JEE Main 2023 April 12 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) 
 
 
  
12/04/2023 
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) 
 
 
  
12/04/2023 
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 five-digit numbers, greater than 
40000 and divisible by 5, which can be formed 
using the digits 0, 1, 3, 5, 7 and 9 without repetition, 
is equal to  
 (1) 132 (2) 120 
 (3) 72 (4) 96 
Answer (2) 
Sol. Case I : Numbers start with 5 
  
 Case II : 
  
 Case III : 
  
 Total numbers = 120 
  
2. Let a, b, c be three distinct real numbers, none 
equal to one. If the vectors 
ˆ ˆ ˆ ˆ ˆ ˆ
,  + + + + ai j k i bj k 
and 
ˆ ˆ ˆ
++ i j ck are coplanar, then 
1 1 1
1 1 1
++
- - - a b c
 is equal to  
 (1) 2 (2) –1 
 (3) –2 (4) 1 
Answer (4) 
Sol. 
11
1 1 0
11
a
b
c
= 
 
1 1 0
0 1 1 0
11
ab
bc
c
--
- - = 
 R1 ? R1 – R2 
 R2 ? R2 – R3 
 ? (a – 1) [c(b – 1) – (1 – c)] + 1 [(1 – b) (1 – c)] = 0 
 ? c(a – 1) (b – 1) – (a – 1) (1 – c) + (1 – b) (1 – c) = 0 
 ? 
1 1 1
1
1 1 1 a b c
+ + =
- - -
 
3. Let <a n> be a sequence such that 
( )( )
2
12
3
...
12
+
+ + + =
++
n
nn
a a a
nn
. If 
 
10
1 2 3
1
1
28 ...
m
k
k
p p p p
a
=
=
?
, where p1, p2,…p m are 
the first m prime numbers, then m is equal to 
 (1) 5 
 (2) 8 
 (3) 6 
 (4) 7 
Answer (3) 
Sol. 
( )( )
( ) ( )
( )
2
2
1 3 1
3
1 2 1
n
nn
nn
a
n n n n
- + -
+
=-
+ + +
 
 
( )( )
4
12 n n n
=
++
 
 
10 10
11
11
( 1)( 2)
4
k
kk
k k k
a
==
= + +
??
 
 = 
10
1
1
( 1)( 2)( 3) ( 1) ( 1)( 2)
16
k
k k k k k k k k
=
+ + + - - + +
?
 
 ( ) ( )
1
1·2 · 3 · 4. 0 2·3·4 ·5 1·2·3·4
16
? - + -
?
 
 ( ) 10·11·12·13 – 9·10·11·12 ?
?
 
 ? ?
1
10·11·12·13
16
== 
 ? 
10
1
1 28 5 11 3 13
28
2
k
k
a
=
? ? ? ?
=
?
 
 = 2·3·5·7·11·13 
 ? 6 m = 
Page 3


   
 
  
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
12/04/2023 
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 five-digit numbers, greater than 
40000 and divisible by 5, which can be formed 
using the digits 0, 1, 3, 5, 7 and 9 without repetition, 
is equal to  
 (1) 132 (2) 120 
 (3) 72 (4) 96 
Answer (2) 
Sol. Case I : Numbers start with 5 
  
 Case II : 
  
 Case III : 
  
 Total numbers = 120 
  
2. Let a, b, c be three distinct real numbers, none 
equal to one. If the vectors 
ˆ ˆ ˆ ˆ ˆ ˆ
,  + + + + ai j k i bj k 
and 
ˆ ˆ ˆ
++ i j ck are coplanar, then 
1 1 1
1 1 1
++
- - - a b c
 is equal to  
 (1) 2 (2) –1 
 (3) –2 (4) 1 
Answer (4) 
Sol. 
11
1 1 0
11
a
b
c
= 
 
1 1 0
0 1 1 0
11
ab
bc
c
--
- - = 
 R1 ? R1 – R2 
 R2 ? R2 – R3 
 ? (a – 1) [c(b – 1) – (1 – c)] + 1 [(1 – b) (1 – c)] = 0 
 ? c(a – 1) (b – 1) – (a – 1) (1 – c) + (1 – b) (1 – c) = 0 
 ? 
1 1 1
1
1 1 1 a b c
+ + =
- - -
 
3. Let <a n> be a sequence such that 
( )( )
2
12
3
...
12
+
+ + + =
++
n
nn
a a a
nn
. If 
 
10
1 2 3
1
1
28 ...
m
k
k
p p p p
a
=
=
?
, where p1, p2,…p m are 
the first m prime numbers, then m is equal to 
 (1) 5 
 (2) 8 
 (3) 6 
 (4) 7 
Answer (3) 
Sol. 
( )( )
( ) ( )
( )
2
2
1 3 1
3
1 2 1
n
nn
nn
a
n n n n
- + -
+
=-
+ + +
 
 
( )( )
4
12 n n n
=
++
 
 
10 10
11
11
( 1)( 2)
4
k
kk
k k k
a
==
= + +
??
 
 = 
10
1
1
( 1)( 2)( 3) ( 1) ( 1)( 2)
16
k
k k k k k k k k
=
+ + + - - + +
?
 
 ( ) ( )
1
1·2 · 3 · 4. 0 2·3·4 ·5 1·2·3·4
16
? - + -
?
 
 ( ) 10·11·12·13 – 9·10·11·12 ?
?
 
 ? ?
1
10·11·12·13
16
== 
 ? 
10
1
1 28 5 11 3 13
28
2
k
k
a
=
? ? ? ?
=
?
 
 = 2·3·5·7·11·13 
 ? 6 m = 
 
   
   
4. Let 
1
1
 51
0 1
A
??
??
=
??
??
??
. If 
12
 
–1 –1
??
=
??
??
B
–1 –2
 
11
A
??
=
??
??
, then 
the sum of all the elements of the matrix 
50
1 =
?
n
n
B is 
equal to 
 (1) 75 
 
 (2) 125 
 (3) 50 (4) 100 
Answer (4) 
Sol. 
( ) ( )
1 2 –1 –2 1 0
–1 –1 1 1 0 1
MN
? ? ? ? ? ?
=
? ? ? ? ? ?
? ? ? ? ? ?
 
 ? MN = I = NM 
 B = MAN 
 B
n
 = (MAN)
n
 = (MAN) (MAN) … 
 = (MA
2
N) MAN 
 = MA
n
N 
 Now 
1
10 0
51
01
00
A I E
??
??
??
= + = +
??
??
??
??
??
 
 E
2
 = 0 
 
2
2
0
()
n n n
A I E I nE C E ? = + = + + 
 = I + nE 
 
1
51
01
n ??
??
=
??
??
??
 
 So, 
1
51 51
1
51 51
nn
nn
B MA N
nn
??
+
??
?? ==
-
??
-
??
??
 
 
50
1
50·51 50·51
50
2·51 2·51 75 25
– 50·51 50·51 –25 25
50 –
2·51 2·51
n
n
B
=
??
+
??
??
??
==
??
??
??
??
??
?
 
 ? Sum = 100 
5. If 
1 1 0
1 1 1 1023
  ...  
1 2 10
-
+ + + + =
+
n n n n
nn
C C C C
nn
then n is equal to 
 (1) 9 (2) 8 
 (3) 7 (4) 6 
Answer (1) 
Sol. 
1
1
0
1 1023 1
1 10 1
n
n n n
r r r
r
n
C C C
rr
+
+
=
+ ??
= ? =
??
++
??
?
 
 
1
1
0
1 1023
1 10
n
n
r
r
C
n
+
+
=
?=
+
?
 
 
1 1 1
1 2 1
1 1023
...
1 10
n n n
n
C C C
n
+ + +
+
??
? + + + =
??
+
 
 
1 10
2 1 1023 2 1
1 10 10
n
n
+
--
? = =
+
 
 ? n + 1 = 10 
 ? n = 9 
6. The area of the region enclosed by the curve 
3
yx = and its tangent at the point (–1, –1) is 
 (1) 
19
4
 (2) 
23
4
 
 (3) 
31
4
 (4) 
27
4
 
Answer (4) 
Sol.  
 
 Given y = x
3
 …(i) 
 
2
3
dy
x
dx
?= 
 
(–1, –1)
3
dy
dx
??
=
??
??
 
P(–1, –1)
Q(2, 8)
Page 4


   
 
  
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
12/04/2023 
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 five-digit numbers, greater than 
40000 and divisible by 5, which can be formed 
using the digits 0, 1, 3, 5, 7 and 9 without repetition, 
is equal to  
 (1) 132 (2) 120 
 (3) 72 (4) 96 
Answer (2) 
Sol. Case I : Numbers start with 5 
  
 Case II : 
  
 Case III : 
  
 Total numbers = 120 
  
2. Let a, b, c be three distinct real numbers, none 
equal to one. If the vectors 
ˆ ˆ ˆ ˆ ˆ ˆ
,  + + + + ai j k i bj k 
and 
ˆ ˆ ˆ
++ i j ck are coplanar, then 
1 1 1
1 1 1
++
- - - a b c
 is equal to  
 (1) 2 (2) –1 
 (3) –2 (4) 1 
Answer (4) 
Sol. 
11
1 1 0
11
a
b
c
= 
 
1 1 0
0 1 1 0
11
ab
bc
c
--
- - = 
 R1 ? R1 – R2 
 R2 ? R2 – R3 
 ? (a – 1) [c(b – 1) – (1 – c)] + 1 [(1 – b) (1 – c)] = 0 
 ? c(a – 1) (b – 1) – (a – 1) (1 – c) + (1 – b) (1 – c) = 0 
 ? 
1 1 1
1
1 1 1 a b c
+ + =
- - -
 
3. Let <a n> be a sequence such that 
( )( )
2
12
3
...
12
+
+ + + =
++
n
nn
a a a
nn
. If 
 
10
1 2 3
1
1
28 ...
m
k
k
p p p p
a
=
=
?
, where p1, p2,…p m are 
the first m prime numbers, then m is equal to 
 (1) 5 
 (2) 8 
 (3) 6 
 (4) 7 
Answer (3) 
Sol. 
( )( )
( ) ( )
( )
2
2
1 3 1
3
1 2 1
n
nn
nn
a
n n n n
- + -
+
=-
+ + +
 
 
( )( )
4
12 n n n
=
++
 
 
10 10
11
11
( 1)( 2)
4
k
kk
k k k
a
==
= + +
??
 
 = 
10
1
1
( 1)( 2)( 3) ( 1) ( 1)( 2)
16
k
k k k k k k k k
=
+ + + - - + +
?
 
 ( ) ( )
1
1·2 · 3 · 4. 0 2·3·4 ·5 1·2·3·4
16
? - + -
?
 
 ( ) 10·11·12·13 – 9·10·11·12 ?
?
 
 ? ?
1
10·11·12·13
16
== 
 ? 
10
1
1 28 5 11 3 13
28
2
k
k
a
=
? ? ? ?
=
?
 
 = 2·3·5·7·11·13 
 ? 6 m = 
 
   
   
4. Let 
1
1
 51
0 1
A
??
??
=
??
??
??
. If 
12
 
–1 –1
??
=
??
??
B
–1 –2
 
11
A
??
=
??
??
, then 
the sum of all the elements of the matrix 
50
1 =
?
n
n
B is 
equal to 
 (1) 75 
 
 (2) 125 
 (3) 50 (4) 100 
Answer (4) 
Sol. 
( ) ( )
1 2 –1 –2 1 0
–1 –1 1 1 0 1
MN
? ? ? ? ? ?
=
? ? ? ? ? ?
? ? ? ? ? ?
 
 ? MN = I = NM 
 B = MAN 
 B
n
 = (MAN)
n
 = (MAN) (MAN) … 
 = (MA
2
N) MAN 
 = MA
n
N 
 Now 
1
10 0
51
01
00
A I E
??
??
??
= + = +
??
??
??
??
??
 
 E
2
 = 0 
 
2
2
0
()
n n n
A I E I nE C E ? = + = + + 
 = I + nE 
 
1
51
01
n ??
??
=
??
??
??
 
 So, 
1
51 51
1
51 51
nn
nn
B MA N
nn
??
+
??
?? ==
-
??
-
??
??
 
 
50
1
50·51 50·51
50
2·51 2·51 75 25
– 50·51 50·51 –25 25
50 –
2·51 2·51
n
n
B
=
??
+
??
??
??
==
??
??
??
??
??
?
 
 ? Sum = 100 
5. If 
1 1 0
1 1 1 1023
  ...  
1 2 10
-
+ + + + =
+
n n n n
nn
C C C C
nn
then n is equal to 
 (1) 9 (2) 8 
 (3) 7 (4) 6 
Answer (1) 
Sol. 
1
1
0
1 1023 1
1 10 1
n
n n n
r r r
r
n
C C C
rr
+
+
=
+ ??
= ? =
??
++
??
?
 
 
1
1
0
1 1023
1 10
n
n
r
r
C
n
+
+
=
?=
+
?
 
 
1 1 1
1 2 1
1 1023
...
1 10
n n n
n
C C C
n
+ + +
+
??
? + + + =
??
+
 
 
1 10
2 1 1023 2 1
1 10 10
n
n
+
--
? = =
+
 
 ? n + 1 = 10 
 ? n = 9 
6. The area of the region enclosed by the curve 
3
yx = and its tangent at the point (–1, –1) is 
 (1) 
19
4
 (2) 
23
4
 
 (3) 
31
4
 (4) 
27
4
 
Answer (4) 
Sol.  
 
 Given y = x
3
 …(i) 
 
2
3
dy
x
dx
?= 
 
(–1, –1)
3
dy
dx
??
=
??
??
 
P(–1, –1)
Q(2, 8)
 
   
   
 Equation of tangent at (–1, –1) 
 (y + 1) = 3(x + 1) 
 y = 3x + 2 …(ii) 
 Solving (i) and (ii) 
 x
3
 = 3x + 2 
  
 Q(2, 8) 
 Required area 
( )
2
3
–1
3 2 – x x dx =+
?
 
 
31
(4 1) 2(2 1) (16 1)
24
= - + + - - 
 
27
4
= 
7. If the total maximum value of the function 
2
sin
3
( ) ,
2sin
x
e
fx
x
??
=??
??
??
0,
2
x
? ??
?
??
??
, is ,
k
e
then 
8
8
8
5
kk
k
e
e
??
++
??
??
 is equal to 
 (1) 
3 6 11
e e e ++ 
 (2) 
5 6 11
e e e ++ 
 (3) 
3 6 10
e e e ++ 
 (4) 
3 5 11
eee ++ 
Answer (1) 
Sol. ( )
2
sin
3
2sin
x
e
fx
x
??
=
??
??
??
 
 
( ) ( )
'
2
2
3
2sin cos ln
2sin
1 2sin 3 1
sin cos
2
3 sin
e
xx
x
f x f x
xe
xx
e x
?? ??
?+ ?? ??
??
??
??
=
??
?
??
? - ?
??
??
 
 ( )
3
sin2 ln sin cos 0
2sin
e
f x x x x
x
?? ??
= - = ?? ??
??
??
?? ??
 
 Sin 2x = 0 (not possible) 
 
31
ln
2sin 2
e
x
??
=+
??
??
??
 
 
31
2sin 2
e
e
x
=+ 
 
3
sin
2
x = 
 
( )
11
17
3
8
8
8
max
c
f e k e
e
= = ? = 
 
8
8
8 3 6 11
5
kk
k e e e
e
e
??
+ + = + +
??
??
 
  
8. Let , ?? be the roots of the quadratic equation 
2
6 3 0 xx + + = . Then 
23 23 14 14
15 15 10 10
? + ? + ? + ?
? + ? + ? + ?
is 
equal to 
 (1) 81 
 (2) 9 
 (3) 72 
 (4) 729 
Answer (2) 
Sol. 
23 23 14 14
15 15 10 10
? + ? + ? + ?
? + ? + ? + ?
 
 Let a n = ?
n
 + ?
n
 
 
23 14
15 10
aa
aa
+
=
+
 
  
 
66
2
x
- ? -
= 
 
1
6
2
i -? ??
=
??
??
 
 
1
3
2
i -? ??
=
??
??
 
 
3
4
3
i
e
?
?= 
x x
3
 – 3 – 2 = 0
–1
–1
? = 2
Page 5


   
 
  
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
12/04/2023 
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 five-digit numbers, greater than 
40000 and divisible by 5, which can be formed 
using the digits 0, 1, 3, 5, 7 and 9 without repetition, 
is equal to  
 (1) 132 (2) 120 
 (3) 72 (4) 96 
Answer (2) 
Sol. Case I : Numbers start with 5 
  
 Case II : 
  
 Case III : 
  
 Total numbers = 120 
  
2. Let a, b, c be three distinct real numbers, none 
equal to one. If the vectors 
ˆ ˆ ˆ ˆ ˆ ˆ
,  + + + + ai j k i bj k 
and 
ˆ ˆ ˆ
++ i j ck are coplanar, then 
1 1 1
1 1 1
++
- - - a b c
 is equal to  
 (1) 2 (2) –1 
 (3) –2 (4) 1 
Answer (4) 
Sol. 
11
1 1 0
11
a
b
c
= 
 
1 1 0
0 1 1 0
11
ab
bc
c
--
- - = 
 R1 ? R1 – R2 
 R2 ? R2 – R3 
 ? (a – 1) [c(b – 1) – (1 – c)] + 1 [(1 – b) (1 – c)] = 0 
 ? c(a – 1) (b – 1) – (a – 1) (1 – c) + (1 – b) (1 – c) = 0 
 ? 
1 1 1
1
1 1 1 a b c
+ + =
- - -
 
3. Let <a n> be a sequence such that 
( )( )
2
12
3
...
12
+
+ + + =
++
n
nn
a a a
nn
. If 
 
10
1 2 3
1
1
28 ...
m
k
k
p p p p
a
=
=
?
, where p1, p2,…p m are 
the first m prime numbers, then m is equal to 
 (1) 5 
 (2) 8 
 (3) 6 
 (4) 7 
Answer (3) 
Sol. 
( )( )
( ) ( )
( )
2
2
1 3 1
3
1 2 1
n
nn
nn
a
n n n n
- + -
+
=-
+ + +
 
 
( )( )
4
12 n n n
=
++
 
 
10 10
11
11
( 1)( 2)
4
k
kk
k k k
a
==
= + +
??
 
 = 
10
1
1
( 1)( 2)( 3) ( 1) ( 1)( 2)
16
k
k k k k k k k k
=
+ + + - - + +
?
 
 ( ) ( )
1
1·2 · 3 · 4. 0 2·3·4 ·5 1·2·3·4
16
? - + -
?
 
 ( ) 10·11·12·13 – 9·10·11·12 ?
?
 
 ? ?
1
10·11·12·13
16
== 
 ? 
10
1
1 28 5 11 3 13
28
2
k
k
a
=
? ? ? ?
=
?
 
 = 2·3·5·7·11·13 
 ? 6 m = 
 
   
   
4. Let 
1
1
 51
0 1
A
??
??
=
??
??
??
. If 
12
 
–1 –1
??
=
??
??
B
–1 –2
 
11
A
??
=
??
??
, then 
the sum of all the elements of the matrix 
50
1 =
?
n
n
B is 
equal to 
 (1) 75 
 
 (2) 125 
 (3) 50 (4) 100 
Answer (4) 
Sol. 
( ) ( )
1 2 –1 –2 1 0
–1 –1 1 1 0 1
MN
? ? ? ? ? ?
=
? ? ? ? ? ?
? ? ? ? ? ?
 
 ? MN = I = NM 
 B = MAN 
 B
n
 = (MAN)
n
 = (MAN) (MAN) … 
 = (MA
2
N) MAN 
 = MA
n
N 
 Now 
1
10 0
51
01
00
A I E
??
??
??
= + = +
??
??
??
??
??
 
 E
2
 = 0 
 
2
2
0
()
n n n
A I E I nE C E ? = + = + + 
 = I + nE 
 
1
51
01
n ??
??
=
??
??
??
 
 So, 
1
51 51
1
51 51
nn
nn
B MA N
nn
??
+
??
?? ==
-
??
-
??
??
 
 
50
1
50·51 50·51
50
2·51 2·51 75 25
– 50·51 50·51 –25 25
50 –
2·51 2·51
n
n
B
=
??
+
??
??
??
==
??
??
??
??
??
?
 
 ? Sum = 100 
5. If 
1 1 0
1 1 1 1023
  ...  
1 2 10
-
+ + + + =
+
n n n n
nn
C C C C
nn
then n is equal to 
 (1) 9 (2) 8 
 (3) 7 (4) 6 
Answer (1) 
Sol. 
1
1
0
1 1023 1
1 10 1
n
n n n
r r r
r
n
C C C
rr
+
+
=
+ ??
= ? =
??
++
??
?
 
 
1
1
0
1 1023
1 10
n
n
r
r
C
n
+
+
=
?=
+
?
 
 
1 1 1
1 2 1
1 1023
...
1 10
n n n
n
C C C
n
+ + +
+
??
? + + + =
??
+
 
 
1 10
2 1 1023 2 1
1 10 10
n
n
+
--
? = =
+
 
 ? n + 1 = 10 
 ? n = 9 
6. The area of the region enclosed by the curve 
3
yx = and its tangent at the point (–1, –1) is 
 (1) 
19
4
 (2) 
23
4
 
 (3) 
31
4
 (4) 
27
4
 
Answer (4) 
Sol.  
 
 Given y = x
3
 …(i) 
 
2
3
dy
x
dx
?= 
 
(–1, –1)
3
dy
dx
??
=
??
??
 
P(–1, –1)
Q(2, 8)
 
   
   
 Equation of tangent at (–1, –1) 
 (y + 1) = 3(x + 1) 
 y = 3x + 2 …(ii) 
 Solving (i) and (ii) 
 x
3
 = 3x + 2 
  
 Q(2, 8) 
 Required area 
( )
2
3
–1
3 2 – x x dx =+
?
 
 
31
(4 1) 2(2 1) (16 1)
24
= - + + - - 
 
27
4
= 
7. If the total maximum value of the function 
2
sin
3
( ) ,
2sin
x
e
fx
x
??
=??
??
??
0,
2
x
? ??
?
??
??
, is ,
k
e
then 
8
8
8
5
kk
k
e
e
??
++
??
??
 is equal to 
 (1) 
3 6 11
e e e ++ 
 (2) 
5 6 11
e e e ++ 
 (3) 
3 6 10
e e e ++ 
 (4) 
3 5 11
eee ++ 
Answer (1) 
Sol. ( )
2
sin
3
2sin
x
e
fx
x
??
=
??
??
??
 
 
( ) ( )
'
2
2
3
2sin cos ln
2sin
1 2sin 3 1
sin cos
2
3 sin
e
xx
x
f x f x
xe
xx
e x
?? ??
?+ ?? ??
??
??
??
=
??
?
??
? - ?
??
??
 
 ( )
3
sin2 ln sin cos 0
2sin
e
f x x x x
x
?? ??
= - = ?? ??
??
??
?? ??
 
 Sin 2x = 0 (not possible) 
 
31
ln
2sin 2
e
x
??
=+
??
??
??
 
 
31
2sin 2
e
e
x
=+ 
 
3
sin
2
x = 
 
( )
11
17
3
8
8
8
max
c
f e k e
e
= = ? = 
 
8
8
8 3 6 11
5
kk
k e e e
e
e
??
+ + = + +
??
??
 
  
8. Let , ?? be the roots of the quadratic equation 
2
6 3 0 xx + + = . Then 
23 23 14 14
15 15 10 10
? + ? + ? + ?
? + ? + ? + ?
is 
equal to 
 (1) 81 
 (2) 9 
 (3) 72 
 (4) 729 
Answer (2) 
Sol. 
23 23 14 14
15 15 10 10
? + ? + ? + ?
? + ? + ? + ?
 
 Let a n = ?
n
 + ?
n
 
 
23 14
15 10
aa
aa
+
=
+
 
  
 
66
2
x
- ? -
= 
 
1
6
2
i -? ??
=
??
??
 
 
1
3
2
i -? ??
=
??
??
 
 
3
4
3
i
e
?
?= 
x x
3
 – 3 – 2 = 0
–1
–1
? = 2
 
   
   
 
5
4
3
i
e
?
?= 
 
23 23 14 14
15 15 10 10
? + ? + ? + ?
? + ? + ? + ?
 
 
( )
( )
35
23 23
44
9
35
14 14
44
35
15 15
44
5
35
10 10
44
3
9
3
ii
ii
ii
ii
ee
ee
ee
ee
??
??
??
??
??
??
??
? ? ?
??
??
??
??
??
+
??
??
??
??
??
?? ??
?? ??
++
?? ??
??
??
=
??
??
??
??
+
??
??
?? ??
??
??
??
??
++
??
??
??
??
??
??
 
 
( )
( )
9
5
11
30
2
9
11
30
2
ii
ii
?? +-+ ??
+
?? ??
??
??
=?
??
+-+ ??
+
??
??
?? ??
 
 = 81 
9. Let the lines 
1
54
:
3 1 2
x y z
l
+ + - ?
==
-
and 
2
: 3 2 2 0 l x y z + + - = 3 2 13 x y z = - + - be 
coplanar. If the point ( , , ) P a b c on 
1
l is nearest to 
the point ( 4, 3, 2), Q-- then | | | | | | a b c ++ is equal 
to 
 (1) 12 (2) 14 
 (3) 8 (4) 10 
Answer (4) 
Sol. 
22
for : 3 2 1
1 3 2
i j k
Pn =
-
 
 ( ) ( ) ( ) 7 5 11 i j k = - + - 
 Let R lies on l2 
 Let z = 0  
 ? 3x + 2y = 2 
  x – 3y = 13   
 ? 11y = –39 + 2 = –37 
 ? 
37
11
y
-
= 
 ? x = 13 + 3y 
  
37 3
13
11
?
=- 
  
143 111 32
11 11
-
== 
 ? 
32 37
, ,0
11 11
R
- ??
??
??
 
  l1 & l2 are coplanar  
 ? 
32 37
54
11 11
3 1 2 0
7 5 11
??
- - - + ?
??
??
-=
??
??
--
??
??
 
  
87 7
11 11
3 1 2 0
7 5 11
--
?
= - =
--
 
 ? ? = 7 
 ? P(3k –5, k –4, –2k + 7) is nearest to (–4, –3, 2) 
  
 
1
0 QP n ?= 
 ( ) ( ) ( )
( ) ( )
3 1 1 2 5 3 2 0 k i k j k k i j k - + - + - + ? + - = 
 ? 9k –3 + k –1 + 4k – 10 = 0  
 = 14k = 14 ? k = 1  
 ? P(–2, –3, 5) ? (a, b, c) 
  |a| + |b| + |c| = 2 + 3 + 5 = 10 
 ? option (4) is correct