KCET EXAMINATION – 2024

SUBJECT : MATHEMATICS VERSION : D4

DATE :- 18-04-2024 TIME : 02.30 PM TO 03.50 PM
 KCET - 2024 (CODE – D4) 2
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1. The value of C in (0, 2) satisfying the mean 4. The function f  x   cos x is

2
value theorem for the function f  x   x  x  1 , (A) Everywhere continuous and differentiable
x  0,2 is equal to (B) Everywhere continuous but not

3 4 1 2 differentiable at odd multiples of
(A) (B) (C) (D) 2
4 3 3 3
(C) Neither continuous nor differentiable at
Ans. B

Sol. f(2)  2, f(0)  0  2n  1 , n  Z
2
20
f '(c)  1 (D) Not differentiable everywhere
20
Ans. B
f '(c)  2c(c  1)  (c  1)2 Sol. Conceptual
option verification
dy
5. If y  2x 3x , then at x  1 is
d  2
 1 2 x  dx
2. cos  cot   is
dx  2x (A) 2 (B) 6 (C) 3 (D) 1
  
Ans. B
3
(A)  Sol. log y  log  2.x 3x 
4
1  log 2  3x log x
(B) 
2 1 dy  1 
 3  x.  log x 
1 y dx  x 
(C)
2 dy
 2.x 3x .3 1  log x 
1 dx
(D)
4  dy 
Ans. D   6
 dx  x 1
Sol. x  2cos 
2  x 2 1  cos    6. Let the function satisfy the equation
  cot 2
2  x 2 1  cos   2 f x  y  f x f y for all x, y  R , where
d 1 1 d  x f  0   0 . If f  5   3 and f '  0   2 , then f '  5 
 1  cos    1  
dx 2 2 dx  2
is
1 (A) 6 (B) 0 (C) 5 (D) -6

4 Ans. A
Sol. f  x  kx
3. For the function f  x   x 3  6x 2  12x  3; x  2
f 5  3
is
(A) a point of minimum k5  3
(B) a point of inflexion  f  x   3x/5
(C) not a critical point
1
(D) a point of maximum f '  x   3x/5 log e 3.
5
Ans. B
f ' 0  2
Sol. f '  x   3x 2  12x  12
log e 3
f ''  x   6x  12 2
5
f ''  2  0 log e 3  10
f '''  2  0 f '  x   2 3
x/5

f ' 5  2  3  6
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5
1
7.  x 6  log x  2
 7 log x  2
dx  9.   x  3  1  x  dx 
1
 
5
1 2 log x  1 (A) 12 (B) (C) 21 (D) 10
(A) log C 6
2 3 log x  2
Ans. A
2 log x  1 5
(B) log C
3 log x  2 Sol.   x  3  x  1  dx
1
3 log x  2 3 5
(C) log C
2 log x  1   2dx   2x  4dx
1 3
1 3 log x  2 5
(D) log C  2  2    x 2  4x 
2 2 log x  1 3

Ans. B  4   25  20    9  12    4  8  12
1
Sol. Put logx  t  dx  dt
x  n n n 1 
1 10. lim  2  2  2  ........  
n n  12 n  22
n  32
5n
I dt  
 3t  2 2t  1  
(A) (B) tan1 3 (C) tan1 2 (D)
1 A B 4 2
 
 3t  2  2t  1 3t  2 2t 1 Ans. C
2n
After solving A  3, B  2  n 
Sol. lim   2 2 
r 1  n  r 
n
3 2
I   dt   dt 2n
3t  2 2t  1 n
 lim  2
n
r 1  r 
  log 3t  2  log 2t  1  C n2 1    
 n
 
2
5x 1
sin  2
dx  tan 1 2
2 1  x
8.  x dx  0

sin
2
11. The area of the region bounded by the line
(A) 2x  sin x  2sin2x  C
y  3x and the curve y  x 2 in sq. units is
(B) x  2sin x  2sin2 x  C
(A) 10
(C) x  2sin x  sin2 x  C
9
(D) 2x  sin x  sin2x  C (B)
2
Ans. C
(C) 9
 5x  x (D) 5
2sin   cos
 2  2 Ans. B
Sol.  dx
x x 3
2sin cos 3
 x2  x3  9
2 2 Sol. A   (3x  x 2 )dx  3     
sin3 x  sin2 x 0
2
  3 0 2
 dx
sin x
  3  4sin2 x  2cos xdx

  1  2cos 2x  2cos x dx
 x  sin2x  2sin x  C
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12. The area of the region bounded by the line 15. The distance between the two planes
y  x and the curve y  x 3 is 2x  3y  4z  4 and 4x  6y  8z  12 is
(A) 0.2 sq. units (A) 2 units (B) 8 units
(B) 0.3 sq. units 2
(C) units (D) 4 units
(C) 0.4 sq. units 29
(D) 0.5 sq. units Ans. C
Ans. D 64 2
1 2 4 1 Sol. d  
x x  4  9  16 29
Sol. A  2 (x  x 3 )dx  2     0.5 sq.units
0  2 4 0
16. The sine of the angle between the straight line
x 2 y 3 4z
  and the plane
3 4 5
2x  2y  z  5 is
1 2 3 3
(A) (B) (C) (D)
5 2 5 2 50 50
Ans. A
3(2)  4(2)  5(1)
Sol. sin  
9  16  25 4  4  1

17. The equation xy  0 in three-dimensional

   space represents
13. If a, b,c are three non-coplanar vectors and p,
  (A) a pair of straight lines
 bc
q, r are vectors defined by p    , (B) a plane
[a b c]
(C) a pair of planes at right angles
   
 ca  ab (D) a pair of parallel planes
q     , r     , then
[a b c] [a b c] Ans. B
         Sol. Conceptual
(a  b).p  (b  c).q  (c  a).r is
(A) 0 (B) 1 (C) 2 (D) 3 18. The plane containing the point (3, 2, 0) and the
Ans. D x 3 y 6 z 4
      line   is
Sol. Since a.(b  c)  a b c  , 1 5 4
         (A) x  y  z  1
(a  b).p  (b  c).q  (c  a).r  1  1  1  3
(B) x  y  z  5
x 1 y  2 z  3 (C) x  2y  z  1
14. If lines   and
3 2k 2 (D) 2x  y  z  5
x 1 y  5 z  6 Ans. A
  are mutually
3k 1 5 i j k
perpendicular, then k is equal to 
Sol. n 0 4 4
10 1 5 4
(A) 
7
7  ˆi(16  20)  ˆj(0  4)  k(0
ˆ  4)
(B) 
10  4iˆ  4ˆj  4k
ˆ
(C) 10  Eq. of plane is
(D) 7 4(x  3)  4(y  2)  4(z  0)  0
Ans. A  4x  12  4y  8  4z  0
Sol. (3)(3k)  (2k)(1)  2(5)  0
 x  y  z 1
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19. Corner points of the feasible region for an LPP 22. If a random variable X follows the binomial
are (0, 2), (3,0), (6, 0), (6, 8) and (0, 5). Let distribution with parameters n  5 , p and
z  4x  6y be the objective function. The P  X  2   9P  X  3  , then p is equal to
minimum value of z occurs at
1 1
(A) Only (0, 2) (A) 10 (B) (C) 5 (D)
10 5
(B) Only (3, 0)
Ans. B
(C) The mid-point of the line segment joining
the points (0, 2) and (3, 0) Sol. Given n  5 , P  X  2   9P  X  3 
(D) Any point on the line segment joining the  n c 2 .q n 2 .p2  9.n c 3 .q n  3 .p3
points (0, 2) and (3, 0)
 q  3  n  2  .p
Ans. D
Sol. Conceptual  1  p  3  3  .p
1
20. A die is thrown 10 times. The probability that p
10
an odd number will come up at least once is
11 1013 1023 1 23. Two finite sets have m and n elements
(A) (B) (C) (D)
1024 1024 1024 1024 respectively. The total number of subsets of
Ans. C the first set is 56 more than the total number
1 1 of subsets of the second set. The values of m
Sol. n  10, p  ; q 
2 2 and n respectively are
P(x  1)  1  P(x  0) (A) 7, 6 (B) 5, 1 (C) 6, 3 (D) 8, 7
0 10 Ans. C
1 1
 1 10 C0     Sol. 2m  56  2n then verification m  6, n  3
2 2

21. A random variable X has the following 24. If  x 2  5  x   6  0 , where [x] denotes the
probability distribution : greatest integer function, then
X 0 1 2 (A) x  3, 4 (B) x  2, 4 
25 1
P X k (C) x  2,3 (D) x   2,3
36 36
Ans. B
1
If the mean of the random variable X is ,
3 Sol.  x   2  x   3   0
then the variance is   x   2 or  x   3
1
(A)  x   2, 4 
18
5
(B) 25. If in two circles, arcs of the same length
18
subtend angles 300 and 780 at the centre,
7
(C) then the ratio of their radii is
18
5 13 13 4
11 (A) (B) (C) (D)
(D) 13 5 4 13
18
Ans. B
Ans. B
Sol. l1  l 2 , 1  300 , 2  780
1 2 1 1 1 5
Sol.  pi x i  3  0  k  36  3  K  3  18  18 l1 r11 r  780 13
Then   1  2  
4 l 2 r2 2 r2 1 300 5
2 06
2   2 
p x
i i
 36  14
p i 1 36

14 14 1 5
 2   2   
36 36 9 18
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26. If ABC is right angled at C , then the value 29. The value of
49
of tanA  tanB is C3  48 C3  47 C3  46 C3  45 C3  45 C4 is
a2 (A) 50
C4
(A) a  b (B)
bc 50
(B) C3
c2 b2 50
(C) (D) (C) C2
ab ac
50
Ans. C (D) C1
A Ans. A
Sol. Since ncr  ncr 1  (n  1)cr ,
b c
Sol.
49
C3  48 C3  47 C3  46 C3   45
C3  45 C4 
B  49 C3  48 C3  47 C3  46 C3  46 C4 ……
C a
Since C is a right angle then  50 C4
a b a 2  b2 c 2
Tan A  Tan B    
b a ab ab 30. In the expansion of (1  x)n
C1 C C C
 2 2  3 3    n n is equal to
1  isin C0 C1 C2 Cn 1
27. The real value of '  ' for which is
1  2isin n  n  1
purely real is (A)
2

(A)  n  1 , n   n
2 (B)
2

(B)  2n  1 , n   n 1
2 (C)
2
(C) n, n  
(D) 3n  n  1

(D)  2n  1 , n   Ans. A
2
C1 C C C
Ans. C Sol.  2 2  3 3    n n
C0 C1 C2 Cn 1
1  isin
Sol. z  , after simplify n.  n  1
1  2isin  n   n  1   n  2  ........  1 
2
z
1  2sin    i  3 sin  
2

and z is purely real

1  4 sin2  31. If Sn stands for sum to n-terms of a G.P. with
then Im  z   0 ‘a’ as the first term and ‘r’ as the common ratio
 sin   0 then Sn : S2n is
   n, n  N
(A) r n  1
28. The length of a rectangle is five times the 1
breadth. If the minimum perimeter of the (B) n
r 1
rectangle is 180 cm , then
(C) r n  1
(A) Breadth  15 cm
1
(B) Breadth  15 cm (D) n
r 1
(C) Length  15 cm
Ans. B
(D) Length = 15 cm
Ans. B 
a. rn  1 
Sol. Given l  5b, P  180 Sn r 1 rn  1 1
Sol.   
 2 l  b   180
S2n a. r2n  1  n n
r 1 r 1  n
r 1   
r 1
 b  15
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32. If A.M. and G.M. of roots of a quadratic 36. The negation of the statement
equation are 5 and 4 respectively, then the “For every real number x; x 2  5 is positive” is
quadratic equation is
(A) For every real number x; x 2  5 is not
2 2
(A) x  10x  16  0 (B) x  10x  16  0
positive.
2
(C) x  10x  16  0 (D) x 2  10x  16  0
(B) For every real number x; x 2  5 is negative
Ans. D
(C) There exists at least one real number x

Sol. Given A.M   5      10 , such that x 2  5 is not positive
2
(D) There exists at least one real number x
G.M  .  4    16
such that x 2  5 is positive
 The quadratic equation is
Ans. C
x 2       x    0 Sol. Conceptual

33. The angle between the line x  y  3 and the 37. Let a, b, c, d and e be the observations with
mean m and standard deviation S. The
line joining the points 1,1 and  3, 4  is
standard deviation of the observations
 1 a  k, b  k,c  k, d  k and e  k is
(A) tan1  7  (B) tan1   
 7
S
(A) kS (B) S  k (C) (D) S
1  1  1  2  k
(C) tan   (D) tan  
7 7 Ans. D
Ans. C Sol. adding constant each observation of S.D does
Sol. Slope of x  y  3 is m1  1 and not effect.
Slope of line joining the points 1,1 ,  3, 4  is
38. Let f : R  R be given by f  x   tan x . Then
3
m2  
4 f 1 1 is
m1  m2 1 
and tan      tan1 (A)
1  m1m2 7 4
  
(B) n  : n  Z 
34. The equation of parabola whose focus is (6,0)  4 
and directrix is x  6 is 
2 2
(C)
(A) y  24x (B) y  24x 3
(C) x 2  24y (D) x 2  24y   
(D) n  : n  Z 
 3 
Ans. A
Ans. B
Sol. Focus =  F   a,0    6,0 

Equation of directrix is x  6 then equation of Sol. tan x  1  x  n 
4
parabola is of the form y 2  4ax,a  6
39. Let f : R  R be defined by f  x   x 2  1 . Then
2 cos x  1 the pre images of 17 and -3 respectively are
35. lim is equal to
x
 cot x  1 (A) , 4, 4
4

1 1 (B) 3, 3 , 

(A) 2 (B) 2 (C) (D)
2 2 (C) 4, 4 , 
Ans. C
(D) 4, 4 , 2, 2
Sol. By L.H. Rule
1 Ans. C
2
 2  sin x  0 2 1 Sol. f  x   x 2  1  17  x  4
Lt  
 cosec2 x 2
2
x
4  2 x 2  1  3 is not possible.
No preimage of -3
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2
44. If A is square matrix such that A 2  A , then
40. Let  gof  x   sin x and  fog  x   sin x   .
Then
 I  A 3 is equal to
(A) 7A-I (B) 7A (C) 7A+I (D) I-7A
(A) f  x   sin2 x, g  x   x
Ans. C
(B) f  x   sin x, g  x   x Sol.  I  A 3  I  3A  3A  A  7A  I
2
(C) f  x   sin x, g  x   x

(D) f  x   sin x, g  x   x 2 1 1 10
45. If A    , then A is equal to
Ans. C 1 1
(A) 28 A (B) 29 A (C) 210 A (D) 211 A
Sol.  
g  f  x    g sin2 x  sin2 x  sin x
Ans. B
2

fog  x   f  x   sin x  Sol. A2  2' A, A4  A3 A  A10  29 A

41. Let A  2,3,4,5,.......16,17,18, . Let R be the x  3 2x 2  18 2x 3  81

relation on the set A of ordered pairs of positive 46. If f  x   x  5 2x 2  50 4x 3  500 , then
integers defined by (a, b) R (c, d) if and only if 1 2 3
ad=bc for all (a, b), (c, d) in A  A . Then the
number of ordered pairs of the equivalence f 1 .f  3   f  3  .f 5   f  5  .f 1 is
class of (8,2) is (A) -1 (B) 0 (C) 1 (D) 2
(A) 4 (B) 5 (C) 6 (D) 7 Ans. No option
Ans. C Sol.
Sol. 6 Pairs
 3,2 ,  6,4  , 9,6  , 12,8  , 18,12 ,(15,10) cos x x 1
f x
47. Let f  x   2 sin x x 2x . Then lim 2 
x 0 x
sin x x x
42. If cos 1 x  cos 1 y  cos 1 z  3, then x  y  z 
(A) -1
 y  z  x   z  x  y  equal to
(B) 0
(A) 0 (B) 1 (C) 6 (D) 12 (C) 3
Ans. C (D) 2
Sol. x  y  z  1 Ans. B
 x  y  z  y z  x   z  x  y   6 Sol. f  x    x 2 cos x  x sin x
f x
1 1 1  lim 0
43. If 2 sin x  3 cos x  4, x   1,1 then 2sin x 0 x2
x  3 cos 1 x is equal to
48. Which one of the following observations is
4  6
(A) correct for the features of logarithm function to
5
and base b>1?
6  4 (A) The domain of the logarithm function is R,
(B)
5 the set of real numbers
3 (B) The range of the logarithm function is R  ,
(C)
2 the set of all positive real numbers.
(D) 0 (C) The point (1,0) is always on the graph of the
Ans. B logarithm function
 (D) The graph of the logarithm function is
Sol. sin1 x  cos 1 x 
2 decreasing as we move from left to right.
4 Ans. C
cos 1 x 
5 Sol. log 1  0
3  8
sin1 x 
10
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
1  3 
 1  x  sinx  cos xdx 
2 2
49. If P  1 3 3  is the adjoint of a 3  3 matrix 53.

2 4 4 
2
A and A  4, then  is equal to (A)   (B) 2  3
3
(A) 4 (B) 5 (C)11 (D) 0
3
Ans. C (C)   (D) 0
2
2
Sol. P  A  A  A  16  2  6  16    11 Ans. D
Sol. f(x) is Odd function, then I  0
x 1 1 54. The function x x ; x  0 is strictly increasing at
x 1 dB
50. If A  and B  1 x 1 , then is 1
1 x dx (A) x   (B) x 
1 1 x e
(A) 3A (B) -3B (C) 3B+1 (D) 1-3A 1
Ans. A (C) x  (D) x  0
e
dB
Sol. A  x 2  1,
dx

 3 x 2  1  3A  Ans. C
1
Sol. f 1  x   x x 1  log x   f 1  x   0  x 
e
If f  x   xe   then f  x  is
x 1 x
51.
(A) increasing in  55. The maximum volume of the right circular
(B) decreasing in  cone with slant height 6 units is
 1  (A) 4 3 cubic units
(C) decreasing in   ,1
 2  (B) 16 3 cubic units
 1  (C) 3 3 cubic units
(D) increasing in   ,1
 2  (D) 6 3 cubic units
Ans. D Ans. B
2
Sol. f1  x   ex x  x  2x  2
1
Sol. V  r2h,
3
 1 
 f is increasing in   ,1 l 2  r 2  h2 ,
 2 
r2  36  h2 ,
sinx  x  16 3
52.  dx 
3  4cos2 x Vmax  16 3
1  2cosx 
(A)  tan 1  C  
2 3  3  56. The vectors AB  3ˆi  4k
ˆ and AC  5ˆi  2ˆj  4k
ˆ
1  cosx 
(B) tan1  C are the sides of a ABC . The length of the
3  3  median through A is
1  cosx 
(C) tan1  C (A) 18 (B) 72 (C) 33 (D) 288
2 3  3  Ans. C
1 1  2cosx  1 1 ^ ^ ^ ^ ^ ^
(D)  tan  C Sol. AB  AC  8 i  2 j 8 k  4 i  j 4 k  33
3  3  2 2
Ans. A
Sol. Put
dt 1 1  2t 
cos x  t   3   2t  2
   tan1  
3 2  3
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57. The volume of the parallelopiped whose co-
terminous edges are ˆj  k,
ˆ ˆi  kˆ and ˆi  ˆj is
(A) 6 cu.units (B) 2 cu.units
(C) 4 cu.units (D) 3 cu.units
Ans. B
0 1 1
^ ^ ^ ^ ^ ^ 
Sol.  j k i  k i  j   1 0 1  2 cub units
 
1 1 0

 
58. Let a and b be two unit vectors and  is the
 
angle between them. Then a  b is a unit
vector if
 
(A)   (B)  
4 3
2 
(C)   (D)  
3 2
Ans. C
2
Sol. ab  1  1  1  2 a b cos   1
1 2
 cos    
2 3

dy
59. The solution of e dx  x  1, y  0   3 is
(A) y  2  xlogx  x
(B) y  x  3  xlogx
(C) y  x  3   x  1 log  x  1
(D) y  x  3   x  1 log  x  1
Ans. D
dy
Sol.  log  x  1  dy  log  x  1 dx
 
dx
And y  0   3 then
 y  x  3   x  1 log  x  1

60. The family of curves whose x and y intercepts

of a tangent at any point are respectively
double the x and y coordinates of that point
is
(A) xy  C (B) x 2  y 2  C
y
(C) x 2  y 2  C (D) C
x
Ans. A
x y
Sol.   1  xy  c
2x1 2y1
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