JEE Main 2023 April 10 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) 
 
 
  
10/04/2023 
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) 
 
 
  
10/04/2023 
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. Let f be a differentiable function such that 
( ) ( )
2
0
4
x
x f x x tf t dt -=
?
, 
( )
2
1
3
f = . Then 18 f(3) is 
equal to 
 (1) 210 
 (2) 160 
 (3) 150 
 (4) 180 
Answer (2) 
Sol. 
2
0
( ) 4 ( )
x
x f x x t f t dt -=
?
 
 Diff. w.r.t. x 
 
2
( ) 2 ( ) 1 4 ( ) x f x x f x xf x ? + - = 
 ? 
2
( ) 1 2 ( ) x f x x f x ? - = 
 ? 
2
21 dy
y
dx x
x
-= Let ( ) ( )
dy
y f x f x
dx
??
= = ?
??
??
 
 ? 
2
2ln
2
1
I.F
dx
x
x
ee
x
-
-
?
= = = 
 ? 
24
1 y
dx C
xx
=+
?
 
 ? 
23
1
3
y
C
xx
= - + Now, 
2
(1)
3
f = 
  
21
33
C = - + 
 ? 1 C = 
 ? 
2
1
()
3
f x x
x
= - + 
 ? 
1
18 (3) 18 9 160
9
f
??
= - + =
??
??
 
2. An arc PQ of a circle subtends a right angle at its 
centre O. The mid point of the arc PQ is R. If 
OP u = , OR v = and OQ u v = ? + ? , then ?, ?
2
 are 
the roots of the equation 
 (1) x
2
 + x – 2 = 0 
 (2) x
2
 – x – 2 = 0 
 (3) 3x
2
 – 2x – 1 = 0 
 (4) 3x
2
 + 2x – 1 = 0 
Answer (2) 
Sol.  
 OQ u v = ? + ? 
 OR v = 
 OP u = 
 ? R will lie on angle bisector of and OQ OP 
 0 OQ OP ?= 
  ( )
2
0 v v u = ? + ? ? ? = 
 ? ? + ? ? cos45° = 0 
 ? 
2
-?
?= 
 
2
1
2
OQ OR r ?= 
 ? ( ) ( )
2
2
r
v v v ? + ? ? = 
  
22
2
22
rr
r = ? ? + ? ? = 
  
1
22
?
= + ? = 
Page 3


   
 
  
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
10/04/2023 
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. Let f be a differentiable function such that 
( ) ( )
2
0
4
x
x f x x tf t dt -=
?
, 
( )
2
1
3
f = . Then 18 f(3) is 
equal to 
 (1) 210 
 (2) 160 
 (3) 150 
 (4) 180 
Answer (2) 
Sol. 
2
0
( ) 4 ( )
x
x f x x t f t dt -=
?
 
 Diff. w.r.t. x 
 
2
( ) 2 ( ) 1 4 ( ) x f x x f x xf x ? + - = 
 ? 
2
( ) 1 2 ( ) x f x x f x ? - = 
 ? 
2
21 dy
y
dx x
x
-= Let ( ) ( )
dy
y f x f x
dx
??
= = ?
??
??
 
 ? 
2
2ln
2
1
I.F
dx
x
x
ee
x
-
-
?
= = = 
 ? 
24
1 y
dx C
xx
=+
?
 
 ? 
23
1
3
y
C
xx
= - + Now, 
2
(1)
3
f = 
  
21
33
C = - + 
 ? 1 C = 
 ? 
2
1
()
3
f x x
x
= - + 
 ? 
1
18 (3) 18 9 160
9
f
??
= - + =
??
??
 
2. An arc PQ of a circle subtends a right angle at its 
centre O. The mid point of the arc PQ is R. If 
OP u = , OR v = and OQ u v = ? + ? , then ?, ?
2
 are 
the roots of the equation 
 (1) x
2
 + x – 2 = 0 
 (2) x
2
 – x – 2 = 0 
 (3) 3x
2
 – 2x – 1 = 0 
 (4) 3x
2
 + 2x – 1 = 0 
Answer (2) 
Sol.  
 OQ u v = ? + ? 
 OR v = 
 OP u = 
 ? R will lie on angle bisector of and OQ OP 
 0 OQ OP ?= 
  ( )
2
0 v v u = ? + ? ? ? = 
 ? ? + ? ? cos45° = 0 
 ? 
2
-?
?= 
 
2
1
2
OQ OR r ?= 
 ? ( ) ( )
2
2
r
v v v ? + ? ? = 
  
22
2
22
rr
r = ? ? + ? ? = 
  
1
22
?
= + ? = 
 
   
   
 ? 2 ?= 
 ? ? = –1 
 ? ? = –1, ?
2
 = 2 
 ? x
2
 – x – 2 = 0 
3. A line segment AB of length ? moves such that the 
points A and B remain on the periphery of a circle 
of radius ?. Then the locus of the point, that divides 
the line segment AB in the ratio 2 : 3, is a circle of 
radius 
 (1) 
3
5
? (2) 
2
3
? 
 (3) 
19
5
? (4) 
19
7
? 
Answer (3) 
Sol.  
 Let O be the origin and radius of circle is ? and AB = ? 
 ? ( )
12
1 2 2cos = - ? - ? 
 ? ( )
12
1
cos
2
? - ? = 
 ? 
12
2 cos 3 cos
5
h
? ? + ? ?
= 
  
12
2 sin 3 sin
5
k
? ? + ? ?
= 
 ? ( ) ( )
2
22
12
4 9 12 cos
25
hk
?
?? + = + + ? - ?
??
 
   
2
19
25
?
=? 
 ? Radius 19
5
?
= 
4. If the coefficient of x
7
 in 
13
2
1
ax
bx
??
-
??
??
and the 
coefficient of x
– 5
 in 
13
2
1
ax
bx
??
+
??
??
 are equal, then 
a
4
b
4
 is equal to: 
 (1) 11 (2) 44 
 (3) 22 (4) 33 
Answer (3) 
Sol. Coefficient of x
7
 in 
13
2
1
ax
bx
??
-
??
??
 
 ( )
13
13
1
2
1
9
r
r
rr
T C x
bx
-
+
??
=-
??
??
 
 ? 2 r = 
 ? Coeff. 
11
13
2
2
a
C
b
= 
 Similarly coeff. of x
–5
 in 
13
2
1
ax
bx
??
+
??
??
 
 ? 6 r = 
 ? Coeff. 
7
13
6
6
a
C
b
= 
 Now, 
11 7
13 13
26
26
aa
CC
bb
= 
 ? a
4
 b
4
 = 22 
5. Let O be the origin and the position vector of the 
point P be – 2 3 i j k -+ . If the position vectors of the 
points A, B and C are –2 3 , 2 4 2 i j k i j k + - + - and 
42 i j k - + - respectively, then the projection of the 
vector OP on a vector perpendicular to the vectors 
AB and AC is 
 (1) 3 (2) 
8
3
 
 (3) 
7
3
 (4) 
10
3
 
Answer (1) 
Page 4


   
 
  
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
10/04/2023 
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. Let f be a differentiable function such that 
( ) ( )
2
0
4
x
x f x x tf t dt -=
?
, 
( )
2
1
3
f = . Then 18 f(3) is 
equal to 
 (1) 210 
 (2) 160 
 (3) 150 
 (4) 180 
Answer (2) 
Sol. 
2
0
( ) 4 ( )
x
x f x x t f t dt -=
?
 
 Diff. w.r.t. x 
 
2
( ) 2 ( ) 1 4 ( ) x f x x f x xf x ? + - = 
 ? 
2
( ) 1 2 ( ) x f x x f x ? - = 
 ? 
2
21 dy
y
dx x
x
-= Let ( ) ( )
dy
y f x f x
dx
??
= = ?
??
??
 
 ? 
2
2ln
2
1
I.F
dx
x
x
ee
x
-
-
?
= = = 
 ? 
24
1 y
dx C
xx
=+
?
 
 ? 
23
1
3
y
C
xx
= - + Now, 
2
(1)
3
f = 
  
21
33
C = - + 
 ? 1 C = 
 ? 
2
1
()
3
f x x
x
= - + 
 ? 
1
18 (3) 18 9 160
9
f
??
= - + =
??
??
 
2. An arc PQ of a circle subtends a right angle at its 
centre O. The mid point of the arc PQ is R. If 
OP u = , OR v = and OQ u v = ? + ? , then ?, ?
2
 are 
the roots of the equation 
 (1) x
2
 + x – 2 = 0 
 (2) x
2
 – x – 2 = 0 
 (3) 3x
2
 – 2x – 1 = 0 
 (4) 3x
2
 + 2x – 1 = 0 
Answer (2) 
Sol.  
 OQ u v = ? + ? 
 OR v = 
 OP u = 
 ? R will lie on angle bisector of and OQ OP 
 0 OQ OP ?= 
  ( )
2
0 v v u = ? + ? ? ? = 
 ? ? + ? ? cos45° = 0 
 ? 
2
-?
?= 
 
2
1
2
OQ OR r ?= 
 ? ( ) ( )
2
2
r
v v v ? + ? ? = 
  
22
2
22
rr
r = ? ? + ? ? = 
  
1
22
?
= + ? = 
 
   
   
 ? 2 ?= 
 ? ? = –1 
 ? ? = –1, ?
2
 = 2 
 ? x
2
 – x – 2 = 0 
3. A line segment AB of length ? moves such that the 
points A and B remain on the periphery of a circle 
of radius ?. Then the locus of the point, that divides 
the line segment AB in the ratio 2 : 3, is a circle of 
radius 
 (1) 
3
5
? (2) 
2
3
? 
 (3) 
19
5
? (4) 
19
7
? 
Answer (3) 
Sol.  
 Let O be the origin and radius of circle is ? and AB = ? 
 ? ( )
12
1 2 2cos = - ? - ? 
 ? ( )
12
1
cos
2
? - ? = 
 ? 
12
2 cos 3 cos
5
h
? ? + ? ?
= 
  
12
2 sin 3 sin
5
k
? ? + ? ?
= 
 ? ( ) ( )
2
22
12
4 9 12 cos
25
hk
?
?? + = + + ? - ?
??
 
   
2
19
25
?
=? 
 ? Radius 19
5
?
= 
4. If the coefficient of x
7
 in 
13
2
1
ax
bx
??
-
??
??
and the 
coefficient of x
– 5
 in 
13
2
1
ax
bx
??
+
??
??
 are equal, then 
a
4
b
4
 is equal to: 
 (1) 11 (2) 44 
 (3) 22 (4) 33 
Answer (3) 
Sol. Coefficient of x
7
 in 
13
2
1
ax
bx
??
-
??
??
 
 ( )
13
13
1
2
1
9
r
r
rr
T C x
bx
-
+
??
=-
??
??
 
 ? 2 r = 
 ? Coeff. 
11
13
2
2
a
C
b
= 
 Similarly coeff. of x
–5
 in 
13
2
1
ax
bx
??
+
??
??
 
 ? 6 r = 
 ? Coeff. 
7
13
6
6
a
C
b
= 
 Now, 
11 7
13 13
26
26
aa
CC
bb
= 
 ? a
4
 b
4
 = 22 
5. Let O be the origin and the position vector of the 
point P be – 2 3 i j k -+ . If the position vectors of the 
points A, B and C are –2 3 , 2 4 2 i j k i j k + - + - and 
42 i j k - + - respectively, then the projection of the 
vector OP on a vector perpendicular to the vectors 
AB and AC is 
 (1) 3 (2) 
8
3
 
 (3) 
7
3
 (4) 
10
3
 
Answer (1) 
 
   
   
Sol. 23 OP i j k = - - + 
 43 AB i j k = + + 
 22 AC i j k = - + + 
 4 3 1
2 1 2
i j k
AB AC ?=
-
 
   5 10 10 i j k = - + 
 Projection of OP on  
  
( )
OP AB AC
AB AC
AB AC
??
?=
?
 
   
( )( )
5 2 3 2 2
5 1 4 4
i j k i j k - - + - +
=
++
 
   = 3 
6. Let the first term a and the common ratio r of a 
geometric progression be positive integers. If the 
sum of squares of its first three terms is 33033, then 
the sum of these three terms is equal to  
 (1) 241 (2) 231 
 (3) 210 (4) 220 
Answer (2) 
Sol. a
2
 + a
2
 r
2
 + a
2
r
4
 = 33033 
 a
2
(1 + r
2
 + r
4
) = 33033 
 a
2
 (1 + r
2
 + r
4
) = 3 × 7 × (11)
2
 × 13 
 ? a
2
 = (11)
2
 
 11 a = 
 ? 1 +r
2
 + r
4
 = 273 
  r
4
 + r
2
 – 272 = 0 
 ? r
2
 = 16  
  4 r = 
 a = 11 
 ar = 44 
 ar
2
 = 176 
 a + ar + ar
2
 = 231 
7. The slope of tangent at any point (x, y) on a curve 
y = y(x) is 
22
, 0.
2
xy
x
xy
+
? If y(2) = 0, then a value 
of y(8) is 
 (1) 42 - (2) 23 
 (3) 23 - (4) 43 
Answer (4) 
Sol. 
22
2
dy x y
dx xy
+
= 
 Put y = vx 
 
dy dv
vx
dx dx
=+ 
 
11
2
dv
v x v
dx v
??
+ = +
??
??
 
 ? 
2
11
2
dv v
x
dx v
??
-
=??
??
??
 
 ? 
2
2
2
log 1 ln lnc
1
v dx
dv v x
x
v
= ? - - = +
-
??
 
 ? 
2 2 2
2
1 y x y
k x k
x
x
??
--
= ? = ??
??
??
 
  y(2) = 0 
 ? k = 2 
 ? 
22
2
xy
x
-
= 
 Put x = 8 
 
2
2
64
2 48
8
y
y
-
= ? = 
   43 y = 
8. The negation of the statement 
 ( ) ( ) ( )
~ p q q r ? ? ? is  
 (1) ( ) ( ) ~ p r q ?? 
 (2) ( ) ( ) ( ) ~ ) ~ p r q ?? 
 (3) ( ) ( ) ( ) ~ ) (~ ~ p q r ?? 
 (4) ( ) ( ) ( ) ~ ) (~ ~ p q r ?? 
Answer (2) 
Page 5


   
 
  
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
10/04/2023 
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. Let f be a differentiable function such that 
( ) ( )
2
0
4
x
x f x x tf t dt -=
?
, 
( )
2
1
3
f = . Then 18 f(3) is 
equal to 
 (1) 210 
 (2) 160 
 (3) 150 
 (4) 180 
Answer (2) 
Sol. 
2
0
( ) 4 ( )
x
x f x x t f t dt -=
?
 
 Diff. w.r.t. x 
 
2
( ) 2 ( ) 1 4 ( ) x f x x f x xf x ? + - = 
 ? 
2
( ) 1 2 ( ) x f x x f x ? - = 
 ? 
2
21 dy
y
dx x
x
-= Let ( ) ( )
dy
y f x f x
dx
??
= = ?
??
??
 
 ? 
2
2ln
2
1
I.F
dx
x
x
ee
x
-
-
?
= = = 
 ? 
24
1 y
dx C
xx
=+
?
 
 ? 
23
1
3
y
C
xx
= - + Now, 
2
(1)
3
f = 
  
21
33
C = - + 
 ? 1 C = 
 ? 
2
1
()
3
f x x
x
= - + 
 ? 
1
18 (3) 18 9 160
9
f
??
= - + =
??
??
 
2. An arc PQ of a circle subtends a right angle at its 
centre O. The mid point of the arc PQ is R. If 
OP u = , OR v = and OQ u v = ? + ? , then ?, ?
2
 are 
the roots of the equation 
 (1) x
2
 + x – 2 = 0 
 (2) x
2
 – x – 2 = 0 
 (3) 3x
2
 – 2x – 1 = 0 
 (4) 3x
2
 + 2x – 1 = 0 
Answer (2) 
Sol.  
 OQ u v = ? + ? 
 OR v = 
 OP u = 
 ? R will lie on angle bisector of and OQ OP 
 0 OQ OP ?= 
  ( )
2
0 v v u = ? + ? ? ? = 
 ? ? + ? ? cos45° = 0 
 ? 
2
-?
?= 
 
2
1
2
OQ OR r ?= 
 ? ( ) ( )
2
2
r
v v v ? + ? ? = 
  
22
2
22
rr
r = ? ? + ? ? = 
  
1
22
?
= + ? = 
 
   
   
 ? 2 ?= 
 ? ? = –1 
 ? ? = –1, ?
2
 = 2 
 ? x
2
 – x – 2 = 0 
3. A line segment AB of length ? moves such that the 
points A and B remain on the periphery of a circle 
of radius ?. Then the locus of the point, that divides 
the line segment AB in the ratio 2 : 3, is a circle of 
radius 
 (1) 
3
5
? (2) 
2
3
? 
 (3) 
19
5
? (4) 
19
7
? 
Answer (3) 
Sol.  
 Let O be the origin and radius of circle is ? and AB = ? 
 ? ( )
12
1 2 2cos = - ? - ? 
 ? ( )
12
1
cos
2
? - ? = 
 ? 
12
2 cos 3 cos
5
h
? ? + ? ?
= 
  
12
2 sin 3 sin
5
k
? ? + ? ?
= 
 ? ( ) ( )
2
22
12
4 9 12 cos
25
hk
?
?? + = + + ? - ?
??
 
   
2
19
25
?
=? 
 ? Radius 19
5
?
= 
4. If the coefficient of x
7
 in 
13
2
1
ax
bx
??
-
??
??
and the 
coefficient of x
– 5
 in 
13
2
1
ax
bx
??
+
??
??
 are equal, then 
a
4
b
4
 is equal to: 
 (1) 11 (2) 44 
 (3) 22 (4) 33 
Answer (3) 
Sol. Coefficient of x
7
 in 
13
2
1
ax
bx
??
-
??
??
 
 ( )
13
13
1
2
1
9
r
r
rr
T C x
bx
-
+
??
=-
??
??
 
 ? 2 r = 
 ? Coeff. 
11
13
2
2
a
C
b
= 
 Similarly coeff. of x
–5
 in 
13
2
1
ax
bx
??
+
??
??
 
 ? 6 r = 
 ? Coeff. 
7
13
6
6
a
C
b
= 
 Now, 
11 7
13 13
26
26
aa
CC
bb
= 
 ? a
4
 b
4
 = 22 
5. Let O be the origin and the position vector of the 
point P be – 2 3 i j k -+ . If the position vectors of the 
points A, B and C are –2 3 , 2 4 2 i j k i j k + - + - and 
42 i j k - + - respectively, then the projection of the 
vector OP on a vector perpendicular to the vectors 
AB and AC is 
 (1) 3 (2) 
8
3
 
 (3) 
7
3
 (4) 
10
3
 
Answer (1) 
 
   
   
Sol. 23 OP i j k = - - + 
 43 AB i j k = + + 
 22 AC i j k = - + + 
 4 3 1
2 1 2
i j k
AB AC ?=
-
 
   5 10 10 i j k = - + 
 Projection of OP on  
  
( )
OP AB AC
AB AC
AB AC
??
?=
?
 
   
( )( )
5 2 3 2 2
5 1 4 4
i j k i j k - - + - +
=
++
 
   = 3 
6. Let the first term a and the common ratio r of a 
geometric progression be positive integers. If the 
sum of squares of its first three terms is 33033, then 
the sum of these three terms is equal to  
 (1) 241 (2) 231 
 (3) 210 (4) 220 
Answer (2) 
Sol. a
2
 + a
2
 r
2
 + a
2
r
4
 = 33033 
 a
2
(1 + r
2
 + r
4
) = 33033 
 a
2
 (1 + r
2
 + r
4
) = 3 × 7 × (11)
2
 × 13 
 ? a
2
 = (11)
2
 
 11 a = 
 ? 1 +r
2
 + r
4
 = 273 
  r
4
 + r
2
 – 272 = 0 
 ? r
2
 = 16  
  4 r = 
 a = 11 
 ar = 44 
 ar
2
 = 176 
 a + ar + ar
2
 = 231 
7. The slope of tangent at any point (x, y) on a curve 
y = y(x) is 
22
, 0.
2
xy
x
xy
+
? If y(2) = 0, then a value 
of y(8) is 
 (1) 42 - (2) 23 
 (3) 23 - (4) 43 
Answer (4) 
Sol. 
22
2
dy x y
dx xy
+
= 
 Put y = vx 
 
dy dv
vx
dx dx
=+ 
 
11
2
dv
v x v
dx v
??
+ = +
??
??
 
 ? 
2
11
2
dv v
x
dx v
??
-
=??
??
??
 
 ? 
2
2
2
log 1 ln lnc
1
v dx
dv v x
x
v
= ? - - = +
-
??
 
 ? 
2 2 2
2
1 y x y
k x k
x
x
??
--
= ? = ??
??
??
 
  y(2) = 0 
 ? k = 2 
 ? 
22
2
xy
x
-
= 
 Put x = 8 
 
2
2
64
2 48
8
y
y
-
= ? = 
   43 y = 
8. The negation of the statement 
 ( ) ( ) ( )
~ p q q r ? ? ? is  
 (1) ( ) ( ) ~ p r q ?? 
 (2) ( ) ( ) ( ) ~ ) ~ p r q ?? 
 (3) ( ) ( ) ( ) ~ ) (~ ~ p q r ?? 
 (4) ( ) ( ) ( ) ~ ) (~ ~ p q r ?? 
Answer (2) 
 
   
   
Sol. ( ) ( ) ( )
~ (~ ) p q q r ? ? ? 
 = ( ) ( )
~ ) ~ (~ p q q r ? ? ? 
 ( ) ( ) ~ ~ ~ p q q r = ? ? ? 
 = ( ) ( ) ( ) ~ ~ ~ ~ ~ p q q p q r ? ? ? ? ? 
 = ( ) ( ) ~ ~ ~ q p r q r ?? ? ? ? ?
??
 
 = ( ) ( ) ( ) ( )
~ ~ ~ ~ q p r q q r ? ? ? ? ? 
 = ( ) ( )
~ ~ ~ q p r q ? ? ? 
 = ( ) ( )
~ (~ ) p r q ?? 
9. Let two vertices of a triangle ABC be (2, 4, 6) and 
(0, –2, –5), and its centroid be (2, 1, –1). If the 
image of the third vertex in the plane x + 2y + 4z = 
11 is (?, ?, ?), then ?? + ?? + ?? is equal to 
 (1) 70 (2) 76 
 (3) 74 (4) 72 
Answer (3) 
Sol. Let the vertex ‘C’ be (a, b, c) 
 
20
24
3
a
a
++
= ? = 
 
42
11
3
b
b
-+
= ? = 
 
65
1 –4
3
c
c
-+
- = ? = 
 ? C(4, 1, –4) 
 Image of C in x + 2y + 4z = 11 
 
( ) 4 2 16 11
4 1 4
22
1 2 4 16 4 1
+ - -
? - ? - ? +
= = = - =
++
 
 ? 
4
26
1
?-
= ? ? = 
  
1
25
2
?-
= ? ? = 
 
4
24
4
?+
= ? ? = 
 ? ?? + ?? + ?? = 30 + 20 + 24 
  = 74 
10. Let N denote the sum of the numbers obtained 
when two dice are rolled. If the probability that 
2
N
 < N! is ,
m
n
 where m and n are coprime, then 
4m – 3n is equal to  
 (1) 6 (2) 12 
 (3) 10 (4) 8 
Answer (4) 
Sol. 2N < N! 
 N = 1 (not possible) ? 0 
 N = 2 (not possible) ? 1 
 N = 3 (not possible) ? 2 
 N = 4 (possible) 
 ? Required probability 
( ) 36 1 2
11
36 12
-+
== 
 ? 4m – 3n 
 = 44 – 36 
 = 8 
 Option (4) is correct 
11. A square piece of tin of side 30 cm is to be made 
into a box without top by cutting a square from each 
corner and folding up the flaps to form a box. If the 
volume of the box is maximum, then its surface 
area (in cm
2
) is equal to 
 (1) 800 (2) 675 
 (3) 1025 (4) 900 
Answer (1) 
Sol.  
 
 Volume = (30 – 2x)
2
·x = V(x) 
 
( ) ( )( )
2
30 2 2 30 2 2 0
dV
x x x
dx
= - + - - = 
 ? x = 5, x = 15 (not possible) 
 ? Surface area  = (30 – 2x) × 4 + (30 – 2x)
2
 
   = 800 cm
2
  (x = 5) 
 ? Option (1) is correct 
(30 – 2 ) cm x
x cm