JEE Main 2022 Mathematics June 29 Shift 1 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


 
   
 
MATHEMATICS 
SECTION - A 
Multiple Choice Questions: This section contains 20 
multiple choice questions. Each question has 4 choices 
(1), (2), (3) and (4), out of which ONLY ONE is correct. 
Choose the correct answer : 
1. The probability that a randomly chosen 2 × 2 matrix 
with all the entries from the set of first 10 primes, is 
singular, is equal to : 
 (A) 
4
133
10
 (B) 
3
18
10
 
 (C) 
3
19
10
 (D) 
4
271
10
 
Answer (C) 
Sol. Let A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} 
 Let E be the event that matrix of order 2 × 2 is singular 
 Case-I 
 All entries are same example 
22
22
??
??
??
 
 = 
10
C
1
 
 Case-II 
 Matrix with two prime numbers only example 
35
35
??
??
??
 
 = 
10
C
2
 × 2! × 2! 
 
10 10
12
4 4 3
2! 2! 190 19
()
10 10 10
CC
PE
+ ? ?
= = = 
2. Let the solution curve of the differential equation 
22
16 ,
dy
x y y x
dx
- = + y(1) = 3 be y = y(x). Then 
y(2) is equal to : 
 (A) 15 (B) 11 
 (C) 13 (D) 17 
Answer (A) 
Sol. 
22
16
dy
x y y x
dx
- = + 
 y = 4x tan ? 
 
2
4tan 4 sec
dy d
x
dx dx
?
= ? + ? 
 
22
4 tan 4 sec 4 tan 4 sec
d
x x x x
dx
?
? + ? - ? = ? 
 sec
dx
d
x
? ? =
??
 
 log |sec? + tan?| = log |x| + C 
 y(1) = 3  ? 3 = 4 tan? 
 
35
tan sec
44
= ? = ? ? = 
 
8
ln ln 1
4
C =+ 
 ? C = ln 2 
 ? |sec? + tan?| = 2|x| 
 To find y(2) put x = 2 
 ? tan
8
y
?= 
 (sec? + tan?)
2
 = 16 
 
sec tan 4
1
sec tan
4
15
2tan 2
48
y
? + ? = ?
? - ? = ?
? = = ?
 
 ? 15 y = 
3. If the mirror image of the point (2, 4, 7) in the plane 
3x – y + 4z = 2 is (a, b, c), then 2a + b + 2c is equal 
to : 
 (A) 54 (B) 50 
 (C) –6 (D) –42 
Answer (C) 
Sol. Mirror image of (2, 4, 7) in 3x – y + 4z = 2 is 
(a, b, c) then 
 
2 2 2
2 4 7 2(6 4 28 2)
3 1 4
3 ( 1) 4
a b c - - - - - + -
= = =
-
+ - +
 
 
2 4 7 28
3 1 4 13
a b c - - - -
= = =
-
 
 
58 80 21
13 13 13
a b c
--
= = = 
 
116 80 42
22
13
a b c
- + -
+ + = 
    = –6 
Page 2


 
   
 
MATHEMATICS 
SECTION - A 
Multiple Choice Questions: This section contains 20 
multiple choice questions. Each question has 4 choices 
(1), (2), (3) and (4), out of which ONLY ONE is correct. 
Choose the correct answer : 
1. The probability that a randomly chosen 2 × 2 matrix 
with all the entries from the set of first 10 primes, is 
singular, is equal to : 
 (A) 
4
133
10
 (B) 
3
18
10
 
 (C) 
3
19
10
 (D) 
4
271
10
 
Answer (C) 
Sol. Let A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} 
 Let E be the event that matrix of order 2 × 2 is singular 
 Case-I 
 All entries are same example 
22
22
??
??
??
 
 = 
10
C
1
 
 Case-II 
 Matrix with two prime numbers only example 
35
35
??
??
??
 
 = 
10
C
2
 × 2! × 2! 
 
10 10
12
4 4 3
2! 2! 190 19
()
10 10 10
CC
PE
+ ? ?
= = = 
2. Let the solution curve of the differential equation 
22
16 ,
dy
x y y x
dx
- = + y(1) = 3 be y = y(x). Then 
y(2) is equal to : 
 (A) 15 (B) 11 
 (C) 13 (D) 17 
Answer (A) 
Sol. 
22
16
dy
x y y x
dx
- = + 
 y = 4x tan ? 
 
2
4tan 4 sec
dy d
x
dx dx
?
= ? + ? 
 
22
4 tan 4 sec 4 tan 4 sec
d
x x x x
dx
?
? + ? - ? = ? 
 sec
dx
d
x
? ? =
??
 
 log |sec? + tan?| = log |x| + C 
 y(1) = 3  ? 3 = 4 tan? 
 
35
tan sec
44
= ? = ? ? = 
 
8
ln ln 1
4
C =+ 
 ? C = ln 2 
 ? |sec? + tan?| = 2|x| 
 To find y(2) put x = 2 
 ? tan
8
y
?= 
 (sec? + tan?)
2
 = 16 
 
sec tan 4
1
sec tan
4
15
2tan 2
48
y
? + ? = ?
? - ? = ?
? = = ?
 
 ? 15 y = 
3. If the mirror image of the point (2, 4, 7) in the plane 
3x – y + 4z = 2 is (a, b, c), then 2a + b + 2c is equal 
to : 
 (A) 54 (B) 50 
 (C) –6 (D) –42 
Answer (C) 
Sol. Mirror image of (2, 4, 7) in 3x – y + 4z = 2 is 
(a, b, c) then 
 
2 2 2
2 4 7 2(6 4 28 2)
3 1 4
3 ( 1) 4
a b c - - - - - + -
= = =
-
+ - +
 
 
2 4 7 28
3 1 4 13
a b c - - - -
= = =
-
 
 
58 80 21
13 13 13
a b c
--
= = = 
 
116 80 42
22
13
a b c
- + -
+ + = 
    = –6 
 
   
 
4. Let ƒ : R ? R be a function defined by : 
 
3
2
max{ 3 } ; 2
2 6 ; 2 3 ƒ( )
[ 3] 9 ; 3 5
2 1 ; 5
tx
t t x
x x x x
xx
xx
?
?
-?
?
?
?
+ - ? ? =
?
?
- + ? ?
?
+? ?
?
 
 where [t] is the greatest integer less than or equal 
to t. Let m be the number of points where ƒ is not 
differentiable and 
2
2
ƒ( ) . I x dx
-
=
?
 Then the ordered 
pair (m, I) is equal to : 
 (A) 
27
3,
4
??
??
??
 (B) 
23
3,
4
??
??
??
 
 (C) 
27
4,
4
??
??
??
 (D) 
23
4,
4
??
??
??
 
Answer (C) 
Sol. 
3
max{ 3 } ; 2
tx
t t x
?
-? 
 g(t) = t
3
 –3t  ? g? (t) = 3t
2
 –3 = 3(t – 1)(t + 1) 
  
  
 
3
2
31
2 1 2
2 6 2 3
ƒ( )
9 3 4
10 4 5
11 5
2 1 5
x x x
x
x x x
x
x
x
x
xx
?
- ? -
?
- ? ?
?
?
+ - ? ?
?
?
=
?
??
?
??
?
?
=
?
+? ?
?
 
 
 Points of non-differentiability = {2, 3, 4, 5} 
 ? m = 4 
 
2 1 2
3
2 2 1
ƒ( ) ( 3 ) 2 I x dx x x dx dx
-
- - -
= = - +
? ? ?
 
 
1
42
2
3 1 3
2(2 1) (4 6) 6
4 2 4 2
xx
-
-
??
??
= - + + = - - - + ??
??
??
??
??
 
    = 
27
4
 
5. Let 
ˆ ˆ ˆ ˆ ˆ ˆ
3 , 3 4 a i j k b i j k = ? + - = - ? + and 
ˆ ˆ ˆ
22 c i j k = + - where ?, ? ? R, be three vectors. If 
the projection of a on c is 
10
3
 and 
ˆ ˆ ˆ
6 10 7 , b c i j k ? = - + + then the value of ? + ? is 
equal to : 
 (A) 3 (B) 4 
 (C) 5 (D) 6 
Answer (A) 
Sol. 
ˆ ˆ ˆ
3 a i j k = ? + - 
 
ˆ ˆ ˆ
34 b i j k = - ? + 
 
ˆ ˆ ˆ
22 c i j k = + - 
 Projection of a on c is 
 
10
3
ac
b
?
= 
 
2 2 2
6 2 8 10
33
1 2 ( 2)
? + + ? +
==
+ + -
 
 ? 2 ?= 
 
ˆ ˆ ˆ
6 10 7 b c i j k ? = - + + 
 
ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
3 4 (2 8) 10 (6 ) 6 10 7
1 2 2
i j k
i j k i j k -? = ? - + + + ? = - + +
-
 
 2 8 6 & 6 7 ? - = - + ? = 
 ? 1 ?= 
 ? + ? = 2 + 1 = 3 
Page 3


 
   
 
MATHEMATICS 
SECTION - A 
Multiple Choice Questions: This section contains 20 
multiple choice questions. Each question has 4 choices 
(1), (2), (3) and (4), out of which ONLY ONE is correct. 
Choose the correct answer : 
1. The probability that a randomly chosen 2 × 2 matrix 
with all the entries from the set of first 10 primes, is 
singular, is equal to : 
 (A) 
4
133
10
 (B) 
3
18
10
 
 (C) 
3
19
10
 (D) 
4
271
10
 
Answer (C) 
Sol. Let A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} 
 Let E be the event that matrix of order 2 × 2 is singular 
 Case-I 
 All entries are same example 
22
22
??
??
??
 
 = 
10
C
1
 
 Case-II 
 Matrix with two prime numbers only example 
35
35
??
??
??
 
 = 
10
C
2
 × 2! × 2! 
 
10 10
12
4 4 3
2! 2! 190 19
()
10 10 10
CC
PE
+ ? ?
= = = 
2. Let the solution curve of the differential equation 
22
16 ,
dy
x y y x
dx
- = + y(1) = 3 be y = y(x). Then 
y(2) is equal to : 
 (A) 15 (B) 11 
 (C) 13 (D) 17 
Answer (A) 
Sol. 
22
16
dy
x y y x
dx
- = + 
 y = 4x tan ? 
 
2
4tan 4 sec
dy d
x
dx dx
?
= ? + ? 
 
22
4 tan 4 sec 4 tan 4 sec
d
x x x x
dx
?
? + ? - ? = ? 
 sec
dx
d
x
? ? =
??
 
 log |sec? + tan?| = log |x| + C 
 y(1) = 3  ? 3 = 4 tan? 
 
35
tan sec
44
= ? = ? ? = 
 
8
ln ln 1
4
C =+ 
 ? C = ln 2 
 ? |sec? + tan?| = 2|x| 
 To find y(2) put x = 2 
 ? tan
8
y
?= 
 (sec? + tan?)
2
 = 16 
 
sec tan 4
1
sec tan
4
15
2tan 2
48
y
? + ? = ?
? - ? = ?
? = = ?
 
 ? 15 y = 
3. If the mirror image of the point (2, 4, 7) in the plane 
3x – y + 4z = 2 is (a, b, c), then 2a + b + 2c is equal 
to : 
 (A) 54 (B) 50 
 (C) –6 (D) –42 
Answer (C) 
Sol. Mirror image of (2, 4, 7) in 3x – y + 4z = 2 is 
(a, b, c) then 
 
2 2 2
2 4 7 2(6 4 28 2)
3 1 4
3 ( 1) 4
a b c - - - - - + -
= = =
-
+ - +
 
 
2 4 7 28
3 1 4 13
a b c - - - -
= = =
-
 
 
58 80 21
13 13 13
a b c
--
= = = 
 
116 80 42
22
13
a b c
- + -
+ + = 
    = –6 
 
   
 
4. Let ƒ : R ? R be a function defined by : 
 
3
2
max{ 3 } ; 2
2 6 ; 2 3 ƒ( )
[ 3] 9 ; 3 5
2 1 ; 5
tx
t t x
x x x x
xx
xx
?
?
-?
?
?
?
+ - ? ? =
?
?
- + ? ?
?
+? ?
?
 
 where [t] is the greatest integer less than or equal 
to t. Let m be the number of points where ƒ is not 
differentiable and 
2
2
ƒ( ) . I x dx
-
=
?
 Then the ordered 
pair (m, I) is equal to : 
 (A) 
27
3,
4
??
??
??
 (B) 
23
3,
4
??
??
??
 
 (C) 
27
4,
4
??
??
??
 (D) 
23
4,
4
??
??
??
 
Answer (C) 
Sol. 
3
max{ 3 } ; 2
tx
t t x
?
-? 
 g(t) = t
3
 –3t  ? g? (t) = 3t
2
 –3 = 3(t – 1)(t + 1) 
  
  
 
3
2
31
2 1 2
2 6 2 3
ƒ( )
9 3 4
10 4 5
11 5
2 1 5
x x x
x
x x x
x
x
x
x
xx
?
- ? -
?
- ? ?
?
?
+ - ? ?
?
?
=
?
??
?
??
?
?
=
?
+? ?
?
 
 
 Points of non-differentiability = {2, 3, 4, 5} 
 ? m = 4 
 
2 1 2
3
2 2 1
ƒ( ) ( 3 ) 2 I x dx x x dx dx
-
- - -
= = - +
? ? ?
 
 
1
42
2
3 1 3
2(2 1) (4 6) 6
4 2 4 2
xx
-
-
??
??
= - + + = - - - + ??
??
??
??
??
 
    = 
27
4
 
5. Let 
ˆ ˆ ˆ ˆ ˆ ˆ
3 , 3 4 a i j k b i j k = ? + - = - ? + and 
ˆ ˆ ˆ
22 c i j k = + - where ?, ? ? R, be three vectors. If 
the projection of a on c is 
10
3
 and 
ˆ ˆ ˆ
6 10 7 , b c i j k ? = - + + then the value of ? + ? is 
equal to : 
 (A) 3 (B) 4 
 (C) 5 (D) 6 
Answer (A) 
Sol. 
ˆ ˆ ˆ
3 a i j k = ? + - 
 
ˆ ˆ ˆ
34 b i j k = - ? + 
 
ˆ ˆ ˆ
22 c i j k = + - 
 Projection of a on c is 
 
10
3
ac
b
?
= 
 
2 2 2
6 2 8 10
33
1 2 ( 2)
? + + ? +
==
+ + -
 
 ? 2 ?= 
 
ˆ ˆ ˆ
6 10 7 b c i j k ? = - + + 
 
ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
3 4 (2 8) 10 (6 ) 6 10 7
1 2 2
i j k
i j k i j k -? = ? - + + + ? = - + +
-
 
 2 8 6 & 6 7 ? - = - + ? = 
 ? 1 ?= 
 ? + ? = 2 + 1 = 3 
 
   
 
6. The area enclosed by y
2
 = 8x and 2 yx = that lies 
outside the triangle formed by 
2 , 1 , 2 2, y x x y = = = is equal to :  
 (A) 
16 2
6
 (B) 
11 2
6
 
 (C) 
13 2
6
 (C) 
52
6
 
Answer (C) 
Sol.  
  
 
( ) ( ) ( )
2, 2 2 , 1 , 2 2 , 1 , 2 A B C 
 Area = ( )
42
2
0
area
8
2
yy
dy BAC
??
- - ? ??
??
??
?
 
 
42
23
0
1
24 2
22
yy
AB BC
??
= - - ? ? ??
??
??
 
 
32 4 2 1
8 2 1 2
24 2
?
= - - ? ? 
 
16 2 2
82
32
= - - 
 ( )
2 13 2
48 32 3
66
= - - = 
7. If the system of linear equations 
 2x + y – z = 7 
 x – 3y + 2z = 1 
 x + 4y + ?z = k, where ?, k ? R 
 has infinitely many solutions, then ? + k is equal to: 
 (A) –3 (B) 3 
 (C) 6 (D) 9 
Answer (B) 
Sol. 2x + y – z = 7 
 x – 3y + 2z = 1 
 x + 4y + ?z = k 
  
2 1 1
1 3 2 7 21 0
14
-
? = - = - ? - =
?
 
 3 ? = - 
 
1
7 1 1
1 3 2
43 k
-
? = -
-
 
 ? 6 – k = 0 ? k = 6 
 3 6 3 k ? + = - + = 
8. Let ? and ? be the roots of the equation x
2
 + 
(2i – 1) = 0. Then, the value of 
88
? + ? is equal to: 
 (A) 50 (B) 250 
 (C) 1250 (D) 1500 
Answer (A) 
Sol. x
2
 + 2i – 1 = 0 
 ?
2
 = ?
2
 = 1 – 2i 
 ?
4
 = (1 – 2i)
2
 = 1 + (2i)
2
 – 4i = –3 – 4i 
 ?
8
 = (–3 – 4i)
2
 = 9 – 16 + 24i = –7 + 24i 
 ( ) ( )
22
88
2 7 24 2 7 24 50 i ? + ? = - + = - + = 
9. Let { , , , } ? ? ? ? ? ? be such that 
( ) ( ) ( )
p q p q q ? ? ? ? is a tautology. Then ? is 
equal to : 
 (A) ? (B) ? 
 (C) ? (D) ? 
Answer (C) 
Sol. ( ) p q q ?? 
 ( ) ~ p q q ?? 
 ( ) ~~ p q q = ? ? 
 ( ) ( ) ~~ = ? ? ? p q q q 
 ( ) ~ p q T = ? ? 
 ~pq =? 
 Now ( ) ( ) ~ p q p q ? ? ? 
 
( ) ( ) ~ ~ ~ p q p p q p q p q p q
T T F T T T
T F F F F T
F T T F T T
F F T F T T
? ? ? ? ?
 
 ? ? = ? 
Page 4


 
   
 
MATHEMATICS 
SECTION - A 
Multiple Choice Questions: This section contains 20 
multiple choice questions. Each question has 4 choices 
(1), (2), (3) and (4), out of which ONLY ONE is correct. 
Choose the correct answer : 
1. The probability that a randomly chosen 2 × 2 matrix 
with all the entries from the set of first 10 primes, is 
singular, is equal to : 
 (A) 
4
133
10
 (B) 
3
18
10
 
 (C) 
3
19
10
 (D) 
4
271
10
 
Answer (C) 
Sol. Let A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} 
 Let E be the event that matrix of order 2 × 2 is singular 
 Case-I 
 All entries are same example 
22
22
??
??
??
 
 = 
10
C
1
 
 Case-II 
 Matrix with two prime numbers only example 
35
35
??
??
??
 
 = 
10
C
2
 × 2! × 2! 
 
10 10
12
4 4 3
2! 2! 190 19
()
10 10 10
CC
PE
+ ? ?
= = = 
2. Let the solution curve of the differential equation 
22
16 ,
dy
x y y x
dx
- = + y(1) = 3 be y = y(x). Then 
y(2) is equal to : 
 (A) 15 (B) 11 
 (C) 13 (D) 17 
Answer (A) 
Sol. 
22
16
dy
x y y x
dx
- = + 
 y = 4x tan ? 
 
2
4tan 4 sec
dy d
x
dx dx
?
= ? + ? 
 
22
4 tan 4 sec 4 tan 4 sec
d
x x x x
dx
?
? + ? - ? = ? 
 sec
dx
d
x
? ? =
??
 
 log |sec? + tan?| = log |x| + C 
 y(1) = 3  ? 3 = 4 tan? 
 
35
tan sec
44
= ? = ? ? = 
 
8
ln ln 1
4
C =+ 
 ? C = ln 2 
 ? |sec? + tan?| = 2|x| 
 To find y(2) put x = 2 
 ? tan
8
y
?= 
 (sec? + tan?)
2
 = 16 
 
sec tan 4
1
sec tan
4
15
2tan 2
48
y
? + ? = ?
? - ? = ?
? = = ?
 
 ? 15 y = 
3. If the mirror image of the point (2, 4, 7) in the plane 
3x – y + 4z = 2 is (a, b, c), then 2a + b + 2c is equal 
to : 
 (A) 54 (B) 50 
 (C) –6 (D) –42 
Answer (C) 
Sol. Mirror image of (2, 4, 7) in 3x – y + 4z = 2 is 
(a, b, c) then 
 
2 2 2
2 4 7 2(6 4 28 2)
3 1 4
3 ( 1) 4
a b c - - - - - + -
= = =
-
+ - +
 
 
2 4 7 28
3 1 4 13
a b c - - - -
= = =
-
 
 
58 80 21
13 13 13
a b c
--
= = = 
 
116 80 42
22
13
a b c
- + -
+ + = 
    = –6 
 
   
 
4. Let ƒ : R ? R be a function defined by : 
 
3
2
max{ 3 } ; 2
2 6 ; 2 3 ƒ( )
[ 3] 9 ; 3 5
2 1 ; 5
tx
t t x
x x x x
xx
xx
?
?
-?
?
?
?
+ - ? ? =
?
?
- + ? ?
?
+? ?
?
 
 where [t] is the greatest integer less than or equal 
to t. Let m be the number of points where ƒ is not 
differentiable and 
2
2
ƒ( ) . I x dx
-
=
?
 Then the ordered 
pair (m, I) is equal to : 
 (A) 
27
3,
4
??
??
??
 (B) 
23
3,
4
??
??
??
 
 (C) 
27
4,
4
??
??
??
 (D) 
23
4,
4
??
??
??
 
Answer (C) 
Sol. 
3
max{ 3 } ; 2
tx
t t x
?
-? 
 g(t) = t
3
 –3t  ? g? (t) = 3t
2
 –3 = 3(t – 1)(t + 1) 
  
  
 
3
2
31
2 1 2
2 6 2 3
ƒ( )
9 3 4
10 4 5
11 5
2 1 5
x x x
x
x x x
x
x
x
x
xx
?
- ? -
?
- ? ?
?
?
+ - ? ?
?
?
=
?
??
?
??
?
?
=
?
+? ?
?
 
 
 Points of non-differentiability = {2, 3, 4, 5} 
 ? m = 4 
 
2 1 2
3
2 2 1
ƒ( ) ( 3 ) 2 I x dx x x dx dx
-
- - -
= = - +
? ? ?
 
 
1
42
2
3 1 3
2(2 1) (4 6) 6
4 2 4 2
xx
-
-
??
??
= - + + = - - - + ??
??
??
??
??
 
    = 
27
4
 
5. Let 
ˆ ˆ ˆ ˆ ˆ ˆ
3 , 3 4 a i j k b i j k = ? + - = - ? + and 
ˆ ˆ ˆ
22 c i j k = + - where ?, ? ? R, be three vectors. If 
the projection of a on c is 
10
3
 and 
ˆ ˆ ˆ
6 10 7 , b c i j k ? = - + + then the value of ? + ? is 
equal to : 
 (A) 3 (B) 4 
 (C) 5 (D) 6 
Answer (A) 
Sol. 
ˆ ˆ ˆ
3 a i j k = ? + - 
 
ˆ ˆ ˆ
34 b i j k = - ? + 
 
ˆ ˆ ˆ
22 c i j k = + - 
 Projection of a on c is 
 
10
3
ac
b
?
= 
 
2 2 2
6 2 8 10
33
1 2 ( 2)
? + + ? +
==
+ + -
 
 ? 2 ?= 
 
ˆ ˆ ˆ
6 10 7 b c i j k ? = - + + 
 
ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
3 4 (2 8) 10 (6 ) 6 10 7
1 2 2
i j k
i j k i j k -? = ? - + + + ? = - + +
-
 
 2 8 6 & 6 7 ? - = - + ? = 
 ? 1 ?= 
 ? + ? = 2 + 1 = 3 
 
   
 
6. The area enclosed by y
2
 = 8x and 2 yx = that lies 
outside the triangle formed by 
2 , 1 , 2 2, y x x y = = = is equal to :  
 (A) 
16 2
6
 (B) 
11 2
6
 
 (C) 
13 2
6
 (C) 
52
6
 
Answer (C) 
Sol.  
  
 
( ) ( ) ( )
2, 2 2 , 1 , 2 2 , 1 , 2 A B C 
 Area = ( )
42
2
0
area
8
2
yy
dy BAC
??
- - ? ??
??
??
?
 
 
42
23
0
1
24 2
22
yy
AB BC
??
= - - ? ? ??
??
??
 
 
32 4 2 1
8 2 1 2
24 2
?
= - - ? ? 
 
16 2 2
82
32
= - - 
 ( )
2 13 2
48 32 3
66
= - - = 
7. If the system of linear equations 
 2x + y – z = 7 
 x – 3y + 2z = 1 
 x + 4y + ?z = k, where ?, k ? R 
 has infinitely many solutions, then ? + k is equal to: 
 (A) –3 (B) 3 
 (C) 6 (D) 9 
Answer (B) 
Sol. 2x + y – z = 7 
 x – 3y + 2z = 1 
 x + 4y + ?z = k 
  
2 1 1
1 3 2 7 21 0
14
-
? = - = - ? - =
?
 
 3 ? = - 
 
1
7 1 1
1 3 2
43 k
-
? = -
-
 
 ? 6 – k = 0 ? k = 6 
 3 6 3 k ? + = - + = 
8. Let ? and ? be the roots of the equation x
2
 + 
(2i – 1) = 0. Then, the value of 
88
? + ? is equal to: 
 (A) 50 (B) 250 
 (C) 1250 (D) 1500 
Answer (A) 
Sol. x
2
 + 2i – 1 = 0 
 ?
2
 = ?
2
 = 1 – 2i 
 ?
4
 = (1 – 2i)
2
 = 1 + (2i)
2
 – 4i = –3 – 4i 
 ?
8
 = (–3 – 4i)
2
 = 9 – 16 + 24i = –7 + 24i 
 ( ) ( )
22
88
2 7 24 2 7 24 50 i ? + ? = - + = - + = 
9. Let { , , , } ? ? ? ? ? ? be such that 
( ) ( ) ( )
p q p q q ? ? ? ? is a tautology. Then ? is 
equal to : 
 (A) ? (B) ? 
 (C) ? (D) ? 
Answer (C) 
Sol. ( ) p q q ?? 
 ( ) ~ p q q ?? 
 ( ) ~~ p q q = ? ? 
 ( ) ( ) ~~ = ? ? ? p q q q 
 ( ) ~ p q T = ? ? 
 ~pq =? 
 Now ( ) ( ) ~ p q p q ? ? ? 
 
( ) ( ) ~ ~ ~ p q p p q p q p q p q
T T F T T T
T F F F F T
F T T F T T
F F T F T T
? ? ? ? ?
 
 ? ? = ? 
 
   
 
10. Let A = [a ij] be a square matrix of order 3 such that 
a ij = 2
j–i
, for all i, j = 1, 2, 3. Then, the matrix A
2
 + A
3
 
+ … + A
10
 is equal to : 
 (A) 
10
33
2
A
??
-
??
??
??
 (B) 
10
31
2
A
??
-
??
??
??
 
 (C) 
10
31
2
A
??
+
??
??
??
 (D) 
10
33
2
A
??
+
??
??
??
 
Answer (A) 
Sol. 
0 1 2
11 12 13
1 0 1
21 22 23
2 1 0
31 32 33
2 2 2
2 2 2
2 2 2
a a a
A a a a
a a a
-
--
??
??
??
??
??
==
??
??
??
?? ??
??
 
 
2
1 2 4 1 2 4 3 6 12
1 1 3
1 2 1 2 3 6 3
2 2 2
1 1 1 1 3 3
1 1 3
4 2 4 2 4 2
AA
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
= = =
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
 
 A
2
 = 3A 
 A
3
 = A.A
2
 = A(3A) = 3A
2
 = 3
2
A 
 A
4
 = 3
3
A 
 Now 
 A
2
 + A
3
 + … + A
10
 
 A[3
1
 + 3
2
 + 3
3
 + … + 3
9
] 
 
9
3 3 1
31
A
??
-
??
=
-
 
 
( )
10
33
2
A
-
= 
11. Let a set A = A1 ?A2 ?… ?A k, where A i ? A j = ? 
for i ? j, 1 ? i, j ? k. Define the relation R from A to 
A by R = {(x, y)  : y ?A i if and only if x?A i, 1 ? i ? 
k}. Then, R is : 
 (A) reflexive, symmetric but not transitive 
 (B) reflexive, transitive but not symmetric 
 (C) reflexive but not symmetric and transitive 
 (D) an equivalence relation  
Answer (D) 
Sol. R = {(x, y) : y?A i, iff x?Ai, 1 ? i ? k} 
 (1) Reflexive  
   (a, a)  ?  a?Ai  iff  a?Ai 
 (2) Symmetric  
   (a, b)  ?  a?Ai  iff  b?Ai 
   (b, a) ?R  as b?Ai  iff a?Ai 
 (3) Transitive  
   (a, b)?R & (b, c) ?R. 
  ? a?Ai   iff   b?A i  &  b?A i  iff   c?A i 
  ? a?Ai   iff   c?A i 
  ? (a, c)?R. 
  ? Relation is equivalence 
12. Let ? ?
0
n
n
a
?
=
 be a sequence such that a0 = a1 = 0 
and a n + 2 = 2a n + 1 – a n + 1 for all n ? 0. 
 Then 
2
7
n
n
n
a
?
=
?
 is equal to : 
 (A) 
6
343
 (B) 
7
216
 
 (C) 
8
343
 (D) 
49
216
 
Answer (B) 
Sol. 
2 1 0 1
2 1 & 0
n n n
a a a a a
++
= - + = = 
 
2 1 0
2 1 1 a a a = - + = 
 
3 2 1
2 1 3 a a a = - + = 
 
4 3 2
2 1 6 a a a = - + = 
 
5 4 3
2 1 10 a a a = - + = 
 
3 24
2 3 4
2
7 7 7 7
n
n
n
a a a a
?
=
= + + +?
?
 
  
2 3 4 5
3 4 5
2 3 4
34
2 3 4
1 3 6 10
7 7 7 7
1 1 3 6
7
7 7 7
6 1 2 3
7
777
6 1 2
49
77
36 1 1 1
49
777
s
s
s
s
s
= + + + + ?
= + + + ?
= + + + ?
= + + ?
= + + + ?
 
Page 5


 
   
 
MATHEMATICS 
SECTION - A 
Multiple Choice Questions: This section contains 20 
multiple choice questions. Each question has 4 choices 
(1), (2), (3) and (4), out of which ONLY ONE is correct. 
Choose the correct answer : 
1. The probability that a randomly chosen 2 × 2 matrix 
with all the entries from the set of first 10 primes, is 
singular, is equal to : 
 (A) 
4
133
10
 (B) 
3
18
10
 
 (C) 
3
19
10
 (D) 
4
271
10
 
Answer (C) 
Sol. Let A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29} 
 Let E be the event that matrix of order 2 × 2 is singular 
 Case-I 
 All entries are same example 
22
22
??
??
??
 
 = 
10
C
1
 
 Case-II 
 Matrix with two prime numbers only example 
35
35
??
??
??
 
 = 
10
C
2
 × 2! × 2! 
 
10 10
12
4 4 3
2! 2! 190 19
()
10 10 10
CC
PE
+ ? ?
= = = 
2. Let the solution curve of the differential equation 
22
16 ,
dy
x y y x
dx
- = + y(1) = 3 be y = y(x). Then 
y(2) is equal to : 
 (A) 15 (B) 11 
 (C) 13 (D) 17 
Answer (A) 
Sol. 
22
16
dy
x y y x
dx
- = + 
 y = 4x tan ? 
 
2
4tan 4 sec
dy d
x
dx dx
?
= ? + ? 
 
22
4 tan 4 sec 4 tan 4 sec
d
x x x x
dx
?
? + ? - ? = ? 
 sec
dx
d
x
? ? =
??
 
 log |sec? + tan?| = log |x| + C 
 y(1) = 3  ? 3 = 4 tan? 
 
35
tan sec
44
= ? = ? ? = 
 
8
ln ln 1
4
C =+ 
 ? C = ln 2 
 ? |sec? + tan?| = 2|x| 
 To find y(2) put x = 2 
 ? tan
8
y
?= 
 (sec? + tan?)
2
 = 16 
 
sec tan 4
1
sec tan
4
15
2tan 2
48
y
? + ? = ?
? - ? = ?
? = = ?
 
 ? 15 y = 
3. If the mirror image of the point (2, 4, 7) in the plane 
3x – y + 4z = 2 is (a, b, c), then 2a + b + 2c is equal 
to : 
 (A) 54 (B) 50 
 (C) –6 (D) –42 
Answer (C) 
Sol. Mirror image of (2, 4, 7) in 3x – y + 4z = 2 is 
(a, b, c) then 
 
2 2 2
2 4 7 2(6 4 28 2)
3 1 4
3 ( 1) 4
a b c - - - - - + -
= = =
-
+ - +
 
 
2 4 7 28
3 1 4 13
a b c - - - -
= = =
-
 
 
58 80 21
13 13 13
a b c
--
= = = 
 
116 80 42
22
13
a b c
- + -
+ + = 
    = –6 
 
   
 
4. Let ƒ : R ? R be a function defined by : 
 
3
2
max{ 3 } ; 2
2 6 ; 2 3 ƒ( )
[ 3] 9 ; 3 5
2 1 ; 5
tx
t t x
x x x x
xx
xx
?
?
-?
?
?
?
+ - ? ? =
?
?
- + ? ?
?
+? ?
?
 
 where [t] is the greatest integer less than or equal 
to t. Let m be the number of points where ƒ is not 
differentiable and 
2
2
ƒ( ) . I x dx
-
=
?
 Then the ordered 
pair (m, I) is equal to : 
 (A) 
27
3,
4
??
??
??
 (B) 
23
3,
4
??
??
??
 
 (C) 
27
4,
4
??
??
??
 (D) 
23
4,
4
??
??
??
 
Answer (C) 
Sol. 
3
max{ 3 } ; 2
tx
t t x
?
-? 
 g(t) = t
3
 –3t  ? g? (t) = 3t
2
 –3 = 3(t – 1)(t + 1) 
  
  
 
3
2
31
2 1 2
2 6 2 3
ƒ( )
9 3 4
10 4 5
11 5
2 1 5
x x x
x
x x x
x
x
x
x
xx
?
- ? -
?
- ? ?
?
?
+ - ? ?
?
?
=
?
??
?
??
?
?
=
?
+? ?
?
 
 
 Points of non-differentiability = {2, 3, 4, 5} 
 ? m = 4 
 
2 1 2
3
2 2 1
ƒ( ) ( 3 ) 2 I x dx x x dx dx
-
- - -
= = - +
? ? ?
 
 
1
42
2
3 1 3
2(2 1) (4 6) 6
4 2 4 2
xx
-
-
??
??
= - + + = - - - + ??
??
??
??
??
 
    = 
27
4
 
5. Let 
ˆ ˆ ˆ ˆ ˆ ˆ
3 , 3 4 a i j k b i j k = ? + - = - ? + and 
ˆ ˆ ˆ
22 c i j k = + - where ?, ? ? R, be three vectors. If 
the projection of a on c is 
10
3
 and 
ˆ ˆ ˆ
6 10 7 , b c i j k ? = - + + then the value of ? + ? is 
equal to : 
 (A) 3 (B) 4 
 (C) 5 (D) 6 
Answer (A) 
Sol. 
ˆ ˆ ˆ
3 a i j k = ? + - 
 
ˆ ˆ ˆ
34 b i j k = - ? + 
 
ˆ ˆ ˆ
22 c i j k = + - 
 Projection of a on c is 
 
10
3
ac
b
?
= 
 
2 2 2
6 2 8 10
33
1 2 ( 2)
? + + ? +
==
+ + -
 
 ? 2 ?= 
 
ˆ ˆ ˆ
6 10 7 b c i j k ? = - + + 
 
ˆ ˆ ˆ
ˆ ˆ ˆ ˆ ˆ ˆ
3 4 (2 8) 10 (6 ) 6 10 7
1 2 2
i j k
i j k i j k -? = ? - + + + ? = - + +
-
 
 2 8 6 & 6 7 ? - = - + ? = 
 ? 1 ?= 
 ? + ? = 2 + 1 = 3 
 
   
 
6. The area enclosed by y
2
 = 8x and 2 yx = that lies 
outside the triangle formed by 
2 , 1 , 2 2, y x x y = = = is equal to :  
 (A) 
16 2
6
 (B) 
11 2
6
 
 (C) 
13 2
6
 (C) 
52
6
 
Answer (C) 
Sol.  
  
 
( ) ( ) ( )
2, 2 2 , 1 , 2 2 , 1 , 2 A B C 
 Area = ( )
42
2
0
area
8
2
yy
dy BAC
??
- - ? ??
??
??
?
 
 
42
23
0
1
24 2
22
yy
AB BC
??
= - - ? ? ??
??
??
 
 
32 4 2 1
8 2 1 2
24 2
?
= - - ? ? 
 
16 2 2
82
32
= - - 
 ( )
2 13 2
48 32 3
66
= - - = 
7. If the system of linear equations 
 2x + y – z = 7 
 x – 3y + 2z = 1 
 x + 4y + ?z = k, where ?, k ? R 
 has infinitely many solutions, then ? + k is equal to: 
 (A) –3 (B) 3 
 (C) 6 (D) 9 
Answer (B) 
Sol. 2x + y – z = 7 
 x – 3y + 2z = 1 
 x + 4y + ?z = k 
  
2 1 1
1 3 2 7 21 0
14
-
? = - = - ? - =
?
 
 3 ? = - 
 
1
7 1 1
1 3 2
43 k
-
? = -
-
 
 ? 6 – k = 0 ? k = 6 
 3 6 3 k ? + = - + = 
8. Let ? and ? be the roots of the equation x
2
 + 
(2i – 1) = 0. Then, the value of 
88
? + ? is equal to: 
 (A) 50 (B) 250 
 (C) 1250 (D) 1500 
Answer (A) 
Sol. x
2
 + 2i – 1 = 0 
 ?
2
 = ?
2
 = 1 – 2i 
 ?
4
 = (1 – 2i)
2
 = 1 + (2i)
2
 – 4i = –3 – 4i 
 ?
8
 = (–3 – 4i)
2
 = 9 – 16 + 24i = –7 + 24i 
 ( ) ( )
22
88
2 7 24 2 7 24 50 i ? + ? = - + = - + = 
9. Let { , , , } ? ? ? ? ? ? be such that 
( ) ( ) ( )
p q p q q ? ? ? ? is a tautology. Then ? is 
equal to : 
 (A) ? (B) ? 
 (C) ? (D) ? 
Answer (C) 
Sol. ( ) p q q ?? 
 ( ) ~ p q q ?? 
 ( ) ~~ p q q = ? ? 
 ( ) ( ) ~~ = ? ? ? p q q q 
 ( ) ~ p q T = ? ? 
 ~pq =? 
 Now ( ) ( ) ~ p q p q ? ? ? 
 
( ) ( ) ~ ~ ~ p q p p q p q p q p q
T T F T T T
T F F F F T
F T T F T T
F F T F T T
? ? ? ? ?
 
 ? ? = ? 
 
   
 
10. Let A = [a ij] be a square matrix of order 3 such that 
a ij = 2
j–i
, for all i, j = 1, 2, 3. Then, the matrix A
2
 + A
3
 
+ … + A
10
 is equal to : 
 (A) 
10
33
2
A
??
-
??
??
??
 (B) 
10
31
2
A
??
-
??
??
??
 
 (C) 
10
31
2
A
??
+
??
??
??
 (D) 
10
33
2
A
??
+
??
??
??
 
Answer (A) 
Sol. 
0 1 2
11 12 13
1 0 1
21 22 23
2 1 0
31 32 33
2 2 2
2 2 2
2 2 2
a a a
A a a a
a a a
-
--
??
??
??
??
??
==
??
??
??
?? ??
??
 
 
2
1 2 4 1 2 4 3 6 12
1 1 3
1 2 1 2 3 6 3
2 2 2
1 1 1 1 3 3
1 1 3
4 2 4 2 4 2
AA
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
= = =
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
 
 A
2
 = 3A 
 A
3
 = A.A
2
 = A(3A) = 3A
2
 = 3
2
A 
 A
4
 = 3
3
A 
 Now 
 A
2
 + A
3
 + … + A
10
 
 A[3
1
 + 3
2
 + 3
3
 + … + 3
9
] 
 
9
3 3 1
31
A
??
-
??
=
-
 
 
( )
10
33
2
A
-
= 
11. Let a set A = A1 ?A2 ?… ?A k, where A i ? A j = ? 
for i ? j, 1 ? i, j ? k. Define the relation R from A to 
A by R = {(x, y)  : y ?A i if and only if x?A i, 1 ? i ? 
k}. Then, R is : 
 (A) reflexive, symmetric but not transitive 
 (B) reflexive, transitive but not symmetric 
 (C) reflexive but not symmetric and transitive 
 (D) an equivalence relation  
Answer (D) 
Sol. R = {(x, y) : y?A i, iff x?Ai, 1 ? i ? k} 
 (1) Reflexive  
   (a, a)  ?  a?Ai  iff  a?Ai 
 (2) Symmetric  
   (a, b)  ?  a?Ai  iff  b?Ai 
   (b, a) ?R  as b?Ai  iff a?Ai 
 (3) Transitive  
   (a, b)?R & (b, c) ?R. 
  ? a?Ai   iff   b?A i  &  b?A i  iff   c?A i 
  ? a?Ai   iff   c?A i 
  ? (a, c)?R. 
  ? Relation is equivalence 
12. Let ? ?
0
n
n
a
?
=
 be a sequence such that a0 = a1 = 0 
and a n + 2 = 2a n + 1 – a n + 1 for all n ? 0. 
 Then 
2
7
n
n
n
a
?
=
?
 is equal to : 
 (A) 
6
343
 (B) 
7
216
 
 (C) 
8
343
 (D) 
49
216
 
Answer (B) 
Sol. 
2 1 0 1
2 1 & 0
n n n
a a a a a
++
= - + = = 
 
2 1 0
2 1 1 a a a = - + = 
 
3 2 1
2 1 3 a a a = - + = 
 
4 3 2
2 1 6 a a a = - + = 
 
5 4 3
2 1 10 a a a = - + = 
 
3 24
2 3 4
2
7 7 7 7
n
n
n
a a a a
?
=
= + + +?
?
 
  
2 3 4 5
3 4 5
2 3 4
34
2 3 4
1 3 6 10
7 7 7 7
1 1 3 6
7
7 7 7
6 1 2 3
7
777
6 1 2
49
77
36 1 1 1
49
777
s
s
s
s
s
= + + + + ?
= + + + ?
= + + + ?
= + + ?
= + + + ?
 
 
   
 
 
2
1
36
7
1
49
1
7
s
=
-
 
 
36 7
49 49 6
s
=
?
 
 
7
216
s = 
13. The distance between the two points A and A? 
which lie on y = 2 such that both the line segments 
AB and A?B (where B is the point (2, 3)) subtend 
angle 
4
?
 at the origin, is equal to 
 (A) 10 (B) 
48
5
 
 (C) 
52
5
 (D) 3 
Answer (C) 
Sol. Let A(?, 2)    Given B(2, 3) 
 
23
&
2
OA OB
mm ==
?
 
 
23
43
2
tan 1
23
4 2 6
1.
2
-
? - ?
?
= ? = ?
?+
+
?
 
 4 – 3? = 2? + 6  &  4 – 3? = -2? - 6 
  
2
& 10
5
-
? = ? = 
 
2
, 2 & (10,2) and (2, 3)
5
A A B
??
-?
??
??
 
 
2 52
10
55
AA? = + = 
14. A wire of length 22 m is to be cut into two pieces. 
One of the pieces is to be made into a square and 
the other into an equilateral triangle. Then, the 
length of the side of the equilateral triangle, so that 
the combined area of the square and the equilateral 
triangle is minimum, is 
 (A) 
22
9 4 3 +
 (B) 
66
9 4 3 +
 
 (C) 
22
4 9 3 +
 (D) 
66
4 9 3 +
 
Answer (B) 
Sol.  
 4a + 3b = 22 
 Total area =
22
3
4
A a b =+ 
   
2
2
22 3 3
44
b
Ab
- ??
=+
??
??
 
  
22 3 3 3
2 .2 0
4 4 4
dA b
b
dB
-- ? ?? ?
= + =
? ?? ?
? ?? ?
 
 ? 
33
(22 3 )
28
b
b =- 
  4 3 66 9 bb =- 
   
66
9 4 3
b =
+
 
15. The domain of the function  
 
1
2
1
1
2sin
41
cos
x
-
-
?? ??
?? ??
? - ?
??
??
?
??
??
 is : 
 (A) 
11
,
22
??
--
??
??
R 
 (B) ( ? ? ) ? ? , 1 1 , 0 -? - ? ? ? 
 (C) ? ?
11
, , 0
22
- ? ? ? ?
-? ? ? ?
? ? ? ?
? ? ? ?
 
 (D) ? ?
11
, , 0
22
- ? ? ? ?
-? ? ? ?
?? ??
? ? ? ?
 
Answer (D) 
Sol. 
1
2
21
1 sin 1
41 x
-
??
- ? ?
??
?
? - ?
 
 
1
2
1
sin
22
41 x
-
??
- ? ?
-
 
 
2
1
11
41 x
- ? ?
-
 
 
2
2
14
10
(2 1)(2 1)
41
x
nx
x
? - ? ?
+-
-