JEE Main 2023 April 11 Shift 1 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) 
 
 
  
11/04/2023 
Morning 
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) 
 
 
  
11/04/2023 
Morning 
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. The number of elements in the set 
? ?
? ?
4 2 6
0,2 : 3cos 5cos 2sin 2 0 = ? ? ? ? - ? - ? + = S
is 
 (1) 10 (2) 8 
 (3) 12 (4) 9 
Answer (4) 
Sol. 3cos
4
? – 5cos
2
? – 2sin
6
? + 2 = 0  
 ? cos
2
?[3cos
2
? – 5] – 2sin
6
?
 
+ 2 = 0 
 ? (1 – sin
2
?) (3 – 3sin
2
? – 5) – 2sin
6
?
 
+ 2 = 0 
 ? (sin
2
? – 1) (3sin
2
? + 2) – 2sin
6
?
 
+ 2 = 0 
 Let sin
2
? = t 
 (t – 1) (3t + 2) – 2t
3
 + 2 = 0 
 (t – 1) [3t + 2 – 2 (t
2
 + t + 1)] = 0 
 (t – 1) [2t
2
 – t] = 0 
 
1
0, 1 , 
2
t = 
 ? sin
2
? = 0 ? 3 solution   
  sin
2
? = 1 ? 2 solution   
  sin
2
? = 
1
2
 ? 4 solution  
 ? Total solution = 9 
2. Let ? ? : 2, 4 ? f be a differentiable function such 
that 
( ) ( ) ( ) ( ) ( ) log log 1 ,   [2, 4] ? + + ? ?
ee
x x f x x f x f x x 
with ( )
1
2
2
= f and ( )
1
4
2
= f . 
 Consider the following two statements : 
 (A) ( ) 1 , for all [2, 4] ?? f x x 
 (B) ( )
1
, for all [2, 4]
8
?? f x x 
 Then, 
 (1) Neither statement (A) nor statement (B) is true 
 (2) Only statement (B) is true 
 (3) Both the statements (A) and (B) are true 
 (4) Only statement (A) is true 
Answer (3*) 
Sol. f : [2, 4] ? ? 
 (x log e x) f?(x) + (log e x) f(x) + f(x) ? 1, x ? [2, 4] 
 ? d [x lnx f(x) – x] ? 0 OR d (x lnx ? f(x)) ? 1 
 ? h(x) = x lnx f(x) – x ? 
 ? h(x) ? h(2) , x ? [2, 4] 
 x lnx f(x) – x ? 2 ln2f(2) – 2  
 ? x lnx f(x) – x ? ln2 – 2, x lnx f(x) – x ? ln4 – 4 
 So, 
 
ln2 2 1 ln4 4 1
()
ln ln ln ln
fx
x x x x x x
--
+ ? ? + 
 f(x) ? 1 
 & 
1
()
8
fx ? 
 Hence both A & B are correct. 
 But LMVT on f(x) ? x lnx can’t be satisfied. Hence no 
such f(x) exist. 
3. Let R be a rectangle given by the lines x = 0, x = 2, 
y = 0 and y = 5. Let A(?, 0) and B(0, ?), ? ? [0, 2] 
and ? ? [0, 5], be such that the line segment AB 
divides the area of the rectangle R in the ratio 4 : 1. 
Then, the mid-point of AB lies on a  
 (1) straight line (2) parabola 
 (3) hyperbola (4) circle 
Answer (3) 
Sol.
 
 
Page 3


   
 
  
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
11/04/2023 
Morning 
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. The number of elements in the set 
? ?
? ?
4 2 6
0,2 : 3cos 5cos 2sin 2 0 = ? ? ? ? - ? - ? + = S
is 
 (1) 10 (2) 8 
 (3) 12 (4) 9 
Answer (4) 
Sol. 3cos
4
? – 5cos
2
? – 2sin
6
? + 2 = 0  
 ? cos
2
?[3cos
2
? – 5] – 2sin
6
?
 
+ 2 = 0 
 ? (1 – sin
2
?) (3 – 3sin
2
? – 5) – 2sin
6
?
 
+ 2 = 0 
 ? (sin
2
? – 1) (3sin
2
? + 2) – 2sin
6
?
 
+ 2 = 0 
 Let sin
2
? = t 
 (t – 1) (3t + 2) – 2t
3
 + 2 = 0 
 (t – 1) [3t + 2 – 2 (t
2
 + t + 1)] = 0 
 (t – 1) [2t
2
 – t] = 0 
 
1
0, 1 , 
2
t = 
 ? sin
2
? = 0 ? 3 solution   
  sin
2
? = 1 ? 2 solution   
  sin
2
? = 
1
2
 ? 4 solution  
 ? Total solution = 9 
2. Let ? ? : 2, 4 ? f be a differentiable function such 
that 
( ) ( ) ( ) ( ) ( ) log log 1 ,   [2, 4] ? + + ? ?
ee
x x f x x f x f x x 
with ( )
1
2
2
= f and ( )
1
4
2
= f . 
 Consider the following two statements : 
 (A) ( ) 1 , for all [2, 4] ?? f x x 
 (B) ( )
1
, for all [2, 4]
8
?? f x x 
 Then, 
 (1) Neither statement (A) nor statement (B) is true 
 (2) Only statement (B) is true 
 (3) Both the statements (A) and (B) are true 
 (4) Only statement (A) is true 
Answer (3*) 
Sol. f : [2, 4] ? ? 
 (x log e x) f?(x) + (log e x) f(x) + f(x) ? 1, x ? [2, 4] 
 ? d [x lnx f(x) – x] ? 0 OR d (x lnx ? f(x)) ? 1 
 ? h(x) = x lnx f(x) – x ? 
 ? h(x) ? h(2) , x ? [2, 4] 
 x lnx f(x) – x ? 2 ln2f(2) – 2  
 ? x lnx f(x) – x ? ln2 – 2, x lnx f(x) – x ? ln4 – 4 
 So, 
 
ln2 2 1 ln4 4 1
()
ln ln ln ln
fx
x x x x x x
--
+ ? ? + 
 f(x) ? 1 
 & 
1
()
8
fx ? 
 Hence both A & B are correct. 
 But LMVT on f(x) ? x lnx can’t be satisfied. Hence no 
such f(x) exist. 
3. Let R be a rectangle given by the lines x = 0, x = 2, 
y = 0 and y = 5. Let A(?, 0) and B(0, ?), ? ? [0, 2] 
and ? ? [0, 5], be such that the line segment AB 
divides the area of the rectangle R in the ratio 4 : 1. 
Then, the mid-point of AB lies on a  
 (1) straight line (2) parabola 
 (3) hyperbola (4) circle 
Answer (3) 
Sol.
 
 
 
   
   
 
1
10
4
2
20 4
1
1
2
- ??
= ? - ?? = ??
??
 
   4 ? ?? = 
 Let , 
22
k
h
?
= ? = 
 ? 4hk = 4 
 ? 1 xy = 
4. Let 
? ?
? ? ? ? , 0, 1, 2 , 1  ,  2 ?? = = ? ? ?
??
ij ij
S M a a i j
be a sample space and ? ? : is invertible ? A M S M 
be an even. Then P(A) is equal to 
 (1) 
16
27
 (2) 
47
81
 
 (3) 
49
81
 (4) 
50
81
 
Answer (4) 
Sol. If M is invertible, then |M| ? 0 
 For |M| = 0 
 1. 
1 1 0 0 2 2
  or   or  
1 1 0 0 2 2
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? Total matrix = 3 
 2.  Two 1’s and Two 0’s ? Total matrix = 4 
 3.  Two 2’s and Two 0’s ? Total matrix = 4 
 4.  Two 1’s and Two 2’s ? Total matrix = 4 
 5.  One 1 and three 0’s ? Total matrix = 4 
 6.  One 2 and Three 0’s ? Total matrix = 4 
 7.  One 1 and one 2 and two 0’s ? Total matrix = 8 
  ( )
31 50
1
81 81
PA = - = 
5. The number of integral solution x of 
2
7
2
7
log 0
23
??
+
??
??
- ??
?
??
-
?? x
x
x
 is  
 (1) 7 
 (2) 8 
 (3) 6 
 (4) 5 
Answer (3) 
Sol. 
2
7
2
7
log 0
23 x
x
x
??
+
??
??
- ??
?
??
-
??
  
 Domain  
 
7
0
2
x+? 
 
–7
2
x ? 
 
7
1
2
x+? 
 
5
2
x
-
? 
 
7
0
23
x
x
-
?
-
 
 x ? 7 
 
3
2
x ? 
 Domain : 
7 5 3
, , 0, 
2 2 2
- ? ? ? ?
- ? -
??
??
? ? ? ?
 
 Case I : 
7
01
2
x ? + ? 
 
75
22
x - ? ? - 
 
2
7
1
23
x
x
- ??
?
??
-
??
 
 
7
11
23
x
x
-
- ? ?
-
 
 
7 2 3
0
23
xx
x
- + -
?
-
 
 
3 10
0
23
x
x
-
?
-
 
  
 
7 2 3
0
23
xx
x
- - +
?
-
 
 
4
0
23
x
x
--
?
-
 
 
4
0
23
x
x
+
?
-
 
Page 4


   
 
  
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
11/04/2023 
Morning 
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. The number of elements in the set 
? ?
? ?
4 2 6
0,2 : 3cos 5cos 2sin 2 0 = ? ? ? ? - ? - ? + = S
is 
 (1) 10 (2) 8 
 (3) 12 (4) 9 
Answer (4) 
Sol. 3cos
4
? – 5cos
2
? – 2sin
6
? + 2 = 0  
 ? cos
2
?[3cos
2
? – 5] – 2sin
6
?
 
+ 2 = 0 
 ? (1 – sin
2
?) (3 – 3sin
2
? – 5) – 2sin
6
?
 
+ 2 = 0 
 ? (sin
2
? – 1) (3sin
2
? + 2) – 2sin
6
?
 
+ 2 = 0 
 Let sin
2
? = t 
 (t – 1) (3t + 2) – 2t
3
 + 2 = 0 
 (t – 1) [3t + 2 – 2 (t
2
 + t + 1)] = 0 
 (t – 1) [2t
2
 – t] = 0 
 
1
0, 1 , 
2
t = 
 ? sin
2
? = 0 ? 3 solution   
  sin
2
? = 1 ? 2 solution   
  sin
2
? = 
1
2
 ? 4 solution  
 ? Total solution = 9 
2. Let ? ? : 2, 4 ? f be a differentiable function such 
that 
( ) ( ) ( ) ( ) ( ) log log 1 ,   [2, 4] ? + + ? ?
ee
x x f x x f x f x x 
with ( )
1
2
2
= f and ( )
1
4
2
= f . 
 Consider the following two statements : 
 (A) ( ) 1 , for all [2, 4] ?? f x x 
 (B) ( )
1
, for all [2, 4]
8
?? f x x 
 Then, 
 (1) Neither statement (A) nor statement (B) is true 
 (2) Only statement (B) is true 
 (3) Both the statements (A) and (B) are true 
 (4) Only statement (A) is true 
Answer (3*) 
Sol. f : [2, 4] ? ? 
 (x log e x) f?(x) + (log e x) f(x) + f(x) ? 1, x ? [2, 4] 
 ? d [x lnx f(x) – x] ? 0 OR d (x lnx ? f(x)) ? 1 
 ? h(x) = x lnx f(x) – x ? 
 ? h(x) ? h(2) , x ? [2, 4] 
 x lnx f(x) – x ? 2 ln2f(2) – 2  
 ? x lnx f(x) – x ? ln2 – 2, x lnx f(x) – x ? ln4 – 4 
 So, 
 
ln2 2 1 ln4 4 1
()
ln ln ln ln
fx
x x x x x x
--
+ ? ? + 
 f(x) ? 1 
 & 
1
()
8
fx ? 
 Hence both A & B are correct. 
 But LMVT on f(x) ? x lnx can’t be satisfied. Hence no 
such f(x) exist. 
3. Let R be a rectangle given by the lines x = 0, x = 2, 
y = 0 and y = 5. Let A(?, 0) and B(0, ?), ? ? [0, 2] 
and ? ? [0, 5], be such that the line segment AB 
divides the area of the rectangle R in the ratio 4 : 1. 
Then, the mid-point of AB lies on a  
 (1) straight line (2) parabola 
 (3) hyperbola (4) circle 
Answer (3) 
Sol.
 
 
 
   
   
 
1
10
4
2
20 4
1
1
2
- ??
= ? - ?? = ??
??
 
   4 ? ?? = 
 Let , 
22
k
h
?
= ? = 
 ? 4hk = 4 
 ? 1 xy = 
4. Let 
? ?
? ? ? ? , 0, 1, 2 , 1  ,  2 ?? = = ? ? ?
??
ij ij
S M a a i j
be a sample space and ? ? : is invertible ? A M S M 
be an even. Then P(A) is equal to 
 (1) 
16
27
 (2) 
47
81
 
 (3) 
49
81
 (4) 
50
81
 
Answer (4) 
Sol. If M is invertible, then |M| ? 0 
 For |M| = 0 
 1. 
1 1 0 0 2 2
  or   or  
1 1 0 0 2 2
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? Total matrix = 3 
 2.  Two 1’s and Two 0’s ? Total matrix = 4 
 3.  Two 2’s and Two 0’s ? Total matrix = 4 
 4.  Two 1’s and Two 2’s ? Total matrix = 4 
 5.  One 1 and three 0’s ? Total matrix = 4 
 6.  One 2 and Three 0’s ? Total matrix = 4 
 7.  One 1 and one 2 and two 0’s ? Total matrix = 8 
  ( )
31 50
1
81 81
PA = - = 
5. The number of integral solution x of 
2
7
2
7
log 0
23
??
+
??
??
- ??
?
??
-
?? x
x
x
 is  
 (1) 7 
 (2) 8 
 (3) 6 
 (4) 5 
Answer (3) 
Sol. 
2
7
2
7
log 0
23 x
x
x
??
+
??
??
- ??
?
??
-
??
  
 Domain  
 
7
0
2
x+? 
 
–7
2
x ? 
 
7
1
2
x+? 
 
5
2
x
-
? 
 
7
0
23
x
x
-
?
-
 
 x ? 7 
 
3
2
x ? 
 Domain : 
7 5 3
, , 0, 
2 2 2
- ? ? ? ?
- ? -
??
??
? ? ? ?
 
 Case I : 
7
01
2
x ? + ? 
 
75
22
x - ? ? - 
 
2
7
1
23
x
x
- ??
?
??
-
??
 
 
7
11
23
x
x
-
- ? ?
-
 
 
7 2 3
0
23
xx
x
- + -
?
-
 
 
3 10
0
23
x
x
-
?
-
 
  
 
7 2 3
0
23
xx
x
- - +
?
-
 
 
4
0
23
x
x
--
?
-
 
 
4
0
23
x
x
+
?
-
 
 
   
   
  
 No intersection, no solution  
 Case II :  
 
7
1
2
x+? 
 
5
2
x ?- 
 
2
7
1
23
x
x
- ??
?
??
-
??
 
 
7
1
23
x
x
-
?-
-
 
 
3 10
, 
23
x
??
?
? ?
??
 
 
7
1
23
x
x
-
?
-
 
 
3
4, 
2
x
??
?-
? ?
??
 
 
3 3 3 10
, , 
2 2 2 3
x
? ? ? ?
? - ?
? ? ? ?
? ? ? ?
 
 Total 6 integers 
6. Let A be a 2 × 2 matrix with real entries such that 
'1 AA = ? + , where { 1 , 1} ? ? - - ., If det 
2
( ) 4 AA -= , the sum of all possible values of ? is 
equal to 
 (1) 0 
 (2) 
3
2
 
 (3) 2 
 (4) 
5
2
 
Answer (4) 
Sol. Let 
ab
A
cd
??
=
??
??
 
 A A I ? = ? + 
 ? 
1
1
a c a b
b d c d
? + ? ? ? ? ?
=
? ? ? ?
? ? +
? ? ? ?
 
 1 aa = ? + ? 
1
1
a =
-?
 .... (i) 
 bc =? .... (ii) 
 cb =? .... (iii) 
 (ii) and (iii) 0 c = or 1 ? = ? (not possible) 
 ? 0 c = 
 Also 
1
1
1
d d d = ? + ? =
-?
 
 ? 0, 0 cb == 
  
2
| | 4 AA -= 
  | || | 4 A A I -= 
  
22
11
14
11
? ? ? ?
-=
? ? ? ?
- ? - ?
? ? ? ?
 
 ? 
1
,2
2
?=  
7. The value of the integral 
2
log 2
2
log 2
log 1
e
x x x
e
e e e dx
-
?? ??
++
?? ??
?? ??
?
is equal to 
 (1) 
( )
2
2 2 5
5
log
2
15
e
??
+
??
-
??
+ ??
??
 
 (2) 
( )
2
25
5
log
2
15
e
??
+
??
+
??
+ ??
??
  
 (3) 
( )
2 2 5
5
log
2
15
e
??
+
??
-
??
??
+
??
  
 (4) 
( )
2
2 3 5
5
log
2
15
e
??
-
??
+
??
+ ??
??
 
Answer (1) 
Page 5


   
 
  
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
11/04/2023 
Morning 
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. The number of elements in the set 
? ?
? ?
4 2 6
0,2 : 3cos 5cos 2sin 2 0 = ? ? ? ? - ? - ? + = S
is 
 (1) 10 (2) 8 
 (3) 12 (4) 9 
Answer (4) 
Sol. 3cos
4
? – 5cos
2
? – 2sin
6
? + 2 = 0  
 ? cos
2
?[3cos
2
? – 5] – 2sin
6
?
 
+ 2 = 0 
 ? (1 – sin
2
?) (3 – 3sin
2
? – 5) – 2sin
6
?
 
+ 2 = 0 
 ? (sin
2
? – 1) (3sin
2
? + 2) – 2sin
6
?
 
+ 2 = 0 
 Let sin
2
? = t 
 (t – 1) (3t + 2) – 2t
3
 + 2 = 0 
 (t – 1) [3t + 2 – 2 (t
2
 + t + 1)] = 0 
 (t – 1) [2t
2
 – t] = 0 
 
1
0, 1 , 
2
t = 
 ? sin
2
? = 0 ? 3 solution   
  sin
2
? = 1 ? 2 solution   
  sin
2
? = 
1
2
 ? 4 solution  
 ? Total solution = 9 
2. Let ? ? : 2, 4 ? f be a differentiable function such 
that 
( ) ( ) ( ) ( ) ( ) log log 1 ,   [2, 4] ? + + ? ?
ee
x x f x x f x f x x 
with ( )
1
2
2
= f and ( )
1
4
2
= f . 
 Consider the following two statements : 
 (A) ( ) 1 , for all [2, 4] ?? f x x 
 (B) ( )
1
, for all [2, 4]
8
?? f x x 
 Then, 
 (1) Neither statement (A) nor statement (B) is true 
 (2) Only statement (B) is true 
 (3) Both the statements (A) and (B) are true 
 (4) Only statement (A) is true 
Answer (3*) 
Sol. f : [2, 4] ? ? 
 (x log e x) f?(x) + (log e x) f(x) + f(x) ? 1, x ? [2, 4] 
 ? d [x lnx f(x) – x] ? 0 OR d (x lnx ? f(x)) ? 1 
 ? h(x) = x lnx f(x) – x ? 
 ? h(x) ? h(2) , x ? [2, 4] 
 x lnx f(x) – x ? 2 ln2f(2) – 2  
 ? x lnx f(x) – x ? ln2 – 2, x lnx f(x) – x ? ln4 – 4 
 So, 
 
ln2 2 1 ln4 4 1
()
ln ln ln ln
fx
x x x x x x
--
+ ? ? + 
 f(x) ? 1 
 & 
1
()
8
fx ? 
 Hence both A & B are correct. 
 But LMVT on f(x) ? x lnx can’t be satisfied. Hence no 
such f(x) exist. 
3. Let R be a rectangle given by the lines x = 0, x = 2, 
y = 0 and y = 5. Let A(?, 0) and B(0, ?), ? ? [0, 2] 
and ? ? [0, 5], be such that the line segment AB 
divides the area of the rectangle R in the ratio 4 : 1. 
Then, the mid-point of AB lies on a  
 (1) straight line (2) parabola 
 (3) hyperbola (4) circle 
Answer (3) 
Sol.
 
 
 
   
   
 
1
10
4
2
20 4
1
1
2
- ??
= ? - ?? = ??
??
 
   4 ? ?? = 
 Let , 
22
k
h
?
= ? = 
 ? 4hk = 4 
 ? 1 xy = 
4. Let 
? ?
? ? ? ? , 0, 1, 2 , 1  ,  2 ?? = = ? ? ?
??
ij ij
S M a a i j
be a sample space and ? ? : is invertible ? A M S M 
be an even. Then P(A) is equal to 
 (1) 
16
27
 (2) 
47
81
 
 (3) 
49
81
 (4) 
50
81
 
Answer (4) 
Sol. If M is invertible, then |M| ? 0 
 For |M| = 0 
 1. 
1 1 0 0 2 2
  or   or  
1 1 0 0 2 2
? ? ? ? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
? Total matrix = 3 
 2.  Two 1’s and Two 0’s ? Total matrix = 4 
 3.  Two 2’s and Two 0’s ? Total matrix = 4 
 4.  Two 1’s and Two 2’s ? Total matrix = 4 
 5.  One 1 and three 0’s ? Total matrix = 4 
 6.  One 2 and Three 0’s ? Total matrix = 4 
 7.  One 1 and one 2 and two 0’s ? Total matrix = 8 
  ( )
31 50
1
81 81
PA = - = 
5. The number of integral solution x of 
2
7
2
7
log 0
23
??
+
??
??
- ??
?
??
-
?? x
x
x
 is  
 (1) 7 
 (2) 8 
 (3) 6 
 (4) 5 
Answer (3) 
Sol. 
2
7
2
7
log 0
23 x
x
x
??
+
??
??
- ??
?
??
-
??
  
 Domain  
 
7
0
2
x+? 
 
–7
2
x ? 
 
7
1
2
x+? 
 
5
2
x
-
? 
 
7
0
23
x
x
-
?
-
 
 x ? 7 
 
3
2
x ? 
 Domain : 
7 5 3
, , 0, 
2 2 2
- ? ? ? ?
- ? -
??
??
? ? ? ?
 
 Case I : 
7
01
2
x ? + ? 
 
75
22
x - ? ? - 
 
2
7
1
23
x
x
- ??
?
??
-
??
 
 
7
11
23
x
x
-
- ? ?
-
 
 
7 2 3
0
23
xx
x
- + -
?
-
 
 
3 10
0
23
x
x
-
?
-
 
  
 
7 2 3
0
23
xx
x
- - +
?
-
 
 
4
0
23
x
x
--
?
-
 
 
4
0
23
x
x
+
?
-
 
 
   
   
  
 No intersection, no solution  
 Case II :  
 
7
1
2
x+? 
 
5
2
x ?- 
 
2
7
1
23
x
x
- ??
?
??
-
??
 
 
7
1
23
x
x
-
?-
-
 
 
3 10
, 
23
x
??
?
? ?
??
 
 
7
1
23
x
x
-
?
-
 
 
3
4, 
2
x
??
?-
? ?
??
 
 
3 3 3 10
, , 
2 2 2 3
x
? ? ? ?
? - ?
? ? ? ?
? ? ? ?
 
 Total 6 integers 
6. Let A be a 2 × 2 matrix with real entries such that 
'1 AA = ? + , where { 1 , 1} ? ? - - ., If det 
2
( ) 4 AA -= , the sum of all possible values of ? is 
equal to 
 (1) 0 
 (2) 
3
2
 
 (3) 2 
 (4) 
5
2
 
Answer (4) 
Sol. Let 
ab
A
cd
??
=
??
??
 
 A A I ? = ? + 
 ? 
1
1
a c a b
b d c d
? + ? ? ? ? ?
=
? ? ? ?
? ? +
? ? ? ?
 
 1 aa = ? + ? 
1
1
a =
-?
 .... (i) 
 bc =? .... (ii) 
 cb =? .... (iii) 
 (ii) and (iii) 0 c = or 1 ? = ? (not possible) 
 ? 0 c = 
 Also 
1
1
1
d d d = ? + ? =
-?
 
 ? 0, 0 cb == 
  
2
| | 4 AA -= 
  | || | 4 A A I -= 
  
22
11
14
11
? ? ? ?
-=
? ? ? ?
- ? - ?
? ? ? ?
 
 ? 
1
,2
2
?=  
7. The value of the integral 
2
log 2
2
log 2
log 1
e
x x x
e
e e e dx
-
?? ??
++
?? ??
?? ??
?
is equal to 
 (1) 
( )
2
2 2 5
5
log
2
15
e
??
+
??
-
??
+ ??
??
 
 (2) 
( )
2
25
5
log
2
15
e
??
+
??
+
??
+ ??
??
  
 (3) 
( )
2 2 5
5
log
2
15
e
??
+
??
-
??
??
+
??
  
 (4) 
( )
2
2 3 5
5
log
2
15
e
??
-
??
+
??
+ ??
??
 
Answer (1) 
 
   
   
Sol. 
log 2
2
log 2
log 1
e
e
x x x
e
I e e e dx
-
??
??
= + +
??
??
??
??
?
 
 Put 
x
et = 
 
x
e dx dt = 
 
2
2
I
1
II
2
1 log 1
e
I t t dt
??
= ? + +
??
??
?
 
 
2
2
2
1
2
1/2
2
ln( 1 )
1
t
t t x dt
t
??
= + + -
??
??
+
?
 
 
2
22
1
2
ln 1 1 t t t t
??
= + + - +
??
??
 
 
1 5 1 5
2ln 5 2 5 ln
2 2 2 2
??
??
= + - - + -
??
??
??
??
 
 
2
2(2 5) 5
ln
2
15
??
+
??
=-
??
+
??
  
8. Let sets A and B have 5 elements each. Let the 
mean of the elements in sets A and B 5 and 8 
respectively and the variance of the elements in 
sets A and B be 12 and 20 respectively. A new set 
C of 10 elements is formed by subtracting 3 from 
each element of A and adding 2 to each element of 
B. Then the sum of the mean and variance of the 
elements of C is ______ .  
 (1) 40 (2) 32 
 (3) 38 (4) 36 
Answer (3) 
Sol. A 
 
1 2 5
mean ( , .... ) 5 x x x = 
 
1 2 5
( 3, 3.... 3) 2 x x x ? - - - = 
 
1 2 5
Var( , ,.... ) 12 x x x = 
 
1 2 5
Var( 3, 3,.... 3) 12 x x x - - - = 
 
2
( 3)
4 12
5
i
x -
-=
?
 
 B 
 
1 2 5
mean( , .... ) 8 y y y = 
 
1 2 5
mean( 2, 2,.... 2) 10 y y y ? + + + = 
 
1 2 5
Var( , .... ) 20 y y y = 
 
1 2 5
Var( 2, 2.... 2) 20 y y y + + + = 
 
2
1
( 2)
100 20
5
y +
-=
?
  
 Combined mean 
5
1
( 3) ( 2)
10
ii
i
xy
=
- + +
??
 
 
10 50
6
10
+
== 
 Combined variance  
  
2
22
( 3) ( 2) 6
10
ii
xy - + + -
=
??
 
  
80 120 5
36 32
10
+?
= - =  
9. Let ( , , ) ? ? ? be the image of point P (2,3,5) in the 
plane 2 3 6 x y z + - = . Then ? + ? + ? is equal to 
 (1) 5 (2) 10 
 (3) 12 (4) 9 
Answer (2) 
Sol. 
2 3 5
2 1 3
? - ? - ? -
==
-
(4 3 15 6)
22
14
+ - -
= - = 
 ? 
2
2
2
?-
=  ? 6 ?= 
  
3
2
1
?-
=  ? 5 ?= 
  
5
2
3
?-
=
-
 ? 1 ? = - 
 ? 6 5 1 10 ? + ? + ? = + - = 
  Option (2) is correct.  
10. Let 
2
( ) – [ ] f x x x x x
??
= - + +
??
, where x ? and [t] 
denotes the greatest integer less than or equal to t. 
Then, f is 
 (1) continuous at x = 0, but not continuous at x = 1
 (2) continuous at x = 1, but not continuous at x = 0 
 (3) continuous at x = 0 and x = 1 
 (4) not continuous at x = 0 and x = 1 
Answer (2)