0,1 roots = 2 2 3 t x = + (1) 0 t 1 x = gives min distance minimum distance 4 1 5 d = + = 8. 8 10 10 2 5 5 5 x x x dx − − = + (A) 3 (B) 5 (C) 6 (D) 4 Ans: A) 8 10 10 2 5 8 2 6 3 2 2 5 5 x x x dx − − − = = = + 9. cos sin ecx xdx − = (A) 2 sin C x + (B) sin x C + (C) sin 2 x C + (D) 2 sin x C + Ans: D) 2 1 sin cos sin sin sin x cosx ecx xdx dx dx x x − − = = 2 sin x C = + 10. If f(x) and g(x) are two functions with 1 ( ) g x x x = − and 3 3 1 ( ) fog x x x = − then f`(x) = (A) 2 1 1 x − (B) 3x 2 + 3 (C) 2 4 3 3 x x + (D) 2 2 1 x x − Ans: B) ( )( ) 3 3 1 fog x x x = − ( ) ( ) ( ) ( ) 3 3 1 1 1 1 1 3 3 x x x x x x x x x x = − + − = − + − ( ) ( ) ( ) 3 ( ) 3 g x g x = + 3 ( ) 3 f x x x = + ( ) 2 ' 3 3 f x x = + 11. A circular plat e of radius 5 cm is heated. Due to expansion, its radius increases at the rate of 0.05 cm/sec. The rate which its area is increasing when the radius is 5.2 cm is (A) 2 0.52 / sec cm (B) 2 5.2 / sec cm (C) 2 27.4 / sec cm (D) 2 5.05 / sec cm Ans: A) Given 0.05 / sec dr cm dt = , 5.2 ? r cm dA dt = = 2 A r = 2 dA dr r dt dt = 5.2 2 (5.2)(0.05) r cm dA dt = = 2 0.52 / sec cm = 12. The distance ‘s’ inmetres travelled by a particle in ‘t’ seconds is given by 3 2 5 18 3 3 t s t = − + . The acceleration when the particle comes to rest is (A) 18 m 2 /sec (B) 3 m 2 /sec (C) 10 m 2 /sec (D) 12 m 2 /sec Ans: D) 3 2 5 18 3 3 t S t = − + Velocity, 2 2 18 ds v t dt = = − , Since 2 2 18 0 t − = 2 2 18 t = 2 9 3 t t = = Acceleration a 4 dv t dt = = 2 3 4(3) 12 / sec t dv m dt = = = 13. A particle moves along the curve 2 2 1 16 4 x y + = . When the rate of change of abscissa is 4 times that of its ordinate, then the quadrant in which the particle lies is (A) II or III (B) I or III (C) II or IV (D) III or IV A ns: C ) 4 dx dy dt dt = --- (1) 2 2 0 16 4 x dx y dy dt dt + = 2 2 1 16 4 x x + = 2 8 4 x dx y dy dt dt − = 2 2 4 1 16 x x + = (4) 8 2 x y x y − = = − 5x 2 = 16 2 16 5 x = 2 4 1 , 1 4 5 5 y x = = − 2 2 4 16 4 5 4 y y = = Points 4 4 , 5 5 x y = 14. 2 5 2 x x dx − + = (A) 2 2 1 5 2 2 log | ( 1) 5 2 | 2 x x x x x x C − − + + − + − + + (B) 2 2 1 5 2 2 log | ( 1) 2 5 | 2 x x x x x x C − − + + + + + + + (C) 2 2 5 2 4 log | ( 1) 2 5 | 2 x x x x x x C − + + + + − + + (D) 2 2 1 5 2 2 log | ( 1) 5 2 | 2 x x x x x x C − + + + − + + + + Ans: . A) 2 2 2 2 2 5 2( )(1) 1 1 5 x x x x − + = − + − + 2 2 ( 1) 2 x = − + ( ) ( ) 2 2 2 2 1 4 1 2 5 2 log 1 5 2 2 2 x x x x x x x c − − + = − + + − + − + + 15. 2 2 1 1 3sin 8cos dx x x = + + (A) 1 2 tan 6 tan 3 x C − + (B) 1 1 tan (2 tan ) 6 x C − + (C) 1 2 tan tan 3 x C − + (D) 1 1 2 tan tan 6 3 x C − + Ans: D) Divide by cos 2 x 2 2 2 2 2 sec sec 1 tan 3tan 8 4 tan 9 xdx xdx x x x = + + + + 2 1 2 2 1 sec 1 2 2 tan tan 4 4 3 3 3 tan 2 xdx x C x − = = + + 1 1 2 tan tan 6 3 x C − = + 16. 0 3 2 2 ( 3 3 3 ( 1) cos( 1) x x x x x x dx − + + + + + + = (A) 1 (B) 0 (C) 3 (D) 4 Ans: D ) 0 4 2 3 2 3 3 4 2 x x x x − + + + ( ) 0 2 ( 1)sin( 1) cos( 1) x x x − + + + + + = ( ) ( ) (4 8 6 6) (sin1 cos1) (sin1 cos1) − − + − + + − + = 4 17. 0 tan sec .cos x x dx x ecx = (A) 2 / 2 (B) / 4 (C) 2 / 4 (D) / 2 Ans: C) Let 0 .tan sec .cos x x I dx x ecx = ------- 1) Replace x by x − ( )( ) 0 tan sec .cos x x x ecx − − − -------- ( 2) By adding (1) & (2) we get, 0 ( ) tan 2 sec .cos x x x I x ecx + − = 0 sin cos 2 1 1 cos sin x x I x x = 2 0 sin 2 I x dx = 0 1 cos 2 . 2 2 x dx − = 0 sin 2 4 2 x x = − 2 ( 0) (0 4 4 = − − = 18. If | | | | a b a b + = − then (A) Inclined to each other at 60 ° (B) a and b are perpendicular (C) a and b are parallel (D) a and b are coincident Ans: B) | | | | a b a b + = − Squaring on b.s 2 2 | | | | a b a b + = − 2 2 | | | | a b + 2 2 2 . | | | | a b a b + = + 2 . a b − 4 . 0 a b = 0 a b = a and b are perpendicular 19. The component of i in the direction of the vector 2 i j k + + is (A) 6 6 (B) 6 (C) 6 (D) 6 6 Ans: A) 2 | | 1 1 4 a i j k a a + + = = + + 1 1 2 6 6 6 i j k = + + Component of 1 6 6 6 i = = 20. In the interval ( ) 0, / 2 , area lying between the curves tan cot y x and y x and = = the x - axis is (A) l og 2 sq. units (B) 3 log 2 sq. units (C) 2 log 2 sq. units (D) 4 log 2 sq. units Ans: A) 4 0 / 4 tan cot A x dx x dx = + / 4 / 4 0 0 2 tan 2(log(sec )) 2log 2 log 2 xdx x = = = = 21. The area of the region bounded by the line 1 y x = + , and the lines 3 5 x and x = = is (A) 7 sq. units (B) 10 sq. units (C) 7 2 sq. units (D) 11 2 sq. units Ans: B) 5 5 2 3 3 25 9 16 ( 1) 5 3 2 10 2 2 2 2 x x dx x = + = + = + − + = + = 22. If a curve passes through the point (1,1) and at any point (x, y) on the curve, the product of the slope of its tangent and x co - ordinate of the point is equal to the y co - ordinate of the point, then the curve also passes through the point. (A) ( ) 3, 0 (B) (2, 2) (C) (3,0) (D) ( - 1, 2) Ans: B) log log( ) dy dy dx x y y xc dx y x = = = y xc = --- (1) Put x = 1 and y = 1 we get, c = 1 From equation (1) we get, y x = 23. The degree of the differential equation 2 2 2 2 3 2 2 1 1 dy d y d y dx dx dx + + = + is (A) 2 (B) 6 (C) 3 (D) 1 Ans: B) 3 2 2 2 2 2 2 1 1 dy d y d y dx dx dx + + = + degree is 6 24. The equation of the plane through the points (2, 1, 0), (3, 2, - 2) and (3, 1, 7) is A) 7 9 5 0 x y z − − − = B) 3 2 6 27 0 x y z − + − = C) 2 3 4 27 0 x y z − + − = D) 6 3 2 7 0 x y z − + − = Ans: A) 2 1 0 1 1 2 0 1 0 7 x y z − − − − = ( 2)7 (1 )(9) ( 1) 0 x y z − + − + − = 7 14 9 9 0 x y z − + − − = 7 9 5 x y z − − = 25. The point of intersection of the line 3 2 1 3 2 y z x + − + + = = with the plane 3 4 5 10 x y z + + = is A) (2, 6, 4) B) ( - 2, 6, - 4) C) (2, - 6, - 4) D) (2, 6, - 4) Ans: D) Put 3 2 1 3 2 y z x t + − + + = = = 1, 3 3, 2 2 x t y t z t = − = − − = − 2 2 z t = − 3( 1) 4(3 3) 5(2 2 ) 10 t t t − + − + − = 5 15 3 t t = = P oint of intersection is (2, 6, - 4) 26. If (2, 3, - 1) is the foot of the perpendicular from (4, 2, 1) to a plane, then the equation of the plane is A) 2 2 5 0 x y z + + − = B) 2 2 1 0 x y z − + + = C) 2 2 1 0 x y z + + − = D) 2 2 0 x y z − + = Ans: B) Direction rat io of P Q = ( - 2, 1, - 2) Eq uation of plane is 2( 2) 1( 3) 2( 1) 0 x y z − − + − − + = 2( 2) ( 3) 2( 1) 0 x y z − − − + + = 2 2 1 0 x y z − + + = 27. 2 2 144 a b a b + = and 4 a then b = is equal to A) 4 B) 12 C) 3 D) 8 Ans: C) ( ) 2 2 144 a b a b + = 2 2 2 2 ( . ) ( . ) 144 a b a b a b − + = 2 2 16| | 144 | | 9 | | 3 b b b = = = 28. If 2 3 0 a b c + + = and ( ) ( ) ( ) ( ) a b b c c a b c + + = Then the value of is equal to A ) 6 B) 2 C) 3 D) 4 Ans: D) 2 3 a b c + = − 2 0 a c b c + = 2( ) ( ) 2( ) b c a c c a b c = = − --- (1) 2 3 a b c + = − 3( ) 3( ) a b c b b c = − = --- (2) 3( ) ( ) 2( ) b c b c b c + − 4( ) 4 b c = = 29. If a line makes an angle of 3 with each X and Y axis then the acute angle made by Z – axis is A) 4 B) 6 C) 3 D) 2 Ans: A) 2 2 2 1 l m n + + = 2 2 2 cos cos cos 1 3 3 + + = 2 1 1 cos 1 4 4 + + = 2 1 cos 2 = 1 cos 2 = 4 = 30. The length of perpendicular drawn from the point (3, - 1, 11) to the line 2 3 2 3 4 x y z − − = = is A) 53 B) 66 C) 29 D) 33 Ans: A) (2 ,3 2, 4 3), (3, 1,1) P t t t Q + + − Directions ratio’s (2 3,3 3, 4 8) PQ t t t = − + − 2(2 3) 3(3 3) 4(4 8) 0 t t t − + + + − = 29 29 1, (2,5, 7) t t P = = Di stance 1 36 16 53 PQ = + + = 31. The shaded region in the figure given is the solution of which of the inequations? A) 7, 2 3 6 0, 0, 0 x y x y x y + − + B) 7, 2 3 6 0, 0, 0 x y x y x y + − + C) 7, 2 3 6 0, 0, 0 x y x y x y + − + D) 7, 2 3 6 0, 0, 0 x y x y x y + − + Ans: B ) 1 7 7 x y + = 1 3 2 x y + = − 0 x 7 x y + = 2 3 1 6 x y − = − 0 y 7 x y + 2 3 6 x y − = − 2 3 6 x y − − 32. If A and B are events such that ( ) ( ) 1 1 2 , / ( / ) ( ) 4 2 3 P A P A B and P B A then P B = = = is A) 1 2 B) 1 6 C) 1 3 D) 2 3 Ans: C) P ( ) ( | ) ( ) P A B A B P B = ( ) ( | ) ( ) P A B P A B P B = --- (1) ( ) ( | ) ( ) P B A P B A P A = ( ) ( | ) ( ) P A B P B A P A = ( ) ( | ) ( ) P A B P B A P A = --- (2) From (1) and (2) ( | ) ( ) ( | ) ( ) P A B P B P B A P A = 1 2 1 ( ) 2 3 4 P B = 2 1 ( ) 2 3 4 P B = 1 ( ) 3 P B = 33. A bag contains 2n +1 coins. It is known that n of these coins have head on both sides whereas the other n + 1 coins are fair. One coin is selected at random and tossed. If the probability that toss results in heads is 31 42 , then the value of n is A) 10 B) 5 C) 6 D) 8 Ans: A) 1 ( ) 2 1 n P E n = + 2 1 ( ) 2 1 n P E n + = + 1 1 ( ) 1 2 1 2 1 2 n n P Head n n + = + + + 2 ( 1) 3 1 2(2 1) 2(2 1) n n n n n + + + = = + + 3 1 31 2(2 1) 42 n n + = + 21(3 1) 31(2 1) n n + = + 63 21 62 31 n n + = + 63 62 10 n n − = n = 10 34. Let , , , , , A x y z u and B a b = = A function : f A B → is selected randomly. The probability that the function is an onto function is A) 1 35 B) 7 8 C) 1 8 D) 5 8 Ans: B) Total number of functions ( ) ( ) ( ) n A n B = Total number of onto 1 2 ( 1) ( 2) n r n r n r C r C r = − − + − 2 n – 2 2 4 – 2 = 14 ( ) 14 ( ) 16 n E n S = = ( ) 14 7 ( ) ( ) 16 8 n E P E n S = = = 35. The modulus of the complex number ( ) ( ) ( )( ) 2 1 1 3 2 6 2 2 i i i i + + − − is (A) 2 4 (B) 4 2 (C) 2 2 (D) 1 2 Ans : A) 2 (1 ) (1 3 ) (2 6 )(2 2 ) i i i i + + − − = 2 2 (1 2 )(1 3 ) (2 6 )(2 2 ) i i i i i + + + − − 2 (1 3 ) 2 6 (2 2 ) i i i i + = − − Apply modulus, we get 2 2 2 2 2 2 2 2 0 2 1 3 2 ( 6) 2 ( 2) + + + = + − + − 4 10 4 36 4 4 = + + = 4 10 40 8 = 2 5 2 2 10 2 2 5 = 2 5 2 = 1 2 2 4 2 2 2 = 36. Given that a, b and x are real numbers and a < b, x < 0 then (A) a b x x (B) a b x x (C) a b x x (D) a b x x Ans : B) a b x x 37. Ten chairs are numbers as 1 to 10. Three women and two men wish to occupy one chair each. First the women choose the chairs marked 1 to 6, then the men choose the chairs from the remaining. The number of possible ways is (A) 6 4 3 2 P C (B) 6 4 3 2 C C (C) 6 4 3 2 P P (D) 6 4 3 2 C P Ans : C ) 6 4 3 2 P P 38. Which of the following is an empty set ? (A) 2 : 2, x x x x R = + (B) 2 : 1 0, x x x R − = (C) 2 : 1 0, x x x R + = (D) 2 : 9 0, x x x R − = Ans: C) 2 2 1 0, 1 x x + = = − 1 x = − x i = R 39. If f(x) = ax + b, where a and b are integers, f( - 1) = - 5 and f(3) = 3 and a and b are respectively (A) 2, 3 (B) - 3, - 1 (C) 2, - 3 (D) 0, 2 Ans : C) ( ) f x ax b = + 5 ( 1) a b − = − + 3 (3) a b = + 5 a b − + = − --- (1) 3 3 a b + = ----- (2) 3 a b + 3 a b = − + 5 = − 6 3 b + = 4 8 a = 3 6 b = − a=2 3 b = − 40. The value of 0 0 0 0 10 10 10 10 log tan 1 log tan 2 log tan 3 ...... log tan 89 e + + + + is ( A) 1 (B) 0 (C) 3 (D) 1 e Ans : A) Since tan89 tan(90 1) = − = cot1 = 1 tan1 , 1 tan 88 tan 2 = and so on tan1 tan 2 ....tan 89 1 = 10 log tan1 tan2 tan3 ....tan89 e = 10 log (1) 1 e e = = 41. The value of 2 0 2 0 0 2 0 0 2 0 0 2 0 2 0 sin 14 sin 66 tan135 sin 66 tan135 sin 14 tan135 sin 14 sin 66 is (A) 2 (B) - 1 (C) 0 (D) 1 Ans : Can’t be determined 42. The distance between the foci of a hyperbola is 16 and its eccentricity is 2 . Its equation is (A) 2 2 32 y x − = (B) 2 2 32 x y − = (C) 2 2 1 4 9 x y − = (D) 2 2 2 3 7 x y − = Ans : B) 2 16 8 ae ae = = 2 e = 8 4 2 2 a = = 2 2 2 ( 1) b a e = − 32(2 1) 32 = − = Eq 2 2 2 2 1 32 32 32 x y x y − = − = 43. If ( ) 0 sin 2 sin 2 lim cos x x x A B x → + − − = , then the values of A and B respectively are (A) 1, 1 (B) 2, 2 (C) 1, 2 (D) 2, 1 Ans : B) 0 limcos(2 ) cos(2 ) cos x x x A B → + + − = cos 2 cos 2 cos A B + = 2cos 2 cos A B 2, 2 A B = = 44. If n is even and the middle term in the expansion of 2 1 n x x + is 924 x 6 , then n is equal to (A) 8 (B) 10 (C) 14 (D) 12 Ans : D) 2 / 2 2 6 2 1 2 1 ( ) 924 n n n n n c T n x x x − + = = 2 2 6 2 ( ) n n x x x = 6 2 12 n x x n = = 45. n th term of series 2 3 5 1 ....... 7 7 + + + is (A) 1 2 1 7 n n − + (B) 1 2 1 7 n n − − (C) 2 1 7 n n + (D) 2 1 7 n n − Ans : B) By inspection method if n = 1 then 1 2 1 7 n n − − = 1 1 2(1) 1 1 7 − − = If n=2 , 2 1 2(2) 1 3 7 7 − − = and so on 46. If 1 1 1 1 1 1 , , p q r q r r p p q + + + are in A.P., then p, q, r (A) are not in G.P. (B) are not in A.P. (C) are in G.P. (D) are in A.P. Ans : D ) By taking p = 1, q = 2, r = 3 1 1 1 1 1 , 1 ,3 1 2 3 3 2 + + + 5 16 27 , , 6 6 6 are in AP 47. A line passes through (2, 2) and is perpendicular to the line 3x + y = 3. Its y - intercept is (A) 4 3 (B) 1 3 (C) 2 3 (D) 1 Ans : A) ( ) 2, 2 3 3 x y + = r ⊥ line 3 0 x y k − + = 2 3(2) 0 k − + = 2 6 0 k − + = 4 k = 3 4 0 x y − + = y - intercept c b − = 4 4 3 3 − = = − 48. Let the r elation R be defined in N by aRb if 3a + 2b = 27 then R is (A) ( ) ( ) ( ) ( ) 2, 1 , 9, 3 , 6, 5 , 3, 7 (B) ( ) ( ) ( ) ( ) 1, 12 , 3, 9 , 5, 6 , 7, 3 (C) ( ) ( ) ( ) ( ) 27 0, , 1, 12 , 3, 9 , 5, 6 , 7, 3 2 (D) ( ) ( ) ( ) ( ) ( ) 1, 12 , 3, 9 , 5, 6 , 7, 3 , 9, 0 Ans : B) 3 2 27 a b + = (2,1) R (1,12) , (3,9) , 15 12 27 R R + = (5, 6) R (7,3) 21 6 27 + = 49. Let f(x) = sin 2x + cos 2x and g(x) = x 2 – 1, then g(f(x)) is invertible in the domain (A) 0, 4 x (B) , 4 4 x − (C) , 8 8 x − (D) , 2 2 x − Ans : C) ( ( )) g f x = (sin 2 cos 2 ) g x x + ( ) 2 sin 2 cos 2 1 x x = + − 2 2 sin 2 cos 2 2sin 2 cos2 1 x x x x = + + − 1 = sin 4 1 x + − sin 4 x 4 2 2 x − 8 8 x − 50. The contrapositive of the statement “ If two lines do not intersect in the same plane then they are parallel.” Is (A) If two lines are parallel then they do not intersect in the same plane. (B) If two lines are not parallel then they intersect in the same plane. (C) If two lines are parallel the n they intersect in the same plane. (D) If two lines are not parallel then they do not intersect in the same plane. Ans : B) 51. The mean of 100 observations is 50 and their standard deviation is 5. Then the sum of squares of all observations is (A) 255000 (B) 50000 (C) 252500 (D) 250000 Ans : (C) 100 i X n = 2 2 2 i i X X n n = − ( ) 2 2 25 50 100 i X = − 2 25 2500 100 i X = − 2 2525 100 i X = 2 252500 i X = 52. : f R R → and ) : 0, g R → are defined by f(x) = x 2 and g(x) = x Which one of the following is not true? (A) ( ) ( ) 2 2 gof − = (B) (gof) (4) = 4 (C) ( ) ( ) 4 4 fog − = (D) (fog) (2) = 2 Ans : C) 2 ( ) : f x x f R R = → ( ) :[0, ) g x x g R = → ( ) ( ( ) g of x g f x = , ( ) ( ) f og x f x = ( ) 2 x = = x ( ( )) f g x x = At 4, (4) 4 x g of = = 2, (2) 2 x fog = = 4, ( 4) 4 x fog = − − = − but given ( 4) 4 fog − = 53. Let : f R R → be defined by f(x) = 3x 2 – 5 and : g R R → by ( ) 2 1 x g x x = + then gof is (A) 2 4 2 3 9 30 2 x x x + − (B) 2 4 2 3 5 9 30 26 x x x − − + (C) 2 4 2 3 5 9 6 26 x x x − − + (D) 2 4 2 3 2 4 x x x + − Ans : B) 2 ( ) 3 5 f x x = − 2 ( ) 1 x g x x = + ( ) ( ( )) g of x g f x = 2 (3 5) g x = − 2 2 2 3 5 (3 5) 1 x x − = − + = 2 4 2 3 5 9 25 30 1 x x x − + − + 2 4 2 3 5 9 30 26 x x x − = − + 54. If A = 2 2 1 3 k k − − is singular matrix, then the value of 5k – k 2 is equal to (A) 6 (B) 4 (C) 6 − (D) 4 − Ans : B) 2 2 0 1 3 k k − = − (2 )(3 ) 2 0 k k − − − = 2 6 2 3 2 0 k k k − − + − = 2 5 4 0 k k − + = ( 4)( 1) 0 k k − − = 4,1 k = 2 5 5(1) 1 k k − = − =4 55. The area of a triangle with vertices ( - 3, 0), (3, 0) and ( 0, k) is 9 sq. units, the value of k is (A) 3 (B) 9 (C) 9 − (D) 6 Ans : A) 3 0 1 1 9 3 0 1 2 0 1 k − = + ( ) ( ) ( ) 18 3 0 0 3 0 1 3 k k = − − − − − + + 18 3 3 k k = + 18 = 6k k = 3 56. If 2 2 2 1 1 1 a a b b c c = and 1 1 1 1 bc ca ab a b c = then (A) 1 =− (B) 1 = (C) 1 3 = (D) 1 Ans : A) 2 2 2 1 1 1 a a b b c c = ( )( )( ) a b b c c a = − − − 1 1 1 1 bc ca ab a b c = = ( )( )( ) a b c a b c − − − − 1 = − 57. If 2 1 1 1 2 2 2 2 1 2 sin cos tan 1 1 1 a a x a a x − − − − + = + + − where ( ) , 0, 1 a x then the value of x is (A) 2 2 1 a a − (B) 0 (C) 2 a (D) 2 2 1 a a + Ans : A) 2 1 1 1 2 2 2 2 1 2 sin cos tan 1 1 1 a a x a a x − − − − + = + + − 1 1 1 2tan 2tan 2tan a a x − − − + = 1 1 2tan tan a x − − = 2 2 1 a x a = − 58. The value of 1 1 sin 1 sin cot 1 sin 1 sin x x x x − − + + − − + where 0, 4 x is (A) 2 x − (B) 2 x (C) 2 x − (D) 3 x − Ans : A ) ( ) 1 sin cos cos sin 2 2 2 2 ot sin cos cos sin 2 2 2 2 x x x x c x x x x − + + − + − − 1 2 cos 2 cot 2 2sin 2 x x x − = 59. If 3 1 15 2 1 5 x y + = − then the value of x and y are ( A) x = - 4, y = 3 (B) x = 4, y = 3 (C) x = 4, y = - 3 (D) x = - 4, y = - 3 Ans : B) 3 15 2 5 x y x y + = − 3x + y = 15 2x – y = 5 ________ 5x = 20 x = 4 y = 3 60. If A and B are two matrices such that AB = B and BA = A then A 2 + B 2 = (A) 2 BA (B) A + B (C) 2 AB (D) AB Ans : B) AB = B & BA = A A 2 + B 2 = ? BAB = B 2 ABA = A 2 AB = B 2 BA = A 2 B = B 2 A = A 2 A 2 + B 2 = A + B *******