JEE Main 2023 April 11 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) 
 
 
  
11/04/2023 
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) 
 
 
  
11/04/2023 
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. The domain of the function  
 ( )
? ? ? ?
2
1
3 10
=
--
fx
xx
 is (where [x] denotes the 
greatest integer less than or equal to x) 
 (1) ( –?, –3] ? (5, ?)  (2) ( –?, –2) ? [6, ?) 
 (3) ( –?, –2) ? (5, ?) (4) ( –?, –3] ? [6, ?) 
Answer (2) 
Sol. ( )
? ? ? ?
2
1
3 10
fx
xx
=
--
 
 For Domain [x]
2
 – 3[x] – 10 > 0 
 ? ([x] – 5) ([x] + 2) > 0 
 ? [x] ?( –?, –2) ? (5, ?)  
 ? x ?( –?, –2) ? [6, ?) 
2. If the system of linear equations 
 7x + 11y + ?z = 13 
 5x + 4y + 7z = ? 
 175x + 194y + 57z = 361 
 has infinitely many solutions, then ? + ? + 2 is equal 
to 
 (1) 4 (2) 3 
 (3) 5 (4) 6 
Answer (1) 
Sol. 7x + 11y + ?z = 13 
 5x + 4y + 7z = ? 
 175x + 194y + 57z = 361 
 For infinite solution, 
7 11
5 4 7 0
175 194 57
?
= 
 ? 
7 11
5 4 7 0
0 81 57 25
?
=
- - ?
 
 ? 81(49 – 5?) + (57 – 25?) ( –27) = 0 
 ? 270? = –2430 ? –9 ?= 
 and ?1 = 0 
 
1 3 1 1 – 9
 4 7 0
361 194 57
?= 
 ? 11 ?= 
 ? ? + ? + 2 = 4 
3. Let A = {1, 3, 4, 6, 9} and B = {2, 4, 5, 8, 10}. Let R 
be a relation defined on A × B such that R = {((a1, 
b1), (a2, b2)) : a1 ? b2 and b1 ? a2}. Then the number 
of elements in the set R is 
 (1) 160 (2) 52 
 (3) 26 (4) 180 
Answer (1) 
Sol. ? ? ? ? 1 , 3, 4, 6, 9 2, 4, 5, 8, 10 AB == 
 ( ) ( ) ( ) ? ?
1 1 2 2 1 2 1 2
, , ,  :  and R a b a b a b b a = ? ? 
For  Similarly for (b1, a2) we 
will have 
a1 = 1 b2 ? 5 b1 ? 2 a2 = 4 
a1 = 3 b2 ? 4 b1 ? 4 a2 = 3 
a1 = 4 b2 ? 4 b1 ? 5 a2 = 2 
a1 = 6 b2 ? 2 b1 ? 8 a2 = 1 
a1 = 9 b2 ? 1 b1 ? 10 a2 = 0 
 16 cases  10 cases 
 ? Total elements in relation = 16 × 10 = 160 
4. Let the mean of 6 observations 1, 2, 4, 5, x and y 
be 5 and their variance be 10. Then their mean 
deviation about the mean is equal to 
 (1) 
7
3
 (2) 3 
 (3) 
8
3
 (4) 
10
3
 
Answer (3) 
Page 3


   
 
 
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
11/04/2023 
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. The domain of the function  
 ( )
? ? ? ?
2
1
3 10
=
--
fx
xx
 is (where [x] denotes the 
greatest integer less than or equal to x) 
 (1) ( –?, –3] ? (5, ?)  (2) ( –?, –2) ? [6, ?) 
 (3) ( –?, –2) ? (5, ?) (4) ( –?, –3] ? [6, ?) 
Answer (2) 
Sol. ( )
? ? ? ?
2
1
3 10
fx
xx
=
--
 
 For Domain [x]
2
 – 3[x] – 10 > 0 
 ? ([x] – 5) ([x] + 2) > 0 
 ? [x] ?( –?, –2) ? (5, ?)  
 ? x ?( –?, –2) ? [6, ?) 
2. If the system of linear equations 
 7x + 11y + ?z = 13 
 5x + 4y + 7z = ? 
 175x + 194y + 57z = 361 
 has infinitely many solutions, then ? + ? + 2 is equal 
to 
 (1) 4 (2) 3 
 (3) 5 (4) 6 
Answer (1) 
Sol. 7x + 11y + ?z = 13 
 5x + 4y + 7z = ? 
 175x + 194y + 57z = 361 
 For infinite solution, 
7 11
5 4 7 0
175 194 57
?
= 
 ? 
7 11
5 4 7 0
0 81 57 25
?
=
- - ?
 
 ? 81(49 – 5?) + (57 – 25?) ( –27) = 0 
 ? 270? = –2430 ? –9 ?= 
 and ?1 = 0 
 
1 3 1 1 – 9
 4 7 0
361 194 57
?= 
 ? 11 ?= 
 ? ? + ? + 2 = 4 
3. Let A = {1, 3, 4, 6, 9} and B = {2, 4, 5, 8, 10}. Let R 
be a relation defined on A × B such that R = {((a1, 
b1), (a2, b2)) : a1 ? b2 and b1 ? a2}. Then the number 
of elements in the set R is 
 (1) 160 (2) 52 
 (3) 26 (4) 180 
Answer (1) 
Sol. ? ? ? ? 1 , 3, 4, 6, 9 2, 4, 5, 8, 10 AB == 
 ( ) ( ) ( ) ? ?
1 1 2 2 1 2 1 2
, , ,  :  and R a b a b a b b a = ? ? 
For  Similarly for (b1, a2) we 
will have 
a1 = 1 b2 ? 5 b1 ? 2 a2 = 4 
a1 = 3 b2 ? 4 b1 ? 4 a2 = 3 
a1 = 4 b2 ? 4 b1 ? 5 a2 = 2 
a1 = 6 b2 ? 2 b1 ? 8 a2 = 1 
a1 = 9 b2 ? 1 b1 ? 10 a2 = 0 
 16 cases  10 cases 
 ? Total elements in relation = 16 × 10 = 160 
4. Let the mean of 6 observations 1, 2, 4, 5, x and y 
be 5 and their variance be 10. Then their mean 
deviation about the mean is equal to 
 (1) 
7
3
 (2) 3 
 (3) 
8
3
 (4) 
10
3
 
Answer (3) 
 
   
  
Sol. 12 + x + y = 30 
 ? x + y = 18 
 and 
22
46
25 10
6
xy ++
-= 
 ? x = 10, y = 8 
 Now, mean deviation about mean  
 
4 3 1 0 5 3 8
63
+ + + + +
== 
5. If four distinct points with position vectors 
, ,  and a b c d are coplanar, then   
??
??
a b c is equal 
to   
 (1)       
? ? ? ? ? ?
++
? ? ? ? ? ?
d b a a c d d b c 
 (2)         
? ? ? ? ? ?
++
? ? ? ? ? ?
a d b d c a d b c 
 (3) 
? ?
        
? ? ? ?
++
? ? ? ?
d c a b d a c d b 
 (4) 
? ?
     c  
? ? ? ?
++
? ? ? ?
b c d d a d b a 
Answer (3) 
Sol.     0 b a c a d a
??
- - - =
??
 
 
( ) ( ) ( )
0 b a c a d a
??
- ? - ? - =
??
 
 
( ) ( )
0 b a c d c a a d - ? ? - ? - ? = 
         0 b c d b c a b a d a c d
? ? ? ? ? ? ? ?
- - - =
? ? ? ? ? ? ? ?
 
 ?          a b c b c d a b d a d c
? ? ? ? ? ? ? ?
= + +
? ? ? ? ? ? ? ?
 
        d c a b d a c d b
? ? ? ? ? ?
= + +
? ? ? ? ? ?
 
6. Let () y y x = be the solution of the differential 
equation 
 
( )
2
5
57
1
5
,0
( 1)
x
dy
yx
dx
x x x
+
+ = ?
+
. If (1) 2, y = then 
(2) y is equal to 
 (1) 
637
128
 (2) 
679
128
 
 (3) 
693
128
 (4) 
697
128
 
Answer (3) 
Sol. I.F. =
5
5
(1 )
dx
xx
e
+
?
 
 = 
6
5
5
1
1
dx
x
x
e
??
+
??
??
?
 
 Put 
5
1
1 t
x
+= 
 
6
5
dx dt
x
-
= 
 
5
ln
5
1
1
dt
t
t
x
ee
t
x
-
-
?
= = =
+
 
 
5 5 5 2
5 5 7
(1 )
11
x x x
y dx c
x x x
??
+
=+ ??
??
++
??
?
 
 
54
5
1
4
1
xx
yc
x
x
??
-
= + + ??
??
+
??
 
 Now (1) 2 y = , then 
7
4
c = 
 ? 
693
(2)
128
y =  
7. Let P be the plane passing through the points (5, 3, 
0), (13, 3, –2) and (1, 6, 2). For ?? , if the 
distances of the points A (3,4, ) ? and B (2, , ) a ? from 
the plane P are 2 and 3 respectively, then the 
positive value of a is 
 (1) 6 (2) 3 
 (3) 5 (4) 4 
Answer (4) 
Sol.   
 
ˆˆ
82 AB i k =- 
 
ˆ ˆ ˆ
4 3 2 AC i j k = - + + 
 
ˆ ˆ ˆ
8 0 2
4 3 2
i j k
AB AC ? = -
-
 
  
ˆ ˆ ˆ
6 8 24 i j k = - + 
Page 4


   
 
 
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
11/04/2023 
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. The domain of the function  
 ( )
? ? ? ?
2
1
3 10
=
--
fx
xx
 is (where [x] denotes the 
greatest integer less than or equal to x) 
 (1) ( –?, –3] ? (5, ?)  (2) ( –?, –2) ? [6, ?) 
 (3) ( –?, –2) ? (5, ?) (4) ( –?, –3] ? [6, ?) 
Answer (2) 
Sol. ( )
? ? ? ?
2
1
3 10
fx
xx
=
--
 
 For Domain [x]
2
 – 3[x] – 10 > 0 
 ? ([x] – 5) ([x] + 2) > 0 
 ? [x] ?( –?, –2) ? (5, ?)  
 ? x ?( –?, –2) ? [6, ?) 
2. If the system of linear equations 
 7x + 11y + ?z = 13 
 5x + 4y + 7z = ? 
 175x + 194y + 57z = 361 
 has infinitely many solutions, then ? + ? + 2 is equal 
to 
 (1) 4 (2) 3 
 (3) 5 (4) 6 
Answer (1) 
Sol. 7x + 11y + ?z = 13 
 5x + 4y + 7z = ? 
 175x + 194y + 57z = 361 
 For infinite solution, 
7 11
5 4 7 0
175 194 57
?
= 
 ? 
7 11
5 4 7 0
0 81 57 25
?
=
- - ?
 
 ? 81(49 – 5?) + (57 – 25?) ( –27) = 0 
 ? 270? = –2430 ? –9 ?= 
 and ?1 = 0 
 
1 3 1 1 – 9
 4 7 0
361 194 57
?= 
 ? 11 ?= 
 ? ? + ? + 2 = 4 
3. Let A = {1, 3, 4, 6, 9} and B = {2, 4, 5, 8, 10}. Let R 
be a relation defined on A × B such that R = {((a1, 
b1), (a2, b2)) : a1 ? b2 and b1 ? a2}. Then the number 
of elements in the set R is 
 (1) 160 (2) 52 
 (3) 26 (4) 180 
Answer (1) 
Sol. ? ? ? ? 1 , 3, 4, 6, 9 2, 4, 5, 8, 10 AB == 
 ( ) ( ) ( ) ? ?
1 1 2 2 1 2 1 2
, , ,  :  and R a b a b a b b a = ? ? 
For  Similarly for (b1, a2) we 
will have 
a1 = 1 b2 ? 5 b1 ? 2 a2 = 4 
a1 = 3 b2 ? 4 b1 ? 4 a2 = 3 
a1 = 4 b2 ? 4 b1 ? 5 a2 = 2 
a1 = 6 b2 ? 2 b1 ? 8 a2 = 1 
a1 = 9 b2 ? 1 b1 ? 10 a2 = 0 
 16 cases  10 cases 
 ? Total elements in relation = 16 × 10 = 160 
4. Let the mean of 6 observations 1, 2, 4, 5, x and y 
be 5 and their variance be 10. Then their mean 
deviation about the mean is equal to 
 (1) 
7
3
 (2) 3 
 (3) 
8
3
 (4) 
10
3
 
Answer (3) 
 
   
  
Sol. 12 + x + y = 30 
 ? x + y = 18 
 and 
22
46
25 10
6
xy ++
-= 
 ? x = 10, y = 8 
 Now, mean deviation about mean  
 
4 3 1 0 5 3 8
63
+ + + + +
== 
5. If four distinct points with position vectors 
, ,  and a b c d are coplanar, then   
??
??
a b c is equal 
to   
 (1)       
? ? ? ? ? ?
++
? ? ? ? ? ?
d b a a c d d b c 
 (2)         
? ? ? ? ? ?
++
? ? ? ? ? ?
a d b d c a d b c 
 (3) 
? ?
        
? ? ? ?
++
? ? ? ?
d c a b d a c d b 
 (4) 
? ?
     c  
? ? ? ?
++
? ? ? ?
b c d d a d b a 
Answer (3) 
Sol.     0 b a c a d a
??
- - - =
??
 
 
( ) ( ) ( )
0 b a c a d a
??
- ? - ? - =
??
 
 
( ) ( )
0 b a c d c a a d - ? ? - ? - ? = 
         0 b c d b c a b a d a c d
? ? ? ? ? ? ? ?
- - - =
? ? ? ? ? ? ? ?
 
 ?          a b c b c d a b d a d c
? ? ? ? ? ? ? ?
= + +
? ? ? ? ? ? ? ?
 
        d c a b d a c d b
? ? ? ? ? ?
= + +
? ? ? ? ? ?
 
6. Let () y y x = be the solution of the differential 
equation 
 
( )
2
5
57
1
5
,0
( 1)
x
dy
yx
dx
x x x
+
+ = ?
+
. If (1) 2, y = then 
(2) y is equal to 
 (1) 
637
128
 (2) 
679
128
 
 (3) 
693
128
 (4) 
697
128
 
Answer (3) 
Sol. I.F. =
5
5
(1 )
dx
xx
e
+
?
 
 = 
6
5
5
1
1
dx
x
x
e
??
+
??
??
?
 
 Put 
5
1
1 t
x
+= 
 
6
5
dx dt
x
-
= 
 
5
ln
5
1
1
dt
t
t
x
ee
t
x
-
-
?
= = =
+
 
 
5 5 5 2
5 5 7
(1 )
11
x x x
y dx c
x x x
??
+
=+ ??
??
++
??
?
 
 
54
5
1
4
1
xx
yc
x
x
??
-
= + + ??
??
+
??
 
 Now (1) 2 y = , then 
7
4
c = 
 ? 
693
(2)
128
y =  
7. Let P be the plane passing through the points (5, 3, 
0), (13, 3, –2) and (1, 6, 2). For ?? , if the 
distances of the points A (3,4, ) ? and B (2, , ) a ? from 
the plane P are 2 and 3 respectively, then the 
positive value of a is 
 (1) 6 (2) 3 
 (3) 5 (4) 4 
Answer (4) 
Sol.   
 
ˆˆ
82 AB i k =- 
 
ˆ ˆ ˆ
4 3 2 AC i j k = - + + 
 
ˆ ˆ ˆ
8 0 2
4 3 2
i j k
AB AC ? = -
-
 
  
ˆ ˆ ˆ
6 8 24 i j k = - + 
 
   
  
 Equation of plane : 6x – 8y + 24z = d passes 
through (5, 3, 0) 
 6 × 5 – 8 × 3 + 24 × 0 = d 
 d = 6 
  6x – 8y + 24z = 6 ? 3x – 4y + 12z = 3 
 Distance of point (3, 4, ?)  
 
9 16 12 3
23
9 16 144
- + ? -
= ? ? =
++
 
 Distance of point (2, ?, a) 
 
3 2 4 3 12 3
3
13
a ? - ? + ? -
= 
 12a – 9 = 39 
 12a = 48 
 a = 4 
8. The converse of ((~ ) ) p q r ?? is 
 (1) ((~ ) ) p q r ?? (2) (~ ) r p q ?? 
 (3) (~ ) ((~ ) ) r p q ?? (4) ( (~ )) (~ ) p q r ?? 
Answer (4) 
Sol. Converse of (?p ? q) ? r is  
 r ? (~p ? q) 
 (p ? (~q)) ? (~r) 
9. Let the function 
? ?
:  0, 2 ? f be defined as 
  
? ?
2
min , [ ]
log ,
, [0, 1)
()
[1 , 2]
e
x x x
xx
ex
fx
ex
-
-??
??
?
?
?
=
?
?
?
?
 
 where [t] denotes the greatest integer less than or 
equal to t. Then the value of the integral 
2
0
() xf x dx
?
 
is 
 (1) 
3
1
2
e
+ (2) 
2
1
( 1)
2
ee
??
-+
??
??
 
 (3) 21 e - (4) 
1
2
2
e - 
Answer (4) 
Sol. 
? ?
2
min ,
[0, 1)
()
[1 , 2]
xx
ex
fx
ex
?
?
?
=
?
?
?
?
 
 
2
[0, 1)
[1 , 2]
x
ex
ex
?
?
?
=
?
? ?
?
 
  ln [1 , 2) xx -? for [1 , 2] x ? 
 ? [ ln ] 1 xx -= 
 
2
2 1 2
0 0 1
( ) . .
x
x f x x e dx x edx =+
? ? ?
 
 
2
xt = ? 2xdx dt = 
 
2
1
2
0
1
1
22
t
x
e dt e =+
?
 
 ?
13
22
ee - ??
+
??
??
?
1
2
2
e - 
 Option (4) is correct. 
10. If : f ? be a continuous function satisfying 
24
00
(sin2 )sin (cos2 )cos 0 f x xdx f x x dx
??
+ ? =
??
, then 
the value of ? is 
 (1) 2 (2) 3 - 
 (3) 3 (4) 2 - 
Answer (4) 
Sol. 
24
00
(sin2 )sin (cos2 ).cos 0 f x xdx f x xdx
??
+ ? =
??
 
 Let 
2
0
(sin2 ).sin I f x xdx
?
=
?
 
 
42
0
4
(sin2 )sin (sin2 ).sin f x xdx f x xdx
??
?
=+
??
 
 
4
0
(cos2 )sin
4
f x x dx
?
???
=-
??
??
?
 
  
4
0
sin2 sin
44
f x x dx
?
?? ?? ? ? ? ?
+ + +
??
? ? ? ?
? ? ? ? ??
?
 
  
4
0
11
(cos2 ) cos sin
22
f x x x dx
?
??
=-
??
??
?
 
Page 5


   
 
 
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
11/04/2023 
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. The domain of the function  
 ( )
? ? ? ?
2
1
3 10
=
--
fx
xx
 is (where [x] denotes the 
greatest integer less than or equal to x) 
 (1) ( –?, –3] ? (5, ?)  (2) ( –?, –2) ? [6, ?) 
 (3) ( –?, –2) ? (5, ?) (4) ( –?, –3] ? [6, ?) 
Answer (2) 
Sol. ( )
? ? ? ?
2
1
3 10
fx
xx
=
--
 
 For Domain [x]
2
 – 3[x] – 10 > 0 
 ? ([x] – 5) ([x] + 2) > 0 
 ? [x] ?( –?, –2) ? (5, ?)  
 ? x ?( –?, –2) ? [6, ?) 
2. If the system of linear equations 
 7x + 11y + ?z = 13 
 5x + 4y + 7z = ? 
 175x + 194y + 57z = 361 
 has infinitely many solutions, then ? + ? + 2 is equal 
to 
 (1) 4 (2) 3 
 (3) 5 (4) 6 
Answer (1) 
Sol. 7x + 11y + ?z = 13 
 5x + 4y + 7z = ? 
 175x + 194y + 57z = 361 
 For infinite solution, 
7 11
5 4 7 0
175 194 57
?
= 
 ? 
7 11
5 4 7 0
0 81 57 25
?
=
- - ?
 
 ? 81(49 – 5?) + (57 – 25?) ( –27) = 0 
 ? 270? = –2430 ? –9 ?= 
 and ?1 = 0 
 
1 3 1 1 – 9
 4 7 0
361 194 57
?= 
 ? 11 ?= 
 ? ? + ? + 2 = 4 
3. Let A = {1, 3, 4, 6, 9} and B = {2, 4, 5, 8, 10}. Let R 
be a relation defined on A × B such that R = {((a1, 
b1), (a2, b2)) : a1 ? b2 and b1 ? a2}. Then the number 
of elements in the set R is 
 (1) 160 (2) 52 
 (3) 26 (4) 180 
Answer (1) 
Sol. ? ? ? ? 1 , 3, 4, 6, 9 2, 4, 5, 8, 10 AB == 
 ( ) ( ) ( ) ? ?
1 1 2 2 1 2 1 2
, , ,  :  and R a b a b a b b a = ? ? 
For  Similarly for (b1, a2) we 
will have 
a1 = 1 b2 ? 5 b1 ? 2 a2 = 4 
a1 = 3 b2 ? 4 b1 ? 4 a2 = 3 
a1 = 4 b2 ? 4 b1 ? 5 a2 = 2 
a1 = 6 b2 ? 2 b1 ? 8 a2 = 1 
a1 = 9 b2 ? 1 b1 ? 10 a2 = 0 
 16 cases  10 cases 
 ? Total elements in relation = 16 × 10 = 160 
4. Let the mean of 6 observations 1, 2, 4, 5, x and y 
be 5 and their variance be 10. Then their mean 
deviation about the mean is equal to 
 (1) 
7
3
 (2) 3 
 (3) 
8
3
 (4) 
10
3
 
Answer (3) 
 
   
  
Sol. 12 + x + y = 30 
 ? x + y = 18 
 and 
22
46
25 10
6
xy ++
-= 
 ? x = 10, y = 8 
 Now, mean deviation about mean  
 
4 3 1 0 5 3 8
63
+ + + + +
== 
5. If four distinct points with position vectors 
, ,  and a b c d are coplanar, then   
??
??
a b c is equal 
to   
 (1)       
? ? ? ? ? ?
++
? ? ? ? ? ?
d b a a c d d b c 
 (2)         
? ? ? ? ? ?
++
? ? ? ? ? ?
a d b d c a d b c 
 (3) 
? ?
        
? ? ? ?
++
? ? ? ?
d c a b d a c d b 
 (4) 
? ?
     c  
? ? ? ?
++
? ? ? ?
b c d d a d b a 
Answer (3) 
Sol.     0 b a c a d a
??
- - - =
??
 
 
( ) ( ) ( )
0 b a c a d a
??
- ? - ? - =
??
 
 
( ) ( )
0 b a c d c a a d - ? ? - ? - ? = 
         0 b c d b c a b a d a c d
? ? ? ? ? ? ? ?
- - - =
? ? ? ? ? ? ? ?
 
 ?          a b c b c d a b d a d c
? ? ? ? ? ? ? ?
= + +
? ? ? ? ? ? ? ?
 
        d c a b d a c d b
? ? ? ? ? ?
= + +
? ? ? ? ? ?
 
6. Let () y y x = be the solution of the differential 
equation 
 
( )
2
5
57
1
5
,0
( 1)
x
dy
yx
dx
x x x
+
+ = ?
+
. If (1) 2, y = then 
(2) y is equal to 
 (1) 
637
128
 (2) 
679
128
 
 (3) 
693
128
 (4) 
697
128
 
Answer (3) 
Sol. I.F. =
5
5
(1 )
dx
xx
e
+
?
 
 = 
6
5
5
1
1
dx
x
x
e
??
+
??
??
?
 
 Put 
5
1
1 t
x
+= 
 
6
5
dx dt
x
-
= 
 
5
ln
5
1
1
dt
t
t
x
ee
t
x
-
-
?
= = =
+
 
 
5 5 5 2
5 5 7
(1 )
11
x x x
y dx c
x x x
??
+
=+ ??
??
++
??
?
 
 
54
5
1
4
1
xx
yc
x
x
??
-
= + + ??
??
+
??
 
 Now (1) 2 y = , then 
7
4
c = 
 ? 
693
(2)
128
y =  
7. Let P be the plane passing through the points (5, 3, 
0), (13, 3, –2) and (1, 6, 2). For ?? , if the 
distances of the points A (3,4, ) ? and B (2, , ) a ? from 
the plane P are 2 and 3 respectively, then the 
positive value of a is 
 (1) 6 (2) 3 
 (3) 5 (4) 4 
Answer (4) 
Sol.   
 
ˆˆ
82 AB i k =- 
 
ˆ ˆ ˆ
4 3 2 AC i j k = - + + 
 
ˆ ˆ ˆ
8 0 2
4 3 2
i j k
AB AC ? = -
-
 
  
ˆ ˆ ˆ
6 8 24 i j k = - + 
 
   
  
 Equation of plane : 6x – 8y + 24z = d passes 
through (5, 3, 0) 
 6 × 5 – 8 × 3 + 24 × 0 = d 
 d = 6 
  6x – 8y + 24z = 6 ? 3x – 4y + 12z = 3 
 Distance of point (3, 4, ?)  
 
9 16 12 3
23
9 16 144
- + ? -
= ? ? =
++
 
 Distance of point (2, ?, a) 
 
3 2 4 3 12 3
3
13
a ? - ? + ? -
= 
 12a – 9 = 39 
 12a = 48 
 a = 4 
8. The converse of ((~ ) ) p q r ?? is 
 (1) ((~ ) ) p q r ?? (2) (~ ) r p q ?? 
 (3) (~ ) ((~ ) ) r p q ?? (4) ( (~ )) (~ ) p q r ?? 
Answer (4) 
Sol. Converse of (?p ? q) ? r is  
 r ? (~p ? q) 
 (p ? (~q)) ? (~r) 
9. Let the function 
? ?
:  0, 2 ? f be defined as 
  
? ?
2
min , [ ]
log ,
, [0, 1)
()
[1 , 2]
e
x x x
xx
ex
fx
ex
-
-??
??
?
?
?
=
?
?
?
?
 
 where [t] denotes the greatest integer less than or 
equal to t. Then the value of the integral 
2
0
() xf x dx
?
 
is 
 (1) 
3
1
2
e
+ (2) 
2
1
( 1)
2
ee
??
-+
??
??
 
 (3) 21 e - (4) 
1
2
2
e - 
Answer (4) 
Sol. 
? ?
2
min ,
[0, 1)
()
[1 , 2]
xx
ex
fx
ex
?
?
?
=
?
?
?
?
 
 
2
[0, 1)
[1 , 2]
x
ex
ex
?
?
?
=
?
? ?
?
 
  ln [1 , 2) xx -? for [1 , 2] x ? 
 ? [ ln ] 1 xx -= 
 
2
2 1 2
0 0 1
( ) . .
x
x f x x e dx x edx =+
? ? ?
 
 
2
xt = ? 2xdx dt = 
 
2
1
2
0
1
1
22
t
x
e dt e =+
?
 
 ?
13
22
ee - ??
+
??
??
?
1
2
2
e - 
 Option (4) is correct. 
10. If : f ? be a continuous function satisfying 
24
00
(sin2 )sin (cos2 )cos 0 f x xdx f x x dx
??
+ ? =
??
, then 
the value of ? is 
 (1) 2 (2) 3 - 
 (3) 3 (4) 2 - 
Answer (4) 
Sol. 
24
00
(sin2 )sin (cos2 ).cos 0 f x xdx f x xdx
??
+ ? =
??
 
 Let 
2
0
(sin2 ).sin I f x xdx
?
=
?
 
 
42
0
4
(sin2 )sin (sin2 ).sin f x xdx f x xdx
??
?
=+
??
 
 
4
0
(cos2 )sin
4
f x x dx
?
???
=-
??
??
?
 
  
4
0
sin2 sin
44
f x x dx
?
?? ?? ? ? ? ?
+ + +
??
? ? ? ?
? ? ? ? ??
?
 
  
4
0
11
(cos2 ) cos sin
22
f x x x dx
?
??
=-
??
??
?
 
 
   
  
  
4
0
11
(cos2 ) cos sin
22
f x x x dx
?
??
++
??
??
?
 
 
4
0
(cos2 )( 2 cos ) f x x dx
?
=
?
 
 ? 2 ? = - 
 Option (4) is correct. 
11. If the radius of the largest circle with centre (2, 0) 
inscribed in the ellipse x
2
 + 4y
2
 = 36 is r, then 12r
2
 
is equal to  
 (1) 115  (2) 92  
 (3) 69 (4) 72 
Answer (2) 
Sol. Equation of normal at P (6cos?, 3sin?) is  
 6sec 3cos 27 x ec y 
 If passes through (2, 0) 
 ? 12sec? = 27 
 ? 
4 65
cos , sin
99
 
  
8 65
,
33
P 
 r = OP  (O = (2, 0)) 
   = 
2
2
8 65
2
33
 
   = 
69
3
 
 ? 
2
69
12 12 92
9
r 
 ? Option (2) is correct. 
12. Let the line passing through the points P(2, –1, 2) 
and Q(5, 3, 4) meet the plane x – y + z = 4 at the 
point R. Then the distance of the point R from the 
plane x + 2y + 3z + 2 = 0 measured parallel to the 
line 
7 3 2
2 2 1
- + -
==
x y z
 is  
 (1) 61 (2) 189 
 (3) 31 (4) 3 
Answer (4) 
Sol. Equation of line PQ. 
 
2 1 2
3 4 2
x y z
 
 Let R be (3? + 2, 4? – 1, 2? + 2) 
 R lies on plane x – y + z = 4  
 ? 3? + 2 – 4? + 1 + 2? + 2 = 4 
 ? ? = –1 
 ? R( –1, –5, 0) 
 Let SR be : 
15
2 2 1
x y z
k 
 
 S : (2k – 1, 2k – 5, k) 
 S lies on plane : x + 2y + 3z + 2 = 0 
 ? (2k – 1) + (4k – 10) + 3k + 2 = 0 
 ? 9k – 9 = 0 ? k = 1 
  S(1, –3, 1)   ? 4 4 1 3 SR 
 ? Option (4) is correct. 
  
13. If ( )
2
1
9
103 81 ,
8
x x x
x x x x
x x x
+
+ ? = +
+?
 then ,
3
?
? 
are the roots of the equation  
 (1) 4x
2
 + 24x – 27 = 0 (2) 4x
2
 – 24x – 27 = 0 
 (3) 4x
2
 + 24x + 27 = 0 (4) 4x
2
 – 24x + 27 = 0 
Answer (4) 
Sol. ( )
2
1
9
103 81
8
x x x
x x x x
x x x
+
+ ? = +
+?
 
 Put x = 0 
 
2
1 0 0
9
0 0 81
8
00
 
 
6
3
3
3
2