HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI 1. Find the mean number o f heads in three tosses of a fair coin: (A) 1.5 (B) 4.5 (C) 2.5 (D) 3.5 Ans: A) , , , , , , , S HHH HHT HTH THH HTT THT TTH TTT = n( S ) = 8 X = number of heads 3,2,1,0 X = Mean 1 1 3 3 1 ( ) 6 1 2 3 8 8 8 8 n i i i X p X = = = + + + 0 3 6 3 8 + + + = 12 8 = = 1.5 2. If A and B are two events such that 1 1 ( ) , ( ) 2 3 P A P B = = and 1 ( | ) 4 P A B = , then ( ) P A B is (A) 1 4 (B) 3 16 (C) 1 12 (D) 3 4 Ans: A) 1 ( ) 2 P A = , 1 ( ) 3 P B = , 1 ( ) 4 P A B = , 1 ( ) 12 P A B = ( ) ( ) ( ) 1 P A B P A B P A B = = − 9 1 12 = − 1 4 = 3. A pandemic has been spreading all over the world. The probabilities are 0.7 that there will be a lockdown, 0.8 that the pandemic is controlled in one month if there is a lockdown and 0.3 that it is controlled in one month if there is no lockdown. The probability that the pandemic will be controlled in one month is (A) 0.65 (B) 1.65 (C) 1.46 (D) 0.46 Ans: A) 1 : E Event that the Lockdown 2 E : not a lockdown A: Controlled in one month 1 ( ) 0.7 P E = X X= 0 X = 1 X= 2 X = 3 P(X) 1 8 3 8 3 8 1 8 CREATIVE LEARNING CLASSES, KARKALA K - CET KEY ANSWER WITH DETAILED SOLUTION - 2022 DEPARTMENT OF MATHEMATICS VERSION CODE – A3 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI 2 ( ) 0.3 P E = 1 ( / ) 0.8 P A E = 2 ( / ) 0.3 P A E = Using T heorem of total probability 1 1 2 2 ( ) ( ). ( / ) ( ). ( / ) P A P E P A E P E P A E = + 7 8 3 3 10 10 10 10 = + 56 9 100 100 = + 65 0.65 100 = = 4. If A and B are two independent events such that ( ) 0.75, ( ) 0.65 P A P A B = = and ( ) P B x = , then find the value of x : (A) 5 14 (B) 8 15 (C) 9 14 (D) 7 15 Ans: D) ( ) 0.75 P A = A & B are independent events ( ) 0.25 P A = ( ) ( ) ( ) P A B P A P B = ( ) P B x = ( ) 0.65 P A B = ( ) ( ) ( ) ( ) P A B P A P B P A B = + − 0.65 0.25 ( ) ( ) x P A P B = + − 0.35 0.25 x x = − 0.35 (1 0.25) x = − 0.35 (0.75) x = 0.35 7 0.75 15 x = = 5. Suppose that the number of elements in set A is p, the number of elements in set B is q and the number of elements in A×B is 7 then 2 2 p q + = ___________. (A) 50 (B) 51 (C) 42 (D) 49 Ans: A) ( ) , ( ) n A p n B q = = ( ) 7 n A B = ( ). ( ) 7 n A n B = 7 p q = This is possible only when 1 & 7 p q = = or p = 7 & q = 1 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI 2 2 1 49 50 p q + = + = 6. The domain of the function 10 1 ( ) 2 log (1 ) f x x x = + + − is (A) ) 2,0 (0,1) − (B) ) 2,1 − (C) ) 2,0 − (D) ) 2,0 (0,1) − Ans: D) 1 2 log(1 ) x x + + − 2 0 x + & 1 – x > 0 and 0 x 2 x − 1 > x x < 1 ) ( ) 2,0 0,1 x − 7. The trigonometric function tan y x = in the II quadrant (A) Decrease from 0 to (B) decrease from − to 0 (C) Increases from 0 to (D) increases from − to 0 Ans: D ) 8. The degree measure of 32 is equal to (A) 0 5 30 20 (B) 0 5 37 20 (C) 0 5 37 30 (D) 0 4 30 30 Ans: C) D.M 180 ( ) R M = 180 = 32 180 32 = 0 20 5 60 32 = 0 5 37 30 = 9. The value of 5 sin sin 12 12 is (A) 0 (B) 1 (C) 1 2 (D) 1 4 Ans: D) HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI 5 sin sin 12 12 = 1 5 .sin 2 2 12 = 1 5 .sin 2 6 = 1 1 1 2 2 4 = 10. 2 2 2 2 cos8 + + + = (A) sin 2 (B) 2 cos (C) 2 sin (D) 2 cos 2 Ans: B) WKT, 2 2 2 2cos 2 2cos n + + + = Required Answer = 2 cos 11. If 1,2,3,....10 A = then number of subsets of A containing only odd numbers is (A) 31 (B) 27 (C) 32 (D) 30 Ans: A) {1,3,5,7,9} B = ( ) 5 n B = Required answer 5 2 1 = − 32 1 31 = − = 12. If all permutations of the letters of the word MASK are arranged in the order as in dictionary with or without meaning, which one of the following is 19 th word? (A) KAMS (B) SAMK (C) AKMS (D) AMSK Ans: Incorrect option 13. If 1 2 3 10 , , ,..., a a a a is a geometric progression and 3 1 25, a a = then 9 5 a a equals (A) 3(5 2 ) (B) 5 4 (C) 5 3 (D) 2(5 2 ) Ans: B) 3 1 25 a a = 2 25 ar a = 2 25 r = 5 r = 8 9 4 5 a ar a ar = 4 r = = 5 4 14. If the straight line 2 3 17 0 x y − + = is perpendicular to the line passing through points (7, 17) and (15, ), then equals (A) - 5 (B) 5 (C) 29 (D) - 29 Ans: B) 2 3 17 0 x y − + = 3 2 17 y x = + HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI 2 17 3 3 y x = + Slope of the line 2 3 17 0 x y − + = is 1 2 3 m = Slope of line joining (7,17) and (15, ) is 2 17 15 7 m − = − 17 8 − = Since 1 2 1 m m = − 2 17 1 3 8 − = − 17 12 − = − 5 = 15. The octant in which the point (2, - 4, - 7) lies is (A) Eighth (B) Third (C) Fourth (D) Fifth Ans: A ) (2, - 4, - 7) = Eight h octant 16. If 2 1, 0 2 ( ) 2 3, 2 3 x x f x x x − = + , the quadratic equation whose roots are 2 lim ( ) x f x − → and 2 lim ( ) x f x + → is (A) 2 14 49 0 x x − + = (B) 2 10 21 0 x x − + = (C) 2 6 9 0 x x − + = (D) 2 7 8 0 x x − + = Ans: B) 2 2 2 2 2 lim ( ) lim 1 3 lim ( ) lim 2 3 7 x x x x f x x a f x x b − − + + → → → → = − = = = + = = Required Quadratic equation ( ) 2 2 0 10 21 0 x a b x ab x x − + + = − + = 17. If 3 (4 ) 6 x i x y i + − = − where x are real numbers, then the values of x and y are respectively, (A) 3, 9 (B) 2, 4 (C) 2, 9 (D) 3, 4 Ans: C) 3 6 2 x x = = and 4 1 8 1 9 x y y y − = − − = − = 18. If the standard deviation of the numbers - 1, 0, 1 k is 5 where k > 0, then k is equal to (A) 5 4 3 (B) 6 (C) 10 2 3 (D) 2 6 Ans: D) 2 2 2 i x k = + and i x k = HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI ( ) 2 2 i x x S D − = 2 2 2 5 4 4 k k + = − 2 2 2 5 4 16 k k + = − 2 2 8 4 5 16 k k + − = 2 80 8 3 k = + 72 = 3k 2 k 2 = 24 2 6 k = 19. If the se t x contains 7 elements and set y contains 8 elements, then the number of bijections from x to y is (A) 0 (B) 8 P 7 (C) 7 ! (D) 8 ! Ans: A) Zero because ( ) ( ) n x n y 20. If : f R R → be defined by 2 2 : 3 ( ) : 1 3 3 : 1 x x f x x x x x = then ( 1) (2) (4) f f f − + + is (A) 5 (B) 10 (C) 9 (D) 14 Ans: C) When x = - 1, f(x) = 3x 2 2, ( ) x f x x = = 4, ( ) 2 x f x x = = ( 1) (2) (4) f f f − + + = 2 3( 1) 2 2(4) − + + 3 4 8 = − + + =9 21. Let the relation R is defined in N by a R b, if 3a + 2b = 27 then R is (A) ( )( )( )( ) 1,12 3,9 5,6 7,3 (B) ( )( )( )( ) 27 0, 1,12 3,9 5, 6 7,3 2 (C) ( )( )( )( )( ) 1,12 3,9 5,6 7,3 9,0 (D) ( )( )( )( ) 2,1 9,3 6,5 3,7 Ans : A) By inspection method. 22. 3 3 0 3 3 lim y y y → + − = (A) 1 2 3 (B) 1 3 2 (C) 2 3 (D) 3 2 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI Ans: A) ( ) ( ) 12 12 3 3 0 1 3 2 3 0 3 3 0 lim 0 3 3 lim 3 3 1 1 3 2 2 3 y y y form y y y − → → + − = + − + − = = 23. If A is a matrix of order 3 3, the (A 2 ) - 1 is equal to (A) ( - A 2 ) 2 (B) (A - 1 ) 2 (C) A 2 (D) ( - A) - 2 Ans: B) ( ) 2 1 2 2 ( ) adj A adjA adjA A A A A − = = ( ) 2 2 adjA A = 2 adjA A = 1 2 ( ) A − = 24. If 2 1 , 3 2 A − = − then the inverse of the matrix A 3 is (A) A (B) - 1 (C) 1 (D) - A Ans: A ) 2 1 3 2 A − = − 2 1 0 0 1 A = 3 2 1 3 2 A − = − 3 4 3 1 A = − + = − ( ) ( ) 3 1 3 3 2 1 2 1 ( ) 1 3 2 3 2 1 adj A A A − − − = = = − − − 25. If A is a skew symmetric matrix, then A 2021 is (A) Row matrix (B) Column matrix (C) Symmetric matrix (D) Skew symmetric matrix Ans: D) Skew symmetric If A is skew symmetric then the odd power of matrix A is also skew symmetric. 26. If 0 1 0 0 A = then ( ) n aI bA + is (where I is the identity matrix of order 2) (A) 2 1 n a I a b A − + (B) 2 1 n a I n a b A − + (C) 2 n a I n a b A + (D) n n a I b A + Ans: B) 0 1 0 0 A = ( ) 1 n n n aI bA a I n a b A − + = + By P.M.I HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI 27. If A is a 3 3 matric such that 5. 5 adj A = then A is equal to (A) 1 (B) 1/ 25 (C) 1/ 5 (D) 5 Ans: C) 5. 5 adj A = n k adjA k adjA = 3 5 5 adjA = 1 n adjA A − = 3 1 3 5 5 A − = 25 2 5 A = 2 1 25 A = 1 5 A = 28. If there are two values of ‘a’ which makes determinant 1 2 5 2 1 86 0 4 2 a a − = − = . Then the sum of these numbers is (A) - 4 (B) 9 (C) 4 (D) 5 Ans: A) 1 2 5 2 1 0 4 2 a a − = − 2 1 2 4 2 4 20 a a = + − − − 2 2 8 44 a a = + + 2 2 8 44 86 a a + + = 2 2 8 42 0 a a + − = 2 4 21 0 a a + − = ( 7) 3( 7) 0 a a a + − + = ( 7)( 3) 0 a a + − = A = 3 or a = - 7 Sum = 3 – 7 = - 4 29. If the vertices of a triangle are ( - 2, 6) (3, - 6) and (1, 5) then the area of the triangle is (A) 40 sq. units (B) 15.5 sq. units (C) 30 sq. units (D) 35 sq. units Ans: B) A rea of triangle 1 1 2 2 3 3 1 1 1 2 1 x y x y x y = 2 6 1 1 3 6 1 2 1 5 1 − = − 1 31 2 = 31 15.5 2 sq units = = HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI 30. Domain of cos - 1 [x] is, where [ ] denotes a greatest integer function (A) ( - 1, 2] (B) ( - 1, 2) (C) [ - 1, 2) (D) [ - 1, 2) Ans: D) [ - 1, 2) 31. If y = ( ) 2 1 1 tan dy x x x then dx − + − is (A) 2 x tan - 1 x (B) 1 tan x x − (C) 2 1 tan x x − (D) 1 tan x x − Ans: A) ( ) 2 1 2 1 1 1 1 tan .(2 ) 1 1 1 2 tan 1 2 tan dy x x x dx x x x x x − − − = + + − + = + − = 32. If sin , x e y e = = cos where is a parameter, then dy dx at (1,1) is equal to (A) 0 (B) 1 2 (C) 1 2 − (D) 1 4 − Ans: A) ( sin ) cos (cos ) sin sin cos cos sin dy dy e e d dx dx e e d dy dx − + = = + − + = + (1,1) , sin cos tan 1 4 at e e = = = , 4 1 1 2 2 0 1 1 2 2 At dy dx = − + = = + 33. If y = 2 ..... 3 2 1 log x x x e d y e x then at x dx = is (A) 3 (B) 5 (C) 0 (D) 1 Ans: A) HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI ( ) ........ 2 2 2 log ........ log log log log log (1) 1 1 x x x y e y x x x y x y y x y y x dy y dx dy y dx d y dy y dx dx = = = = = → = = = = 3 log (1) log 3 e x From y x y e y e y = = = = 34. If f(1) = 1, f(1) = 3 then the derivative of f(f(f(x))) + (f(x)) 2 at x =1 is (A) 10 (B) 33 (C) 35 (D) 12 Ans: B) ( ( ( )). ( ( )). ( ) 2 ( ). ( ) f f f x f f x f x f x f n + At 1 x = ( ) ( ) (1) (1) . (1) 2 (1). (1) f f f f f f f f + (1) . (1).(3) 2(1).(3) f f f + (1).(3)(3) 6 f + (3)(3)(3) 6 + 27 6 33 = + = 35. If sin (sin ) 2 x x dy y x x then at x dx = + = is (A) 4 (B) log 2 (C) 1 (D) 2 2 Ans: C) HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI ( ) ( ) ( ) sin sin 1 sin , log sin .(1) log .cos sin .cos log sin .(1) sin , 2 2 log .(0) 1 .0 log1 2 2 2 2 0 1 2 x x v v x x y x x d v du dv By u u u dx u dx dx dy x x x x x x x x dx x x at x = + = + = + + + = = + + + = + = 36. If 1 1 n n n A n n − = − then 1 2 2021 ....... A A A + + + = (A) - 2021 (B) – (2021) 2 (C) (2021) 2 (D) 4042 Ans: B) 1 1 n n n A n n − = − 1 0 1 1 1 0 A = = − 2 1 2 3 2 1 A − = = − − 3 2 3 5 3 2 A − = = − − 2021 2021 A = − 1 2 2021 ... A A A + + + 1 3 5 7..... 2021 = − − − − − ( ) 1 3 5 7 .... 2021 = − + + + + + 2 (2021) = − 37. The function 2 ( ) log(1 ) 2 x f x x x = + − + is increasing on (A) ( , ) − (B) ( , 1) − (C) ( 1, ) − (D) ( ,0) − Ans: C) 2 ( ) log(1 ) (2 ) x f x x x = + − + HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI 2 2 ( ) (1 )(2 ) x f x x x = + + Since f(x) is increasing function. So, 1 0 x + 1 x − i.e., x ( 1, ) − 38. The co - ordinates of the point on the 6 x y + = at which the tangent is equal inclined to the axes is (A) (4,4) (B) (1,1) (C) (9,9) (D) (6,6) Ans: C) The only point (9, 9) satisfies the given curve 6 x y + = 39. The function f( x ) = 4 sin 3 x – 6 sin 2 x +12 sin x +100 is strictly (A) Decreasing in , 2 2 − (B) decreasing in 0, 2 (C) increasing in 3 , 2 (D) Decreasing in , 2 Ans: D) 3 2 ( ) 4sin 6sin 12sin 100 f x x x x = − + + ( ) 2 ( ) 12sin 12sin 12 cos x f x x x = − + 2 ( ) 12(sin sin 1) cos f x x x x = − + Since , 2 sin sin 1 0 x x − + ( ) 0 f x cos 0 x So , Decreasing in , 2 40. If [x] is the greatest integer function not greater than x then 8 0 x dx is equal to (A) 28 (B) 30 (C) 29 (D) 20 Ans: D) 8 0 [ ]. x dx 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 0. 1. 2. 3. 5. 5. 6. 7. dx dx dx dx dx dx dx dx = + + + + + + + 2 3 4 5 6 7 8 1 2 3 4 5 6 7 0 ] 2 ] 3 ] 4 ] 5 ] 6 ] 7 ] x x x x x x x = + + + + + + + (2 1) (6 4) (12 9) (20 16) (30 25) (42 36) (56 49) = − + − + − + − + − + − + − 1 2 3 ..... 7 = + + + + 7(7 1) 7(8) 28 2 2 + = = = 41. /2 0 sin cos 3 θ dθ is equal to (A) 8 23 (B) 7 23 (C) 8 21 (D) 7 21 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI Ans: C) / 2 2 0 sin .cos .cos d ( ) / 2 2 0 sin . 1 sin .cos d = − Put sin t = Diff. w.r.t cos d dt = When 0 0 t = = 1 2 t = = ( ) 1 5/ 2 0 t t dt − 1 3 7 2 2 0 2 2 . 3 7 t t = − 2 2 14 6 8 3 7 21 21 − = − = = 42. If e y +xy=e the order ed pair 2 2 , dy d y dx dx at x=0 is equal to (A) 2 1 1 , e e (B) 2 1 1 , e e − − (C) 2 1 1 , e e − (D) 2 1 1 , e e − Ans: D ) At x = 0 e y = e y = 1 y e xy e + = 0 y e xy e + − = By partial differentiation ( ) y dy y dx e x − = + 1 0 dy dx e − = + x = 0, y = 1 dy dx 1 e − = Diff. w.r.t ‘x’ ( ) ( ) ( ) 2 2 2 1 y y y dy dy e x y e d y dx dx dx e x + − − − + = + At x = 0, y = 1 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI ( ) 2 1 1 0 (1) 1 e e e e e − + + + = 1 1 − = 1 + 2 2 1 e e = 43. cos 2 cos 2 cos cos x x − − dx is equal (A) 2(sin x - x cos )+c (B) 2(sin x - x cos )+c (C) 2(sin x - 2x cos )+c (D) 2(sin x - 2x cos )+c Ans: B) ( ) 2 2 2 2 2 cos 1 2 cos 1 cos cos 2(cos cos ) cos cos cos cos 2 1 2(sin cos ) x dx x x dx x x dx x x c − − − − = − + = = + + 44. 1 3 0 (2 ) x xe dx x + is equal to (A) 1 1 27 8 e − (B) 1 1 27 8 e + ( C) 1 1 9 4 e + (D) 1 1 9 4 e − Ans: D) 1 3 0 (2 ) 2 . (2 ) x x e dx x + − + 1 2 3 0 1 2 (2 ) (2 ) x e dx x x − = + + + Here 2 1 ( ) (2 ) f x x = + and 3 2 ( ) (2 ) f x x − = + = 1 2 0 1 . (2 ) x e x + 1 9 4 e = − 45. If 2 ( 2)( 1) dx x x = + + a log 2 |1 | x + +b tan - 1 x+ 1 5 log | 2 | , x c + + then (A) 1 2 , 10 5 a b − = = (B) 1 2 , 10 5 a b = = (C) 1 2 , 10 5 a b − − = = (D) 1 2 , 10 5 a b − = = Ans: A) 2 2 1 ( 2)( 1) 2 1 A Bx C x n x x + = + + + + + 2 1 ( 1) ( )( 2) A x Bx C x = + + + + HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI Put 2 1 1 5 5 x A A = − = = Put 0 1 2 1 1 2 5 4 2 5 2 5 x A C C C C = = + − = = = Put 1 1 2 3 3 2 6 1 3 5 5 8 1 3 5 8 3 1 5 3 3 5 1 5 x A B C B B B B B = = + + = + + = + = − = − = − 2 2 1 1 2 5 5 5 2 1 1 x dx x x x − + + + + + 2 1 1 1 2 log( 2) log( 1) tan ( ) 5 10 5 x x x C − = + − + + + 1 2 , 10 5 a b − = = 46. Area of the region bounded by the curve y= tan x, the x - axis and the line x = 3 is (A) l og 1 2 (B) log 2 (C) 0 (D) - log 2 Ans: B) Required Ans 3 0 tan x dx = 3 0 log(sec ) x = log(2) log(1) = − log 2 = 47. Evaluate 3 2 2 x dx as the limit of a su m HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI (A) 72 6 (B) 53 9 (C) 25 7 (D) 19 3 Ans: D) 3 2 2 x .dx = 3 3 2 3 x 1 19 27 8 3 3 = − = 48. /2 0 cos sin 1 sin x x x + dx is equal to (A) l og 2 - 1 (B) log 2 (C) – log 2 (D) 1 - log 2 Ans: D) 2 0 cos .sin 1 sin x x dx x + ( ) / 2 0 cos 1 sin 1 1 sin x x x + − = + ( ) / 2 0 cos 1 sin cos 1 sin 1 sin x x x dx x x + = − + + / 2 0 cos cos 1 sin x x dx x = − + ( ) / 2 0 sin log 1 sin x x = − + ( ) 1 log 2 (0 0) = − − − 1 log 2 = − 49. If 2 dy y x dx x + = , then 2y(2) - y(1)= (A) 11 4 (B) 15 4 (C) 9 4 (D) 13 4 Ans: B) ( ) ( ) ( ) ( ) 2 1 log 2 3 4 3 1 , 4 3 dx x x P Q x x IF e e x The solution is y IF Q IF dx c y x x x dx c yx x dx c x xy c x c y x = = = = = = + = + = + = + = + HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI 8 1 (2) (1) 4 2 4 2 2 1 2 (2) (1) 2 2 2 4 1 1 15 4 4 4 4 4 c y and y c c c y y c c c = + = + = + − = + − − = + − − = − = 50. The solution of the differential equation 2 ( ) dy x y dx = + is (A) tan - 1 (x+y) = x+c (B) tan - 1 (x+y) = 0 (C) cot - 1 (x+y) = c (D) cot - 1 (x+y) = x+ c Ans: A) ( ) ( ) 2 2 2 . . 1 1 1 1 Put x y t Diff w r t x dy dt dx dx dy dt dx dx dy x y dx dt t dx dt t dx + = + = = − = + − = = + 2 2 1 1 1 1 1 1 tan ( ) dt dx t Integrate on both sides dt dx t t x c − = + = + = + 51. If ( ) y x be the solution of differential equation log 2 log , ( ) dy x x y x x y e dx + = is equal to (A) e (B) 0 (C) 2 (D) 2e Ans: D) log 2 .log dy x x y x x dx + = log x x 1 2 log dy y dx x x + = dy py Q dx + = 1 log P x x = and Q = 2 pdx I F e = log dx x x e = dt t I F e = logt e = log x = HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI Sol ution is ( ) ( ) Y I F Q I F dx C = + log 2.log y x x dx C = + log 2 (log 1) y x x x C = − + At x = e log 2 (log 11) y e e e C = − + y C = log 2 (log 1) y x x x y = − + 2 y x = ( ) 2 y e e = 52. If | | 2 and | | 3 a b = = and the angle between and a b is 120 0 , then the length of the vector 2 2 3 a b − is (A) 2 (B) 3 (C) 1 6 (D) 1 Ans: B) 0 | | 2,| | 3, 120 a b = = = 0 1 cos(120 ) 2 = − 2 2 2 0 2 .cos120 2 3 2 3 2 3 a b a b a b − = + − 1 1 1 1 (4) (9) (2)(3) 4 9 3 2 − = + − 1 1 1 3 = + + = 53. 2 36 a b a b + = and | | 3 a = then b is equal to (A) 9 (B) 36 (C) 4 (D) 2 Ans: Insufficient data 54. If ˆ ˆ ˆ ˆ ˆ 3 , 2 i j i j k = − = + − then express in the form 1 2 = + where 1 is parallel to 2 and is perpendicular to 1 and is given by (A) 5 ˆ ˆ ( 3 ) 8 i j − (B) 5 ˆ ˆ ( 3 ) 8 i j + (C) ˆ ˆ 3 i j − (D) ˆ ˆ 3 i j + Ans: Insufficient data 55. The sum of the degree and order of the differential equation 2 2 / 3 1 2 (1 ) y y + = is (A) 4 (B) 6 (C) 5 (D) 7 Ans: C) ( ) 2 / 3 2 1 2 1 y y + = Take cube on both side ( ) 2 2 3 1 2 1 y y + = order = 2 degree = 3 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI Ans = 2 + 3 =5 56. The co - ordinates of foot of the perpendicular drawn from the origin to the plane 2 3 4 29 x y z − + = are (A) (2,3,4) (B) (2, - 3, - 4) (C) (2, - 3,4) (D) ( - 2, - 3,4) Ans: C) 2 3 4 29 x y z − + = : m : n 2 : 3 : 4 l = − 0 0 0 2 3 4 x y z − − − = = = − 2 3 4 x y z = = − = ( ) ( ) ( ) 2 2 3 3 4 4 29 − − + = 4 9 16 29 + + = 29 29 = 1 = 2 : 3: 4 x y z = = − = Required answer = (2, - 3, 4) 57. The angle between the pair of lines 3 1 3 3 5 4 x y z + − + = = and 1 4 5 1 4 2 x y z + − − = = is (A) 1 27 cos 5 − = (B) 1 8 3 cos 15 − = (C) 1 19 cos 21 − = (D) 1 5 3 cos 16 − = Ans: No Option 3 1 3 3 5 4 x y z + − + = = 1 4 5 1 4 2 x y z + − − = = 1 2 1 2 1 2 2 2 2 2 2 2 1 2 3 1 2 3 cos a a b b c c a a a b b b + + = + + + + 3 1 5 4 4 2 cos 9 25 16 1 16 4 + + = + + + + 3 20 8 31 cos 50 21 1050 + + = = 58. The corner points of the feasible region of an LPP are (0,2), (3,0), (6,0), (6,8) and (0,5), then the minimum value of 4 6 z x y = + occurs at (A) Finite number of points (B) infinite number of points HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE , UDUPI (C) Only one point (D) only two points Ans: B ) Corner point Z = 4x + 6y (0, 5) 30 (0, 2) 12 (6, 8) 72 (3, 0) 12 (6, 0) 24 59. A dietician has to develop a special diet using two foods X and Y. Each packet (containing 30 g) of food. X contains 12 units of calcium, 4 units of iron, 6 units of cholesterol and 6 units of vitamin A Each packet of the same quantity of food Y contains 3 units of calcium, 20 units of iron, 4 units of cholesterol and 3 units of vitamin A. The diet requires atleast 240 units of calcium, atleast 460 un its of iron and atmost 300 units of cholesterol. The corner points of the feasible region are (A) (2,72), (40,15), (15,20) (B) (2,72), (15,20), (0,23) (C) (0,23), (40,15), (2,72) (D) (2,72), (40,15), (115,0) Ans: A ) X Y Availability Calcium 12 3 240 Iron 4 20 460 Cholesterol 6 4 300 Subject to constraints 12 3 240 x y + 4 20 460 x y + 6 4 300 x y + 60. The distance of the point whose position vector is ˆ ˆ ˆ (2 ) i j k + − from the plane ˆ ˆ ˆ .( 2 4 ) 4 r i j k − + = is (A) 8 21 (B) 8 21 (C) 8 21 − (D) 8 21 − Ans: A) r n d = a n d p n − = 2 a i j k = + + 2 p = 2 − 4 4 1 4 16 − − + + 2 4 n i j k = − + d = 4 x 0 20 y 80 0 x 0 115 y 23 0 x 0 0 y 75 0