Updated Maths K-cet Key answers and Solution.pdf - CREATIVE P.U COLLEGE

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CREATIVE LEARNING CLASSES, KARKALA K-CET KEY ANSWER WITH DETAILED SOLUTION - 2022 DEPARTMENT OF MATHEMATICS VERSION CODE – A3 1. Find the mean number of 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 X = 3, 2,1,0 X X= 0 X = 1 X= 2 X = 3 P(X) 1 3 3 1 8 8 8 8 n 1 3 3 1 Mean =  X i p( X i ) = 6  + 1 + 2  + 3  i =1 8 8 8 8 0 + 3 + 6 + 3 12 = = = 1.5 8 8 1 1 1 2. If A and B are two events such that P ( A) = , P ( B ) = and P ( A | B ) = , then P( A  B) is 2 3 4 1 3 1 3 (A) (B) (C) (D) 4 16 12 4 Ans: A) 1 1 1 1 P ( A) = , P( B) = , P( A B) = , P( A  B ) = 2 3 4 12 9 1 P ( A  B) = P ( A  B ) = 1 − P ( A  B ) = 1 − = 12 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) E1 : Event that the Lockdown E 2 : not a lockdown A: Controlled in one month P( E1 ) = 0.7 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI P( E2 ) = 0.3 P( A / E1 ) = 0.8 P( A / E2 ) = 0.3 Using Theorem of total probability  P( A) = P( E1 ).P( A / E1 ) + P( E2 ).P( A / E2 ) 7 8 3 3 56 9 65 = . + . = + = = 0.65 10 10 10 10 100 100 100 4. If A and B are two independent events such that P( A) = 0.75, P( A  B) = 0.65 and P( B) = x , then find the value of x : 5 8 9 7 (A) (B) (C) (D) 14 15 14 15 Ans: D) ( ) P A = 0.75 A & B are independent events P( A) = 0.25 P ( A  B ) = P( A) P( B) P( B) = x P( A  B) = 0.65 P( A  B) = P( A) + P( B) − P( A  B) 0.65 = 0.25 + x − P( A) P( B) 0.35 = x − 0.25 x 0.35 = x(1 − 0.25) 0.35 = x(0.75) 0.35 7 x= = 0.75 15 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 p 2 + q 2 = ___________. (A) 50 (B) 51 (C) 42 (D) 49 Ans: A) n( A) = p, n( B) = q n( A  B) = 7 n( A).n( B) = 7 p.q = 7 This is possible only when p = 1 & q = 7 or p = 7 & q = 1 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI  p 2 + q 2 = 1 + 49 = 50 1 6. The domain of the function f ( x) = + x + 2 is log10 (1 − x) (A) −2,0)  (0,1) (B)  −2,1) (C)  −2,0) (D) −2,0)  (0,1) 1 Ans: D) + x+2 log(1 − x) x+20 & 1–x>0 and x  0 x  −2 1>x x<1  x −2,0)  ( 0,1) 7. The trigonometric function y = tan 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 is equal to 32 (A) 503020 (B) 503720 (C) 503730 (D) 403030 Ans: C) 180 D.M =  ( R.M )  180  180  20  =  = = 50   60  = 503730  32 32  32  5  9. The value of sin sin is 12 12 1 1 (A) 0 (B) 1 (C) (D) 2 4 Ans: D) HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI 5  1  5  1 5 1 1 1 sin sin = .sin 2   = .sin = . = 12 12 2  12  2 6 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 2n = 2cos   Required Answer = 2 cos  11. If  A = 1, 2,3,....10 then number of subsets of A containing only odd numbers is (A) 31 (B) 27 (C) 32 (D) 30 Ans: A) B = {1,3,5,7,9}  n( B) = 5 Required answer = 25 −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 19th word? (A) KAMS (B) SAMK (C) AKMS (D) AMSK Ans: Incorrect option a a 13. If a1 , a2 , a3 ,..., a10 is a geometric progression and 3 = 25, then 9 equals a1 a5 (A) 3(52) (B) 54 (C) 53 (D) 2(52) Ans: B) a3 = 25 a1 ar 2 = 25 a r 2 = 25 r = 5 a9 ar 8  = 4 = r 4 = 54 a5 ar 14. If the straight line 2 x − 3 y + 17 = 0 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 x − 3 y + 17 = 0 3 y = 2 x + 17 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI 2 17 y= x+ 3 3 2  Slope of the line 2 x − 3 y + 17 = 0 is = m1 3 Slope of line joining (7,17) and (15,  ) is  − 17  − 17 m2 = = 15 − 7 8 Since m1m2 = −1  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) = Eighth octant  x 2 − 1, 0  x  2 16. If f ( x) =  , the quadratic equation whose roots are lim− f ( x) and lim+ f ( x) is 2 x + 3, 2  x  3 x →2 x →2 (A) x2 − 14 x + 49 = 0 (B) x2 −10x + 21 = 0 (C) x2 − 6 x + 9 = 0 (D) x2 − 7 x + 8 = 0 Ans: B) lim f ( x) = lim− x 2 − 1 = 3 = a x → 2− x →2 lim f ( x) = lim+ 2 x + 3 = 7 = b x → 2+ x →2 Required Quadratic equation  x 2 − ( a + b ) x + ab = 0 x 2 − 10 x + 21 = 0 17. If 3x + i(4x − y) = 6 − 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) 4 x − y = −1 3x = 6 and 8 − y = −1 x=2 y=9 18. If the standard deviation of the numbers -1, 0, 1 k is 5 where k > 0, then k is equal to 5 10 (A) 4 (B) 6 (C) 2 (D) 2 6 3 3 Ans: D) x 2 i = 2 + k 2 and x = k i HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI  x − ( x) 2 2 i  S .D = 2 + k2  k  2 5= −  4 4 2 + k2 k2 5= − 4 16 8 + 4k 2 − k 2 5= 16 80 = 8 + 3k 2 72 = 3k2 k2 = 24 k =2 6 19. If the set x contains 7 elements and set y contains 8 elements, then the number of bijections from x to y is (A) 0 (B) 8 P7 (C) 7 ! (D) 8 ! Ans: A) Zero because n( x)  n( y) 2 x : x3  2 20. If f : R → R be defined by f ( x) =  x : 1  x  3 then f (−1) + f (2) + f (4) is 3x : x 1  (A) 5 (B) 10 (C) 9 (D) 14 Ans: C) When x = -1, f(x) = 3x x = 2, f ( x) = x 2 x = 4, f ( x) = 2x  f (−1) + f (2) + f (4) = 3(−1) + 22 + 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  27   (A) (1,12)(3,9)(5,6)( 7,3) (B)  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. 3 + y3 − 3 22. lim = y →0 y3 1 1 (A) (B) (C) 2 3 (D) 3 2 2 3 3 2 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI Ans: A) 3 + y3 − 3 0 lim = form y →0 y3 0 (3 + y ) − 3 1 1 2 3 2 lim y →0 (3 + y ) − 3 3 1 − 12 1 = 3 = 2 2 3 23. If A is a matrix of order 3  3, the (A2)-1 is equal to (A) (-A2)2 (B) (A-1)2 (C) A2 (D) (-A)-2 Ans: B) adj ( A2 ) adjA.adjA ( A2 ) = −1 = A2 A A ( adjA ) 2  adjA  2 −1 2 = 2 =  = (A ) A  A   2 −1 24. If A =   , then the inverse of the matrix A3 is  3 −2  (A) A (B) -1 (C) 1 (D) -A Ans: A)  2 −1 1 0   2 −1 A=  A2 =   A3 =    3 −2  0 1   3 −2  A3 = −4 + 3 = −1 adj ( A3 ) 1  −2 1   2 −1 (A ) 3 −1 = = = −1  −3 2   3 −2  (A )3 25. If A is a skew symmetric matrix, then A2021 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. 0 1  26. If A =  then ( aI + bA) is (where I is the identity matrix of order 2) n  0 0 (A) a2 I + an−1b.A (B) a2 I + n.an−1b.A (C) a 2 I + n a nb A (D) a n I + bn A 0 1  Ans: B) A =   0 0 ( aI + bA) = a n I + n a n−1b.A n 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.adj A = 5 then A is equal to (A)  1 (B) 1/ 25 (C) 1/ 5 (D)  5 Ans: C) 5.adj.A = 5 k adjA = k n adjA n −1 53 adjA = 5 adjA = A 3−1 53 A =5 25 A = 5 2 1 A = 2 25 1 A = 5 1 −2 5 28. If there are two values of ‘a’ which makes determinant  = 2 a −1 = 86 . Then the sum of these 0 4 2a numbers is (A) -4 (B) 9 (C) 4 (D) 5 1 −2 5 Ans: A)  = 2 a −1 0 4 2a  = 1 2a 2 + 4 − 2  −4a − 20  = 2a2 + 8a + 44 2a2 + 8a + 44 = 86 2a2 + 8a − 42 = 0 a2 + 4a − 21 = 0 a(a + 7) − 3(a + 7) = 0 (a + 7)(a − 3) = 0 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 x1 y1 1 −2 6 1 1 1 1 31 Ans: B) Area of triangle = x2 y2 1 = 3 −6 1 = 31 = = 15.5 sq.units 2 2 2 2 x3 y3 1 1 5 1 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 = (1 + x 2 ) tan −1 x − x then dy is dx tan −1 x (A) 2x tan-1 x (B) (C) x2 tan −1 x (D) x tan −1 x x Ans: A) = (1 + x 2 ) . dy 1 + tan −1 x.(2 x) − 1 dx 1 + x2 = 1 + 2 x tan −1 x − 1 = 2 x tan −1 x dy 32. If x = e sin  , y = e cos  where  is a parameter, then at (1,1) is equal to dx 1 1 1 (A) 0 (B) (C) − (D) − 2 2 4 Ans: A) dy dy d e (− sin  ) + cos  e = =  dx dx e (cos  ) + sin  e d dy − sin  + cos  = dx cos  + sin  at (1,1) , e sin  = e cos  tan  = 1  = 4  At  = , 4 1 1 − + dy 2 2 =0 = dx 1 1 + 2 2 d2y 33. If y = e x x x ..... x  1then 2 at x = log e3 is dx (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 y=e x x x ........ log y = x x x........ log y = x log y ( log y ) = x log y 2 log y = x → (1) 1 dy =1 y dx dy =y dx d 2 y dy = =y dx 2 dx From(1) log y = x y = ex y = e loge 3 y=3 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) f ( f ( f ( x)). f ( f ( x)). f ( x) + 2 f ( x). f (n) At x = 1 f   f ( f (1) ) . f  ( f (1) ) . f (1) + 2 f (1). f (1) f  f (1). f (1).(3) + 2(1).(3) f (1).(3)(3) + 6 (3)(3)(3) + 6 = 27 + 6 = 33 dy  35. If y = x sin x + (sin x) x then at x = is dx 2 4  2 (A) (B)  log (C) 1 (D)  2 2 Ans: C) HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI y = x sin x + ( sin x ) x  v du dv  By , d v dx ( u ) = u v  . + log u.   u dx dx  dy  sin x  x  x  = xsin x  .(1) + log x.cos x. + ( sin x )  .cos x + log sin x.(1)  dx  x   sin x   at x = , 2 1    2     =    + log .(0)  + 1  .0 + log1  2   2  2   2  = + 0 = 1 2    1 − n n  36. If An =   then A1 + A2 + ....... + A2021 =  n 1 − n (A) -2021 (B) –(2021)2 (C) (2021)2 (D) 4042 1 − n n  Ans: B) An =    n 1 − n 0 1 A1 = = −1 1 0 −1 2 A2 = = −3 2 −1 −2 3 A3 = = −5 3 −2 . . . A2021 = −2021  A1 + A2 + ... + A2021 = −1 − 3 − 5 − 7..... − 2021 = − (1 + 3 + 5 + 7 + .... + 2021) = −(2021) 2 2x 37. The function f ( x) = log(1 + x) − is increasing on 2+ x (A) (−, ) (B) (, −1) (C) (−1, ) (D) (−,0) Ans: C) 2x f ( x) = log(1 + x) − (2 + x) HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI x2 f ( x) = (1 + x)(2 + x) 2 Since f(x) is increasing function. So, 1 + x  0  x  −1 i.e., x  (−1, ) 38. The co-ordinates of the point on the x + y = 6 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 x+ y =6 39. The function f(x) = 4 sin3x – 6 sin2 x +12 sin x +100 is strictly  −     (A) Decreasing in  ,  (B) decreasing in 0, 2   2 2  3    (C) increasing in   ,  (D) Decreasing in  ,    2  2  Ans: D) f ( x) = 4sin 3 x − 6sin 2 x + 12sin x + 100 f ( x) = (12sin 2 x − 12sin x + 12 ) cos x f ( x) = 12(sin 2 x − sin x + 1) cos x Since, sin 2 x − sin x + 1  0 f ( x)  0  cos x  0   So, Decreasing in  ,   2  8 40. If [x] is the greatest integer function not greater than x then   x  dx is equal to 0 (A) 28 (B)30 (C)29 (D) 20 Ans: D) 8 1 2 3 4 5 6 7 8  [ x].dx =  0.dx +  1.dx +  2.dx +  3.dx +  5.dx +  5.dx +  6.dx +  7.dx 0 0 1 2 3 4 5 6 7 = 0 + x]12 + 2 x]32 + 3 x]34 + 4 x]54 + 5 x]56 + 6 x]67 + 7 x]87 = (2 −1) + (6 − 4) + (12 − 9) + (20 −16) + (30 − 25) + (42 − 36) + (56 − 49) 7(7 + 1) 7(8) = 1 + 2 + 3 + ..... + 7 = = = 28 2 2  /2 41.  0 sin  cos3θ dθ is equal to 8 7 8 7 (A) (B) (C) (D) 23 23 21 21 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI Ans: C)  /2  /2  sin  .cos2  .cos  .d =  sin  . (1 − sin 2  ) .cos  .d 0 0 Put sin  = t Diff. w.r.t  cos  .d = dt When  = 0  t = 0  =  t =1 2 ( ) 1  t − t 5/ 2 .dt 0 2 3 2 7 1 2 2 14 − 6 8 = t 2 − .t 2  = − = = 3 7  0 3 7 21 21  dy d 2 y  42. If ey+xy=e the ordered pair  , 2  at x=0 is equal to  dx dx  1 1   −1 −1   1 −1   −1 1  (A)  , 2  (B)  , 2  (C)  , 2  (D)  , 2  e e   e e  e e   e e  Ans: D) At x = 0 ey = e y=1 e y + xy = e e y + xy − e = 0 By partial differentiation dy −( y ) = dx e y + x dy  −1   = dx  e + 0 x = 0, y = 1 dy −1 = dx e Diff. w.r.t ‘x’  dy   dy  d y 2 (e y + x) −  − (− y) ey  dx  + 1  dx  = (ey + x) 2 dx 2 At x = 0, y = 1 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI   −1   ( e + 0 )  1  + (1)  e   + 1 e   e   = 2 e 1 −1 + 1 1 = = 2 e2 e cos 2 x − cos 2 43.  cos x − cos  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 cos 2 x − 1( 2 cos 2  − 1)  cos x − cos  dx 2(cos 2 x − cos 2  ) = dx cos x − cos  cos x + cos  = 2 dx 1 = 2(sin x + x cos  ) + c 1 xe x 44. 0 (2 + x)3 dx is equal to 1 1 1 1 1 1 1 1 (A) .e − (B) .e + (C) .e + (D) .e − 27 8 27 8 9 4 9 4 Ans: D) 1  (2 + x) − 2  e x  (2 + x)3  .dx 0   1  1  −2  =  ex  +  3  .dx  (2 + x)  (2 + x)  2 0 1 −2 Here f ( x) = and f ( x) = (2 + x) 2 (2 + x)3 1 1  e 1 =e. x 2 = − (2 + x)  0 9 4 dx 1 45. If  ( x + 2)( x 2 + 1) = a log |1 + x 2 | +b tan-1 x+ log | x + 2 | +c, then 5 −1 2 1 2 −1 −2 1 −2 (A) a = ,b= (B) a = ,b = (C) a = ,b = (D) a = ,b = 10 5 10 5 10 5 10 5 Ans: A) 1 A Bx + C = + 2 ( x + 2)(n + 1) x + 2 x + 1 2 1 = A( x 2 + 1) + ( Bx + C )( x + 2) HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI Put x = −2 1 1 = 5A  A = 5 Put x = 0 1 = A + 2C 1 1 − = 2C 5 4 = 2C 5 2 C= 5 Put x = 1 1 = 2 A + 3B + 3C 2 6 1 = + + 3B 5 5 8 1 = + 3B 5 8 3B = 1 − 5 3 3B = − 5 1 B=− 5  1 1 2   5 − x  1 1 2  + 25 + 25  .dx = log( x + 2) − log( x 2 + 1) + tan −1 ( x) + C  x + 2 x +1 x +1 5 10 5   −1 2 a = ,b = 10 5  46. Area of the region bounded by the curve y= tan x, the x-axis and the line x = is 3 1 (A) log (B) log 2 (C) 0 (D) -log 2 2 Ans: B)  3 Required Ans =  tan x.dx 0  = log(sec x)03 = log(2) − log(1) = log 2 3 x 2 47. Evaluate dx as the limit of a sum 2 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI 72 53 25 19 (A) (B) (C) (D) 6 9 7 3 Ans: D) 3 x3  3 1 19 2 x 2 .dx =  =  27 − 8 = 3 2 3 3  /2 cos x sin x 48.  0 1 + sin x dx is equal to (A) log 2-1 (B) log 2 (C) – log 2 (D) 1- log 2 Ans: D) 2 cos x.sin x 0 1 + sin x dx  /2 cos x (1 + sin x ) − 1 =  0 1 + sin x  /2  cos x (1 + sin x ) cos x  =   0 1 + sin x −  dx 1 + sin x   /2  cos x  =   cos x − 1 + sin x  dx 0  /2 = sin x − log (1 + sin x )  0 = (1 − log 2) − (0 − 0) = 1 − log 2 dy y 49. If + = x 2 , then 2y(2) - y(1)= dx x 11 15 9 13 (A) (B) (C) (D) 4 4 4 4 Ans: B) 1 P= , Q = x2 x 1 IF = e  x = elog x = x dx  The solution is y ( IF ) =  Q. ( IF ) dx + c y ( x ) =  x 2 . ( x ) dx + c yx =  x 3 dx + c x4 xy = + c 4 x3 c y= + 3 x HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI 8 c 1  y (2) = + and y (1) = +c 4 2 4 c = 2+ 2  c 1  2 y (2) − y (1) = 2  2 +  − − c  2 4 1 1 15 = 4+c− −c= 4− = 4 4 4 dy 50. The solution of the differential equation = ( x + y ) 2 is dx -1 -1 (A) tan (x+y) = x+c (B) tan (x+y) = 0 (C) cot-1(x+y) = c (D) cot-1(x+y) = x+c Ans: A) Put x + y = t Diff . w.r.t x dy dt 1+ = dx dx dy dt = −1 dx dx dy  = ( x + y) 2 dx dt −1 = (t ) 2 dx dt = t2 +1 dx 1  2 dt = dx t +1 Integrate on both sides 1  t 2 + 1 dt =  1 dx tan −1 (t ) = x + c dy 51. If y( x) be the solution of differential equation x log x + y = 2 x log x, y (e) is equal to dx (A) e (B) 0 (C) 2 (D) 2e Ans: D) dy x log x + y = 2 x.log x x log x dx dy 1 + y=2 dx x log x dy 1 + py = Q P= and Q = 2 dx x log x I .F = e pdx dx  x log x =e dt I .F = e  t = elogt = log x HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI Solution is Y . ( I .F ) =  Q. ( I .F ) dx + C y log x =  2.log x dx + C y log x = 2 x(log x −1) + C At x = e y log e = 2e(log e −11) + C y =C y log x = 2 x(log x −1) + y  y = 2x y(e) = 2e 2 a b 52. If | a |= 2 and | b |= 3 and the angle between a and b is 120 , then the length of the vector − 0 is 2 3 1 (A) 2 (B) 3 (C) (D) 1 6 Ans: B) | a |= 2,| b |= 3,  = 1200 1 cos(120 0 ) = − 2 2 2 2 a b a b a b − = + − 2 . .cos1200 2 3 2 3 2 3 1 1 1  −1  = (4) + (9) − (2)(3)   = 1 + 1 + 1 = 3 4 9 3  2  2 53. a  b + a.b = 36 and | a |= 3 then b is equal to (A) 9 (B) 36 (C) 4 (D) 2 Ans: Insufficient data 54. If  = iˆ − 3 ˆj ,  = iˆ + 2 ˆj − kˆ then express  in the form  = 1 + 2 where 1 is parallel to  and 2 is perpendicular to  and 1 is given by 5 5 (A) (iˆ − 3 ˆj ) (B) (iˆ + 3 ˆj ) (C) iˆ − 3 ˆj (D) iˆ + 3 ˆj 8 8 Ans: Insufficient data 55. The sum of the degree and order of the differential equation (1 + y12 ) 2 / 3 = y2 is (A) 4 (B) 6 (C) 5 (D) 7 Ans: C) (1 + y ) 2 2/3 1 = y2 Take cube on both side (1 + y ) 2 2 1 = y23 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 x − 3 y + 4 z = 29 are (A) (2,3,4) (B) (2,-3, -4) (C) (2,-3,4) (D) (-2,-3,4) Ans: C) 2 x − 3 y + 4 z = 29 l : m : n = 2 : −3 : 4 x−0 y−0 z −0 = = = 2 −3 4 x = 2 y = −3 z = 4 2 ( 2 ) − 3( −3 ) + 4 ( 4 ) = 29 4 + 9 + 16 = 29 29 = 29  =1  x = 2 : y = −3: z = 4 Required answer = (2, -3, 4) x + 3 y −1 z + 3 x +1 y − 4 z − 5 57. The angle between the pair of lines = = and = = is 3 5 4 1 4 2  27  8 3  (A)  = cos −1   (B)  = cos −1   5  15  19  5 3  (C)  = cos −1   (D)  = cos −1    21   16  Ans: No Option x + 3 y −1 z + 3 = = 3 5 4 x +1 y − 4 z − 5 = = 1 4 2 a1a2 + b1b2 + c1c2 cos  = a + a22 + a32 b12 + b22 + b32 2 1 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 z = 4 x + 6 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 units 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 x 0 20 12 x + 3 y  240 y 80 0 4x + 20 y  460 x 0 115 y 23 0 6 x + 4 y  300 x 0 0 y 75 0 60. The distance of the point whose position vector is (2iˆ + ˆj − kˆ) from the plane r .(iˆ − 2 ˆj + 4kˆ) = 4 is 8 −8 −8 (A) (B) 8 21 (C) (D) 21 21 21 Ans: A) r.n = d a.n − d p= a = 2i + j + k n 2 − 2 −4−4 p= n = i − 2 j + 4k d=4 1 + 4 + 16 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI −8 8 p= p= 4 21 Department of Mathematics ❖ Mr. Ashwath S L ❖ Mr. Ganapathi Bhat K S ❖ Mr. Nandeesh H B ❖ Mr. Aditya Vati K ❖ Mr. Rakshith B S ❖ Mr. Sumanth Damle ❖ Mr. Kiran Kumar ❖ Mr. Shashiraj ❖ Mr. Agraja ❖ Mr. Raviranjan ❖ Mr. Prakash Ganiga CREATIVE EDUCATION FOUNDATION MOODBIDRI (R) Website : www.creativeedu.in Phone No. : 9019844492 HKS P U COLLEGE, HASSAN CREATIVE P U COLLEGE, KARKALA CREATIVE P U COLLEGE, UDUPI