JEE Main 2023 April 8 Shift 2 Paper & Solutions | Mock Tests for JEE Main and Advanced 2025 PDF Download

 Page 1


   
 
 
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
08/04/2023 
Evening 
Time : 3 hrs. M.M. : 300 
IMPORTANT INSTRUCTIONS: 
(1) The test is of 3 hours duration. 
(2) The Test Booklet consists of 90 questions. The maximum marks are 300. 
(3) There are three parts in the question paper consisting of Mathematics, Physics and Chemistry 
having 30 questions in each part of equal weightage. Each part (subject) has two sections. 
 (i) Section-A: This section contains 20 multiple choice questions which have only one correct 
answer. Each question carries 4 marks for correct answer and –1 mark for wrong answer. 
 (ii) Section-B: This section contains 10 questions. In Section-B, attempt any five questions out 
of 10. The answer to each of the questions is a numerical value. Each question carries 4 marks 
for correct answer and –1 mark for wrong answer. For Section-B, the answer should be 
rounded off to the nearest integer. 
Page 2


   
 
 
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
08/04/2023 
Evening 
Time : 3 hrs. M.M. : 300 
IMPORTANT INSTRUCTIONS: 
(1) The test is of 3 hours duration. 
(2) The Test Booklet consists of 90 questions. The maximum marks are 300. 
(3) There are three parts in the question paper consisting of Mathematics, Physics and Chemistry 
having 30 questions in each part of equal weightage. Each part (subject) has two sections. 
 (i) Section-A: This section contains 20 multiple choice questions which have only one correct 
answer. Each question carries 4 marks for correct answer and –1 mark for wrong answer. 
 (ii) Section-B: This section contains 10 questions. In Section-B, attempt any five questions out 
of 10. The answer to each of the questions is a numerical value. Each question carries 4 marks 
for correct answer and –1 mark for wrong answer. For Section-B, the answer should be 
rounded off to the nearest integer. 
 
   
  
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 
( )
1 2 sin
0, 2 : is purely imaginary .
1 sin
i
A
i
+? ??
= ? ? ?
??
-?
??
 
Then the sum of the elements in A is  
 (1) 4? (2) 3? 
 (3) ? (4) 2? 
Answer (1) 
Sol. 
1 2 sin
1 sin
i
i
+?
-?
 is purely imaginary 
 So, 
2
2
1 2 sin
0
1 sin
-?
=
+?
 
 ? 
2
1
sin
2
?= 
 ? 
357
,,,
4 4 4 4
? ? ? ?
?= , for ? ? (0, 2?) 
 ? Sum of all values = 
16
4
4
?
=? 
2. Let the vectors 
1 2 3
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
,  and u i j ak u i bj k u ci j k = + + = + + = + + 
be coplanar. If the vectors ( )
1
ˆ ˆ ˆ
, v a b i cj ck = + + + 
( )
2
ˆ ˆ ˆ
v ai b c j ak = + + + and ( )
3
ˆˆ
v bi bj c a k = + + + 
are also coplanar, then 6 (a + b + c) is equal to 
 (1) 0 (2) 4 
 (3) 12 (4) 6 
Answer (3) 
Sol. Given : 
12
ˆ ˆ ˆ ˆ ˆ ˆ
, u i j ak u i bj k = + + = + + and 
3
ˆ ˆ ˆ
u ci j k = + + 
 and 
1
ˆ ˆ ˆ
( ) , v a b i cj ck = + + + 
2
ˆ ˆ ˆ
() v ai b c j ak = + + + 
and 
3
ˆ ˆ ˆ
() v bi bj c a k = + + + 
 Now 
11
1 1 0
11
a
b
c
= ?  ( 1) (1 ) (1 ) 0 b c a bc - - - + - = 
 2 a b c abc + + = + …(i) 
 and 0
a b c c
a b c a
b b c a
+
+=
+
 
 ? 0
2 2 0
a b c c
a b c a
ac
+
+=
--
   ( )
3 3 1 2
() R R R R ? - + 
 ? abc = 0 
 ? a + b + c = 2 
 ? 6( ) 12 a b c + + = 
3. Let a
n
 be n
th
 term of the series 5 + 8 + 14 + 23 + 35 
+ 50 + …. and 
1
.
n
nk
k
Sa
=
=
?
 Then S
30
 – a
40
 is equal 
to 
 (1) 11310 (2) 11260 
 (3) 11290 (4) 11280 
Answer (3) 
Sol. S
n
 = 5 + 8 + 14 + 23 + 35 + 50 + ….. + a
n
 
 S
n
 = 5 + 8 + 14 + 23 + 35 + ….. + a
n
         
 
( –1) terms
0 5 3 6 9 12 ..... –
n
n
a = + + + + + 
 ? ? ?
–1
5 6 ( – 2)3
2
n
n
an = + + 
  
2
– 1 3 3
5 [3 ] 5 –
2 2 2
n
n n n = + = +
?
 
  
2
1
[3 – 3 10] 2
2
nn = + = 
 ? 
40
2345 a = 
 
2
11
[3 – 3 10 1 ] 3
22
n
S n n x = + =
? ? ?
 
 
30
1 3 30 31 61 30 31
– 3 10 30 13,635
2 6 2
S
? ? ? ? ??
= ? + ? =
??
??
 
 ? S
30
 – a
40
 = 11,290 
4. The value of 36(4cos
2
9° – 1)(4cos
2
27° – 1) 
(4cos
2
81° – 1)(4cos
2
243° – 1) is 
 (1) 54 (2) 18 
 (3) 27 (4) 36 
Answer (4) 
Page 3


   
 
 
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
08/04/2023 
Evening 
Time : 3 hrs. M.M. : 300 
IMPORTANT INSTRUCTIONS: 
(1) The test is of 3 hours duration. 
(2) The Test Booklet consists of 90 questions. The maximum marks are 300. 
(3) There are three parts in the question paper consisting of Mathematics, Physics and Chemistry 
having 30 questions in each part of equal weightage. Each part (subject) has two sections. 
 (i) Section-A: This section contains 20 multiple choice questions which have only one correct 
answer. Each question carries 4 marks for correct answer and –1 mark for wrong answer. 
 (ii) Section-B: This section contains 10 questions. In Section-B, attempt any five questions out 
of 10. The answer to each of the questions is a numerical value. Each question carries 4 marks 
for correct answer and –1 mark for wrong answer. For Section-B, the answer should be 
rounded off to the nearest integer. 
 
   
  
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 
( )
1 2 sin
0, 2 : is purely imaginary .
1 sin
i
A
i
+? ??
= ? ? ?
??
-?
??
 
Then the sum of the elements in A is  
 (1) 4? (2) 3? 
 (3) ? (4) 2? 
Answer (1) 
Sol. 
1 2 sin
1 sin
i
i
+?
-?
 is purely imaginary 
 So, 
2
2
1 2 sin
0
1 sin
-?
=
+?
 
 ? 
2
1
sin
2
?= 
 ? 
357
,,,
4 4 4 4
? ? ? ?
?= , for ? ? (0, 2?) 
 ? Sum of all values = 
16
4
4
?
=? 
2. Let the vectors 
1 2 3
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
,  and u i j ak u i bj k u ci j k = + + = + + = + + 
be coplanar. If the vectors ( )
1
ˆ ˆ ˆ
, v a b i cj ck = + + + 
( )
2
ˆ ˆ ˆ
v ai b c j ak = + + + and ( )
3
ˆˆ
v bi bj c a k = + + + 
are also coplanar, then 6 (a + b + c) is equal to 
 (1) 0 (2) 4 
 (3) 12 (4) 6 
Answer (3) 
Sol. Given : 
12
ˆ ˆ ˆ ˆ ˆ ˆ
, u i j ak u i bj k = + + = + + and 
3
ˆ ˆ ˆ
u ci j k = + + 
 and 
1
ˆ ˆ ˆ
( ) , v a b i cj ck = + + + 
2
ˆ ˆ ˆ
() v ai b c j ak = + + + 
and 
3
ˆ ˆ ˆ
() v bi bj c a k = + + + 
 Now 
11
1 1 0
11
a
b
c
= ?  ( 1) (1 ) (1 ) 0 b c a bc - - - + - = 
 2 a b c abc + + = + …(i) 
 and 0
a b c c
a b c a
b b c a
+
+=
+
 
 ? 0
2 2 0
a b c c
a b c a
ac
+
+=
--
   ( )
3 3 1 2
() R R R R ? - + 
 ? abc = 0 
 ? a + b + c = 2 
 ? 6( ) 12 a b c + + = 
3. Let a
n
 be n
th
 term of the series 5 + 8 + 14 + 23 + 35 
+ 50 + …. and 
1
.
n
nk
k
Sa
=
=
?
 Then S
30
 – a
40
 is equal 
to 
 (1) 11310 (2) 11260 
 (3) 11290 (4) 11280 
Answer (3) 
Sol. S
n
 = 5 + 8 + 14 + 23 + 35 + 50 + ….. + a
n
 
 S
n
 = 5 + 8 + 14 + 23 + 35 + ….. + a
n
         
 
( –1) terms
0 5 3 6 9 12 ..... –
n
n
a = + + + + + 
 ? ? ?
–1
5 6 ( – 2)3
2
n
n
an = + + 
  
2
– 1 3 3
5 [3 ] 5 –
2 2 2
n
n n n = + = +
?
 
  
2
1
[3 – 3 10] 2
2
nn = + = 
 ? 
40
2345 a = 
 
2
11
[3 – 3 10 1 ] 3
22
n
S n n x = + =
? ? ?
 
 
30
1 3 30 31 61 30 31
– 3 10 30 13,635
2 6 2
S
? ? ? ? ??
= ? + ? =
??
??
 
 ? S
30
 – a
40
 = 11,290 
4. The value of 36(4cos
2
9° – 1)(4cos
2
27° – 1) 
(4cos
2
81° – 1)(4cos
2
243° – 1) is 
 (1) 54 (2) 18 
 (3) 27 (4) 36 
Answer (4) 
 
   
  
Sol. 36(4cos
2
9° – 1) (4cos
2
27° – 1) (4cos
28
1° – 1) 
(4cos
2
243° – 1) 
 = 36[(4cos
2
9° – 1) (4sin
2
9° –1) (4cos
2
27° –1) 
(4sin
2
27° –1)] 
 = 36[(4sin
2
18° – 4 + 1) (4sin
2
54° – 4 + 1)] 
 = 36[(4sin
2
18° – 3) (4sin
2
54° – 3)] 
 
( ) ( )
22
5 – 1 5 1
36 – 3 – 3
44
?? ? ?? ?
+
?? ? ?? ?
=
? ?? ?
??
? ?? ?
??
? ?? ? ??
 
 
( ) ( )
6 – 2 5 – 12 6 2 5 – 12
36
44
??
+
??
=
??
??
 
 
( )
2
2 5 – 36
–36 36
16
??
??
==
??
??
??
 
5. Let the mean and variance of 12 observations be 
9
2
 and 4 respectively. Later on, it was observed that 
two observations were considered as 9 and 10 
instead of 7 and 14 respectively. If the correct 
variance is ,
m
n
 where m and n are coprime, then  
m + n are coprime, then m + n is equal to 
 (1) 315 (2) 316 
 (3) 314 (4) 317 
Answer (4) 
Sol. 
1 2 12
..... 9 10 ..... 9
12 2
x x x
x
+ + + + + +
== 
 x
1
 + x
2
 ….. + x
12
 + 19 = 54 
 x
1
 + x
2
 + ….. + x
12
 + 7 + 14 = 54 – 19 + 7 + 14 
 
1 2 12
..... 7 14 56
12 12
x x x + + + + +
= 
 New 
14
3
x = 
 
2 2 2 2 2
1 2 12
..... (9) (10) 9
–4
12 2
x x x + + + + +
= 
 
2 2 2 2 2
1 2 12
..... 7 14 355 x x x + + + + + = 
 New variance 
2
2
– ( )
i
x
x
N
=
?
 
   
2
355 14
–
12 3
??
=
??
??
 
   
281
36
= 
 ? m = 281 
   n = 36 
 ? m + n = 317 
6. Let A = {1, 2, 3, 4, 5, 6, 7}. Then the relation 
( ) ? ?
, : 7 = ? ? + = R x y A A x y is 
 (1) an equivalence relation 
 (2) symmetric but neither reflexive nor transitive 
 (3) transitive but neither symmetric nor reflexive 
 (4) reflexive but neither symmetric nor transitive 
Answer (2) 
Sol. x + y = 7 
 y = 7 – x 
 ( ) ? ?
1, 6 , (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) R = 
 ? ( ) ( ) ,  , a b R b a R ? ? ? 
 Relation is symmetric but not reflexive and not 
transitive 
7. Let O be the origin and OP and OQ be the tangents 
to the circle 
22
6 4 8 0 + - + + = x y x y at the points 
P and Q on it. If the circumcircle of the triangle OPQ 
passes through the point 
1
, ,
2
??
?
??
??
 then a value of ? 
is 
 (1) 
3
2
 (2) 
1
2
- 
 (3) 
5
2
 (4) 1 
Answer (3) 
Sol.  
 (0, 0) and (3, –2) are the diametric end points of the 
circumcircle 
 ? ( )( ) ( )( ) 0 3 0 2 0 x x y y - - + - + = 
   
22
3 2 0 x y x y + - + = 
Page 4


   
 
 
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
08/04/2023 
Evening 
Time : 3 hrs. M.M. : 300 
IMPORTANT INSTRUCTIONS: 
(1) The test is of 3 hours duration. 
(2) The Test Booklet consists of 90 questions. The maximum marks are 300. 
(3) There are three parts in the question paper consisting of Mathematics, Physics and Chemistry 
having 30 questions in each part of equal weightage. Each part (subject) has two sections. 
 (i) Section-A: This section contains 20 multiple choice questions which have only one correct 
answer. Each question carries 4 marks for correct answer and –1 mark for wrong answer. 
 (ii) Section-B: This section contains 10 questions. In Section-B, attempt any five questions out 
of 10. The answer to each of the questions is a numerical value. Each question carries 4 marks 
for correct answer and –1 mark for wrong answer. For Section-B, the answer should be 
rounded off to the nearest integer. 
 
   
  
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 
( )
1 2 sin
0, 2 : is purely imaginary .
1 sin
i
A
i
+? ??
= ? ? ?
??
-?
??
 
Then the sum of the elements in A is  
 (1) 4? (2) 3? 
 (3) ? (4) 2? 
Answer (1) 
Sol. 
1 2 sin
1 sin
i
i
+?
-?
 is purely imaginary 
 So, 
2
2
1 2 sin
0
1 sin
-?
=
+?
 
 ? 
2
1
sin
2
?= 
 ? 
357
,,,
4 4 4 4
? ? ? ?
?= , for ? ? (0, 2?) 
 ? Sum of all values = 
16
4
4
?
=? 
2. Let the vectors 
1 2 3
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
,  and u i j ak u i bj k u ci j k = + + = + + = + + 
be coplanar. If the vectors ( )
1
ˆ ˆ ˆ
, v a b i cj ck = + + + 
( )
2
ˆ ˆ ˆ
v ai b c j ak = + + + and ( )
3
ˆˆ
v bi bj c a k = + + + 
are also coplanar, then 6 (a + b + c) is equal to 
 (1) 0 (2) 4 
 (3) 12 (4) 6 
Answer (3) 
Sol. Given : 
12
ˆ ˆ ˆ ˆ ˆ ˆ
, u i j ak u i bj k = + + = + + and 
3
ˆ ˆ ˆ
u ci j k = + + 
 and 
1
ˆ ˆ ˆ
( ) , v a b i cj ck = + + + 
2
ˆ ˆ ˆ
() v ai b c j ak = + + + 
and 
3
ˆ ˆ ˆ
() v bi bj c a k = + + + 
 Now 
11
1 1 0
11
a
b
c
= ?  ( 1) (1 ) (1 ) 0 b c a bc - - - + - = 
 2 a b c abc + + = + …(i) 
 and 0
a b c c
a b c a
b b c a
+
+=
+
 
 ? 0
2 2 0
a b c c
a b c a
ac
+
+=
--
   ( )
3 3 1 2
() R R R R ? - + 
 ? abc = 0 
 ? a + b + c = 2 
 ? 6( ) 12 a b c + + = 
3. Let a
n
 be n
th
 term of the series 5 + 8 + 14 + 23 + 35 
+ 50 + …. and 
1
.
n
nk
k
Sa
=
=
?
 Then S
30
 – a
40
 is equal 
to 
 (1) 11310 (2) 11260 
 (3) 11290 (4) 11280 
Answer (3) 
Sol. S
n
 = 5 + 8 + 14 + 23 + 35 + 50 + ….. + a
n
 
 S
n
 = 5 + 8 + 14 + 23 + 35 + ….. + a
n
         
 
( –1) terms
0 5 3 6 9 12 ..... –
n
n
a = + + + + + 
 ? ? ?
–1
5 6 ( – 2)3
2
n
n
an = + + 
  
2
– 1 3 3
5 [3 ] 5 –
2 2 2
n
n n n = + = +
?
 
  
2
1
[3 – 3 10] 2
2
nn = + = 
 ? 
40
2345 a = 
 
2
11
[3 – 3 10 1 ] 3
22
n
S n n x = + =
? ? ?
 
 
30
1 3 30 31 61 30 31
– 3 10 30 13,635
2 6 2
S
? ? ? ? ??
= ? + ? =
??
??
 
 ? S
30
 – a
40
 = 11,290 
4. The value of 36(4cos
2
9° – 1)(4cos
2
27° – 1) 
(4cos
2
81° – 1)(4cos
2
243° – 1) is 
 (1) 54 (2) 18 
 (3) 27 (4) 36 
Answer (4) 
 
   
  
Sol. 36(4cos
2
9° – 1) (4cos
2
27° – 1) (4cos
28
1° – 1) 
(4cos
2
243° – 1) 
 = 36[(4cos
2
9° – 1) (4sin
2
9° –1) (4cos
2
27° –1) 
(4sin
2
27° –1)] 
 = 36[(4sin
2
18° – 4 + 1) (4sin
2
54° – 4 + 1)] 
 = 36[(4sin
2
18° – 3) (4sin
2
54° – 3)] 
 
( ) ( )
22
5 – 1 5 1
36 – 3 – 3
44
?? ? ?? ?
+
?? ? ?? ?
=
? ?? ?
??
? ?? ?
??
? ?? ? ??
 
 
( ) ( )
6 – 2 5 – 12 6 2 5 – 12
36
44
??
+
??
=
??
??
 
 
( )
2
2 5 – 36
–36 36
16
??
??
==
??
??
??
 
5. Let the mean and variance of 12 observations be 
9
2
 and 4 respectively. Later on, it was observed that 
two observations were considered as 9 and 10 
instead of 7 and 14 respectively. If the correct 
variance is ,
m
n
 where m and n are coprime, then  
m + n are coprime, then m + n is equal to 
 (1) 315 (2) 316 
 (3) 314 (4) 317 
Answer (4) 
Sol. 
1 2 12
..... 9 10 ..... 9
12 2
x x x
x
+ + + + + +
== 
 x
1
 + x
2
 ….. + x
12
 + 19 = 54 
 x
1
 + x
2
 + ….. + x
12
 + 7 + 14 = 54 – 19 + 7 + 14 
 
1 2 12
..... 7 14 56
12 12
x x x + + + + +
= 
 New 
14
3
x = 
 
2 2 2 2 2
1 2 12
..... (9) (10) 9
–4
12 2
x x x + + + + +
= 
 
2 2 2 2 2
1 2 12
..... 7 14 355 x x x + + + + + = 
 New variance 
2
2
– ( )
i
x
x
N
=
?
 
   
2
355 14
–
12 3
??
=
??
??
 
   
281
36
= 
 ? m = 281 
   n = 36 
 ? m + n = 317 
6. Let A = {1, 2, 3, 4, 5, 6, 7}. Then the relation 
( ) ? ?
, : 7 = ? ? + = R x y A A x y is 
 (1) an equivalence relation 
 (2) symmetric but neither reflexive nor transitive 
 (3) transitive but neither symmetric nor reflexive 
 (4) reflexive but neither symmetric nor transitive 
Answer (2) 
Sol. x + y = 7 
 y = 7 – x 
 ( ) ? ?
1, 6 , (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) R = 
 ? ( ) ( ) ,  , a b R b a R ? ? ? 
 Relation is symmetric but not reflexive and not 
transitive 
7. Let O be the origin and OP and OQ be the tangents 
to the circle 
22
6 4 8 0 + - + + = x y x y at the points 
P and Q on it. If the circumcircle of the triangle OPQ 
passes through the point 
1
, ,
2
??
?
??
??
 then a value of ? 
is 
 (1) 
3
2
 (2) 
1
2
- 
 (3) 
5
2
 (4) 1 
Answer (3) 
Sol.  
 (0, 0) and (3, –2) are the diametric end points of the 
circumcircle 
 ? ( )( ) ( )( ) 0 3 0 2 0 x x y y - - + - + = 
   
22
3 2 0 x y x y + - + = 
 
   
  
 Put 
1
, 
2
??
?
??
??
 in the above equation 
 
2
1
3 1 0
4
? + - ? + = 
 
2
5
30
4
? - ? + = 
 
2
4 12 5 0 ? - ? + = 
 
2
4 10 2 5 0 ? - ? - ? + = 
 ( ) ( ) 2 2 5 1 2 5 0 ? ? - - ? - = 
 
15
, 
22
?=  
8. Let A(0, 1), B(1, 1) and C(1, 0) be the mid-points of 
the sides of a triangle with incentre at the point D. If 
the focus of the parabola y
2
 = 4ax passing through 
D is 
( )
2, 0 , ? + ? where ? and ? are rational 
numbers, then 
2
?
?
 is equal to 
 (1) 8 (2) 12 
 (3) 6 (4) 
9
2
 
Answer (1) 
Sol. Mid-point is (0, 1), (1, 0) and (1, 1) 
 In center (I) = 
44
, 
4 2 2 4 2 2
??
??
++ ??
 
 
2
4 y ax = 
 
2
4 4 1
4  
4 2 2 4 2 2 4 2 2
aa
? ? ? ?
= ? =
? ? ? ?
+ + + ? ? ? ?
 
 Focus (a, 0) 
  
1
, 0
4 2 2
??
=
??
+??
 
  
4 2 2
, 0
8
??
-
=
??
??
??
 
 
4 1 1
, 
8 2 4
? = = ? = - 
 
22
1
16
2
8
2
1
4
?
= = =
?
- ??
??
??
 
9. The absolute difference of the coefficients of x
10
 
and x
7
 in the expansion of 
11
2
1
2
2
??
+
??
??
x
x
 is equal to 
 (1) 13
3
 – 13 (2) 11
3
 – 11 
 (3) 10
3
 – 10 (4) 12
3
 – 12 
Answer (4) 
Sol. 
( )
( )
11
11 2
1
22
r
r
rr
T C x x
-
-
+
= 
 For coefficient of x
7
  
 ? 22 – 2r – r = 7 
 ? r = 5 
 For coefficient of x
10
  
 22 – 3r = 10 
 ? r = 4 
 Absolute difference = 
67
11 11
54
54
22
22
CC ? - ? 
   = 924 2640 1716 -= 
   = 12
3
 – 12 
 Option (4) is correct 
10. For a, b ? Z and |a – b| ? 10, let the angle between 
the plane P : ax + y – z = b and the line l : x – 1 =  
a – y = z + 1 be cos
–1
 
1
.
3
??
??
??
 If the distance of the 
point (6, –6, 4) from the plane P is 3 6, then  
a
4
 + b
2
 is equal to 
 (1) 32 (2) 85 
 (3) 25 (4) 48 
Answer (1) 
Sol. 
11
1 2 2
cos sin
33
--
??
??
? = =
??
??
??
??
??
 
 ( )
( ) ( )( )
2
1 1 1 1
22
cos 90
3
23
a
a
+ - + -
? - ? = =
+?
 
 ? 
2
2 2 2
3
2
a
a
-
=
+
 
 After solving we get 
2
, 2
5
aa = - = - 
 ? a = –2 ( ? a ? Z) 
Page 5


   
 
 
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
08/04/2023 
Evening 
Time : 3 hrs. M.M. : 300 
IMPORTANT INSTRUCTIONS: 
(1) The test is of 3 hours duration. 
(2) The Test Booklet consists of 90 questions. The maximum marks are 300. 
(3) There are three parts in the question paper consisting of Mathematics, Physics and Chemistry 
having 30 questions in each part of equal weightage. Each part (subject) has two sections. 
 (i) Section-A: This section contains 20 multiple choice questions which have only one correct 
answer. Each question carries 4 marks for correct answer and –1 mark for wrong answer. 
 (ii) Section-B: This section contains 10 questions. In Section-B, attempt any five questions out 
of 10. The answer to each of the questions is a numerical value. Each question carries 4 marks 
for correct answer and –1 mark for wrong answer. For Section-B, the answer should be 
rounded off to the nearest integer. 
 
   
  
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 
( )
1 2 sin
0, 2 : is purely imaginary .
1 sin
i
A
i
+? ??
= ? ? ?
??
-?
??
 
Then the sum of the elements in A is  
 (1) 4? (2) 3? 
 (3) ? (4) 2? 
Answer (1) 
Sol. 
1 2 sin
1 sin
i
i
+?
-?
 is purely imaginary 
 So, 
2
2
1 2 sin
0
1 sin
-?
=
+?
 
 ? 
2
1
sin
2
?= 
 ? 
357
,,,
4 4 4 4
? ? ? ?
?= , for ? ? (0, 2?) 
 ? Sum of all values = 
16
4
4
?
=? 
2. Let the vectors 
1 2 3
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ
,  and u i j ak u i bj k u ci j k = + + = + + = + + 
be coplanar. If the vectors ( )
1
ˆ ˆ ˆ
, v a b i cj ck = + + + 
( )
2
ˆ ˆ ˆ
v ai b c j ak = + + + and ( )
3
ˆˆ
v bi bj c a k = + + + 
are also coplanar, then 6 (a + b + c) is equal to 
 (1) 0 (2) 4 
 (3) 12 (4) 6 
Answer (3) 
Sol. Given : 
12
ˆ ˆ ˆ ˆ ˆ ˆ
, u i j ak u i bj k = + + = + + and 
3
ˆ ˆ ˆ
u ci j k = + + 
 and 
1
ˆ ˆ ˆ
( ) , v a b i cj ck = + + + 
2
ˆ ˆ ˆ
() v ai b c j ak = + + + 
and 
3
ˆ ˆ ˆ
() v bi bj c a k = + + + 
 Now 
11
1 1 0
11
a
b
c
= ?  ( 1) (1 ) (1 ) 0 b c a bc - - - + - = 
 2 a b c abc + + = + …(i) 
 and 0
a b c c
a b c a
b b c a
+
+=
+
 
 ? 0
2 2 0
a b c c
a b c a
ac
+
+=
--
   ( )
3 3 1 2
() R R R R ? - + 
 ? abc = 0 
 ? a + b + c = 2 
 ? 6( ) 12 a b c + + = 
3. Let a
n
 be n
th
 term of the series 5 + 8 + 14 + 23 + 35 
+ 50 + …. and 
1
.
n
nk
k
Sa
=
=
?
 Then S
30
 – a
40
 is equal 
to 
 (1) 11310 (2) 11260 
 (3) 11290 (4) 11280 
Answer (3) 
Sol. S
n
 = 5 + 8 + 14 + 23 + 35 + 50 + ….. + a
n
 
 S
n
 = 5 + 8 + 14 + 23 + 35 + ….. + a
n
         
 
( –1) terms
0 5 3 6 9 12 ..... –
n
n
a = + + + + + 
 ? ? ?
–1
5 6 ( – 2)3
2
n
n
an = + + 
  
2
– 1 3 3
5 [3 ] 5 –
2 2 2
n
n n n = + = +
?
 
  
2
1
[3 – 3 10] 2
2
nn = + = 
 ? 
40
2345 a = 
 
2
11
[3 – 3 10 1 ] 3
22
n
S n n x = + =
? ? ?
 
 
30
1 3 30 31 61 30 31
– 3 10 30 13,635
2 6 2
S
? ? ? ? ??
= ? + ? =
??
??
 
 ? S
30
 – a
40
 = 11,290 
4. The value of 36(4cos
2
9° – 1)(4cos
2
27° – 1) 
(4cos
2
81° – 1)(4cos
2
243° – 1) is 
 (1) 54 (2) 18 
 (3) 27 (4) 36 
Answer (4) 
 
   
  
Sol. 36(4cos
2
9° – 1) (4cos
2
27° – 1) (4cos
28
1° – 1) 
(4cos
2
243° – 1) 
 = 36[(4cos
2
9° – 1) (4sin
2
9° –1) (4cos
2
27° –1) 
(4sin
2
27° –1)] 
 = 36[(4sin
2
18° – 4 + 1) (4sin
2
54° – 4 + 1)] 
 = 36[(4sin
2
18° – 3) (4sin
2
54° – 3)] 
 
( ) ( )
22
5 – 1 5 1
36 – 3 – 3
44
?? ? ?? ?
+
?? ? ?? ?
=
? ?? ?
??
? ?? ?
??
? ?? ? ??
 
 
( ) ( )
6 – 2 5 – 12 6 2 5 – 12
36
44
??
+
??
=
??
??
 
 
( )
2
2 5 – 36
–36 36
16
??
??
==
??
??
??
 
5. Let the mean and variance of 12 observations be 
9
2
 and 4 respectively. Later on, it was observed that 
two observations were considered as 9 and 10 
instead of 7 and 14 respectively. If the correct 
variance is ,
m
n
 where m and n are coprime, then  
m + n are coprime, then m + n is equal to 
 (1) 315 (2) 316 
 (3) 314 (4) 317 
Answer (4) 
Sol. 
1 2 12
..... 9 10 ..... 9
12 2
x x x
x
+ + + + + +
== 
 x
1
 + x
2
 ….. + x
12
 + 19 = 54 
 x
1
 + x
2
 + ….. + x
12
 + 7 + 14 = 54 – 19 + 7 + 14 
 
1 2 12
..... 7 14 56
12 12
x x x + + + + +
= 
 New 
14
3
x = 
 
2 2 2 2 2
1 2 12
..... (9) (10) 9
–4
12 2
x x x + + + + +
= 
 
2 2 2 2 2
1 2 12
..... 7 14 355 x x x + + + + + = 
 New variance 
2
2
– ( )
i
x
x
N
=
?
 
   
2
355 14
–
12 3
??
=
??
??
 
   
281
36
= 
 ? m = 281 
   n = 36 
 ? m + n = 317 
6. Let A = {1, 2, 3, 4, 5, 6, 7}. Then the relation 
( ) ? ?
, : 7 = ? ? + = R x y A A x y is 
 (1) an equivalence relation 
 (2) symmetric but neither reflexive nor transitive 
 (3) transitive but neither symmetric nor reflexive 
 (4) reflexive but neither symmetric nor transitive 
Answer (2) 
Sol. x + y = 7 
 y = 7 – x 
 ( ) ? ?
1, 6 , (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) R = 
 ? ( ) ( ) ,  , a b R b a R ? ? ? 
 Relation is symmetric but not reflexive and not 
transitive 
7. Let O be the origin and OP and OQ be the tangents 
to the circle 
22
6 4 8 0 + - + + = x y x y at the points 
P and Q on it. If the circumcircle of the triangle OPQ 
passes through the point 
1
, ,
2
??
?
??
??
 then a value of ? 
is 
 (1) 
3
2
 (2) 
1
2
- 
 (3) 
5
2
 (4) 1 
Answer (3) 
Sol.  
 (0, 0) and (3, –2) are the diametric end points of the 
circumcircle 
 ? ( )( ) ( )( ) 0 3 0 2 0 x x y y - - + - + = 
   
22
3 2 0 x y x y + - + = 
 
   
  
 Put 
1
, 
2
??
?
??
??
 in the above equation 
 
2
1
3 1 0
4
? + - ? + = 
 
2
5
30
4
? - ? + = 
 
2
4 12 5 0 ? - ? + = 
 
2
4 10 2 5 0 ? - ? - ? + = 
 ( ) ( ) 2 2 5 1 2 5 0 ? ? - - ? - = 
 
15
, 
22
?=  
8. Let A(0, 1), B(1, 1) and C(1, 0) be the mid-points of 
the sides of a triangle with incentre at the point D. If 
the focus of the parabola y
2
 = 4ax passing through 
D is 
( )
2, 0 , ? + ? where ? and ? are rational 
numbers, then 
2
?
?
 is equal to 
 (1) 8 (2) 12 
 (3) 6 (4) 
9
2
 
Answer (1) 
Sol. Mid-point is (0, 1), (1, 0) and (1, 1) 
 In center (I) = 
44
, 
4 2 2 4 2 2
??
??
++ ??
 
 
2
4 y ax = 
 
2
4 4 1
4  
4 2 2 4 2 2 4 2 2
aa
? ? ? ?
= ? =
? ? ? ?
+ + + ? ? ? ?
 
 Focus (a, 0) 
  
1
, 0
4 2 2
??
=
??
+??
 
  
4 2 2
, 0
8
??
-
=
??
??
??
 
 
4 1 1
, 
8 2 4
? = = ? = - 
 
22
1
16
2
8
2
1
4
?
= = =
?
- ??
??
??
 
9. The absolute difference of the coefficients of x
10
 
and x
7
 in the expansion of 
11
2
1
2
2
??
+
??
??
x
x
 is equal to 
 (1) 13
3
 – 13 (2) 11
3
 – 11 
 (3) 10
3
 – 10 (4) 12
3
 – 12 
Answer (4) 
Sol. 
( )
( )
11
11 2
1
22
r
r
rr
T C x x
-
-
+
= 
 For coefficient of x
7
  
 ? 22 – 2r – r = 7 
 ? r = 5 
 For coefficient of x
10
  
 22 – 3r = 10 
 ? r = 4 
 Absolute difference = 
67
11 11
54
54
22
22
CC ? - ? 
   = 924 2640 1716 -= 
   = 12
3
 – 12 
 Option (4) is correct 
10. For a, b ? Z and |a – b| ? 10, let the angle between 
the plane P : ax + y – z = b and the line l : x – 1 =  
a – y = z + 1 be cos
–1
 
1
.
3
??
??
??
 If the distance of the 
point (6, –6, 4) from the plane P is 3 6, then  
a
4
 + b
2
 is equal to 
 (1) 32 (2) 85 
 (3) 25 (4) 48 
Answer (1) 
Sol. 
11
1 2 2
cos sin
33
--
??
??
? = =
??
??
??
??
??
 
 ( )
( ) ( )( )
2
1 1 1 1
22
cos 90
3
23
a
a
+ - + -
? - ? = =
+?
 
 ? 
2
2 2 2
3
2
a
a
-
=
+
 
 After solving we get 
2
, 2
5
aa = - = - 
 ? a = –2 ( ? a ? Z) 
 
   
  
 Distance = 36 
 
2
6 6 4
36
2
ab
a
- - -
=
+
 
 22 18 b = + = 
 ? b = –4, –18 –22 = –40 
 ? b = –4 ( ? |a – b| ? 10) 
 ? a
4
 + b
2
 = 16 + 16 = 32 
 Option (1) is correct 
11. The integral 
??
? ? ? ?
?? +
? ? ? ?
? ? ? ? ??
? 2
2
log
2
xx
x
x dx
x
 is equal to 
 (1) 
? ? ? ?
++
? ? ? ?
? ? ? ?
2
2
xx
x
C
x
 (2) 
? ? ? ?
+
? ? ? ?
? ? ? ?
2
–
2
xx
x
C
x
 
 (3) 
2
log
22
x
xx
C
? ? ? ?
+
? ? ? ?
? ? ? ?
 (4) 
2
2
log
2
x
x
C
x
? ? ? ?
+
? ? ? ?
? ? ? ?
 
Answer (*) 
Sol. 
2
2
log
2
xx
x
I x dx
x
??
? ? ? ?
?? =+
? ? ? ?
? ? ? ? ??
?
 
 Let 
2
x
x
t
??
=
??
??
 
 ? 
22
log log
2
x
xt
??
=
??
??
 
 ? 
22
log log log .log
2
ee
x
x e t e
??
?=
??
??
 
 On differentiating both sides we get : 
 
2 1 1
log
22
e
x dt
x
x t dx
??
+ ? ? =
??
??
 
 ? 
1
log 1
2
e
x dt
t dx
??
+=
??
??
 
 Then solution is not possible. 
12. If the number of words, with or without meaning, 
which can be made using all the letters of the word 
MATHEMATICS in which C and S do not come 
together, is (6!)k, then k is equal to 
 (1) 2835 (2) 5670 
 (3) 1890 (4) 945 
Answer (2) 
Sol. Total words = 
11!
2!2!2!
 
 C, S comes together in = 
10!
2!
2!2!2!
? 
 ? required number of words 
  = 
32
11! 10!
–
(2!) (2!)
 
  = 
2
10! 11
–1
2
(2!)
??
??
??
 
  = 
10!
9
8
? 
  = 
10 9 8 7 9
6!
8
????
? 
  = 6! (5670) 
 ? k = 5670  
13. The negation of ( ) ( ) ( ) ?? ~~ p q p is equivalent to 
 (1) ( ) ? ~ pq (2) ? pq 
 (3) ( ) ( ) ?? ~ p q p (4) ( ) ( ) ?? ~ p q p 
Answer (2) 
Sol. ( ) ~ (~ ) (~ ) p q p ?? ??
??
 
 ( ) ~ ~ ( ) p q p ?? ??
??
 
 ( ) ~ ( ) p q p ?? 
 = ( ) ~ ( ) p p q p ? ? ? 
 = () C q p ?? 
 = () pq ? 
14. If ? ? ? ? 0 are the roots of the equation 
+ + =
2
10 ax bx , and  
 
( )
( )
1
2
2
2
1
1 cos 1 1 1
lim –
21
x
x bx a
k
x
?
?
??
- + + ??
?? =
??
??
??
?? -? ??
, then k is 
equal to 
 (1) ? 2 (2) ? 
 (3) ? 2 (4) ? 
Answer (3)