JEE Main 2023 April 13 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) 
 
 
  
13/04/2023 
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) 
 
 
  
13/04/2023 
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 ?, ? be the roots of the equation 
2
2 2 0. xx - + = Then 
14 14
? + ? is equal to   
 (1) –64 (2) 64 2 - 
 (3) –128 (4) 128 2 - 
Answer (3) 
Sol. 
2
2 2 0 xx - + = 
 
26
2
i
x
?
= 
 Let 
2
1 3 1 3
,2
22
ii +-
? = ? = = ? 
 2 =? 
 
14 7 2 14 7
2 , 2 ? = ? ? = ? 
 ? ( )
14 14 7
21 ? + ? = - 
 = –2
7
 
2. The plane, passing through the points (0, –1, 2) and 
(–1, 2, 1) and parallel to the line passing through  
(5, 1, –7) and (1, –1, –1), also passes through the 
point 
 (1) (–2, 5, 0) (2) (1, –2, 1) 
 (3) (2, 0, 1) (4) (0, 5, –2) 
Answer (1) 
Sol. Let A(0, –1, 2) & B(–1, 2, 1) 
 So, normal vector to the plane 
 ( ) ( )
ˆ
ˆˆ
ˆ ˆ ˆ
3 1 1 16 10 (14)
6 42
j
ik
n i j k = - - = - - +
--
 
 = 
ˆ ˆ ˆ
16 10 14 i j k ++ 
 ? Equation of plane 
 16(x – 0) + 10(y + 1) + 14 (z – 2) = 0 
 ? 8x + 5y + 7z = 9 
 ? Option (1) is correct 
3. The statement (p ? (?q)) ? ((?p) ? q) ? ((?p) ? (?q)) 
is equivalent to _____ 
 (1) (?p) ? q (2) (?p) ? (?q) 
 (3) p ? (?q) (4) p ? q 
Answer (2) 
Sol. ( ) ( ) ( ) (~ ) (~ ) (~ ) (~ ) p q p q p q ? ? ? ? ? 
 = ( ) ( ) ~ ( ~ ) ~ p q q p q ? ? ? ? 
 = ( ) ~~ p p q ?? 
 = ( ) ( ) ~ ~ ~ p p p q ? ? ? 
 = ~~ pq ? 
4. Let the centre of a circle C be (?, ?) and its radius r 
< 8. Let 3x + 4y = 24 and 3x – 4y = 32 be two 
tangents and 4x + 3y = 1 be a normal to C. Then  
(? – ? + r) is equal to  
 (1) 7 (2) 5 
 (3) 6 (4) 9 
Answer (1) 
Sol. ? (?, ?) lies on 4x + 3y = 1 
 ? 4? + 3? = 1 …(i) 
 and 
3 4 24 3 4 32
55
? + ? - ? - ? -
= 
 Take (+ve) 
  3? + 4? – 24 = 3? – 4? – 32 
 ? 8? = –8 ? 1 ? = - 
 So 1 ?=  By equation (i) 
 And 
3 4 24
58
5
r
--
= = ? 
 Take (–ve) 
 3? + 4? – 24 = –3? + 4? + 32 
 6? = 56 ? 
56 1 4 28
and 1
6 3 3
? ??
? = ? = -
??
??
 
 For then r > 8 
 ? r = 5, ? = 1, ? = –1 
 ? ? – ? + r = 7 
Page 3


   
 
 
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
13/04/2023 
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 ?, ? be the roots of the equation 
2
2 2 0. xx - + = Then 
14 14
? + ? is equal to   
 (1) –64 (2) 64 2 - 
 (3) –128 (4) 128 2 - 
Answer (3) 
Sol. 
2
2 2 0 xx - + = 
 
26
2
i
x
?
= 
 Let 
2
1 3 1 3
,2
22
ii +-
? = ? = = ? 
 2 =? 
 
14 7 2 14 7
2 , 2 ? = ? ? = ? 
 ? ( )
14 14 7
21 ? + ? = - 
 = –2
7
 
2. The plane, passing through the points (0, –1, 2) and 
(–1, 2, 1) and parallel to the line passing through  
(5, 1, –7) and (1, –1, –1), also passes through the 
point 
 (1) (–2, 5, 0) (2) (1, –2, 1) 
 (3) (2, 0, 1) (4) (0, 5, –2) 
Answer (1) 
Sol. Let A(0, –1, 2) & B(–1, 2, 1) 
 So, normal vector to the plane 
 ( ) ( )
ˆ
ˆˆ
ˆ ˆ ˆ
3 1 1 16 10 (14)
6 42
j
ik
n i j k = - - = - - +
--
 
 = 
ˆ ˆ ˆ
16 10 14 i j k ++ 
 ? Equation of plane 
 16(x – 0) + 10(y + 1) + 14 (z – 2) = 0 
 ? 8x + 5y + 7z = 9 
 ? Option (1) is correct 
3. The statement (p ? (?q)) ? ((?p) ? q) ? ((?p) ? (?q)) 
is equivalent to _____ 
 (1) (?p) ? q (2) (?p) ? (?q) 
 (3) p ? (?q) (4) p ? q 
Answer (2) 
Sol. ( ) ( ) ( ) (~ ) (~ ) (~ ) (~ ) p q p q p q ? ? ? ? ? 
 = ( ) ( ) ~ ( ~ ) ~ p q q p q ? ? ? ? 
 = ( ) ~~ p p q ?? 
 = ( ) ( ) ~ ~ ~ p p p q ? ? ? 
 = ~~ pq ? 
4. Let the centre of a circle C be (?, ?) and its radius r 
< 8. Let 3x + 4y = 24 and 3x – 4y = 32 be two 
tangents and 4x + 3y = 1 be a normal to C. Then  
(? – ? + r) is equal to  
 (1) 7 (2) 5 
 (3) 6 (4) 9 
Answer (1) 
Sol. ? (?, ?) lies on 4x + 3y = 1 
 ? 4? + 3? = 1 …(i) 
 and 
3 4 24 3 4 32
55
? + ? - ? - ? -
= 
 Take (+ve) 
  3? + 4? – 24 = 3? – 4? – 32 
 ? 8? = –8 ? 1 ? = - 
 So 1 ?=  By equation (i) 
 And 
3 4 24
58
5
r
--
= = ? 
 Take (–ve) 
 3? + 4? – 24 = –3? + 4? + 32 
 6? = 56 ? 
56 1 4 28
and 1
6 3 3
? ??
? = ? = -
??
??
 
 For then r > 8 
 ? r = 5, ? = 1, ? = –1 
 ? ? – ? + r = 7 
 
   
  
5. The coefficient of x
5
 in the expansion of 
5
3
2
1
2
3
x
x
??
-
??
??
 is  
 (1) 
80
9
 (2) 9 
 (3) 8 (4) 
26
3
 
Answer (1) 
Sol. 
( )
5–
53
1
2
–1
2
3
r
r
rr
T C x
x
+
??
=
??
??
 
 15 – 3r – 2r = 5 
 ? r = 2 
 Coefficient ( )
3
5
2
1 80
2
99
C
??
==
??
??
 
6. All words, with or without meaning, are made using 
all the letters of the word MONDAY. These words 
are written as in a dictionary with serial numbers. 
The serial number of the word MONDAY is 
 (1) 327 (2) 328 
 (3) 324 (4) 326 
Answer (1) 
Sol.  
 
3 5 4 2 1 6
M O N D A Y
2 3 2 1 0 0
5! 4! 3! 2! 1! 0!
 
 ? Rank = (2 × 5! + 3 × 4! + 2 × 3! + 1 × 2!) + 1 
   = 240 + 72 + 12 + 2 + 1 = 327 
7. The value of 
( )
4
50
4
0
4
49 51
0
tan
tan tan
x
x
e e xdx
e x x dx
?
?
-
-
?
-
+
+
?
?
is 
 (1) 51 (2) 50 
 (3) 25 (4) 49 
Answer (2) 
Sol. 
( )
( )
4
50
4
0
4
49 2
0
tan
tan sec
x
x
e e x dx
e x x dx
?
-?
-
?
-
+
?
?
 
 
( )
4
50
4
0
4 50 50
4
0 0
tan
tan tan
50 50
x
xx
e e x dx
e x e x
dx
?
?
-
-
?
?
--
+
=
??
??
??
??
+ ??
??
??
??
??
??
?
?
 
 = 50 
8. Let for a triangle ABC, 
 
ˆ ˆ ˆ
23 AB i j k = - + + 
 
ˆ ˆ ˆ
CB i j k = ? + ? + ? 
 
ˆ ˆ ˆ
43 CA i j k = + + ? 
 If 0 ?? and the area of the triangle ABC is 5 6,
then CB CA ? is equal to 
 (1) 60 (2) 54 
 (3) 108 (4) 120 
Answer (1) 
Sol.  
 CA + AB = CB 
 +2i + 4j + (? + 3)k = ?i + ?j + ?k  
 ? ? = +2, ? = 4, ? = ? + 3 
 Area 
11
2 1 3 5 6
22
24
i j k
AB BC = ? = - =
+?
 
  ( ) ( )
( )
2
22
12 6 2 100 10 6 = ? - + + ? + = 
  ? 5?
2
 = 320 
Page 4


   
 
 
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
13/04/2023 
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 ?, ? be the roots of the equation 
2
2 2 0. xx - + = Then 
14 14
? + ? is equal to   
 (1) –64 (2) 64 2 - 
 (3) –128 (4) 128 2 - 
Answer (3) 
Sol. 
2
2 2 0 xx - + = 
 
26
2
i
x
?
= 
 Let 
2
1 3 1 3
,2
22
ii +-
? = ? = = ? 
 2 =? 
 
14 7 2 14 7
2 , 2 ? = ? ? = ? 
 ? ( )
14 14 7
21 ? + ? = - 
 = –2
7
 
2. The plane, passing through the points (0, –1, 2) and 
(–1, 2, 1) and parallel to the line passing through  
(5, 1, –7) and (1, –1, –1), also passes through the 
point 
 (1) (–2, 5, 0) (2) (1, –2, 1) 
 (3) (2, 0, 1) (4) (0, 5, –2) 
Answer (1) 
Sol. Let A(0, –1, 2) & B(–1, 2, 1) 
 So, normal vector to the plane 
 ( ) ( )
ˆ
ˆˆ
ˆ ˆ ˆ
3 1 1 16 10 (14)
6 42
j
ik
n i j k = - - = - - +
--
 
 = 
ˆ ˆ ˆ
16 10 14 i j k ++ 
 ? Equation of plane 
 16(x – 0) + 10(y + 1) + 14 (z – 2) = 0 
 ? 8x + 5y + 7z = 9 
 ? Option (1) is correct 
3. The statement (p ? (?q)) ? ((?p) ? q) ? ((?p) ? (?q)) 
is equivalent to _____ 
 (1) (?p) ? q (2) (?p) ? (?q) 
 (3) p ? (?q) (4) p ? q 
Answer (2) 
Sol. ( ) ( ) ( ) (~ ) (~ ) (~ ) (~ ) p q p q p q ? ? ? ? ? 
 = ( ) ( ) ~ ( ~ ) ~ p q q p q ? ? ? ? 
 = ( ) ~~ p p q ?? 
 = ( ) ( ) ~ ~ ~ p p p q ? ? ? 
 = ~~ pq ? 
4. Let the centre of a circle C be (?, ?) and its radius r 
< 8. Let 3x + 4y = 24 and 3x – 4y = 32 be two 
tangents and 4x + 3y = 1 be a normal to C. Then  
(? – ? + r) is equal to  
 (1) 7 (2) 5 
 (3) 6 (4) 9 
Answer (1) 
Sol. ? (?, ?) lies on 4x + 3y = 1 
 ? 4? + 3? = 1 …(i) 
 and 
3 4 24 3 4 32
55
? + ? - ? - ? -
= 
 Take (+ve) 
  3? + 4? – 24 = 3? – 4? – 32 
 ? 8? = –8 ? 1 ? = - 
 So 1 ?=  By equation (i) 
 And 
3 4 24
58
5
r
--
= = ? 
 Take (–ve) 
 3? + 4? – 24 = –3? + 4? + 32 
 6? = 56 ? 
56 1 4 28
and 1
6 3 3
? ??
? = ? = -
??
??
 
 For then r > 8 
 ? r = 5, ? = 1, ? = –1 
 ? ? – ? + r = 7 
 
   
  
5. The coefficient of x
5
 in the expansion of 
5
3
2
1
2
3
x
x
??
-
??
??
 is  
 (1) 
80
9
 (2) 9 
 (3) 8 (4) 
26
3
 
Answer (1) 
Sol. 
( )
5–
53
1
2
–1
2
3
r
r
rr
T C x
x
+
??
=
??
??
 
 15 – 3r – 2r = 5 
 ? r = 2 
 Coefficient ( )
3
5
2
1 80
2
99
C
??
==
??
??
 
6. All words, with or without meaning, are made using 
all the letters of the word MONDAY. These words 
are written as in a dictionary with serial numbers. 
The serial number of the word MONDAY is 
 (1) 327 (2) 328 
 (3) 324 (4) 326 
Answer (1) 
Sol.  
 
3 5 4 2 1 6
M O N D A Y
2 3 2 1 0 0
5! 4! 3! 2! 1! 0!
 
 ? Rank = (2 × 5! + 3 × 4! + 2 × 3! + 1 × 2!) + 1 
   = 240 + 72 + 12 + 2 + 1 = 327 
7. The value of 
( )
4
50
4
0
4
49 51
0
tan
tan tan
x
x
e e xdx
e x x dx
?
?
-
-
?
-
+
+
?
?
is 
 (1) 51 (2) 50 
 (3) 25 (4) 49 
Answer (2) 
Sol. 
( )
( )
4
50
4
0
4
49 2
0
tan
tan sec
x
x
e e x dx
e x x dx
?
-?
-
?
-
+
?
?
 
 
( )
4
50
4
0
4 50 50
4
0 0
tan
tan tan
50 50
x
xx
e e x dx
e x e x
dx
?
?
-
-
?
?
--
+
=
??
??
??
??
+ ??
??
??
??
??
??
?
?
 
 = 50 
8. Let for a triangle ABC, 
 
ˆ ˆ ˆ
23 AB i j k = - + + 
 
ˆ ˆ ˆ
CB i j k = ? + ? + ? 
 
ˆ ˆ ˆ
43 CA i j k = + + ? 
 If 0 ?? and the area of the triangle ABC is 5 6,
then CB CA ? is equal to 
 (1) 60 (2) 54 
 (3) 108 (4) 120 
Answer (1) 
Sol.  
 CA + AB = CB 
 +2i + 4j + (? + 3)k = ?i + ?j + ?k  
 ? ? = +2, ? = 4, ? = ? + 3 
 Area 
11
2 1 3 5 6
22
24
i j k
AB BC = ? = - =
+?
 
  ( ) ( )
( )
2
22
12 6 2 100 10 6 = ? - + + ? + = 
  ? 5?
2
 = 320 
 
   
  
   ?
2
 = 64 
   ? = 8 
   ? = 5 
 ( )( ) 2 4 8 4 3 5 CB CA i j k i j k ? = + + + + 
   = 8 + 12 + 40   
   = 60 
9. Let 
1 2 3
, , ,..... a a a be a G.P. of increasing positive 
numbers. Let the sum of its 6
th
 and 8
th
 terms be 2 
and the product of its 3
rd
 and 5
th
 terms be 
1
9
. Then 
2 4 4 6
6( )( ) a a a a ++ is equal to  
 (1) 3 (2) 33 
 (3) 2 (4) 22 
Answer (1) 
Sol. a6 + a8 = 2 
 ? ar
5
 + ar
7
 = 2 …(i) 
  
2 2 4
35
11
99
a a a r r ? = ? ? ? = 
 ? 
3
1
3
ar = 
  
24
2
33
rr
+= 
 ? r
4
 + r
2
 = 6 
 ? (r
2
 + 3) (r
2
 – 2) = 0 
 ? r
2
 = 2 
 ? 
11
2
36
ar ar ? = ? = 
 Now, 6(a2 + a4) (a4 + a6)  
 = 6(ar + ar
3
) (ar
3
 + ar
5
) 
 
1 1 1 2
6
6 3 3 3
? ?? ?
= + +
? ?? ?
? ?? ?
 
 
1
6 1 3
2
= ? ? = 
10. The range of 
2
1
2
( ) 4sin
1
x
fx
x
-
??
= ??
??
+
??
is  
 (1) [0, 2 ] ? (2) [0, ] ? 
 (3) [0, 2 ) ? (4) [0, ) ? 
Answer (3) 
Sol.  
2
22
1
11
11
x
xx
= - ?
++
 
 ? 
2
2
01
1
x
x
??
+
 
 ? 
2
1
2
0 sin
2
1
x
x
-
??
?
?? ??
??
+
??
 
 ? 
2
1
2
0 4sin 2
1
x
x
-
??
? ? ? ??
??
+
??
 
 ? Option (3) is correct. 
11. The random variable X follows binomial 
distribution B (n, p), for which the difference of the 
mean and the variance is 1.  
 If 2P (X = 2) = 3P (X = 1), then n
2
P (X > 1) is equal 
to   
 (1) 15 (2) 11 
 (3) 12 (4) 16 
Answer (2) 
Sol.  np – npq = 1 
 ? np(1 – q) = 1 
 ? np
2
 = 1 
  2P(X = 2) = 3P(X = 1) 
  
2 2 1
21
23
n n n n
C p q C p q
--
? = ? ? 
 ? 
( ) 1
23
2
nn
p n q
?-
? ? = ? ? 
 ? (n – 1)p = 3(1 – p) 
 ? ( )
2
1
1 3 1 pp
p
??
- = - ??
??
??
 
 ? 
( )( )
( )
11
31
pp
p
p
-+
=- 
 ? 1 + p = 3p 
 ? 
1
2
p = 
 ? n = 4 
 n
2
P(x > 1) = n
2
 (1 – P(x = 1) – P(x = 0)) 
   
44
4
1
11
16 1 11
22
C
??
? ? ? ?
?? = - ? - =
? ? ? ?
??
? ? ? ?
??
 
 Option (2) is correct. 
Page 5


   
 
 
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
13/04/2023 
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 ?, ? be the roots of the equation 
2
2 2 0. xx - + = Then 
14 14
? + ? is equal to   
 (1) –64 (2) 64 2 - 
 (3) –128 (4) 128 2 - 
Answer (3) 
Sol. 
2
2 2 0 xx - + = 
 
26
2
i
x
?
= 
 Let 
2
1 3 1 3
,2
22
ii +-
? = ? = = ? 
 2 =? 
 
14 7 2 14 7
2 , 2 ? = ? ? = ? 
 ? ( )
14 14 7
21 ? + ? = - 
 = –2
7
 
2. The plane, passing through the points (0, –1, 2) and 
(–1, 2, 1) and parallel to the line passing through  
(5, 1, –7) and (1, –1, –1), also passes through the 
point 
 (1) (–2, 5, 0) (2) (1, –2, 1) 
 (3) (2, 0, 1) (4) (0, 5, –2) 
Answer (1) 
Sol. Let A(0, –1, 2) & B(–1, 2, 1) 
 So, normal vector to the plane 
 ( ) ( )
ˆ
ˆˆ
ˆ ˆ ˆ
3 1 1 16 10 (14)
6 42
j
ik
n i j k = - - = - - +
--
 
 = 
ˆ ˆ ˆ
16 10 14 i j k ++ 
 ? Equation of plane 
 16(x – 0) + 10(y + 1) + 14 (z – 2) = 0 
 ? 8x + 5y + 7z = 9 
 ? Option (1) is correct 
3. The statement (p ? (?q)) ? ((?p) ? q) ? ((?p) ? (?q)) 
is equivalent to _____ 
 (1) (?p) ? q (2) (?p) ? (?q) 
 (3) p ? (?q) (4) p ? q 
Answer (2) 
Sol. ( ) ( ) ( ) (~ ) (~ ) (~ ) (~ ) p q p q p q ? ? ? ? ? 
 = ( ) ( ) ~ ( ~ ) ~ p q q p q ? ? ? ? 
 = ( ) ~~ p p q ?? 
 = ( ) ( ) ~ ~ ~ p p p q ? ? ? 
 = ~~ pq ? 
4. Let the centre of a circle C be (?, ?) and its radius r 
< 8. Let 3x + 4y = 24 and 3x – 4y = 32 be two 
tangents and 4x + 3y = 1 be a normal to C. Then  
(? – ? + r) is equal to  
 (1) 7 (2) 5 
 (3) 6 (4) 9 
Answer (1) 
Sol. ? (?, ?) lies on 4x + 3y = 1 
 ? 4? + 3? = 1 …(i) 
 and 
3 4 24 3 4 32
55
? + ? - ? - ? -
= 
 Take (+ve) 
  3? + 4? – 24 = 3? – 4? – 32 
 ? 8? = –8 ? 1 ? = - 
 So 1 ?=  By equation (i) 
 And 
3 4 24
58
5
r
--
= = ? 
 Take (–ve) 
 3? + 4? – 24 = –3? + 4? + 32 
 6? = 56 ? 
56 1 4 28
and 1
6 3 3
? ??
? = ? = -
??
??
 
 For then r > 8 
 ? r = 5, ? = 1, ? = –1 
 ? ? – ? + r = 7 
 
   
  
5. The coefficient of x
5
 in the expansion of 
5
3
2
1
2
3
x
x
??
-
??
??
 is  
 (1) 
80
9
 (2) 9 
 (3) 8 (4) 
26
3
 
Answer (1) 
Sol. 
( )
5–
53
1
2
–1
2
3
r
r
rr
T C x
x
+
??
=
??
??
 
 15 – 3r – 2r = 5 
 ? r = 2 
 Coefficient ( )
3
5
2
1 80
2
99
C
??
==
??
??
 
6. All words, with or without meaning, are made using 
all the letters of the word MONDAY. These words 
are written as in a dictionary with serial numbers. 
The serial number of the word MONDAY is 
 (1) 327 (2) 328 
 (3) 324 (4) 326 
Answer (1) 
Sol.  
 
3 5 4 2 1 6
M O N D A Y
2 3 2 1 0 0
5! 4! 3! 2! 1! 0!
 
 ? Rank = (2 × 5! + 3 × 4! + 2 × 3! + 1 × 2!) + 1 
   = 240 + 72 + 12 + 2 + 1 = 327 
7. The value of 
( )
4
50
4
0
4
49 51
0
tan
tan tan
x
x
e e xdx
e x x dx
?
?
-
-
?
-
+
+
?
?
is 
 (1) 51 (2) 50 
 (3) 25 (4) 49 
Answer (2) 
Sol. 
( )
( )
4
50
4
0
4
49 2
0
tan
tan sec
x
x
e e x dx
e x x dx
?
-?
-
?
-
+
?
?
 
 
( )
4
50
4
0
4 50 50
4
0 0
tan
tan tan
50 50
x
xx
e e x dx
e x e x
dx
?
?
-
-
?
?
--
+
=
??
??
??
??
+ ??
??
??
??
??
??
?
?
 
 = 50 
8. Let for a triangle ABC, 
 
ˆ ˆ ˆ
23 AB i j k = - + + 
 
ˆ ˆ ˆ
CB i j k = ? + ? + ? 
 
ˆ ˆ ˆ
43 CA i j k = + + ? 
 If 0 ?? and the area of the triangle ABC is 5 6,
then CB CA ? is equal to 
 (1) 60 (2) 54 
 (3) 108 (4) 120 
Answer (1) 
Sol.  
 CA + AB = CB 
 +2i + 4j + (? + 3)k = ?i + ?j + ?k  
 ? ? = +2, ? = 4, ? = ? + 3 
 Area 
11
2 1 3 5 6
22
24
i j k
AB BC = ? = - =
+?
 
  ( ) ( )
( )
2
22
12 6 2 100 10 6 = ? - + + ? + = 
  ? 5?
2
 = 320 
 
   
  
   ?
2
 = 64 
   ? = 8 
   ? = 5 
 ( )( ) 2 4 8 4 3 5 CB CA i j k i j k ? = + + + + 
   = 8 + 12 + 40   
   = 60 
9. Let 
1 2 3
, , ,..... a a a be a G.P. of increasing positive 
numbers. Let the sum of its 6
th
 and 8
th
 terms be 2 
and the product of its 3
rd
 and 5
th
 terms be 
1
9
. Then 
2 4 4 6
6( )( ) a a a a ++ is equal to  
 (1) 3 (2) 33 
 (3) 2 (4) 22 
Answer (1) 
Sol. a6 + a8 = 2 
 ? ar
5
 + ar
7
 = 2 …(i) 
  
2 2 4
35
11
99
a a a r r ? = ? ? ? = 
 ? 
3
1
3
ar = 
  
24
2
33
rr
+= 
 ? r
4
 + r
2
 = 6 
 ? (r
2
 + 3) (r
2
 – 2) = 0 
 ? r
2
 = 2 
 ? 
11
2
36
ar ar ? = ? = 
 Now, 6(a2 + a4) (a4 + a6)  
 = 6(ar + ar
3
) (ar
3
 + ar
5
) 
 
1 1 1 2
6
6 3 3 3
? ?? ?
= + +
? ?? ?
? ?? ?
 
 
1
6 1 3
2
= ? ? = 
10. The range of 
2
1
2
( ) 4sin
1
x
fx
x
-
??
= ??
??
+
??
is  
 (1) [0, 2 ] ? (2) [0, ] ? 
 (3) [0, 2 ) ? (4) [0, ) ? 
Answer (3) 
Sol.  
2
22
1
11
11
x
xx
= - ?
++
 
 ? 
2
2
01
1
x
x
??
+
 
 ? 
2
1
2
0 sin
2
1
x
x
-
??
?
?? ??
??
+
??
 
 ? 
2
1
2
0 4sin 2
1
x
x
-
??
? ? ? ??
??
+
??
 
 ? Option (3) is correct. 
11. The random variable X follows binomial 
distribution B (n, p), for which the difference of the 
mean and the variance is 1.  
 If 2P (X = 2) = 3P (X = 1), then n
2
P (X > 1) is equal 
to   
 (1) 15 (2) 11 
 (3) 12 (4) 16 
Answer (2) 
Sol.  np – npq = 1 
 ? np(1 – q) = 1 
 ? np
2
 = 1 
  2P(X = 2) = 3P(X = 1) 
  
2 2 1
21
23
n n n n
C p q C p q
--
? = ? ? 
 ? 
( ) 1
23
2
nn
p n q
?-
? ? = ? ? 
 ? (n – 1)p = 3(1 – p) 
 ? ( )
2
1
1 3 1 pp
p
??
- = - ??
??
??
 
 ? 
( )( )
( )
11
31
pp
p
p
-+
=- 
 ? 1 + p = 3p 
 ? 
1
2
p = 
 ? n = 4 
 n
2
P(x > 1) = n
2
 (1 – P(x = 1) – P(x = 0)) 
   
44
4
1
11
16 1 11
22
C
??
? ? ? ?
?? = - ? - =
? ? ? ?
??
? ? ? ?
??
 
 Option (2) is correct. 
 
   
  
12. The line, that is coplanar to the line 
3 1 5
3 1 5
x y z
, is  
 (1) 
1 2 5
1 2 4
x y z
 
 (2) 
1 2 5
1 2 5
x y z
 
 (3) 
1 2 5
1 2 5
x y z
  
 (4) 
1 2 5
1 2 5
x y z
 
Answer (2) 
Sol. Given line : 
3 1 5
3 1 5
x y z + - -
==
-
 
 No line in given options is parallel to given line. 
 ? If two lines are coplanar, then distance between 
them must be zero. 
 Among the given option, only option (2) satisfies 
above condition. 
  
2 1 0
3 1 5
1 2 5
--
-
-
 
   = –2(–5) + 1 (–15 + 5) = 0. 
 ? Option (2) is correct. 
13.  The area of the region 
22
, : 4 , 1 x y x y x y is  
 (1) 
( )
4
4 2 1
3
- (2) 
( )
4
4 2 1
3
+ 
 (3) 
( )
3
4 2 1
4
+ (4) 
( )
3
4 2 1
4
- 
Answer (1) 
Sol.  
 Required area 
22
14
2 4– ydy ydy
??
??
= +
??
??
??
 
 
2
3/2
4
3/2
2
1
2(4 – )
2
3 –
3
2
y
y
??
?
?
??
? =
?
??
?
?
?
??
??
 
 
( )
4
4 2 – 1
3
= 
14. Let 2, 3 ab and the angle between the 
vectors a and b be 
4
. Then 
( ) ( )
2
2 2 3 a b a b + ? - is equal to 
 (1) 441 (2) 482 
 (3) 841 (4) 882 
Answer (4) 
Sol. 2 a = 
 4 b = 
 
4
ab
?
?= 
 
( ) ( )
2
2 2 3 a b a b +?- 
  
( ) ( )
2
34 a b b a = - ? + ? 
  
( )
2
7ba =? 
  
22
2
49 sin
4
ab
?
= 
  
1
49 4 9
2
? ? ? 
  = 882 
15. If the system of equations  
 2x + y – z = 5 
 2x – 5y + ?z = ? 
 x + 2y – 5z = 7 
 has infinitely many solutions, then 
( ) ( )
22
? + ? + ? - ? is equal to 
 (1) 904 (2) 916 
 (3) 912 (4) 920 
Answer (2)