17/04/2025

Version Code - B2
Subject Code: M

Corporate Office : AESL, 3rd Floor, Incuspaze Campus-2, Plot-13, Sector-18, Udyog Vihar,
Gurugram, Haryana-122015

Answers & Solutions

for
Karnataka Common Entrance Test-2025
Time : 80 Minutes
(MATHEMATICS) M.M. : 60

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 KCET-2025 (17-04-2025)

MATHEMATICS
Choose the correct answer: Answer (3)
 e x + ax , x  0  e1/ x − 1
1. The function f (x) =  is  , x0
b ( x − 1) , x  0
2 Sol. f ( x ) =  e1/ x + 1

 0 , x =0
differentiable at x = 0. Then
(1) a = 3, b = –1 (2) a = 1, b = 1 e1/ x − 1 0 −1
lim = = −1
(3) a = 3, b = 1 (4) a = –3, b = 1 x →0 −
e 1/ x
+1 0 +1

Answer (4) e1/ x − 1 1 − e −1/ x 1− 0

lim = lim = =1
+ 1/ x + −1/ x 1+ 0
 e + ax , x  0
x x →0 e +1 x →0 1+ e
Sol. f ( x ) = 
b ( x − 1) , x  0
2 LHL  RHL
Not continuous at x = 0
For continuity,
dy 3
lim f ( x ) = lim f ( x ) = f ( 0 ) 3. If y = a sin3t, x = a cos3t, then at t = is
− + dx 4
x →0 x →0

e0 + a × 0 = b(0 –1)2 (1) 1 (2) – 1

b=1 (3)
1
(4) − 3
For differentiable 3
Answer (1)
lim e x + a = lim 2b ( x − 1)
− +
x →0 x →0 Sol. y = a sin3t
e0 + a = 2(1)(0 – 1) x = a cos3t
1+a=–2
a=–3
dy
dt
(
= a 3 sin2 t . cos t )
2. A function

 1
dx
dt
(
= a 3cos2 t ( − sin t ) )
e x
−1
 , if x  0
f (x) =  1 , is dy 3a sin2 t . cos t
e x + 1  =
dx −3a cos2 t sin t
 0 , if x = 0

sin t
(1) differentiable at x = 0, but not continuous at x = =− = − tan t
cos t
0
(2) continuous at x = 0 dy  3
 = − tan
dx t = 3 4
(3) not continuous at x = 0 4
(4) differentiable at x = 0 = – (– 1) = 1

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KCET-2025 (17-04-2025)
4. The derivative of sin x with respect to log x is dx
cos x
7. The value of
 ( x + 1) ( x + 2) is
(1) (2) cos x
x
x +1 x –1
cos x (1) log +c (2) log +c
(3) x cos x (4) x+2 x+2
log x
x –1 x+2
Answer (3) (3) log +c (4) log +c
x–2 x +1
Sol. y = sin x
z = log x Answer (1)

dy dz 1 dx  ( x + 2) – ( x + 1) 
dx
= cos x, =
dx x
Sol.  ( x + 1)( x + 2) =   ( x + 1)( x + 2) 
dx

dy cos x
= = x cos x 1 1
dz 1/ x =  x + 1dx –  x + 2dx + C
5. The minimum value of 1 – sin x is
= log|x + 1|– log |x + 2|+ C
(1) 2 (2) 0
x +1
(3) – 1 (4) 1 = log +C
x+2
Answer (2)
1
Sol. y = 1 – sin x
 sin
5
8. The value of x cos4 x dx is
–1  sin x  1 –1

–1  –sin x  1
–
0  1 – sin x  2 (1) 0 (2)
2
Minimum value = 0

6. The function f(x) = tan x – x (3)  (4)
2
(1) neither increases nor decreases
Answer (1)
(2) always increases
1
(3) always decreases
 sin
5
Sol. I = x cos4 xdx
(4) never increases −1

Answer (2) 1
Sol. f(x) = tan x – x (
I =  sin5 x cos4 x − sin5 x cos4 x dx )
0
f (x) = sec2x – 1 = tan2x
I=0
tan2x is always positive
 a a 
 f (x)  0 

 f ( x )dx =  ( f ( x ) + f (− x )) dx 
–a 0 
 f(x) is always increasing.

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 KCET-2025 (17-04-2025)

2 dx
x
9. The value of
 1 + sin  dx is Sol.  x 2 ( x 4 + 1)3/4
0
2

dx
(1) 0 (2) 8 =    1  
3/4
x 2  x3 1 + 
(3) 4 (4) 2   4 
 x  
Answer (2)
dx
2
x x
=  3/4
5 1 
Sol. I =  1 + 2sin cos
4 4 x 1 + 
0  x4 

2 2 2 1
  x    x  x x Let 1 + =t
=   sin x  4   +  cos  4   + 2sin 4 cos 4 x4
0
–4x – 5dx = dt
2 2 2
 x x x x dx – dt
  sin 4 + cos 4  =
 
 sin
4
+ cos dx
4 5
=
4
0 0 x

x  
Let, = t ,  dx = 4dt −dt –1  t −3/4 + 1
4 =  =   +C
4t 3/4 4  −3
+1 
  4 
2
 I =  | sin t + cos t | 4dt 1
0 = –t 4 + C

 1
= 4 ( − cos t + sin t ) 2  1 4
0 = – 1 +  +C
 x4 
= 4[1 –(–1 + 0)] = 8
1
 x 4 + 1 4
dx = –  +C
10. x equals  x4 
(x )  
3/4
2 4
+1
1
1 
 x +1
4 
1
4  x +1
4 
1
4
11.  log  x – 1 dx is
(1) –  4  + c (2)  4  + c 0

 x   x 
 1
(1) loge   (2) 1
1 1 2
(3) (x 4
+1 ) 4 +c (
(4) – x 4 + 1 ) 4 +c
(3) 0 (4) loge 2
Answer (1)
Answer (3)

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KCET-2025 (17-04-2025)
1 13. The area of the region bounded by the curve y = x2
 1− x 
Sol. I2 =  log   dx …(i)
 x  and the line y = 16 is
0
128 32
1
 (1 − (1 − x )  (1) sq. units (2) sq. units
I =  log   dx
3 3
0  (1 − x ) 
256 64
(3) sq. units (4) sq. units
1 3 3
 x 
I =  log   dx …(ii)
0  1− x  Answer (3)

Add (i) and (ii) Sol.

1
  (1 − x )   x 
2I =   log   + log   dx
0  x   1− x  

 I=0

x
12. The area bounded by the curve y = sin   , x axis,
3 4
Required area = 2 (16 − x 2 )dx
0
the lines x = 0 and x = 3 is
4
(1) 3 sq. units  x3 
= 2 ·  16 x − 
(2) 9 sq. units  3 
  0
1
(3) sq. units
3  64  4 256
= 2  64 −  = · 64 = sq. units
 3  3 3
(4) 6 sq. units
Answer (4) 14. General solution of the differential equation
dy
x + y tan x = sec x is
Sol. y = sin   dx
3
(1) x sec x = tan y + c (2) y sec x = tan x + c
3
x
Area =  sin   dx (3) y tan x = sec x + c (4) cosec x = y tan x + c
0 3
Answer (2)
x
Let = t  dx = 3dt dy
3 Sol. + y tan x = sec x
dx

Area =  sin t·3dt I.F. e 
tan xdx
= e
x sec x
= sec x
0

y · sec x =  sec 2 x dx
= 3(– cos t )0
= tan x + c
= 3(1 + 1) = 6

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 KCET-2025 (17-04-2025)

15. If ‘a’ and ‘b’ are the order and degree respectively of 18. A and B are two sets having 3 and 6 elements
the differentiable equation respectively.
2
 d 2 y   dy 3 Consider the following statements.
  + + x 4 = 0, then a – b = ________
 dx 2   dx  Statement (I): Minimum number of elements in A  B
 
is 3
(1) 0 (2) 1
Statement (II): Maximum number of elements in
(3) 2 (4) –1
A  B is 3
Answer (1)
Which of the following is correct?
2
 d 2y   dy  3
(1) Both statements (I) and (II) are false
Sol.   + + x4 = 0
 dx 2   dx 
  (2) Statement (I) is true, statement (II) is false
Order = 2 = a, degree = 2 = b (3) Statement (I) is false, statement (II) is true
a–b=0 (4) Both statements (I) and (II) are true
16. The distance of the point P(–3, 4, 5) from yz plane is Answer (3)
(1) 3 units (2) 4 units Sol. Minimum number of elements in A  B is 6, hence
(3) 5 units (4) –3 units (I) is false.
Answer (1) Maximum number of elements in A  B is 3, hence
Sol. Distance from yz plane = absolute value of x (II) is correct.
coordinate 19. Domain of the function f, given by
= |–3| = 3 1
f (x) = is
17. If A = {x : x is an integer and x2 – 9 = 0} ( x − 2)( x − 5)

B = {x : x is a natural number and 2  x < 5} (1) (–, 3]  (5, ) (2) (–, 2]  [5, )
C = {x : x is a prime number  4} (3) (–, 2)  (5, ) (4) (–, 3)  [5, )
Then (B – C)  A is, Answer (3)
(1) {2, 3, 5} (2) {–3, 3, 4} Sol. (x – 2)(x – 5) > 0
(3) {2, 3, 4} (4) {3, 4, 5}  x(–, 2)  (5, )
Answer (2) 20. If f(x) = sin[2]x – sin[–2]x, where [x] = greatest integer
Sol. A = {–3, 3}  x, then which of the following is not true?
B = {2, 3, 4} (1) f() = –1 (2) f(0) = 0
C = {2, 3}
  1
(3) f   = 1 (4) f   = 1 +
B – C = {4} 2 4 2
(B – C)  A = {–3, 3, 4}  2 Answer (1)

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KCET-2025 (17-04-2025)
Sol. f(x) = sin([2]x) – sin([–2]x) 24. A random experiment has five outcomes w1, w2, w3,
w4 and w5. The probabilities of the occurrence of the
 9 < 2 < 10, –10 < –2 < –9
1
outcomes w1, w2, w4 and w5 are respectively , a, b
 f(x) = sin(9x) – sin(–10x) = sin(9x) + sin(10x) 6
  1 1
 f() = 0, f(0) = 0, f   = 1, f   = +1 and such that 12a + 12b – 1 = 0. Then the
 
2  
4 2 12
21. Which of the following is not correct? probabilities of occurrence of the outcome w3 is

(1) tan45° = tan(–315°) (2) cos5 = cos4 1 2

(1) (2)
12 3
(3) sin2 = sin(–2) (4) sin4 = sin6
Answer (2) 1 1
(3) (4)
3 6
Sol. (1) tan45° = 1, tan(–315°) = –tan(315°) = –tan(360°
– 45°) = 1 Answer (2)
(2) cos(5) = –1, cos(4) = 1 Sol. Let P(w3) = x
(3) sin2 = 0, sin(–2) = 0  P(w1) + P(w2) + P(w3) + P(w4) + P(w5) = 1
(4) sin(4) = 0, sin(6) = 0
1 1
22. If cosx + cos2x = 1, then the value of sin2x + sin4x is  = +a+ x +b+ = 1  2 + 12x + 12(a + b)
6 12
(1) 2 (2) –1 + 1 = 12
(3) 1 (4) 0
 12(a + b) = 1
Answer (3)
2
Sol.  cosx + cos2x = 1  cosx = sin2x  2 + 12x + 1 + 1 = 12  12x = 8  x =
3
 sin2x + sin4x = sin2x + cos2x = 1
25. A die has two faces each with number '1', three faces
23. The mean deviation about the mean for the data 4, 7, each with number '2' and one face with number '3'. If
8, 9, 10, 12, 13, 17 is the die is rolled once, then P(1 or 3) is
(1) 4.03 (2) 10
1 2
(3) 3 (4) 8.5 (1) (2)
6 3
Answer (3)
1 1
(3) (4)
4 + 7 + 8 + 9 + 10 + 12 + 13 + 17 2 3
Sol. mean = x = = 10
8
Answer (3)
 mean deviation about mean
2 3 1
1 Sol. P (1) = , P (2) = , P (3) =
= (|4 – 10| + |7 – 10| + |8 – 10| + |9 – 10| + |10 – 6 6 6
8
10| + |12 – 10| + |13 – 10| + |17 – 10|) 2 1 1
P (1 3) = P (1) + P (3) – (1 3) = + –0=
6 6 2
1
= (6 + 3 + 2 + 1 + 0 + 2 + 3 + 7) = 3 Option (3) is correct.
8

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 KCET-2025 (17-04-2025)

26. Let A = {a, b, c}, then the number of equivalence  

Sol. f : 0,  → R
relations on A containing (b, c) is  2
(1) 4 (2) 1 f(x) = sin x

(3) 3 (4) 2  
g : 0,  → R
 2
Answer (4)
g(x) = cos x
Sol. A = {a, b, c}
 
Smallest equivalence relation containing (b, c) is Statement (I):  f(x) is increasing in 0,  and
 2
R1 = (a, a ), (b, b ), (c, c ), (b, c ), (c, b ) attains the value from [0, 1]

 
Now we are left with (a, b), (a, c), (b, a), (c, a)  f(x) is one-one in 0, 
 2
If we add any one pair [say (a, b)], then for
Also,
symmetry we must add (b, a)
 
Also for transitivity we are required to add g(x) is decreasing in 0,  and take the values
 2
(c, a), (a, c) which make it a universal relation. from [0, 1]
 Only 2 equivalence relation is possible. i.e. R1  
 g(x) is also one-one in 0, 
and universal relation.  2
 Statement (I) is correct.
 
27. Let the functions ‘’f’ ’’ and ‘’g’’ be f : 0,  → R given
 2 Statement (II): f + g  sinx + cosx = h(x)

  h1(x) = cosx – sinx = 0

by f(x) = sinx and g : 0,  → R given by g(x) =
 2 sin x
 = 1  tan x = 1
cosx, where R is the set of real numbers cos x

Consider the following statements: 

 x=
4
Statement (I): f and g are one-one

Statement (II): f + g is one-one h(x) attain its maxima at x =
4
Which of the following are correct? 
 among x = the function  ses or  es
(1) Both statements (I) and (II) are false 4

(2) Statement (I) is true, statement (II) is false  h(x) is not one-one
28. sec2 (tan–1 2) + cosec2 (cot–13) =
(3) Statement (I) is false, statement (II) is true
(1) 10 (2) 1
(4) Both statements (I) and (II) are true
(3) 5 (4) 15
Answer (2)
Answer (4)

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KCET-2025 (17-04-2025)
Sol. sec2 (tan–1 2) + cosec2 (cot–13) 0  cos–1  
2
E = sec(tan–1 2) + cosec(cot –1 3)
2 0  2y  2 …(i)
Also

– 
2
(
 sin–1 2 x 1– x 2  )
2

– 
 sin–1(2y )  …(ii)
Let tan–12 = A 2 2
 tanA = 2 From (i) and (ii)

 sec A = 5 
0  2y 
2
 A = sec –1 5

0 y
4


 0  cos –1 x 
4
B = cot–1 3
1
cot B = 3  x 1
2
cosec B = 10
30. Consider the following statements:
B = cosec –1 10
Statement (I): In a LPP, the objective function is
–1 2 –1 2
 E = (sec(sec 5 )) + (cosec(cosec 10 )) always linear.
Statement (II) : In a LPP, the linear inequalities on
= ( 5 ) + ( 10 )
2 2
variables are called constraints.

= 15 Which of the following is correct?

29. 2cos–1 x = sin–1 2 x 1– x 2( ) is valid for all values

(1) Statement (I) is false, Statement (II) is true
(2) Statement (I) is true, Statement (II) is true
of ‘x’ satisfying
(3) Statement (I) is true, Statement (II) is false
1 1
(1)  x 1 (2) 0  x  (4) Both Statements (I) and (II) are false
2 2
Answer (1)
(3) –1  x  1 (4) 0  x  1
Sol. Objective function can be anything.
Answer (1)
 Statement (I) is incorrect
Sol. 2cos –1
x = sin –1
( 2x 1– x 2 ) Statement (II): In LPP, the linear constant on
variable are called constraints.
Let cos–1 x = y
 Statement (II) is correct

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 KCET-2025 (17-04-2025)

31. The maximum value of z = 3x + 4y, subject to the This statement is correct.
constraints x + y  40, x + 2y  60 and x, y  0 is Statement-II: P(A  B) = 0, but P(A), P(B)  0
(1) 40 (2) 130  P(A) · P(B)  P(A  B)
(3) 120 (4) 140 Statement-II is correct.
Answer (4) 33. If A and B are two non-mutually exclusive events such
Sol. that P(A|B) = P(B|A), then
(1) P(A) = P(B) (2) A  B but A  B
(3) A = B (4) A  B = 
Answer (1)
 A B
Sol. P   = P  
 
B  A

P ( A  B) P (B  A)
 =
P (B ) P ( A)

 P(A) = P(B)

z = 3x + 4y 34. If A and B are two events such that A  B and

P(B)  0, then which of the following is correct?
(0, 0) → 0
(1) P(A) = P(B)
(40, 0) → 120
P (B )
(20, 20) → 140  Max (2) P ( A | B ) =
P ( A)
(0, 30) → 120
(3) P(A|B) < P(A)
32. Consider the following statements.
(4) P(A|B)  P(A)
Statement (I) : If E and F are two independent
events, then E and F are also independent. Answer (4)

Statement (II) : Two mutually exclusive events with Sol. A  B

non-zero probabilities of occurrence cannot be P(A) < P(B) Option 1 in incorrect

 A  P ( A  B)
independent.
P  =
Which of the following is correct? B P (B )
(1) Both the statements are false
P ( A)
=
(2) Statement (I) is true and statement (II) is false P (B )
(3) Statement (I) is false and statement (II) is true
P(B) < 1
(4) Both the statements are true
 A  P ( A)
Answer (4) P  =  P ( A)
 B  P (B )
Sol. Statement-I : If E and F are independent E and F are
also independent. Option 4 is correct.

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KCET-2025 (17-04-2025)
35. Meera visits only one of the two temples A and B in 36. If Z1 and Z2 are two non-zero complex numbers,
2 then which of the following is not true?
her locality. Probability that she visits temple A is
5
(1) Z1 + Z2  Z1 + Z2 (2) Z1 + Z2 = Z1 + Z2
1
. If she visits temple A, is the probability that she
3
(3) Z1 Z2 = Z1 · Z2 (4) Z1·Z2 = Z1·Z2
2
meets her friend, whereas it is if she visits
7 Answer (1)
temple B. Meera met her friend at one of the two
Sol. Z1 + Z2  Z1 + Z2
temples. The probability that she met her at temple
B is Using triangular inequality.
9 7
(1) (2) 37. Consider the following statements:
16 16
Statement (I): The set of all solutions of the linear
5 3
(3) (4) inequalities 3x + 8 < 17 and 2x + 8  12 are x < 3
16 16
Answer (1) and x  2 respectively.

Sol. A : Meera visits temple A Statement (II): The common set of solutions of
B : Meera visits temple B linear inequalities 3x + 8 < 17 and 2x + 8  12 is
F : She meets her friend (2,3)
2 3
P ( A) = ; P (B ) = Which of the following is true?
5 5
(1) Both the statements are false
F  1 F  2
P  = ; P  = (2) Statement (I) is true but statement (II) is false
 A 3 B 7

 B  P (B  F )
(3) Statement (I) is false but statement (II) is true
P  =
F  P (F ) (4) Both the statements are true

F  Answer (2)
P ( B ) ·P  
= B Sol. 3x + 8 < 17
F  F 
P ( A ) · P   + P ( B ) ·P   = 3x < 9
 
A B
3 2 =x<3

= 5 7 2x + 8  12
2 1 3 2
 + 
5 3 5 7 = 2x  4
6 x2
= 7
2 6 Statement-I is true.
+
3 7
Statement-II is false
9
= (–, 3)  [2, ) = [2, 3]
16

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 KCET-2025 (17-04-2025)

38. The number of four digit even number that can be 41. If 4th, 10th and 16th terms of a G.P. are x, y and z
formed using the digits 0, 1, 2 and 3 without respectively, then
repetition is x+z
(1) y = (2) z = xy
(1) 12 (2) 6 2
(3) 10 (4) 4 (3) y = xz (4) x = yz
Answer (3)
Answer (3)
Sol. Case-I (when unit digit is 0)
Sol. Let a be 1st term and r be common ratio,

x = ar 3
y = ar 9  xz = a r = (ar ) = y
2 18 9 2 2
Case-II (when unit digit is 2)
z = ar 15  y = xz

42. If A is a square matrix such that A2 = A, then (I – A)3

is
 Total numbers = 6 + 4 = 10
(1) –I – A (2) I – A
39. The number of diagonals that can be drawn in an
octagon is (3) A – I (4) I + A

(1) 30 (2) 15 Answer (2)

(3) 20 (4) 28 Sol. A 2 = A  A3 = A 2 = A

Answer (3)
(I – A)3 = I3 – 3A + 3A2 – A3
Sol. Number of diagonals in an convex polygon
= I – 3A + 3A – A = I – A
= nC2 – n
43. If A and B are two matrices such that AB is an identity
for n = 8 matrix and the order of matrix B is 3 × 4 then the order
Number of diagonals = 8C2 – 8 of matrix A is

= 28 – 8 = 20 (1) 4 × 4

40. If the number of terms in the binomial expansion of (2) 3 × 4

(2x + 3)3n is 22, then the value of n is (3) 3 × 3
(1) 9 (2) 8 (4) 4 × 3
(3) 6 (4) 7 Answer (4)
Answer (4) Sol. AB = I
Sol. Number of terms in (x + y)n is n = 1. For product to be valid the order of Am × n Bn × m to
for (2n = 3)3n, number of terms = 3n + 1 be an identity matrix of order m × m.
 2n + 1 = 22  n × m  (3 × 4)  Am × n  A4 × 3 order of A will
 n=7 be 4 × 3

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KCET-2025 (17-04-2025)
44. Which of the following statements is not correct? k 2  3
46. If A =   and |A | = 125, then the value of k is
(1) A skew symmetric matrix has all diagonal  2 k 
elements equal to zero
(1) –4 (2) ±2
(2) A row matrix has only one row
(3) ±3 (4) –5
(3) A diagonal matrix has all diagonal elements
equal to zero Answer (3)

(4) A symmetric matrix A is a square matrix k 2 

Sol. A =    |A| = k – 4
2
satisfying A = A 2 k 
Answer (3)
 |A3| = |k2 – 4|3 = 125 = 53
Sol. (2) A row matrix consist of only a now.
 k2 – 4 = 5  k = ±3
(1) For skew-symmetric matrix has all diagonal
elements equal to 0. 47. If A is a square matrix satisfying the equation A2 – 5A
+ 7I = 0, where I is the identity matrix and 0 is null
(3) A diagonal matrix has non-diagonal element
matrix of same order, then A–1 =
zero but there is no condition that diagonal
elements should be zero. 1 1
(1) (7I − A) (2) (5I − A)
Hence, option (C) is not correct. 5 7

(4) AT = A  Symmetrix matrix 1

(3) ( A − 5I ) (4) 7(5I – A)
7
1 1 6
45. If a matrix A =   satisfies A = kA, then the
1 1 Answer (2)
value of k is Sol. If A2 – 5A + 7I = 0
(1) 6 (2) 32 Melting A–1 to RHS both side
1 AT(A2 – 5A + 7I) = A–1 0
(3) 1 (4)
32
A–1 AA – 5A–1A + 7A–1I = 0
Answer (2)
1
1 1 1 1  A – 5I + 7A–1 = 0  A–1 = (5I − A)
Sol. A =   AT =  7
 
1 1 1 1
48. If A is a square matrix of order 3 × 3, det A = 3, then
1−  1 the value of det (3A–1) is
 | A − I | = = 0 using characteristic
1 1− 
1
(1) 9 (2)
equation, (1 – )2 – 1 = 0 3
 2 – 2 = 0 (3) 3 (4) 27
 A2 – 2A = 0  A2 = 2A Answer (A)
 A4 = 4A2 = 4(2A) = 8A Sol. |A| = 3
A6 = A2  A4 = (2A)(8A) = 16A2 1 33
|3A–1| = 33|A–1| = 33  = =9
= 32A | A| 3

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 KCET-2025 (17-04-2025)

1 3  Sol. a + b = iˆ + 2 jˆ + kˆ + (iˆ − jˆ + 4kˆ )

49. If B =   be the adjoint of a matrix A and
1  
= (1 +  )iˆ + (2 −  ) ˆj + (1 + 4 )kˆ
|A| = 2, then the value of  is
(1) 3 (2) 4 (a + b ).c = 0

(3) 5 (4) 2 (1 + ) + (2 – ) + (1 + 4) = 0

Answer (3)  4 + 4= 0

 = –1
Sol. |B| = (1)() – (3)(1) =  – 3
|B| = |A|2–1 = |A| 52. If a = 10 , b = 2 and a.b = 12 , then the value of

 – 3 = 2 a  b is

=5 (1) 16 (2) 5

50. The system of equations 4x + 6y = 5 and (3) 10 (4) 14
8x + 12y = 0 has Answer (1)
(1) Only two solutions
Sol. a = 10 ; b = 2
(2) No solution
a . b = 12
(3) Infinitely many solutions
(4) A unique solution a . b cos  = 12

Answer (4) 12 3
cos  = =
Sol. 4x + 6y = 5 … (i) 20 5
8x + 12y = 0 … (ii) 4 4
 a  b = a b sin  = (10)(2)  = 20  = 16
Multiply eq. (i) by 2 5 5
53. Consider the following statements:
Both are same equations.
 Infinitely many solutions. Statements (I) : If either a = 0 or b = 0 , then

a.b = 0 .
51. If a = iˆ + 2 jˆ + kˆ , b = iˆ − jˆ + 4kˆ and c = iˆ + ˆj + kˆ are

such that a + b is perpendicular to c , then the Statements (II) : If ab = 0 , then a is

value of  is perpendicular to b .

(1) 0 Which of the following is correct?

(1) Both Statement (I) and Statement (II) are false
(2) 1
(2) Statement (I) is true but Statement (II) is false
(3) ±1
(3) Statement (I) is false but Statement (II) is true
(4) –1
(4) Both Statement (I) and Statement (II) are true
Answer (4)
Answer (2)

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KCET-2025 (17-04-2025)
Sol. Statements (I) : 55. The equation of the line through the point (0, 1, 2) and
x −1 y +1 z −1
If a = 0 or b = 0 , then a.b =| a | . | b | cos  = 0 . perpendicular to the line = = is
2 3 −2
x y −1 z − 2 x y −1 z − 2
Statement (I) is true. (1) = = (2) = =
3 −4 3 3 4 −3
Statements (II) :
x y −1 z − 2 x y −1 z − 2
(3) = = (4) = =
If a  b = 0 , then a  b =| a || b | sin  = 0 −3 4 3 3 4 3
Answer (3)
This means sin = 0, a = 0 or | b |= 0 Sol. Let a1 = 2iˆ + 3 ˆj − 2kˆ

If sin = 0,  = 0 or  meaning a and b are parallel b1 = aiˆ + bjˆ + ckˆ

or antiparallel. a1  b1 = 2a + 3b − 2c = 0
Either, a = 0 or | b |= 0 . (i) 2(3) + 3(–4) – 2(3) = 6 – 12 – 6  0
(ii) 2(3) + 3(4) – 2(–3) = 6 + 12 + 6  0
Statement (II) is false.
(iii) 2(–3) + 3(4) – 2(3) = –6 + 12 – 6 = 0
54. If a line makes angles 90°, 60° and  with x, y and z  Option (3) is correct.
axes respectively, where  is acute, then the value of 56. A line passes through (–1, –3) and perpendicular to
 is x + 6y = 5. Its x intercept is

(1) /2 (2) /6 1

(1) 2 (2)
2
(3) /4 (4) /3
1
(3) − (4) –2
Answer (2) 2

Sol. We know that, Answer (4)

Sol. Given x + 6y = 5
cos2 + cos2 + cos2 = 1
 6y = 5 – x
cos290° + cos260° + cos2 =1 5 x
y= − …(i)
6 6
1
0 + + cos2  = 1
4 −1
 m=
6
3
cos2  = Let m1 be the slope perpendicular to (i).
4
 m  m1 = –1
3  −1 
cos  =  m1   = −1
2  6
 m1 = 6
 0
2  Equation of line passing through (–1, –3) having
slope 6 is

 (y + 3) = 6(x + 1)
6

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 KCET-2025 (17-04-2025)

 y + 3 = 6x + 6 1
4−
 y = 6x + 3 = 2
1
For x-intercept, put y = 0. 2
0 = 6x + 3 7
 x = –2 = 2 =7
1
 Option (4) is correct. 2
57. The length of the latus rectum of x2 + 3y2 = 12 is
 Option (3) is correct.
2
(1) 24 units (2) units cos x
3 59. If y = , then
1 + sin x
1 4
(3) units (4) units dy −1
3 3 (a) =
dx 1 + sin x
Answer (4)
dy 1
(b) =
2b2 dx 1 + sin x
Sol. Length of latus rectum =
a
dy 1  x
(c) = − sec 2  − 
2 4 dx 2 4 2
=
12
dy 1  x
(d) = sec 2  − 
=
4 dx 2 4 2
3
(1) Both (b) and (d) are correct
 Option (4) is correct.
(2) Only (b) is correct
x − x
4
(3) Only (a) is correct
58. lim is
x →1 x −1 (4) Both (a) and (c) are correct
1 Answer (4)
(1) (2) 0
2
cos x
(3) 7 (4) Does not exist Sol. y =
1 + sin x
Answer (3)
Differentiable both sides w.r.t. x
x4 − x
Sol. lim dy d  cos x 
=
x −1 dx dx  1 + sin x 
x →1

0/0 form, applying L’Hospital.

d  d 
(1 + sin x ) (cos x ) − cos x  (1 + sin x ) 
d 4 1 dx  dx 
(x − x ) 4x3 − =
lim dx = lim 2 x (1 + sin x )2
x →1 d x →1 1
( x − 1) −0
dx 2 x (1 + sin x )( − sin x ) − cos x(cos x )
=
(1 + sin x )2

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KCET-2025 (17-04-2025)

− sin x − sin2 x − cos2 x (1) a – iii, b – ii, c – iv, d – i

=
(1 + sin x )2 (2) a – i, b – ii, c – iv, d – iii
(3) a – iv, b – iii, c – i, d – ii
− sin x − (sin2 x + cos2 x )
=
(1 + sin x )2 (4) a – ii, b – iv, c – iii, d – i
Answer (4)
− sin x − 1
=
(1 + sin x )2 Sol. (a) x|x| is continuous in (–1, 1) and differentiable in
(–1, 1).
2
−1 −1    
 =  sec  −    (ii) is correct.
(1 + sin x ) 2   4 2 
(b) | x | is not differentiable in (–1, 1).
 Option (4) is correct.
60. Match the following :  (iv) is correct.
In the following, [x] denotes the greatest integer less  x − 1, x  ( −1, 0)
than or equal to x. (c) x + [ x ] = 
 x, x  [0, 1)
Column-I Column-II
 (iii) is correct.
(a) x|x| (i) Continuous in
(–1, 1)

(b) |x| (ii) Differentiable in

(–1, 1)

(c) x + [x] (iii) Strictly increasing in

(–1, 1)
(d) |x – 1| + |x + 1| is continuous in (–1, 1).
(d) |x – 1| + |x + 1| (iv) Not differentiable at,
at least one point in  (i) is correct.
(–1, 1) So, option (4) is correct.

❑ ❑ ❑

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