Page 1 : KARNATAKA CET BOOSTER, , MAXIMUM, MARKS, 60, , TOTAL DURATION, 80 MINUTES, , 1. If A and B are Finite sets and A  B ,then, a) n  A  B   n  A, b) n  A  B   n  B , c) n  A  B   n  B , , MAXIMUM TIME, FOR ANSWERING, 70 MINUTES, , Solution:(b) (i), , d) n  A  B   , ,  , ,  , , 2. The value of cos2 450  sin 2 150 is, (a) 3, , (b), , 2, , 3, 4, ,  , , 2, ,  1   3 1 , , ,   ,  2  2 2 ,  1   3 1 2 3 ,     , , 8, 2 , ,  1  42 3  1  2 3 ,     ,      , , 8, 2 ,  2  4 ,  1  2 3 ,     , ,  2   4 ,  2  2  3 , 3,  ,  , 4, ,  4, 3. 3  5  7  ... to n terms is, 2, (a) n  n  2 , (b)  n  1, , (d) n  n  2 , , (c) n 2, , Solution:(a) Short cut method, By principle of mathematical induction method, Option (a) satisfies the given condition, 3  5  7  ...to n th term  n  n  2   1, , (i) Put n  1  3  11  2   3 Eq (1)true for n  1, Hence Options c ,b& d are not correct, So option (a) is correct, ,  1 i , 4. If ,   1 then the least positive integral,  1 i , m, , value of m is, a)1, b) 4, , c) 2, , [NCERT-DIRECT QUESTION], , KCET/JEE MATHEMATICS BOOSTER, , (ii)  1  i   1,  1 i , m, , m, , 3 1, (d) 3  1, 2 2, 2 2, 2, 0, 2, 0, Solution:(b) cos 45  sin 15, 2, , 1  i  i  1 2 i, , i, 11, 2, ,  i   1  m  4, , (c), ,  , , A, , 1  i 1  i 1  i  1  i  i  i 2, , , 1  i 1  i 1  i , 1 i2, , , Solution:(c) If A  B then n  A  B   n  B , , VERSION, CODE, , d) 3, , 5. If x  2  1 ,then, a) x  1,3, , b) x  1,3 , , c) x   1,3, , d) x   1,3, , Solution:(a) x  2  1, , 1  x  2  1, 1  2  x  2  2  1  2, 1 x  3,  x  1,3, 6. If nC12  n C8 then n is equal to, (a)12, (b)26, (c)6, , (d)20, , [NCERT-DIRECT QUESTION], , Solution:(d) If nC p  nCq  n  p  q, So nC12 nC8  n  12  8  20, 7. The total number of terms in the expansion of, ( x  a)47  ( x  a) 47 after simplification will be, (a) 24, (b) 47, (c) 48, (d) 96, Solution:(a) Short cut method, W.K.T If n is odd, then, , (x  y )n  (x  y )n and (x  y )n  (x  y )n ,both have the, n 1, .,  2 , , same number of terms equal to , , So required number   n  1   47  1  48  24,  2 , , 2, , 2, , 8. Equation of line passes through the point 1, 2  ,and, is perpendicular to the line y  3 x  1 ., (a) x  3 y  7  0, (b) x  3 y  7  0, (c) x  3 y  0, (d) x  3 y  0, Solution:(a) Short cut method, By inspection method option a satisfies the given two, conditions, [email protected], Phone number-8105418762 / 7019881906, 1

Page 2 : KARNATAKA CET-2017 PAPER WITH SOLUTIONS, , (i) 1, 2  satisfies the option a equation x  3 y  7  0, , q : It is odd. We have p  q, The contrapositive of p  q is ~ q  ~ p, i.e., If x is not odd then it is not a prime, 13. If the coefficient of variation and standard, deviation are 60and 24 respectively, the arithmetic, Mean of distribution, , But not satisfies the options b, c & d, 9. The eccentricity of the ellipse, (a) 2 5, , (c) 2 13, , (b) 2 5, , 6, , x2 y 2, , 1,, 36 16, (d) 2 13, , 6, , 4, , 4, , a) 40, , [NCERT-DIRECT QUESTION], ,  x2 y 2, , x2 y 2, Solution:(a) ,  1  2  2  1, 36 16, b,  a, , , b) 1, , c) 7, , d) 20, , 40, , 20, , 7, , [KCET-2017-1M][One option correct type], [NCERT-DIRECT QUESTION], , Solution:(a) W.K.T CV , , W.K.T a  c  b, c 2  a 2  b 2  36  16  20,  c  20, 2, , e, , 2, , 2, , c, 20 2 5, , , a, 6, 6, , 14. The range of the function f  x   9  x 2 is, , 10. The perpendicular distance of the point, P  6,7,8 from XY-plane is, (a)8, (b)7, Solution:(a), , (c)6, , (c)  0,3, , (d)5, , (d)  0,3, , [NCERT-DIRECT QUESTION], , Z, , Solution:(b) y  9  x 2  0  1 Squaring on both sides, y 2  9  x2, , 8, 6, , Y, 7, , 1  cos 4, is,  0 1  cos 6, 4, 9, 9, 3, (a), (b), (c), (d), 9, 4, 3, 4, 1  cos 4, Solution:(a) l  lim, 0 0 Form,   0 1  cos 6, By using L’ Hospital’s rule,, 4sin 4, 0 0 Form, l  lim,   0 6sin 6, By using L’ Hospital’s rule,, 4  4 cos 4, l  lim,   0 6  6 cos 6, 11. The value of lim, , 16 cos 0 16 4, ,   cos 0  1, 36 cos 0 36 9, 12. The contrapositive statement of the statement, “If x is a prime then it is odd.” Is, (a) If x is not a prime then it is not odd, (b) If x is a prime then it is not odd, (c) If x is not a prime then it is odd, (d) If x is not odd then it is not a prime, l, , [NCERT-DIRECT QUESTION], , Solution:(d) p : x is a prime., 2, , (b)  0,3, , (a)  0,3, ,  P  6,7,8, , X, , , , 100, x, , x,  100, C.V, 24, x  100  40, 60, ,  a 2  36, b2  16, ,  Point of inflection, , O,  , , A, , x2  9  y 2, , x  9  y2, ,  9  y2  0,  y2  9  0,  y 2  32  0, ,   y  3 y  3  0, ,   y   3   y  3  0,  3  y  3   2 , , From equations (1&2), 0  y  3  0,3 [ y is the range of a function], 15. Let f : R  R be defined by f  x   x 4 then f is, a) one-one and Onto, b) may be one-one and onto, c) one-one but not onto, d) neither one-one nor onto, [KCET-2014-1m] [Only one option correct], [NCERT-DIRECT QUESTION], , Solution:(d), (i) Injective(one-one),  1 and 1 R  Domain  & 1  1, But f  1  f 1   1  1, 4, , Hence f is not one  one.  1m , , [email protected], Phon number-8105418762 / 7019881906, t.me/mcpmathematicskar_1986, , MCP-MATHEMATICS

Page 3 : KARNATAKA CET BOOSTER, , (ii) Onto [surjective], f  x   x 4  0, x  R, Range of f  0,    R (codomain), Hnce f is not onto,  f is neither one-one nor onto, OR, , y  x4, , O, X, ,  1, 1  x  , sin  x  tan    , 1,  , 19. If A  , ,   1  x , 1,  sin   cot  x  ,  , , , 8, P  6,7,8, , 1, 1  x  ,   cos  x  tan    , 1,    then, , Z &B   , , 1  x , 1,  sin    tan  x  ,  , , , A  B is equal to, Point6 of inflection, Y, a) I, b) O, c) 2I, d) 1 I,  1, 1, sin  x   cos  x , 1, A B  ,   1  x , 1  x ,  sin    sin  , ,  ,  , , , , 0, 1 2,  1    1, A B  ,      ,      2  0, , 0, 2 , , , 16. The range of sec 1 x is,  , (a)  0,     , (b)    ,  ,  2 2, 2,  , , , , , , , (c)   , , (d)  0,  ,  2 2, , 7, , [NCERT-DIRECT QUESTION], , Solution:(a)  0,      , 2, 17. If tan 1 x  tan 1 y  4, equal to, (a) 2, , (b) , , 5, , 5, , then cot 1 x  cot 1 y is, , (c) 3, , 5, , , , (d) , , 5, , Solution:(b) tan x  tan y  4, 1, , 1, , , , 5, ,  cot 1 x   cot 1 y  4, 5, 2, 2, , ,   cot 1 x   cot 1 y   4, 5, 2, 2,  , cot 1 x  cot 1 y   4  , 5 2 2, 4  5, cot 1 x  cot 1 y   4   , 5, 5, , cot 1 x  cot 1 y , 5, 1, , 18. If f  x   8 x3 and g  x   x 3 ,Then f g  x  is, (a) 8x, 3, (c) 8x , , 2, , A Solution: (d), ,  nor onto,  f is neither one-one, , (b) 83 x, (d) 8x 3, , 0, 1 , , 1, A B   I, 2, , , , 1, 1, 1, 1,  sin    cos    2 & cot    tan    2 , , , , 20. If a matrix A is symmetric as well as skew, symmetric, then, a) A is a diagonal matrix, b) A is a null matrix, c) A is a unit matrix, d) A is a triangular matrix, [NCERT-DIRECT QUESTION], , Solution (b) Since A is symmetric A = A ; Since A, is skew symmetric A = – A;, A = –A,  2A = O, A=O,  1 3   y 0  5 6 , 21. If 2 , , ,  then the value of,  0 x   1 2  1 8 , x & y are, a) x  3, y  3, b) x  3, y  3, c) x  3, y  3, d) x  3, y  3, [NCERT-DIRECT QUESTION], , [NCERT-DIRECT QUESTION], 3, ,  , Solution:(a) f g  x   f  g  x    8  x   8 x,  , 1, 3, , KCET/JEE MATHEMATICS BOOSTER, , x,  x , tan 1    tan 1   ,  ,  , , cot 1  x   tan 1  x , , , 1 3   y, , 0 x   1, 2 6   y, 0 2 x    1, ,  , , Solution:(a) 2 , , 0  5 6 , , 2  1 8 , 0  5 6 , , 2  1 8 , , [email protected], Phone number-8105418762 / 7019881906, , 3

Page 4 : KARNATAKA CET-2017 PAPER WITH SOLUTIONS, , 6  5 6 , 2  y, ,  1, 2 x  2  1 8 , , (i) 2  y  5  y  5  2  3, 6, (ii) 2 x  2  8  2 x  8  2  6  x   3, 2, 22. Binary operation  on R  1 defined by, , a, is , b 1, (a) associative and commutative, (b) associative but not commutative, (c) neither associative nor commutative, (d) commutative but not associative, , a b , , [NCERT-DIRECT QUESTION], , Solution:(c)(i) Commutative, 1, 1, 1, 2  R  1 1 2 ,   1, 2 1 3, 2, 2, and 2 1 ,   1   2, 11 2, From1&2  1 2  2 1, Hence  is not commutative, , (ii) Associative, , 1 2   3  , , 1 , 1, 3  3, 3,  2 1 , , 1, 1,  3   1, 3  1 12,  2 , 2, 1, 1  2  3  1  ,   1    1  ,  3 1 , 4, 2, , ,  1  1 2, ,     2, 1  3 3,  1 , 2  2, From1and 2  1 2   3  1  2  3, , Hence  is not associative,   is neither associative nor commutative, , 3 x, x 1, , , , 3 2, 4 1, , (a) 2 2, (c) 8, , then x is equal to, (b) 4, (d) 2, , [NCERT-DIRECT QUESTION], , Solution:(a) 3 x  3 2, x 1, , 4 1, , 3  x  3  8  3  x  5, 2, , 2, ,  8  x  x   8  2 2, 2, , 4, , c) k 3 A, , d) 3k A, , [NCERT-DIRECT QUESTION], , Solution:(c) If A   aij  & k  R  kA  k n A, nn, , kA  k 3 A, 25. The area of triangle with vertices  k ,0  ,  4,0 , , and  0, 2  is area 4 sq. Units. Then value of k is., (a) 0 or 8, (b) 0 or  8, (c) 0, (d) 8, [NCERT-DIRECT QUESTION], , x1, , y1 1, , x3, , y3 1, , Solution:(a) 1 x y 1  A, 2, 2, 2, , k, k 0 1, 1, 4 0 1  4  4, 2, 0, 0 2 1, , 0 1, 0 1  8, 2 1, , 2  k  4   8, , 1, 2,3 , , 23. If, , 24. If A is a square matrix of order 3  3 , then kA is, equal to, a) k A, b) k 2 A, ,  4  k  4, 4k  4  k  0 &, 4  k  4  k  8, , Ax, 26. Let   By, , Cz, , x2 1, , A, , B, , C, , y, , 2, , 1 & 1  x, , y, , z , then, , z, , 2, , 1, , zx, , xy, , zy, , a) 1  , , b) 1  2, , c) 1  , , d) 1  , , A, , B, , C, , Solution:(a) 1  x, , y, , z, , zy, , zx, , xy, , A, xyz, 1 , x, xyz, zy, , B, , C, , y, , z, , zx, , xy, , C1  xC1 , C2  yC2 , C3  zC3, , xA, 1, 1 , x2, xyz, xzy, xA, xyz 2, 1 , x, xyz, 1, , [email protected], Phon number-8105418762 / 7019881906, t.me/mcpmathematicskar_1986, , yB, , zC, , 2, , z2, , y, , xyz, , xyz, , yB, , zC, , 2, , z2, , 1, , 1, , y, , MCP-MATHEMATICS

Page 5 : KARNATAKA CET BOOSTER, , Ax, 1  By, Cz, , y  x  1, 2, , x, , 2, , 1, , y, , 2, , 1, , Diff w.r.t. x, , z, , 2, , 1, , 2y, , 1  , , kx, , 27. If f ( x)  , , 3, , 2, , if x  2 ,  , is continuous at, if x  2 , , Given dy  tan   1, dx, 4, Equation (2)  2 y 1  1, , x  2 , then the value of k is, (a) 4, , (b) 3, , 3, , 4, , y, (d) 4, , (c) 3, , Put y , , [NCERT-DIRECT QUESTION], , Solution:(b) Given f  x  is continuous at x  2 ., , LHL  RHL, , lim f  x   lim f  x , , x  2, , x2, ,    lim  3, 2, , x2, , k  2  3, 2, , 3, 4, , (c) 4, , (b) 2, 7, (d), 2, , [NCERT-DIRECT QUESTION], , Solution:(a) C   2, 4   f   C  , 16  4 12, 2C , , 6, 2, 2, , 2, , 1, 1,    x x, 4, 2, , (c)  1,  , , 28. The value of C in Mean value theorem for the, function f  x   x 2 , x   2, 4 is., (a) 3, , 1, in y 2  x, 2, , increasing in the interval, (a)  1,  , (b)  , 1, , k  4  3, k, , 1, 2, , 1 1,  The required point is  , , 4 2, 30. The function f  x   x 2  2 x  5 is strictly, , x 2, , lim kx, , dy,  1   2, dx, , (d)  , 1, , [NCERT-DIRECT QUESTION], , Solution:(a) f  x   x 2  2 x  5, f   x   2x  2, f  x  is strictly increasing, , f  x  0, f  4  f  2, 42, , 2x  2  0, x 1  0, x  1  x   1,  , , 31. The rate of change of volume of a sphere with, respect to its surface area when the radius is 4cm is, 6, C   3   2, 4 , (a) 4 cm3 / cm2, (b) 2 cm3 / cm2, 2, (c) 6 cm3 / cm2, (d) 8 cm3 / cm2, 29. The point on the curve y 2  x where the tangent, 4,  with X-axis is, Solution:(b) W.K.T V   r 3 , S  4 r 2, makes an angle of, 3, 4, 4, dV, 4, 3, V  r ,   3   r 2  4 r 2, a)  1 , 1 , b)  1 , 1 , 3, dr 3, 4 2, 2 4, dS, 2, c)  4, 2 , d) 1,1, S  41,, r 2,  dr  8 r, Solution:(b), dV, , dV, 4 r 2, r,  dr , , 4, dS 128 r, dS, 2, dr, dV, 4,   2 cm3 / cm2, dS r  4 2, Tangent,  0,1, y2  x,  1,0 , KCET/JEE MATHEMATICS BOOSTER, [email protected], Phone, number-8105418762 / 7019881906, 5

Page 6 : KARNATAKA CET-2017 PAPER WITH SOLUTIONS, , f  x l a, dy,  g  x m b, dx, h  x  n c, ,  sin x  cos x ,  ,then,  cos x  sin x , , 1, 32. If y  tan , , dy, is equal to, dx, , OR, , (a) 1, , (b) 0, , 2, , (c), , , , (d) 1, , 4, , l, , f  x g  x h  x, , 33. If y , to, , l, , m, , n, , a, , b, , c, , dy, ,then, is equal, dx, , a), , l, , m, , n, , a, , b, , c, , l, b) f   x , , b, , d), , 6, , Put t  tan , ,  2 tan , x  sin 1 , 2,  1  tan , x  sin 1  sin 2 , , 2t, 1 t2, , , 1  2 tan  , ,  & y  tan , 2,  1  tan  , , & y  tan 1  tan 2 , y  2, , &, , dy, 2, d, , &, , , , n, , (b), , 1, , (c), , , , l, , m, , n, , a, , b, , c, , f   x  g   x  h  x , l, , m, , n, , a, , b, , c, , u  cos, , 1, , , , 2, x, , (d) 2, , 2 1  x2, , Solution:(d) Let, u  cos1 2 x 2  1, , f  x g  x h  x, , OR, , tan y , , &, , (a) 1  x 2, , [NCERT-DIRECT QUESTION], , dy, , dx, , 2t, 1 t2, , is, , a, b, c, f   x  g   x  h  x , , Solution:(a,c,d) y , , sin x , , (b) 2, (d) 1, , , , c, , m, , to, (a) 0, (c) 1, Solution:(c), , dy, dy d 2, , ,  1, dx dx 2, d, 35. The derivative of cos1 2 x2  1 w. r. t cos1  x , , f  x l a, c) g   x  m b, h  x  n c, l, , m, , n, , 2t, 2t, dy, ,then, is equal, , tan y , 2, 2, 1 t, 1 t, dx, , dx, 2, d, , m, n, g   x  h  x , , a, , l, , 34. If sin x , , x  2, , f   x  g   x  h  x , , n, , a, b, c  a, b, c, f   x  g   x  h  x  f   x  g   x  h  x , , [NCERT-DIRECT QUESTION], , Solution:(d) SHORT CUT METHOD,  a cos f x   b sin f  x   dy,    f   x , W.K.T y  tan 1 ,  b cos f x   a sin f  x   dx, Hence,  sin x  cos x , y  tan 1 ,   f  x  x,  cos x  sin x , dy, ,  f  x  1, dx, , m, , dy,   1 1, dx, , , , &, , v  cos1  x , , Put x  cos, 1, & v  cos  cos  , 2cos   1, , , , 2, , u  cos1  cos 2 , , & v  cos 1  cos  , , u  2, du, 2, d, , & v , dv, &, 1, d, du, du d, 2, , ,  2, dv d v 1, d, , [email protected], Phon number-8105418762 / 7019881906, t.me/mcpmathematicskar_1986, , MCP-MATHEMATICS

Page 7 : KARNATAKA CET BOOSTER, 2, , d y, 36. If y  log  log x  ,then 2 is equal to, dx,  1  log x ,  1  log x , (a), ,  x log x , 1  log x , (c), 2,  x log x , , x log x, , I , , 1  log x , , I , , (b), , 2, , (d), , I , , 2, , 2, , x log x, , , , 39., , 2,  x  1, , dx, , dx, , x 2  2 x  5  2log x  1  x 2  2 x  5  c, , b), , d 2 y  1  log x , , 2, dx 2,  x log x , , 2,  x  1 x 2  2 x  5  log x  1  x 2  2 x  5  c, d), 2, ,  x  3 e x dx,   x  4 2, 1, ,  x  4, , 2, , c, ,  x  4, , 2, , c, , (b), , ex, c,  x  3, , (d), , e, c,  x  4, ,  1, 1 ,   ex , , dx, x  4   x  4 2 ,  , , , 1, c,  x  4, , cos 2 x  cos 2,  cos x  cos  dx is equal to, a) 2  sin x  x cos    C, b) 2  sin x  x cos    C, c) 2  sin x  2 x cos    C, d) 2  sin x  2 x cos    C, , [NCERT-DIRECT QUESTION], , Solution:(a) I  , ,  x  1, , Solution:(a), ,  x  3 e x dx   x  4   1e x dx, Solution:(d) , 2,   x  4 2,  x  4, ,  ex, , x 2  2 x  5  log x  1  x 2  2 x  5  c, , 2, , x 2  2 x  5  2 log x  1  x 2  2 x  5  c, , [NCERT-DIRECT QUESTION], , x, , x, , c), , , ,  x  1, , cos 2 x  cos 2, dx, cos x  cos , , KCET/JEE MATHEMATICS BOOSTER, , , , x 2  2 x  5 dx  , ,  x  1, , 2, ,  22 dx, , x 2  2 x  5  2 log x  1  x 2  2 x  5  c, , 2, , , , tan 7 x,  cot 7 x  tan 7 x dx is, , , , (b), (c), (d), 4, 2, 3, , 40. The value of, (a), , [NCERT-DIRECT QUESTION], , 38., , cos x  cos , , x, , x log x  0   1  log x , d y, x, , , 2, 2, dx,  x log x , 2, , e, , cos x  cos , 2  cos x  cos   cos x  cos  , , x 2  2 x  5 dx is equal to, ,  x  1, , a), , Diff dy w.r.t x, , (c), , dx, , I  2  sin x  x cos    C, , Diff y w.r.t x, dy, 1, , dx x log x, , (a), , , , 2 cos 2 x  cos 2 , , I  2  cos x  cos  dx, , Solution:(a) y  log  log x , , 37., , 2 cos x  1  2 cos 2   1, dx, cos x  cos , 2, , , 6, , 2, 0, , [NCERT-DIRECT QUESTION], , Solution:(b) W.K.T Short cut method, , , , , , 2, 0, , , tan n x, cot n x, , 2, dx, , dx , n, n, n, n, , 0 tan x  cot x, tan x  cot x, 4, , ,  2, 0, , tan 7 x, , dx , 7, 7, cot x  tan x, 4, , 5, , 41., , , , x  2 dx is equal to, , 5, , (a) 29, , (b) 28, , (c) 27, , [NCERT-DIRECT QUESTION], , (d) 30, , Solution:(a), 5, , I   x  2 dx, 5, , , , 2, , 5, , x  2  0  x  2, , I   ( x  2)dx   ( x  2)dx, 5, , I  , , , 2, , 2, ( x  2) 2, 2, , 5, ,    ( x  2) 2 ,  5  2  2, , [email protected], Phone number-8105418762 / 7019881906, , 7

Page 8 : KARNATAKA CET-2017 PAPER WITH SOLUTIONS, , , 1, 1, I    9   49, 2, 2, 9 49 58, I , , 2 2, 2, , 2, , 43., , 0, , I  29, , 2, ,  e, , 42., , , , dx, is equal to, 1, , sin x, , , (c), 2, , (b) 1, , 2, , 1, dx is equal to, sin x  b 2 cos 2 x, 2, , (a)  a, , (b)  a, , 4b, , (c) b, 4a, , (d) , 2ab, , 2b, , [NCERT-DIRECT QUESTION], , 2, , (a) 0, , a, , (d), , , 2, , , 2, , Solution:(d) I  , 0, , , 2, , , , 2, , , , dx,  1, sin x, e 1, , , , Solution:(d) I , , 1, dx, a sin x  b 2 cos 2 x, 2, , 2, , 1, I 2, 2, b, cos, 0, , 2, , , 2, , , , e, , 2, ,   , , sin     x ,  2 2, , , , , I, ,  e, , dx, 1, , , , , , sin   x , , 1, , , , 2, ,  e, , , , , , , , , dx,  sin  x , , 1, , 2, , esin  x , , 2, , , 1, , 2, , , , sin  x , , I, , esin  x , , 2, , , , , , sin x, , 2, , II , , , , , , 2I , , , , , , I, , 1dx, , sin  x , e, 1, , 2, , , 2, , , , 2, , sin  x , , e, 1, dx , sin  x , e, 1, , 2, , e   dx,  1  esin x , sin x, , 2, , , , 2, , , 2, ,  1dx, , , , 2, , , , 1,  x  2, 2 2, , 1      1     1,   , ,   , 2  2  2   2  2 2  2, , I, 2, I, , 44. The area of the region bounded by the curve, y  x 2 and the line y  16 is, a) 128 sq.units, b) 64 sq.units, 3, 3, c) 32 sq.units, d) 256 sq.units, 3, 3, y, [NCERT-DIRECT QUESTION], Solution:(d), 16, , 16, , 0, , 0, , y  16, , A  2  x dy  2  y dy, , y  x2, , y0, , 16, , 3, 2  2  32 , 4, A, y, , 16,  2,  , 3  0 3, , A, 8, , 1  1  a, , a, , tan  tan   tan 1  tan 0  , , ab , 2, b, b, , , I, , Equations (1)+(2), , , , dx  tan 1 f  x   c , , , , 1  1  a, , a, , tan  tan   tan 1  tan 0  , , ab , 2, b, b, , 1,  tan 1     tan 1  0  , I, ab , 1 , , , I,  0 , ab  2, 2, ab, , , e   dx, I ,   2, sin  x ,  1 e, 2, , 2, , 1  1  a,  2, I, tan  tan x  , , ab , b, 0, , dx, ,  1  e, , f  x, ,  1 f  x,  , , , , , , 2, , I, , dx, , 2, , , , 1, , , 2, , , , , a, sec 2 x, 1 2 b, I, dx, 2, ab 0  a, ,  tan x   1, b, , , dx, , , , I, , , , , , 1,  2, dx, 2, x  a sin x,  1 ,  2,  b cos 2 x, , , 4 4 32 4 6 256,  2   3  2  3 sq.units, 3, , [email protected], Phon number-8105418762 / 7019881906, t.me/mcpmathematicskar_1986, , MCP-MATHEMATICS

Page 9 : KARNATAKA CET BOOSTER, , 45. The area of the region bounded by the curve, y  cos x, x  0 & x   is, a) 3sq.units, b) 1sq.units, c) 4 sq.units, d) 2 sq.units, [NCERT-DIRECT QUESTION], , Solution:(d), , –3/2, , 1,  xC, 1 y, 48. The integrating factor of the differential equation, dy, x  2 y  x2  x  0, dx, log, , Y, , a) x, , O, – –/2 /2, , y=1, X, y = –1, , 3/2, 2, , , b) log x, , log x, c) e, , d) x, , [NCERT-DIRECT QUESTION], , dy,  2 y  x2, dx, [Divided by x on both sides], , Solution:(a) x, , Y, , , dy 2,  y  x  dy  Py  Q , dx x,  dx, , , A  2 2 cos x dx, 0, , , , A  2 sin x02  2 1  0, A  2 sq.units, 46. The degree of the differential equation, 2,   dy  2 , d2y, 1      2 is, dx,   dx  , a) 1, b) 2, c) 3, d) 4, [NCERT-DIRECT QUESTION], 2, ,   dy  2 , d2y, Solution:(a) 1      2, dx,   dx  , 4, , 2, , dy,  y  1 y  1 is, dx, 1,  xC, 1 y, , P, , 2, , Pdx,  dx, I .F  e   e x  elog f  x   f  x  , , , ,  e2log x  elog x  x2, 49. If a  2iˆ  ˆj  kˆ & b  ˆi  2jˆ  3kˆ are orthogonal,, then value of  is, 3, 5, (a) 0, (b) 1, (c), (d) , 2, 2, 2, , 2, , a b  0,  2,  ,1  1, 2,3  0, , Solution:(d), , 2  2  3  0, 2  5  0   , , 1,  x  C, 1 y, dy, Solution:(a),  y 1, dx, dy,  1 y, dx, dy,  dx, 1 y, dy,  1  y   1dx,  log 1  y  x  C, , d) log, , KCET/JEE MATHEMATICS BOOSTER, , 5, 2, , 50. If a , b, c are unit vectors such that a  b  c  0 ., Find the value of a.b  b.c  c.a is, , b) log 1  y  x  C, c) log 1  y  x  C, , 2, &Qx, x, , [NCERT direct question], ,  dy ,  dy  d y, 1    2   2, dx,  dx ,  dx ,  order -2and degree-1, 47. The general solution of the differential equation, , a) log, , 2, , (a) 1, , (c) , , (b) 3, , 3, 2, , (d), , 3, 2, , [NCERT direct question], , Solution:(c)W.K.T, 2, , 2, , 2, , 2, ,  , , , ,      , a  b  c  a  b  c  2  a b  b c  c  a , , , , , , 0  12  12  12  2   , 0  3  2    , , 3, 2, , 51. If a & b are unit vectors then what is the angle, between a & b for 3 a  b be unit vector?, (a) 30 0, (b) 450, (c) 90 0, (d) 60 0, Solution:(a) Given a  b  1 and, [email protected], Phone number-8105418762 / 7019881906, , 9

Page 10 : KARNATAKA CET-2017 PAPER WITH SOLUTIONS, , 3 a  b  1 (Squaring on both sides), , sin sin 1    , , 2, , 3a  b  12, 2, , 3 a  b 2, 2, , , , , , 2, , 2 3 6, , , , 7, 7 7, , 4, 9 36, 1, , , 49 49 49, , 3a  b  1, , 2, 2,  7 , 1 7, ,  , 3 a  b  3, , 3 1  1   2 3 a  b  1, 2, , 1, 0, 0   , , a b cos  , , cos  , , 54. The distance of the point  2, 4, 5 from the line, , 3 , a  b  a b cos  , , 2 3 , , x 3 y 4 z 8, is, , , 3, 5, 6, , 3 3, 3, , a  b  1, , , 2, 2 3, ,    30, , 52. Reflection of the point  ,  ,   in XY plane is, (a)  ,  ,0 , , (b)  0,0,  , , (c)   ,  ,  , , (d)  ,  ,  , , Solution:(d), , 37, 10, , (a), , 0, , 37, 10, , (b), , , (c), , 37, 10, , (d), , 37, 10, , , , Solution:(b)Let a   3, 4, 8  ,    2, 4, 5  ,, , , , , , , a     1,0, 3 , b   3,5, 6  ,, , Z, , d, ,  P  ,  ,  , ,   ,  a   b, , , , , b, Y, Reflection point, , X, ,  Q  ,  ,  , , /, , Z, , , , O,  , , 7, , 3, (a), 2, 2, (c), 7, Solution:(c), , , , 2, (b), 3, 3, (d), 7, 2 x  3 y  6 z  11  0  1, , 22  32  62  4  9  36  49  7 , , , Equation(1)  7 ,we get normal form of plane, 2, 3, 6, 11, Equation x  y  z , 7, 7, 7, 7, 2 3 6,  nˆ   ,  , , 7 7 7, Direction cosines of X-axis  bˆ  1, 0, 0 , , , sin  , , , , bn, , , , ,    sin 1  , , , , , , i, , j, , k, , 8, , 1 0 3, , A, , 3, d, , 53. The plane 2 x  3 y  6 z  11  0 makes an angle, , sin 1   With X-axis. The value of  is equal to, , , , d, d, d, , 65, , , , 6, , 9  25  36, , , , , , , , 15 i   6  9  j  5 k, 70, , 15, 3, 5, 70, 225  9  25, , 70, , 259, 37  7, , 70, 10  7, , 37, 10, , 55. A box has 100 pens of which 10 are defective. The, probability that out of a sample of 5 pens drawn, one by one with replacement and at most one is, defective is, , 1 9 , (b)  , 2  10 , , 9, (a), 10, 5, ,  9  1 9 ,    ,  10  2  10 , , (c) , , 4, , (d), , 4, , 1 9 ,  , 2  10 , , 5, , Solution:(c) Let X denote the number of defective, pens, (X=0,1, 2, 3, 4, 5), 10 1, 1 9, n5 , p, , , q  1 p  1  ,, 100 10, 10 10, , b n, , 10, , [email protected], Phon number-8105418762 / 7019881906, t.me/mcpmathematicskar_1986, , MCP-MATHEMATICS

Page 11 : KARNATAKA CET BOOSTER, , 9, P  X  x   nCx q n x p x  5Cx  ,  10 , P  X  1  P  X  0   P  X  1, 9,  5C0  ,  10 , , 5 0, , 5 x, , 0, , 1,  ,  10 , , 1 5 9,    C1  ,  10 ,  10 , , 5, , 51, ,  0,5, , 5, , 56. Two events A and B will be independent if, (a) A and B are mutually exclusive, (b) P( A  B)  1  P( A)   1  P( B) , (c) P( A)  P( B), (d) P ( A)  P ( B )  1, Solution:(b) P( A  B)  P( A  B ),  1  P( A  B),  1  P( A)  P( B)  P( A  B),  1  P( A)  P( B)  P( A) P( B),  1  P( A)   P( B) 1  P( A) ,  1  P( A) 1  P( B) , ,   x  dx is, 3.5, , 0.2, , (a) 3, (b) 3.5, Solution:(c), , (d) 4, , (c) 4.5, ,   x dx    x  dx    x  dx    x  dx    x  dx, , (d) 1, , 3.5, , 1, , 0.2, , 0.2, , 2, , 3, , 1, , 1, , 3.5, , 2, , 2, , 3, , 3, , 3.5, ,   0 dx   1 dx   2dx   3 dx, , [NCERT-DIRECT QUESTION], 3, , Solution:(a), , x, , 59. If an LPP admits optimal solution at two, consecutive vertices of a feasible region, then, a) the required optimal solution is at the midpoint, of the line joining two points., b) the optimal solution occurs at every point on the, line joining these two points, c) the LPP under consideration is not solvable, d) the LPP under consideration must be, reconstructed, Solution:(b) If an LPP admits optimal solution at two, consecutive vertices of a feasible region, then, the optimal solution occurs at every point on the, line joining these two point, 60., , 57. The probability distribution of X is, x, 0, 1 2, 3 Find the value of k, P(x) 0.3 k 2k 2k, (c) 0.7, ,  6, 0 , ,  4, 0 , , 4, , (b) 0.3, , x6, , 5 x  4 y  20, , 4, ,  9   1  9 ,       ,  10   2  10 , , y3, ,  0,3, 1, , 1,  ,  10 , , 9, 9 1,  1   1  5    , 10,  ,  10   10 , , (a) 0.14, , y, , Solution:(c), , x, , 0.2, ,  P( X  x)  1, , 1, , 2, , x 0, , 0.3  k  2k  2k  1, 5k  1  0.3  0.7, 0.7, k,  0.14, 5, 58. The shaded region shown in the fig is given by the, in-equation, , 3, ,  0   x 1  2  x 2  3  x 3, 2, , 3, , 3.5, ,  0   x 1  2  x 2  3  x 3, 2, , 3, , 3.5, ,   2  1   6  4   10.5  9   1  2  1.5, yx, ,  4.5, , y, L.R, ,  0,5,  0,3, ,  4, 0 , (a), (b), (c), (d), ,  6, 0 , , x, , 5 x  4 y  20, x  6, y  3, x  0, y  0, 5 x  4 y  20, x  6, y  3, x  0, y  0, 5 x  4 y  20, x  6, y  3, x  0, y  0, 5 x  4 y  20, x  6, y  3, x  0, y  0, , KCET/JEE MATHEMATICS BOOSTER, , [email protected], Phone number-8105418762 / 7019881906, , 11