JEE Main 2022 Mathematics June 29 Shift 2 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. Let ? be a root of the equation 1 + x
2
 + x
4
 = 0. Then 
the value of ?
1011
 + ?
2022
 – ?
3033
 is equal to 
 (A) 1  (B) ? 
 (C) 1 + ? (D) 1 + 2? 
Answer (A) 
Sol. 1 + x
2
 + x
4
 = 0 
 Root is ?(cube root of unity) 
 ?
1011
 + ?
2022
 – ?
3033
 
 = (?
3
)
337
 + (?
3
)
674
 – (?
3
)
1011 
 
= 1 + 1 – 1 = 1 
2. Let arg(z) represent the principal argument of the 
complex number z. 
 Then, |z| = 3 and arg(z – 1) – arg(z + 1) 
4
?
= 
intersect 
 (A) exactly at one point 
  (B) exactly at two points 
 (C) nowhere 
 (D) at infinitely many points 
Answer (C) 
Sol.  
 |z| = 3 
 arg(z – 1) – arg(z + 1) 
4
?
= 
 ?AKL = ?ACB 
4
?
= 
 ? LK = AL = ? = 1 
 K(0, 1) 
 radius 2 = 
 PL = PK + KL 21 =+ 
 P(0, 1 + 2 ) 
 Number of intersection = 0 
3. Let 
21
.
02
A
- ??
=
??
??
 If B = I – 
5
C1(adjA) + 
5
C2(adjA)
2
 – 
…. – 
5
C5(adjA)
5
, then the sum of all elements of the 
matrix B is 
 (A) –5  (B) –6 
 (C) –7 (D) –8 
Answer (C) 
Sol. A =
21
02
- ??
??
??
 ? adj(A) 
21
02
??
??
??
 
 B = I – 
5
C1(adjA) + 
5
C2(adjA)
2
 + ….. + 
5
C5(adjA)
5 
 
= (I – adjA)
5
 = 
5 5
1 0 2 1 1 1
0 1 0 2 0 1
?? -- ? ? ? ? ? ?
-=
??
? ? ? ? ? ?
? ? ? ? ? ? ??
  
 Let 
11
01
P
-- ??
=
??
-
??
 ? B = P
5 
 
2
1 1 1 1 1 2
0 1 0 1 0 1
P
- - - - ? ? ? ? ? ?
==
? ? ? ? ? ?
--
? ? ? ? ? ?
 
 
3
1 2 1 1 1 3
0 1 0 1 0 1
P
- - - - ? ? ? ? ? ?
==
? ? ? ? ? ?
--
? ? ? ? ? ?
 
 
4
1 3 1 1 1 4
0 1 0 1 0 1
P
- - - - ? ? ? ? ? ?
==
? ? ? ? ? ?
--
? ? ? ? ? ?
 
 
5
1 4 1 1 1 5
0 1 0 1 0 1
PB
- - - - ? ? ? ? ? ?
= = =
? ? ? ? ? ?
-
? ? ? ? ? ?
 
 Sum of elements = –1 –5 – 1 + 0 = –7 
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. Let ? be a root of the equation 1 + x
2
 + x
4
 = 0. Then 
the value of ?
1011
 + ?
2022
 – ?
3033
 is equal to 
 (A) 1  (B) ? 
 (C) 1 + ? (D) 1 + 2? 
Answer (A) 
Sol. 1 + x
2
 + x
4
 = 0 
 Root is ?(cube root of unity) 
 ?
1011
 + ?
2022
 – ?
3033
 
 = (?
3
)
337
 + (?
3
)
674
 – (?
3
)
1011 
 
= 1 + 1 – 1 = 1 
2. Let arg(z) represent the principal argument of the 
complex number z. 
 Then, |z| = 3 and arg(z – 1) – arg(z + 1) 
4
?
= 
intersect 
 (A) exactly at one point 
  (B) exactly at two points 
 (C) nowhere 
 (D) at infinitely many points 
Answer (C) 
Sol.  
 |z| = 3 
 arg(z – 1) – arg(z + 1) 
4
?
= 
 ?AKL = ?ACB 
4
?
= 
 ? LK = AL = ? = 1 
 K(0, 1) 
 radius 2 = 
 PL = PK + KL 21 =+ 
 P(0, 1 + 2 ) 
 Number of intersection = 0 
3. Let 
21
.
02
A
- ??
=
??
??
 If B = I – 
5
C1(adjA) + 
5
C2(adjA)
2
 – 
…. – 
5
C5(adjA)
5
, then the sum of all elements of the 
matrix B is 
 (A) –5  (B) –6 
 (C) –7 (D) –8 
Answer (C) 
Sol. A =
21
02
- ??
??
??
 ? adj(A) 
21
02
??
??
??
 
 B = I – 
5
C1(adjA) + 
5
C2(adjA)
2
 + ….. + 
5
C5(adjA)
5 
 
= (I – adjA)
5
 = 
5 5
1 0 2 1 1 1
0 1 0 2 0 1
?? -- ? ? ? ? ? ?
-=
??
? ? ? ? ? ?
? ? ? ? ? ? ??
  
 Let 
11
01
P
-- ??
=
??
-
??
 ? B = P
5 
 
2
1 1 1 1 1 2
0 1 0 1 0 1
P
- - - - ? ? ? ? ? ?
==
? ? ? ? ? ?
--
? ? ? ? ? ?
 
 
3
1 2 1 1 1 3
0 1 0 1 0 1
P
- - - - ? ? ? ? ? ?
==
? ? ? ? ? ?
--
? ? ? ? ? ?
 
 
4
1 3 1 1 1 4
0 1 0 1 0 1
P
- - - - ? ? ? ? ? ?
==
? ? ? ? ? ?
--
? ? ? ? ? ?
 
 
5
1 4 1 1 1 5
0 1 0 1 0 1
PB
- - - - ? ? ? ? ? ?
= = =
? ? ? ? ? ?
-
? ? ? ? ? ?
 
 Sum of elements = –1 –5 – 1 + 0 = –7 
 
   
 
4. The sum of the infinite series  
 
2 3 4 5 6
5 12 22 35 51 70
1
6
6 6 6 6 6
+ + + + + + + ….. is equal to 
 (A) 
425
216
  (B) 
429
216
 
 (C) 
288
125
 (D) 
280
125
 
Answer (C) 
Sol. 
23
23
2 3 4
2 3 4
2 3 4
5 12 22
1 ....
6
66
1 1 5 12
....
66
66
5 4 7 10 13
1 ....
66
666
5 1 4 7 10
....
36 6
666
25 3 3 3 3
1 ....
36 6
666
= + + + +
= + + +
= + + + + +
= + + + +
= + + + + +
S
S
S
S
S
 
 
3
25
6
1
1
36
1
6
S
=+
-
 
 
25 8
36 5
S
= 
 
288
125
S = 
5. The value of 
( )
( )
22
43
1
1 sin
lim
2 2 1
x
xx
x x x
?
-?
- + -
 is equal to 
 (A) 
2
6
?
  (B) 
2
3
?
 
 (C) 
2
2
?
 (D) ?
2
 
Answer (D) 
Sol. 
( )
22
43
1
1 sin
lim
2 2 1
x
xx
x x x
?
-?
- + -
 
 = 
( )( )
( ) ( )
2
3
1
1 1 sin
lim
11
x
x x x
xx
?
+ - ?
-+
 
 Let x – 1 = t 
 
( )
( )
2
3
0
2 sin
lim
2
?
+?
+
t
t t t
tt
 
 = 
2
2
22
0
sin
lim ·
t
t
t
?
?
=?
?
 
 = ?
2
 
6. Let f : R ? R be a function defined by  
 ( ) ( ) ( )
12
12
3 5 , , . = - - ?
nn
f x x x n n N Then, which 
of the following is NOT true? 
 (A) For 
12
3, 4, nn == there exists ( ) 3, 5 ?? 
where f attains local maxima. 
 (B) For 
12
4, 3, nn == there exists ( ) 3, 5 ?? 
where f attains local minima. 
 (C) For 
12
3, 5, nn == there exists ( ) 3, 5 ?? 
where f attains local maxima. 
 (D) For 
12
4, 6, nn == there exists ( ) 3, 5 ?? 
where f attains local maxima. 
Answer (C) 
Sol. For n2 ? odd, there will be local minima in (3, 5) 
for n2 ? even, there will be local maxima in (3, 5) 
7. Let f be a real valued continuous function on [0, 1] 
and ( ) ( ) ( )
1
0
. f x x x t f t dt = + -
?
 Then, which of the 
following points (x, y) lies on the curve ( ) y f x = ? 
 (A) (2, 4) (B) (1, 2) 
 (C) (4, 17) (D) (6, 8) 
Answer (D) 
Sol. ( ) ( ) ( )
1
0
f x x x t f t dt =-
?
 
 ( ) ( ) ( )
11
00
f x x x f t dt tf t dt = + -
??
 
 ( ) ( ) ( )
11
00
1
??
= + -
??
??
??
??
f x x f t dt tf t dt 
 Let ( ) ( )
11
00
1 and 1 f t dt a tf t dt + = =
??
 
 ( )=- f x ax b 
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. Let ? be a root of the equation 1 + x
2
 + x
4
 = 0. Then 
the value of ?
1011
 + ?
2022
 – ?
3033
 is equal to 
 (A) 1  (B) ? 
 (C) 1 + ? (D) 1 + 2? 
Answer (A) 
Sol. 1 + x
2
 + x
4
 = 0 
 Root is ?(cube root of unity) 
 ?
1011
 + ?
2022
 – ?
3033
 
 = (?
3
)
337
 + (?
3
)
674
 – (?
3
)
1011 
 
= 1 + 1 – 1 = 1 
2. Let arg(z) represent the principal argument of the 
complex number z. 
 Then, |z| = 3 and arg(z – 1) – arg(z + 1) 
4
?
= 
intersect 
 (A) exactly at one point 
  (B) exactly at two points 
 (C) nowhere 
 (D) at infinitely many points 
Answer (C) 
Sol.  
 |z| = 3 
 arg(z – 1) – arg(z + 1) 
4
?
= 
 ?AKL = ?ACB 
4
?
= 
 ? LK = AL = ? = 1 
 K(0, 1) 
 radius 2 = 
 PL = PK + KL 21 =+ 
 P(0, 1 + 2 ) 
 Number of intersection = 0 
3. Let 
21
.
02
A
- ??
=
??
??
 If B = I – 
5
C1(adjA) + 
5
C2(adjA)
2
 – 
…. – 
5
C5(adjA)
5
, then the sum of all elements of the 
matrix B is 
 (A) –5  (B) –6 
 (C) –7 (D) –8 
Answer (C) 
Sol. A =
21
02
- ??
??
??
 ? adj(A) 
21
02
??
??
??
 
 B = I – 
5
C1(adjA) + 
5
C2(adjA)
2
 + ….. + 
5
C5(adjA)
5 
 
= (I – adjA)
5
 = 
5 5
1 0 2 1 1 1
0 1 0 2 0 1
?? -- ? ? ? ? ? ?
-=
??
? ? ? ? ? ?
? ? ? ? ? ? ??
  
 Let 
11
01
P
-- ??
=
??
-
??
 ? B = P
5 
 
2
1 1 1 1 1 2
0 1 0 1 0 1
P
- - - - ? ? ? ? ? ?
==
? ? ? ? ? ?
--
? ? ? ? ? ?
 
 
3
1 2 1 1 1 3
0 1 0 1 0 1
P
- - - - ? ? ? ? ? ?
==
? ? ? ? ? ?
--
? ? ? ? ? ?
 
 
4
1 3 1 1 1 4
0 1 0 1 0 1
P
- - - - ? ? ? ? ? ?
==
? ? ? ? ? ?
--
? ? ? ? ? ?
 
 
5
1 4 1 1 1 5
0 1 0 1 0 1
PB
- - - - ? ? ? ? ? ?
= = =
? ? ? ? ? ?
-
? ? ? ? ? ?
 
 Sum of elements = –1 –5 – 1 + 0 = –7 
 
   
 
4. The sum of the infinite series  
 
2 3 4 5 6
5 12 22 35 51 70
1
6
6 6 6 6 6
+ + + + + + + ….. is equal to 
 (A) 
425
216
  (B) 
429
216
 
 (C) 
288
125
 (D) 
280
125
 
Answer (C) 
Sol. 
23
23
2 3 4
2 3 4
2 3 4
5 12 22
1 ....
6
66
1 1 5 12
....
66
66
5 4 7 10 13
1 ....
66
666
5 1 4 7 10
....
36 6
666
25 3 3 3 3
1 ....
36 6
666
= + + + +
= + + +
= + + + + +
= + + + +
= + + + + +
S
S
S
S
S
 
 
3
25
6
1
1
36
1
6
S
=+
-
 
 
25 8
36 5
S
= 
 
288
125
S = 
5. The value of 
( )
( )
22
43
1
1 sin
lim
2 2 1
x
xx
x x x
?
-?
- + -
 is equal to 
 (A) 
2
6
?
  (B) 
2
3
?
 
 (C) 
2
2
?
 (D) ?
2
 
Answer (D) 
Sol. 
( )
22
43
1
1 sin
lim
2 2 1
x
xx
x x x
?
-?
- + -
 
 = 
( )( )
( ) ( )
2
3
1
1 1 sin
lim
11
x
x x x
xx
?
+ - ?
-+
 
 Let x – 1 = t 
 
( )
( )
2
3
0
2 sin
lim
2
?
+?
+
t
t t t
tt
 
 = 
2
2
22
0
sin
lim ·
t
t
t
?
?
=?
?
 
 = ?
2
 
6. Let f : R ? R be a function defined by  
 ( ) ( ) ( )
12
12
3 5 , , . = - - ?
nn
f x x x n n N Then, which 
of the following is NOT true? 
 (A) For 
12
3, 4, nn == there exists ( ) 3, 5 ?? 
where f attains local maxima. 
 (B) For 
12
4, 3, nn == there exists ( ) 3, 5 ?? 
where f attains local minima. 
 (C) For 
12
3, 5, nn == there exists ( ) 3, 5 ?? 
where f attains local maxima. 
 (D) For 
12
4, 6, nn == there exists ( ) 3, 5 ?? 
where f attains local maxima. 
Answer (C) 
Sol. For n2 ? odd, there will be local minima in (3, 5) 
for n2 ? even, there will be local maxima in (3, 5) 
7. Let f be a real valued continuous function on [0, 1] 
and ( ) ( ) ( )
1
0
. f x x x t f t dt = + -
?
 Then, which of the 
following points (x, y) lies on the curve ( ) y f x = ? 
 (A) (2, 4) (B) (1, 2) 
 (C) (4, 17) (D) (6, 8) 
Answer (D) 
Sol. ( ) ( ) ( )
1
0
f x x x t f t dt =-
?
 
 ( ) ( ) ( )
11
00
f x x x f t dt tf t dt = + -
??
 
 ( ) ( ) ( )
11
00
1
??
= + -
??
??
??
??
f x x f t dt tf t dt 
 Let ( ) ( )
11
00
1 and 1 f t dt a tf t dt + = =
??
 
 ( )=- f x ax b 
 
   
  
 Now, ( )
1
0
1 1 1
22
aa
a at b dt b b = + - = + - ? + =
?
 
 ( )
1
0
32
 
3 2 2 3 9
a b b a a
b t at b dt b = - = - ? = ? =
?
 
  
2
1
29
aa
+=  
 
18 4
 
13 13
ab ? = = 
 ( )
18 4
13
x
fx
-
= 
 (6, 8) lies on f(x) i.e. option (D) 
8. If 
21
2
22
00
2 2 1 1
2
y
x x x dx y dy
??
??
- - = - - - + ??
??
??
??
??
??
 
2
2
1
2
2
y
dy I
??
-+ ??
??
??
?
 then I equal is 
 (A) 
1
2
0
11 y dy
??
+-
??
??
?
 
 (B) 
1
2
2
0
11
2
y
y dy
??
- - + ??
??
??
?
 
 (C) 
1
2
0
11 y dy
??
--
??
??
?
 
 (D) 
1
2
2
0
11
2
y
y dy
??
+ - + ??
??
??
?
 
Answer (C) 
Sol. ( )
2 2 2 1
2
2
2
0 0 0 0
2 1 1 1 1
2
y
xdx x dx dy y dy
??
- - - = - - - ??
??
??
? ? ? ?
 
 + 1 + I 
 ? 
11
22
00
82
2 1 1 1
33
y dy y dy I - - = + - - +
??
 
 ? 
1
2
0
11 I y dy = - -
?
 
 ? 
1
2
0
11 I y dy
??
= - -
??
??
?
 
9. If y = y (x) is the solution of the differential equation 
( ) ( )
22
1 2 1 0
xx
dy
e y e
dx
+ + + = and ( ) 0 0, y = then 
( )
( ) ( )
2
6 0 log 3
e
yy
??
?+
??
??
 is equal to 
 (A) 2 
 (B) –2 
 (C) –4 
 (D) –1 
Answer (C) 
Sol. 
( ) ( )
22
1 2 1 0
xx
dy
e y e
dx
+ + + = 
 
22
2
11
=-
++
??
x
x
dy e
dx
ye
  
=
=
x
x
et
e dx dt
 
 
1
2
tan 2
1
dt
y
t
-
=-
+
?
 
 
( )
11
tan 2tan
x
y e c
--
=+ 
 ( ) 0
2
yc
?
?= 
 
( )
11
tan 2tan
2
x
ye
--
?
= - + 
 
( )
1
cot 2tan
-
=
x
ye 
 
( )
21
2
2
cosec 2tan
1
x
x
x
dy e
e
dx
e
-
??
=- ??
??
+
??
 
 ( )
0
2
01
2
=
-
? = = = -
x
dy
y
dx
 
 
( )
1
cot 2tan
x
ye
-
= 
 
( )
3
1 log
ln 3 cot 2tan
e
ye
-
??
=
??
??
 
 
( )
1
21
cot 2tan 3 cot cot
33
3
-
?? ??
= = = - = -
??
??
 
 ( )
( ) ( )
2
2
1
6 0 ln 3 6 1
3
yy
??
?? ??
??
? + = - + -
?? ??
??
?? ??
??
 
   
1
6 1 4
3
??
= - + = -
??
??
 
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. Let ? be a root of the equation 1 + x
2
 + x
4
 = 0. Then 
the value of ?
1011
 + ?
2022
 – ?
3033
 is equal to 
 (A) 1  (B) ? 
 (C) 1 + ? (D) 1 + 2? 
Answer (A) 
Sol. 1 + x
2
 + x
4
 = 0 
 Root is ?(cube root of unity) 
 ?
1011
 + ?
2022
 – ?
3033
 
 = (?
3
)
337
 + (?
3
)
674
 – (?
3
)
1011 
 
= 1 + 1 – 1 = 1 
2. Let arg(z) represent the principal argument of the 
complex number z. 
 Then, |z| = 3 and arg(z – 1) – arg(z + 1) 
4
?
= 
intersect 
 (A) exactly at one point 
  (B) exactly at two points 
 (C) nowhere 
 (D) at infinitely many points 
Answer (C) 
Sol.  
 |z| = 3 
 arg(z – 1) – arg(z + 1) 
4
?
= 
 ?AKL = ?ACB 
4
?
= 
 ? LK = AL = ? = 1 
 K(0, 1) 
 radius 2 = 
 PL = PK + KL 21 =+ 
 P(0, 1 + 2 ) 
 Number of intersection = 0 
3. Let 
21
.
02
A
- ??
=
??
??
 If B = I – 
5
C1(adjA) + 
5
C2(adjA)
2
 – 
…. – 
5
C5(adjA)
5
, then the sum of all elements of the 
matrix B is 
 (A) –5  (B) –6 
 (C) –7 (D) –8 
Answer (C) 
Sol. A =
21
02
- ??
??
??
 ? adj(A) 
21
02
??
??
??
 
 B = I – 
5
C1(adjA) + 
5
C2(adjA)
2
 + ….. + 
5
C5(adjA)
5 
 
= (I – adjA)
5
 = 
5 5
1 0 2 1 1 1
0 1 0 2 0 1
?? -- ? ? ? ? ? ?
-=
??
? ? ? ? ? ?
? ? ? ? ? ? ??
  
 Let 
11
01
P
-- ??
=
??
-
??
 ? B = P
5 
 
2
1 1 1 1 1 2
0 1 0 1 0 1
P
- - - - ? ? ? ? ? ?
==
? ? ? ? ? ?
--
? ? ? ? ? ?
 
 
3
1 2 1 1 1 3
0 1 0 1 0 1
P
- - - - ? ? ? ? ? ?
==
? ? ? ? ? ?
--
? ? ? ? ? ?
 
 
4
1 3 1 1 1 4
0 1 0 1 0 1
P
- - - - ? ? ? ? ? ?
==
? ? ? ? ? ?
--
? ? ? ? ? ?
 
 
5
1 4 1 1 1 5
0 1 0 1 0 1
PB
- - - - ? ? ? ? ? ?
= = =
? ? ? ? ? ?
-
? ? ? ? ? ?
 
 Sum of elements = –1 –5 – 1 + 0 = –7 
 
   
 
4. The sum of the infinite series  
 
2 3 4 5 6
5 12 22 35 51 70
1
6
6 6 6 6 6
+ + + + + + + ….. is equal to 
 (A) 
425
216
  (B) 
429
216
 
 (C) 
288
125
 (D) 
280
125
 
Answer (C) 
Sol. 
23
23
2 3 4
2 3 4
2 3 4
5 12 22
1 ....
6
66
1 1 5 12
....
66
66
5 4 7 10 13
1 ....
66
666
5 1 4 7 10
....
36 6
666
25 3 3 3 3
1 ....
36 6
666
= + + + +
= + + +
= + + + + +
= + + + +
= + + + + +
S
S
S
S
S
 
 
3
25
6
1
1
36
1
6
S
=+
-
 
 
25 8
36 5
S
= 
 
288
125
S = 
5. The value of 
( )
( )
22
43
1
1 sin
lim
2 2 1
x
xx
x x x
?
-?
- + -
 is equal to 
 (A) 
2
6
?
  (B) 
2
3
?
 
 (C) 
2
2
?
 (D) ?
2
 
Answer (D) 
Sol. 
( )
22
43
1
1 sin
lim
2 2 1
x
xx
x x x
?
-?
- + -
 
 = 
( )( )
( ) ( )
2
3
1
1 1 sin
lim
11
x
x x x
xx
?
+ - ?
-+
 
 Let x – 1 = t 
 
( )
( )
2
3
0
2 sin
lim
2
?
+?
+
t
t t t
tt
 
 = 
2
2
22
0
sin
lim ·
t
t
t
?
?
=?
?
 
 = ?
2
 
6. Let f : R ? R be a function defined by  
 ( ) ( ) ( )
12
12
3 5 , , . = - - ?
nn
f x x x n n N Then, which 
of the following is NOT true? 
 (A) For 
12
3, 4, nn == there exists ( ) 3, 5 ?? 
where f attains local maxima. 
 (B) For 
12
4, 3, nn == there exists ( ) 3, 5 ?? 
where f attains local minima. 
 (C) For 
12
3, 5, nn == there exists ( ) 3, 5 ?? 
where f attains local maxima. 
 (D) For 
12
4, 6, nn == there exists ( ) 3, 5 ?? 
where f attains local maxima. 
Answer (C) 
Sol. For n2 ? odd, there will be local minima in (3, 5) 
for n2 ? even, there will be local maxima in (3, 5) 
7. Let f be a real valued continuous function on [0, 1] 
and ( ) ( ) ( )
1
0
. f x x x t f t dt = + -
?
 Then, which of the 
following points (x, y) lies on the curve ( ) y f x = ? 
 (A) (2, 4) (B) (1, 2) 
 (C) (4, 17) (D) (6, 8) 
Answer (D) 
Sol. ( ) ( ) ( )
1
0
f x x x t f t dt =-
?
 
 ( ) ( ) ( )
11
00
f x x x f t dt tf t dt = + -
??
 
 ( ) ( ) ( )
11
00
1
??
= + -
??
??
??
??
f x x f t dt tf t dt 
 Let ( ) ( )
11
00
1 and 1 f t dt a tf t dt + = =
??
 
 ( )=- f x ax b 
 
   
  
 Now, ( )
1
0
1 1 1
22
aa
a at b dt b b = + - = + - ? + =
?
 
 ( )
1
0
32
 
3 2 2 3 9
a b b a a
b t at b dt b = - = - ? = ? =
?
 
  
2
1
29
aa
+=  
 
18 4
 
13 13
ab ? = = 
 ( )
18 4
13
x
fx
-
= 
 (6, 8) lies on f(x) i.e. option (D) 
8. If 
21
2
22
00
2 2 1 1
2
y
x x x dx y dy
??
??
- - = - - - + ??
??
??
??
??
??
 
2
2
1
2
2
y
dy I
??
-+ ??
??
??
?
 then I equal is 
 (A) 
1
2
0
11 y dy
??
+-
??
??
?
 
 (B) 
1
2
2
0
11
2
y
y dy
??
- - + ??
??
??
?
 
 (C) 
1
2
0
11 y dy
??
--
??
??
?
 
 (D) 
1
2
2
0
11
2
y
y dy
??
+ - + ??
??
??
?
 
Answer (C) 
Sol. ( )
2 2 2 1
2
2
2
0 0 0 0
2 1 1 1 1
2
y
xdx x dx dy y dy
??
- - - = - - - ??
??
??
? ? ? ?
 
 + 1 + I 
 ? 
11
22
00
82
2 1 1 1
33
y dy y dy I - - = + - - +
??
 
 ? 
1
2
0
11 I y dy = - -
?
 
 ? 
1
2
0
11 I y dy
??
= - -
??
??
?
 
9. If y = y (x) is the solution of the differential equation 
( ) ( )
22
1 2 1 0
xx
dy
e y e
dx
+ + + = and ( ) 0 0, y = then 
( )
( ) ( )
2
6 0 log 3
e
yy
??
?+
??
??
 is equal to 
 (A) 2 
 (B) –2 
 (C) –4 
 (D) –1 
Answer (C) 
Sol. 
( ) ( )
22
1 2 1 0
xx
dy
e y e
dx
+ + + = 
 
22
2
11
=-
++
??
x
x
dy e
dx
ye
  
=
=
x
x
et
e dx dt
 
 
1
2
tan 2
1
dt
y
t
-
=-
+
?
 
 
( )
11
tan 2tan
x
y e c
--
=+ 
 ( ) 0
2
yc
?
?= 
 
( )
11
tan 2tan
2
x
ye
--
?
= - + 
 
( )
1
cot 2tan
-
=
x
ye 
 
( )
21
2
2
cosec 2tan
1
x
x
x
dy e
e
dx
e
-
??
=- ??
??
+
??
 
 ( )
0
2
01
2
=
-
? = = = -
x
dy
y
dx
 
 
( )
1
cot 2tan
x
ye
-
= 
 
( )
3
1 log
ln 3 cot 2tan
e
ye
-
??
=
??
??
 
 
( )
1
21
cot 2tan 3 cot cot
33
3
-
?? ??
= = = - = -
??
??
 
 ( )
( ) ( )
2
2
1
6 0 ln 3 6 1
3
yy
??
?? ??
??
? + = - + -
?? ??
??
?? ??
??
 
   
1
6 1 4
3
??
= - + = -
??
??
 
 
   
 
10. Let P : y
2
 = 4ax, a > 0 be a parabola with focus S. 
Let the tangents to the parabola P make an angle 
of 
4
?
 with the line y = 3x + 5 touch the parabola P 
at A and B. Then the value of a for which A, B and 
S are collinear is 
 (A) 8 only (B) 2 only 
 (C) 
1
4
 only (D) any a > 0  
Answer (D) 
Sol. ( )
2
: 4 , 0 , 0 =? P y ax a S a 
 Equation of tangent on parabola =+
a
y mx
m
 
     35 yx =+ 
 
( )
3
tan 3 1 3
4 1 3
m
mm
m
?-
= ? - = ? +
+
 
 
3 1 3
2
- = +
=-
mm
m
 
 
3 1 3
1
2
- = - -
=
mm
m
 
 Equation of one tangent : 2
2
= - -
a
yx 
 Equation of other tangent : 2
2
x
ya =+ 
 Point of contact are 
 
( )
( )
22
22
, and ,
1
2
21
2
2
a a a a
??
??
??
--
??
??
??
?? -
- ??
?? ??
????
????
 
 ( ) , and 4 , 4
4
a
A a B a a
??
-
??
??
 
 Now or (?ABS) = 0 [S is the focus] 
  
1
4
1
4 4 1 0
2
01
a
a
aa
a
-= 
 ? ( ) ( )
( ) ( )
2
4 0 4 1 0 4 0
4
a
a a a a a - - - - + - - = 
  
2 2 2
3 4 0 = - - + = a a a 
  Always true 
11. Let a triangle ABC be inscribed in the circle 
22
– 2( ) 0 + + = x x y y such that .
2
?
?= BAC If 
the length of side AB is 2 , then the area of the 
?ABC is equal to : 
 (A) 
( )
2 6 3 + 
 (B) 
( )
6 3 2 + 
 (C) 
( )
3 3 4 + 
 (D) 
( )
6 2 3 4 + 
Answer (Dropped) 
Sol. 
22
– 2 – 2 0 x x y y += 
 Centre 
11
,
22
??
??
??
 
 Radius = 1 
  
 BC is diameter 
 
1
ar( ) 2 2 1
2
ABC ? = ? ? = 
12. Let 
2 1 3
3 –2 –1
- + +
==
x y z
 lie on the plane px – qy + 
z = 5, for some ,. ? pq The shortest distance of 
the plane from the origin is : 
 (A) 
3
109
 
 (B) 
5
142
 
 (C) 
5
71
 
 (D) 
1
142
 
Answer (B) 
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. Let ? be a root of the equation 1 + x
2
 + x
4
 = 0. Then 
the value of ?
1011
 + ?
2022
 – ?
3033
 is equal to 
 (A) 1  (B) ? 
 (C) 1 + ? (D) 1 + 2? 
Answer (A) 
Sol. 1 + x
2
 + x
4
 = 0 
 Root is ?(cube root of unity) 
 ?
1011
 + ?
2022
 – ?
3033
 
 = (?
3
)
337
 + (?
3
)
674
 – (?
3
)
1011 
 
= 1 + 1 – 1 = 1 
2. Let arg(z) represent the principal argument of the 
complex number z. 
 Then, |z| = 3 and arg(z – 1) – arg(z + 1) 
4
?
= 
intersect 
 (A) exactly at one point 
  (B) exactly at two points 
 (C) nowhere 
 (D) at infinitely many points 
Answer (C) 
Sol.  
 |z| = 3 
 arg(z – 1) – arg(z + 1) 
4
?
= 
 ?AKL = ?ACB 
4
?
= 
 ? LK = AL = ? = 1 
 K(0, 1) 
 radius 2 = 
 PL = PK + KL 21 =+ 
 P(0, 1 + 2 ) 
 Number of intersection = 0 
3. Let 
21
.
02
A
- ??
=
??
??
 If B = I – 
5
C1(adjA) + 
5
C2(adjA)
2
 – 
…. – 
5
C5(adjA)
5
, then the sum of all elements of the 
matrix B is 
 (A) –5  (B) –6 
 (C) –7 (D) –8 
Answer (C) 
Sol. A =
21
02
- ??
??
??
 ? adj(A) 
21
02
??
??
??
 
 B = I – 
5
C1(adjA) + 
5
C2(adjA)
2
 + ….. + 
5
C5(adjA)
5 
 
= (I – adjA)
5
 = 
5 5
1 0 2 1 1 1
0 1 0 2 0 1
?? -- ? ? ? ? ? ?
-=
??
? ? ? ? ? ?
? ? ? ? ? ? ??
  
 Let 
11
01
P
-- ??
=
??
-
??
 ? B = P
5 
 
2
1 1 1 1 1 2
0 1 0 1 0 1
P
- - - - ? ? ? ? ? ?
==
? ? ? ? ? ?
--
? ? ? ? ? ?
 
 
3
1 2 1 1 1 3
0 1 0 1 0 1
P
- - - - ? ? ? ? ? ?
==
? ? ? ? ? ?
--
? ? ? ? ? ?
 
 
4
1 3 1 1 1 4
0 1 0 1 0 1
P
- - - - ? ? ? ? ? ?
==
? ? ? ? ? ?
--
? ? ? ? ? ?
 
 
5
1 4 1 1 1 5
0 1 0 1 0 1
PB
- - - - ? ? ? ? ? ?
= = =
? ? ? ? ? ?
-
? ? ? ? ? ?
 
 Sum of elements = –1 –5 – 1 + 0 = –7 
 
   
 
4. The sum of the infinite series  
 
2 3 4 5 6
5 12 22 35 51 70
1
6
6 6 6 6 6
+ + + + + + + ….. is equal to 
 (A) 
425
216
  (B) 
429
216
 
 (C) 
288
125
 (D) 
280
125
 
Answer (C) 
Sol. 
23
23
2 3 4
2 3 4
2 3 4
5 12 22
1 ....
6
66
1 1 5 12
....
66
66
5 4 7 10 13
1 ....
66
666
5 1 4 7 10
....
36 6
666
25 3 3 3 3
1 ....
36 6
666
= + + + +
= + + +
= + + + + +
= + + + +
= + + + + +
S
S
S
S
S
 
 
3
25
6
1
1
36
1
6
S
=+
-
 
 
25 8
36 5
S
= 
 
288
125
S = 
5. The value of 
( )
( )
22
43
1
1 sin
lim
2 2 1
x
xx
x x x
?
-?
- + -
 is equal to 
 (A) 
2
6
?
  (B) 
2
3
?
 
 (C) 
2
2
?
 (D) ?
2
 
Answer (D) 
Sol. 
( )
22
43
1
1 sin
lim
2 2 1
x
xx
x x x
?
-?
- + -
 
 = 
( )( )
( ) ( )
2
3
1
1 1 sin
lim
11
x
x x x
xx
?
+ - ?
-+
 
 Let x – 1 = t 
 
( )
( )
2
3
0
2 sin
lim
2
?
+?
+
t
t t t
tt
 
 = 
2
2
22
0
sin
lim ·
t
t
t
?
?
=?
?
 
 = ?
2
 
6. Let f : R ? R be a function defined by  
 ( ) ( ) ( )
12
12
3 5 , , . = - - ?
nn
f x x x n n N Then, which 
of the following is NOT true? 
 (A) For 
12
3, 4, nn == there exists ( ) 3, 5 ?? 
where f attains local maxima. 
 (B) For 
12
4, 3, nn == there exists ( ) 3, 5 ?? 
where f attains local minima. 
 (C) For 
12
3, 5, nn == there exists ( ) 3, 5 ?? 
where f attains local maxima. 
 (D) For 
12
4, 6, nn == there exists ( ) 3, 5 ?? 
where f attains local maxima. 
Answer (C) 
Sol. For n2 ? odd, there will be local minima in (3, 5) 
for n2 ? even, there will be local maxima in (3, 5) 
7. Let f be a real valued continuous function on [0, 1] 
and ( ) ( ) ( )
1
0
. f x x x t f t dt = + -
?
 Then, which of the 
following points (x, y) lies on the curve ( ) y f x = ? 
 (A) (2, 4) (B) (1, 2) 
 (C) (4, 17) (D) (6, 8) 
Answer (D) 
Sol. ( ) ( ) ( )
1
0
f x x x t f t dt =-
?
 
 ( ) ( ) ( )
11
00
f x x x f t dt tf t dt = + -
??
 
 ( ) ( ) ( )
11
00
1
??
= + -
??
??
??
??
f x x f t dt tf t dt 
 Let ( ) ( )
11
00
1 and 1 f t dt a tf t dt + = =
??
 
 ( )=- f x ax b 
 
   
  
 Now, ( )
1
0
1 1 1
22
aa
a at b dt b b = + - = + - ? + =
?
 
 ( )
1
0
32
 
3 2 2 3 9
a b b a a
b t at b dt b = - = - ? = ? =
?
 
  
2
1
29
aa
+=  
 
18 4
 
13 13
ab ? = = 
 ( )
18 4
13
x
fx
-
= 
 (6, 8) lies on f(x) i.e. option (D) 
8. If 
21
2
22
00
2 2 1 1
2
y
x x x dx y dy
??
??
- - = - - - + ??
??
??
??
??
??
 
2
2
1
2
2
y
dy I
??
-+ ??
??
??
?
 then I equal is 
 (A) 
1
2
0
11 y dy
??
+-
??
??
?
 
 (B) 
1
2
2
0
11
2
y
y dy
??
- - + ??
??
??
?
 
 (C) 
1
2
0
11 y dy
??
--
??
??
?
 
 (D) 
1
2
2
0
11
2
y
y dy
??
+ - + ??
??
??
?
 
Answer (C) 
Sol. ( )
2 2 2 1
2
2
2
0 0 0 0
2 1 1 1 1
2
y
xdx x dx dy y dy
??
- - - = - - - ??
??
??
? ? ? ?
 
 + 1 + I 
 ? 
11
22
00
82
2 1 1 1
33
y dy y dy I - - = + - - +
??
 
 ? 
1
2
0
11 I y dy = - -
?
 
 ? 
1
2
0
11 I y dy
??
= - -
??
??
?
 
9. If y = y (x) is the solution of the differential equation 
( ) ( )
22
1 2 1 0
xx
dy
e y e
dx
+ + + = and ( ) 0 0, y = then 
( )
( ) ( )
2
6 0 log 3
e
yy
??
?+
??
??
 is equal to 
 (A) 2 
 (B) –2 
 (C) –4 
 (D) –1 
Answer (C) 
Sol. 
( ) ( )
22
1 2 1 0
xx
dy
e y e
dx
+ + + = 
 
22
2
11
=-
++
??
x
x
dy e
dx
ye
  
=
=
x
x
et
e dx dt
 
 
1
2
tan 2
1
dt
y
t
-
=-
+
?
 
 
( )
11
tan 2tan
x
y e c
--
=+ 
 ( ) 0
2
yc
?
?= 
 
( )
11
tan 2tan
2
x
ye
--
?
= - + 
 
( )
1
cot 2tan
-
=
x
ye 
 
( )
21
2
2
cosec 2tan
1
x
x
x
dy e
e
dx
e
-
??
=- ??
??
+
??
 
 ( )
0
2
01
2
=
-
? = = = -
x
dy
y
dx
 
 
( )
1
cot 2tan
x
ye
-
= 
 
( )
3
1 log
ln 3 cot 2tan
e
ye
-
??
=
??
??
 
 
( )
1
21
cot 2tan 3 cot cot
33
3
-
?? ??
= = = - = -
??
??
 
 ( )
( ) ( )
2
2
1
6 0 ln 3 6 1
3
yy
??
?? ??
??
? + = - + -
?? ??
??
?? ??
??
 
   
1
6 1 4
3
??
= - + = -
??
??
 
 
   
 
10. Let P : y
2
 = 4ax, a > 0 be a parabola with focus S. 
Let the tangents to the parabola P make an angle 
of 
4
?
 with the line y = 3x + 5 touch the parabola P 
at A and B. Then the value of a for which A, B and 
S are collinear is 
 (A) 8 only (B) 2 only 
 (C) 
1
4
 only (D) any a > 0  
Answer (D) 
Sol. ( )
2
: 4 , 0 , 0 =? P y ax a S a 
 Equation of tangent on parabola =+
a
y mx
m
 
     35 yx =+ 
 
( )
3
tan 3 1 3
4 1 3
m
mm
m
?-
= ? - = ? +
+
 
 
3 1 3
2
- = +
=-
mm
m
 
 
3 1 3
1
2
- = - -
=
mm
m
 
 Equation of one tangent : 2
2
= - -
a
yx 
 Equation of other tangent : 2
2
x
ya =+ 
 Point of contact are 
 
( )
( )
22
22
, and ,
1
2
21
2
2
a a a a
??
??
??
--
??
??
??
?? -
- ??
?? ??
????
????
 
 ( ) , and 4 , 4
4
a
A a B a a
??
-
??
??
 
 Now or (?ABS) = 0 [S is the focus] 
  
1
4
1
4 4 1 0
2
01
a
a
aa
a
-= 
 ? ( ) ( )
( ) ( )
2
4 0 4 1 0 4 0
4
a
a a a a a - - - - + - - = 
  
2 2 2
3 4 0 = - - + = a a a 
  Always true 
11. Let a triangle ABC be inscribed in the circle 
22
– 2( ) 0 + + = x x y y such that .
2
?
?= BAC If 
the length of side AB is 2 , then the area of the 
?ABC is equal to : 
 (A) 
( )
2 6 3 + 
 (B) 
( )
6 3 2 + 
 (C) 
( )
3 3 4 + 
 (D) 
( )
6 2 3 4 + 
Answer (Dropped) 
Sol. 
22
– 2 – 2 0 x x y y += 
 Centre 
11
,
22
??
??
??
 
 Radius = 1 
  
 BC is diameter 
 
1
ar( ) 2 2 1
2
ABC ? = ? ? = 
12. Let 
2 1 3
3 –2 –1
- + +
==
x y z
 lie on the plane px – qy + 
z = 5, for some ,. ? pq The shortest distance of 
the plane from the origin is : 
 (A) 
3
109
 
 (B) 
5
142
 
 (C) 
5
71
 
 (D) 
1
142
 
Answer (B) 
 
   
  
Sol. 
2 1 3
3 –2 –1
x y z - + +
= = = ? 
 (3 2,–2 – 1 , – – 3) ? + ? ? lies on plane px – qy + z =5 
 (3 2) – (–2 – 1) (– – 3) 5 pq ? + ? + ? = 
 (3 2 – 1) (2 – 8) 0 p q p q ? + + + = 
 
3 2 – 1 0 15
2 – 8 0 –22
p q p
p q q
+ = = ?
?
+ = =
?
 
 Equation of plane 15 22 – 5 0 x y z + + = 
 Shortest distance from origin = 
22
| 0 0 0 – 5 |
15 22 1
++
++
 
 = 
5
710
 
 = 
5
142
 
13. The distance of the origin from the centroid of the 
triangle whose two sides have the equations  
x – 2y + 1 = 0 and 2x – y – 1 = 0 and whose 
orthocenter is 
77
,
33
??
??
??
 is :   
 (A) 2 (B) 2 
 (C) 22 (D) 4 
Answer (C) 
Sol. 
: 2 1 0
(1 , 1)
: 2 – – 1 0
AB x y
A
AC x y
- + = ?
?
=
?
 
 
 Altitude from B is BH = 2 – 7 0 (3, 2) x y B + = ? 
 Altitude from C is CH = 2 – 7 0 (2, 3) x y C + = ? 
 Centroid of ?ABC = (2, 2) 2 2 = G OG 
14. Let Q be the mirror image of the point P(1, 2, 1) with 
respect to the plane x + 2y + 2z = 16. Let T be a 
plane passing through the point Q and contains the 
line 
( )
ˆ ˆ ˆ ˆ
– 2 , . = + ? + + ? ? r k i j k Then, which of 
the following points lies on T? 
 (A) (2, 1, 0) (B) (1, 2, 1) 
 (C) (1, 2, 2) (D) (1, 3, 2) 
Answer (B) 
Sol. P(1, 2, 1) image in plane x + 2y + 2z = 16 
 
( )
222
2 1 2 2 2 1 16
1 2 1
1 2 2
1 2 2
x y z
- + ? + ? -
- - -
= = =
++
 
 
1 2 1
2
1 2 2
x y z - - -
= = = 
 Q(3, 6, 5)  
 
( )
ˆ ˆ ˆ ˆ
2 r k i j k = - + ? + + 
  
 
ˆ ˆ ˆ
3 6 6 AQ i j k = + + 
 
( )
ˆ ˆ ˆ
3 2 2 i j k = + + 
 
( ) ( )
ˆ ˆ ˆ ˆ ˆ ˆ
2 2 2 n i j k i j k = + + ? + + 
 
ˆ ˆ ˆ
1 2 2
1 1 2
i j k
 
 
ˆ ˆ ˆ
20 i j k = - - 
 Equation of plane ? 2(x – 0)  + 0 (y – 0) – 1 (z + 1) 
= 0 
 2x – z = 1 
 Point lying on plane from the option is (1, 2, 1) i.e., 
option (B) 
15. Let A, B, C be three points whose position vectors 
respectively are  
 
ˆ ˆ ˆ
43 = + + a i j k 
 
ˆ ˆ ˆ
2 4 , = + ? + ?? b i j k 
 
ˆ ˆ ˆ
3 – 2 5 =+ c i j k 
 If ? is the smallest positive integer for which ,, a b c
are non collinear, then the length of the median, in 
?ABC, through A is: 
 (A) 
82
2
 (B) 
62
2
 
 (C) 
69
2
 (D) 
66
2
 
Answer (A)