JEE Main 2023 April 15 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) 
 
 
  
15/04/2023 
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) 
 
 
  
15/04/2023 
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 S be the set of all (?, ?) for which the vectors 
ˆ ˆ ˆ ˆ ˆ ˆ
,  2 i j k i j k ? - + + + ? and 
ˆ ˆ ˆ
3 4 5 , i j k -+ where  
? – ? = 5, are coplanar, then 
( )
( )
22
,
80
S ? ? ?
? + ?
?
 is 
equal to 
 (1) 2210 (2) 2130 
 (3) 2290 (4) 2370 
Answer (3) 
Sol. ? 
11
1 2 0
35 4
? -
?
=
-
 
 ? ??10 + 4?] + 1 [5 – 3?? + 1 [–10] = 0 
 ? 4?? + 10? – 3? = 5 …(i) 
 And ? – ? = 5 
 So, by (1) 
 4(5 + ?)? + 10(5 + ?) – 3? = 5 
 ? 4?
2
 + 27? + 45 = 0 
 ? 4?
2
 + 15? + 12? + 45 = 0 
 ? (4? + 15) – (? + 3) = 0 
 ? 
15
3,
4
-
? = - 
 
5
2,
4
?= 
 ? 
( )
( )
22
225 25
80 80 9 4
16 16
s ???
??
? + ? = + + +
??
??
?
 
  
125
80 13 10 229
8
??
= + = ?
??
??
 
  = 2290 
2. Let the determinant of a square matrix A of order m 
be m – n, where m and n satisfy 4m + n = 22 and 
17m + 4n = 93. If det(n adj(adj (mA))) = 3
a
5
b
6
c
, then 
a + b + c is equal to 
 (1) 84 (2) 96 
 (3) 101 (4) 109 
Answer (2) 
Sol. |A| = m – n, where 4m + n = 22     …(i) 
 and 17m + 4n = 93  …(ii) 
 Solving (i) and (ii) 
 m = 5, n = 2 
 Order = 5 
 |A| = 3 
 ? det (n adj (adj(mA))) 
 = |2adj (adj (5A))| 
 = 2
5
 |5A|
16
 
 = 2
5
 5
80
 |A|
16
 = 2
5
· 3
16
· 5
80
 
 = 3
11
 5
80
 6
5
 
 So, a + b + c = 96 
3. If (?, ?) is the orthocenter of the triangle ABC with 
vertices A(3, –7), B(–1, 2) and C(4, 5), then  
9? – 6? + 60 is equal to  
 (1) 25 (2) 35 
 (3) 30 (4) 40 
Answer (1) 
Sol.  
 ( )
5
: 7 3
3
AD y x
-
+ = - 
 ? 3y + 21 = –5x + 15 
  5x + 3y + 6 = 0 …(i) 
 BE : ( )
1
21
12
yx
-
- = + 
 ? 12y – 24 = –x – 1 
 ? x = 23 – 12y 
 by (ii) 115 – 60y + 3y + 6 = 0 
 57y = 121 
 
121 121
, 23 12
57 57
yx = = - ? 
 ? 9? – 6? + 60 = 9 × 23 – 
121 121
108 6 60
57 57
? - ? + 
 = 207 – 242 + 60 = 25 
Page 3


   
 
  
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
15/04/2023 
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 S be the set of all (?, ?) for which the vectors 
ˆ ˆ ˆ ˆ ˆ ˆ
,  2 i j k i j k ? - + + + ? and 
ˆ ˆ ˆ
3 4 5 , i j k -+ where  
? – ? = 5, are coplanar, then 
( )
( )
22
,
80
S ? ? ?
? + ?
?
 is 
equal to 
 (1) 2210 (2) 2130 
 (3) 2290 (4) 2370 
Answer (3) 
Sol. ? 
11
1 2 0
35 4
? -
?
=
-
 
 ? ??10 + 4?] + 1 [5 – 3?? + 1 [–10] = 0 
 ? 4?? + 10? – 3? = 5 …(i) 
 And ? – ? = 5 
 So, by (1) 
 4(5 + ?)? + 10(5 + ?) – 3? = 5 
 ? 4?
2
 + 27? + 45 = 0 
 ? 4?
2
 + 15? + 12? + 45 = 0 
 ? (4? + 15) – (? + 3) = 0 
 ? 
15
3,
4
-
? = - 
 
5
2,
4
?= 
 ? 
( )
( )
22
225 25
80 80 9 4
16 16
s ???
??
? + ? = + + +
??
??
?
 
  
125
80 13 10 229
8
??
= + = ?
??
??
 
  = 2290 
2. Let the determinant of a square matrix A of order m 
be m – n, where m and n satisfy 4m + n = 22 and 
17m + 4n = 93. If det(n adj(adj (mA))) = 3
a
5
b
6
c
, then 
a + b + c is equal to 
 (1) 84 (2) 96 
 (3) 101 (4) 109 
Answer (2) 
Sol. |A| = m – n, where 4m + n = 22     …(i) 
 and 17m + 4n = 93  …(ii) 
 Solving (i) and (ii) 
 m = 5, n = 2 
 Order = 5 
 |A| = 3 
 ? det (n adj (adj(mA))) 
 = |2adj (adj (5A))| 
 = 2
5
 |5A|
16
 
 = 2
5
 5
80
 |A|
16
 = 2
5
· 3
16
· 5
80
 
 = 3
11
 5
80
 6
5
 
 So, a + b + c = 96 
3. If (?, ?) is the orthocenter of the triangle ABC with 
vertices A(3, –7), B(–1, 2) and C(4, 5), then  
9? – 6? + 60 is equal to  
 (1) 25 (2) 35 
 (3) 30 (4) 40 
Answer (1) 
Sol.  
 ( )
5
: 7 3
3
AD y x
-
+ = - 
 ? 3y + 21 = –5x + 15 
  5x + 3y + 6 = 0 …(i) 
 BE : ( )
1
21
12
yx
-
- = + 
 ? 12y – 24 = –x – 1 
 ? x = 23 – 12y 
 by (ii) 115 – 60y + 3y + 6 = 0 
 57y = 121 
 
121 121
, 23 12
57 57
yx = = - ? 
 ? 9? – 6? + 60 = 9 × 23 – 
121 121
108 6 60
57 57
? - ? + 
 = 207 – 242 + 60 = 25 
 
   
   
4. Let ABCD be a quadrilateral. If E and F are the mid 
points of the diagonals AC and BD respectively and 
( ) ( )
– – AB BC AD DC k FE += , then k is equal to 
 (1) 4 (2) –2 
 (3) 2 (4) –4 
Answer (4) 
Sol.  
 
Let position vector of A, B, C and D are , , and a b c d 
respectively 
 ? Position vector of 
22
OC OA c a
E
++
== 
  Position vector of 
2
bd
F
+
= 
 Now, 
( )
AB BC AD DC - + - 
 
( ) ( )
b a c b d a c d ? - - - + - - - 
 2 2 2 2 b a c d ? - - + 
 
( ) ( )
22 b d a c ? + - + 
 44
22
b d a c
OF OE
??
++
??
? - = -
??
??
??
??
 
 44 EF FE ? = - 
 ? k = –4 
5. Let the foot of perpendicular of the point  
P(3, –2, –9) on the plane passing through the points 
(–1, –2, –3), (9, 3, 4), (9, –2, 1) be Q(?, ?, ?). Then 
the distance Q from the origin is  
 (1) 42 (2) 38 
 (3) 35 (4) 29 
Answer (1) 
Sol.  
 10 4 AC i k =+ 
 10 5 7 AB i j k = + + 
 10 0 4
10 5 7
i j k
AC AB ?= 
   = –20i – 30j + 50k 
 Equation of plane 
  2x + 3y – 5z = d 
 Put (–1, –2, –3) 
  –2 – 6 + 15 = d 
  d = 7 
 ? 2x + 3y – 5z = 7 
 Foot of ?
r
 
 
3 2 9 38
2 3 5 38
x y z - + + ??
= = = -
??
-
??
 
 x = 1, y = –5, z = –4 
 Q(1, –5, –4) 
 Distance from origin 1 25 16 = + + 
   42 = 
6. The number of common tangents, to the circles  
 x
2
 + y
2
 – 18x –15y + 131 = 0 and x
2
 + y
2
 – 6x – 6y 
– 7 = 0, is  
 (1) 3 (2) 1 
 (3) 4 (4) 2 
Answer (1) 
Sol. 
22
18 15 131 0 x y x y + - - + = 
 
11
15 225 5
9, , 81 131
2 4 2
Cr
??
= + - =
??
??
 
 
22
6 6 7 0 x y x y + - - - = 
 
22
(3, 3), 9 9 7 5 Cr = + + = 
 
2
2
12
15 81 15
(9 3) 3 36
2 4 2
d C C
??
= = - + - = + =
??
??
 
 
12
5 15
5
22
rr + = + = 
 
1 2 1 2
C C r r ? = + 
 ? Circles touch each other externally, 3 common 
tangents. 
A B
D C
A(–1, –2, –3)
B(9, 3, 4)
C(9, –2, 1)
Page 4


   
 
  
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
15/04/2023 
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 S be the set of all (?, ?) for which the vectors 
ˆ ˆ ˆ ˆ ˆ ˆ
,  2 i j k i j k ? - + + + ? and 
ˆ ˆ ˆ
3 4 5 , i j k -+ where  
? – ? = 5, are coplanar, then 
( )
( )
22
,
80
S ? ? ?
? + ?
?
 is 
equal to 
 (1) 2210 (2) 2130 
 (3) 2290 (4) 2370 
Answer (3) 
Sol. ? 
11
1 2 0
35 4
? -
?
=
-
 
 ? ??10 + 4?] + 1 [5 – 3?? + 1 [–10] = 0 
 ? 4?? + 10? – 3? = 5 …(i) 
 And ? – ? = 5 
 So, by (1) 
 4(5 + ?)? + 10(5 + ?) – 3? = 5 
 ? 4?
2
 + 27? + 45 = 0 
 ? 4?
2
 + 15? + 12? + 45 = 0 
 ? (4? + 15) – (? + 3) = 0 
 ? 
15
3,
4
-
? = - 
 
5
2,
4
?= 
 ? 
( )
( )
22
225 25
80 80 9 4
16 16
s ???
??
? + ? = + + +
??
??
?
 
  
125
80 13 10 229
8
??
= + = ?
??
??
 
  = 2290 
2. Let the determinant of a square matrix A of order m 
be m – n, where m and n satisfy 4m + n = 22 and 
17m + 4n = 93. If det(n adj(adj (mA))) = 3
a
5
b
6
c
, then 
a + b + c is equal to 
 (1) 84 (2) 96 
 (3) 101 (4) 109 
Answer (2) 
Sol. |A| = m – n, where 4m + n = 22     …(i) 
 and 17m + 4n = 93  …(ii) 
 Solving (i) and (ii) 
 m = 5, n = 2 
 Order = 5 
 |A| = 3 
 ? det (n adj (adj(mA))) 
 = |2adj (adj (5A))| 
 = 2
5
 |5A|
16
 
 = 2
5
 5
80
 |A|
16
 = 2
5
· 3
16
· 5
80
 
 = 3
11
 5
80
 6
5
 
 So, a + b + c = 96 
3. If (?, ?) is the orthocenter of the triangle ABC with 
vertices A(3, –7), B(–1, 2) and C(4, 5), then  
9? – 6? + 60 is equal to  
 (1) 25 (2) 35 
 (3) 30 (4) 40 
Answer (1) 
Sol.  
 ( )
5
: 7 3
3
AD y x
-
+ = - 
 ? 3y + 21 = –5x + 15 
  5x + 3y + 6 = 0 …(i) 
 BE : ( )
1
21
12
yx
-
- = + 
 ? 12y – 24 = –x – 1 
 ? x = 23 – 12y 
 by (ii) 115 – 60y + 3y + 6 = 0 
 57y = 121 
 
121 121
, 23 12
57 57
yx = = - ? 
 ? 9? – 6? + 60 = 9 × 23 – 
121 121
108 6 60
57 57
? - ? + 
 = 207 – 242 + 60 = 25 
 
   
   
4. Let ABCD be a quadrilateral. If E and F are the mid 
points of the diagonals AC and BD respectively and 
( ) ( )
– – AB BC AD DC k FE += , then k is equal to 
 (1) 4 (2) –2 
 (3) 2 (4) –4 
Answer (4) 
Sol.  
 
Let position vector of A, B, C and D are , , and a b c d 
respectively 
 ? Position vector of 
22
OC OA c a
E
++
== 
  Position vector of 
2
bd
F
+
= 
 Now, 
( )
AB BC AD DC - + - 
 
( ) ( )
b a c b d a c d ? - - - + - - - 
 2 2 2 2 b a c d ? - - + 
 
( ) ( )
22 b d a c ? + - + 
 44
22
b d a c
OF OE
??
++
??
? - = -
??
??
??
??
 
 44 EF FE ? = - 
 ? k = –4 
5. Let the foot of perpendicular of the point  
P(3, –2, –9) on the plane passing through the points 
(–1, –2, –3), (9, 3, 4), (9, –2, 1) be Q(?, ?, ?). Then 
the distance Q from the origin is  
 (1) 42 (2) 38 
 (3) 35 (4) 29 
Answer (1) 
Sol.  
 10 4 AC i k =+ 
 10 5 7 AB i j k = + + 
 10 0 4
10 5 7
i j k
AC AB ?= 
   = –20i – 30j + 50k 
 Equation of plane 
  2x + 3y – 5z = d 
 Put (–1, –2, –3) 
  –2 – 6 + 15 = d 
  d = 7 
 ? 2x + 3y – 5z = 7 
 Foot of ?
r
 
 
3 2 9 38
2 3 5 38
x y z - + + ??
= = = -
??
-
??
 
 x = 1, y = –5, z = –4 
 Q(1, –5, –4) 
 Distance from origin 1 25 16 = + + 
   42 = 
6. The number of common tangents, to the circles  
 x
2
 + y
2
 – 18x –15y + 131 = 0 and x
2
 + y
2
 – 6x – 6y 
– 7 = 0, is  
 (1) 3 (2) 1 
 (3) 4 (4) 2 
Answer (1) 
Sol. 
22
18 15 131 0 x y x y + - - + = 
 
11
15 225 5
9, , 81 131
2 4 2
Cr
??
= + - =
??
??
 
 
22
6 6 7 0 x y x y + - - - = 
 
22
(3, 3), 9 9 7 5 Cr = + + = 
 
2
2
12
15 81 15
(9 3) 3 36
2 4 2
d C C
??
= = - + - = + =
??
??
 
 
12
5 15
5
22
rr + = + = 
 
1 2 1 2
C C r r ? = + 
 ? Circles touch each other externally, 3 common 
tangents. 
A B
D C
A(–1, –2, –3)
B(9, 3, 4)
C(9, –2, 1)
 
   
   
7. The total number of three-digit numbers, divisible 
by 3, which can be formed using the digits 1, 3, 5, 
8, if repetition of digits is allowed, is 
 (1) 21 (2) 20 
 (3) 22 (4) 18 
Answer (3) 
Sol. Sum of digits 3 : (1, 1, 1) 
 Sum of digits 9 : (1, 3, 5) or (3, 3, 3) 
 Sum of digits 12 : (1, 3, 8) 
 Sum of digits 15 : (5, 5, 5) 
 Sum of digits 18 : (5, 5, 8) 
 Sum of digits 21 : (5, 8, 8)  
 Sum of digits 24 : (8, 8, 8) 
 Possible numbers are 
 
3! 3!
1 3! 1 3! 1 1
2! 2!
= + + + + + + + 
 = 22 
8. If the set : , Re Re
2 3 5
3
z z z z
z
z z
z
 is 
equal to the interval (?, ?], then 24(? – ?) is equal 
to 
 (1) 36 (2) 27 
 (3) 30 (4) 42 
Answer (3) 
Sol. Re
2 3 5
z z zz
zz
??
-+
??
??
-+
??
 
 
( )
( ) ( )
22
Re
2 3 5
x iy x iy x y
x iy x iy
??
+ - - + +
??
?? - + + -
??
 
 
( )
22
2
Re
2 2 8
x y i y
x iy
??
++
??
?? +-
??
 
 
( )
( ) ( )
( ) ( ) ( )
22
2 2
2 2 1 8
Re
2 1 8
x y yi x iy
xy
??
+ + + +
??
??
?? ++
??
 
 
( )
( )
( ) ( )
2 2 2
22
2 1 16
4 1 8
x y x y
xy
+ + -
=
++
 
 Put x = 3 
 
( )
22
2
8 9 16
64 64
yy
y
+-
=
+
 
 
( )
( )
2
2
9
1
8
1
y
fy
y
-
=
+
 
 Range of f(y) = (– 0.125, 1.125] 
  ? = – 0.125 
  ? = 1.125  
  ? – ? = 1.25 
  24(? – ?) = 30 
9. Let 
( )
20
10
2
10
, , ,
i
i
i
a bx cx p x a b c
=
+ + = ?
?
. If  
p1 = 20 and p2 = 210, then 2(a + b + c) is equal to 
 (1) 6 (2) 15 
 (3) 12 (4) 8 
Answer (3) 
Sol. General term : ( ) ( )
( )
3
12
2
1 2 3
10!
!!!
r
rr
a bx cx
r r r
 
 For coeff. of x : r2 + 2r3 = 1 
  
1 2 3
910
r r r
   
 ? coff of 
91
10!
20
9!
x a b == 
 ? a
9
 ? b = 2 …(i) 
  Coff of x
2
 : 
8 2 9
10! 10!
210
8!2!0! 9!0!1!
a b a c ? + ? ? = 
 ? 45a
8
 ? b
2
 + 10 ? a
9
 ? c = 210 
 ? 9a
8
b
2
 + 2a
9
 ? c = 42 
  as a, b, c ?N 
 ? a = 1, b = 2, c = 3 
  2(a + b + c) = 2(3 + 2 + 1) = 12 
10. The number of real roots of the equation 
5 2 6 0, x x x is  
 (1) 5 (2) 4 
 (3) 6 (4) 3 
Answer (4) 
Page 5


   
 
  
Answers & Solutions 
for 
JEE (Main)-2023 (Online) Phase-2 
(Mathematics, Physics and Chemistry) 
 
 
  
15/04/2023 
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 S be the set of all (?, ?) for which the vectors 
ˆ ˆ ˆ ˆ ˆ ˆ
,  2 i j k i j k ? - + + + ? and 
ˆ ˆ ˆ
3 4 5 , i j k -+ where  
? – ? = 5, are coplanar, then 
( )
( )
22
,
80
S ? ? ?
? + ?
?
 is 
equal to 
 (1) 2210 (2) 2130 
 (3) 2290 (4) 2370 
Answer (3) 
Sol. ? 
11
1 2 0
35 4
? -
?
=
-
 
 ? ??10 + 4?] + 1 [5 – 3?? + 1 [–10] = 0 
 ? 4?? + 10? – 3? = 5 …(i) 
 And ? – ? = 5 
 So, by (1) 
 4(5 + ?)? + 10(5 + ?) – 3? = 5 
 ? 4?
2
 + 27? + 45 = 0 
 ? 4?
2
 + 15? + 12? + 45 = 0 
 ? (4? + 15) – (? + 3) = 0 
 ? 
15
3,
4
-
? = - 
 
5
2,
4
?= 
 ? 
( )
( )
22
225 25
80 80 9 4
16 16
s ???
??
? + ? = + + +
??
??
?
 
  
125
80 13 10 229
8
??
= + = ?
??
??
 
  = 2290 
2. Let the determinant of a square matrix A of order m 
be m – n, where m and n satisfy 4m + n = 22 and 
17m + 4n = 93. If det(n adj(adj (mA))) = 3
a
5
b
6
c
, then 
a + b + c is equal to 
 (1) 84 (2) 96 
 (3) 101 (4) 109 
Answer (2) 
Sol. |A| = m – n, where 4m + n = 22     …(i) 
 and 17m + 4n = 93  …(ii) 
 Solving (i) and (ii) 
 m = 5, n = 2 
 Order = 5 
 |A| = 3 
 ? det (n adj (adj(mA))) 
 = |2adj (adj (5A))| 
 = 2
5
 |5A|
16
 
 = 2
5
 5
80
 |A|
16
 = 2
5
· 3
16
· 5
80
 
 = 3
11
 5
80
 6
5
 
 So, a + b + c = 96 
3. If (?, ?) is the orthocenter of the triangle ABC with 
vertices A(3, –7), B(–1, 2) and C(4, 5), then  
9? – 6? + 60 is equal to  
 (1) 25 (2) 35 
 (3) 30 (4) 40 
Answer (1) 
Sol.  
 ( )
5
: 7 3
3
AD y x
-
+ = - 
 ? 3y + 21 = –5x + 15 
  5x + 3y + 6 = 0 …(i) 
 BE : ( )
1
21
12
yx
-
- = + 
 ? 12y – 24 = –x – 1 
 ? x = 23 – 12y 
 by (ii) 115 – 60y + 3y + 6 = 0 
 57y = 121 
 
121 121
, 23 12
57 57
yx = = - ? 
 ? 9? – 6? + 60 = 9 × 23 – 
121 121
108 6 60
57 57
? - ? + 
 = 207 – 242 + 60 = 25 
 
   
   
4. Let ABCD be a quadrilateral. If E and F are the mid 
points of the diagonals AC and BD respectively and 
( ) ( )
– – AB BC AD DC k FE += , then k is equal to 
 (1) 4 (2) –2 
 (3) 2 (4) –4 
Answer (4) 
Sol.  
 
Let position vector of A, B, C and D are , , and a b c d 
respectively 
 ? Position vector of 
22
OC OA c a
E
++
== 
  Position vector of 
2
bd
F
+
= 
 Now, 
( )
AB BC AD DC - + - 
 
( ) ( )
b a c b d a c d ? - - - + - - - 
 2 2 2 2 b a c d ? - - + 
 
( ) ( )
22 b d a c ? + - + 
 44
22
b d a c
OF OE
??
++
??
? - = -
??
??
??
??
 
 44 EF FE ? = - 
 ? k = –4 
5. Let the foot of perpendicular of the point  
P(3, –2, –9) on the plane passing through the points 
(–1, –2, –3), (9, 3, 4), (9, –2, 1) be Q(?, ?, ?). Then 
the distance Q from the origin is  
 (1) 42 (2) 38 
 (3) 35 (4) 29 
Answer (1) 
Sol.  
 10 4 AC i k =+ 
 10 5 7 AB i j k = + + 
 10 0 4
10 5 7
i j k
AC AB ?= 
   = –20i – 30j + 50k 
 Equation of plane 
  2x + 3y – 5z = d 
 Put (–1, –2, –3) 
  –2 – 6 + 15 = d 
  d = 7 
 ? 2x + 3y – 5z = 7 
 Foot of ?
r
 
 
3 2 9 38
2 3 5 38
x y z - + + ??
= = = -
??
-
??
 
 x = 1, y = –5, z = –4 
 Q(1, –5, –4) 
 Distance from origin 1 25 16 = + + 
   42 = 
6. The number of common tangents, to the circles  
 x
2
 + y
2
 – 18x –15y + 131 = 0 and x
2
 + y
2
 – 6x – 6y 
– 7 = 0, is  
 (1) 3 (2) 1 
 (3) 4 (4) 2 
Answer (1) 
Sol. 
22
18 15 131 0 x y x y + - - + = 
 
11
15 225 5
9, , 81 131
2 4 2
Cr
??
= + - =
??
??
 
 
22
6 6 7 0 x y x y + - - - = 
 
22
(3, 3), 9 9 7 5 Cr = + + = 
 
2
2
12
15 81 15
(9 3) 3 36
2 4 2
d C C
??
= = - + - = + =
??
??
 
 
12
5 15
5
22
rr + = + = 
 
1 2 1 2
C C r r ? = + 
 ? Circles touch each other externally, 3 common 
tangents. 
A B
D C
A(–1, –2, –3)
B(9, 3, 4)
C(9, –2, 1)
 
   
   
7. The total number of three-digit numbers, divisible 
by 3, which can be formed using the digits 1, 3, 5, 
8, if repetition of digits is allowed, is 
 (1) 21 (2) 20 
 (3) 22 (4) 18 
Answer (3) 
Sol. Sum of digits 3 : (1, 1, 1) 
 Sum of digits 9 : (1, 3, 5) or (3, 3, 3) 
 Sum of digits 12 : (1, 3, 8) 
 Sum of digits 15 : (5, 5, 5) 
 Sum of digits 18 : (5, 5, 8) 
 Sum of digits 21 : (5, 8, 8)  
 Sum of digits 24 : (8, 8, 8) 
 Possible numbers are 
 
3! 3!
1 3! 1 3! 1 1
2! 2!
= + + + + + + + 
 = 22 
8. If the set : , Re Re
2 3 5
3
z z z z
z
z z
z
 is 
equal to the interval (?, ?], then 24(? – ?) is equal 
to 
 (1) 36 (2) 27 
 (3) 30 (4) 42 
Answer (3) 
Sol. Re
2 3 5
z z zz
zz
??
-+
??
??
-+
??
 
 
( )
( ) ( )
22
Re
2 3 5
x iy x iy x y
x iy x iy
??
+ - - + +
??
?? - + + -
??
 
 
( )
22
2
Re
2 2 8
x y i y
x iy
??
++
??
?? +-
??
 
 
( )
( ) ( )
( ) ( ) ( )
22
2 2
2 2 1 8
Re
2 1 8
x y yi x iy
xy
??
+ + + +
??
??
?? ++
??
 
 
( )
( )
( ) ( )
2 2 2
22
2 1 16
4 1 8
x y x y
xy
+ + -
=
++
 
 Put x = 3 
 
( )
22
2
8 9 16
64 64
yy
y
+-
=
+
 
 
( )
( )
2
2
9
1
8
1
y
fy
y
-
=
+
 
 Range of f(y) = (– 0.125, 1.125] 
  ? = – 0.125 
  ? = 1.125  
  ? – ? = 1.25 
  24(? – ?) = 30 
9. Let 
( )
20
10
2
10
, , ,
i
i
i
a bx cx p x a b c
=
+ + = ?
?
. If  
p1 = 20 and p2 = 210, then 2(a + b + c) is equal to 
 (1) 6 (2) 15 
 (3) 12 (4) 8 
Answer (3) 
Sol. General term : ( ) ( )
( )
3
12
2
1 2 3
10!
!!!
r
rr
a bx cx
r r r
 
 For coeff. of x : r2 + 2r3 = 1 
  
1 2 3
910
r r r
   
 ? coff of 
91
10!
20
9!
x a b == 
 ? a
9
 ? b = 2 …(i) 
  Coff of x
2
 : 
8 2 9
10! 10!
210
8!2!0! 9!0!1!
a b a c ? + ? ? = 
 ? 45a
8
 ? b
2
 + 10 ? a
9
 ? c = 210 
 ? 9a
8
b
2
 + 2a
9
 ? c = 42 
  as a, b, c ?N 
 ? a = 1, b = 2, c = 3 
  2(a + b + c) = 2(3 + 2 + 1) = 12 
10. The number of real roots of the equation 
5 2 6 0, x x x is  
 (1) 5 (2) 4 
 (3) 6 (4) 3 
Answer (4) 
 
   
   
Sol. x|x| – 5|x + 2| + 6 = 0 
 Case-I : 
  x < –2 
  –x
2
 + 5(x + 2) + 6 = 0 
 ? x
2
 – 5x – 16 = 0 
 ? 
5 25 64
2
x 
 ? 
5 89
2
x is accepted 
 Case-II : 
  –2 ? x < 0 
  –x
2
 – 5(x + 2) + 6 = 0 
 ? x
2
 + 5x + 4 = 0 
 ? (x + 1) (x + 4) = 0 
  x = –1 is accepted 
 Case-III : 
  x ? 0 
  x
2
 – 5(x + 2) + 6 = 0 
 ? x
2
 – 5x – 4 = 0 
  
5 25 16
2
x 
   
5 41
2
 
  
5 41
2
x is accepted 
 ? 3 real roots are possible. 
 Option (4) is correct. 
11. Let A1 and A2 be two arithmetic means and G1, G2 
and G3 be three geometric means of two distinct 
positive numbers. Then 
4 4 4 2 2
1 2 3 1 3
G G G G G is 
equal to   
 (1) (A1 + A2)
2
 G1G3 
 (2) 2(A1 + A2) G1G3 
 (3) ( )
22
1 2 1 3
A A G G + 
 (4) ( )
22
1 2 1 3
2 A A G G + 
Answer (1) 
Sol. Let the two numbers are a, b. 
  
1
2
33
ba
ab
Aa 
  
2
2
2
33
ba
ab
Aa 
  
1
4
1
b
Ga
a
??
=
??
??
 
  
2
4
2
b
Ga
a
??
=
??
??
 
  
3
4
3
b
Ga
a
??
=
??
??
 
  ( ) ( ) ( ) ( ) ( )
42 4 4 2
1 2 3 1 3
G G G G G + + + ? 
 ? 
2 3 2
4 4 4 4
2 3 2
b b b b
a a a a
a
aaa
? + ? + ? + ? 
   = ba
3
 + b
2
a
2
 + b
3
a + a
2
b
2
 
   = ab(a
2
 + b
2
 + 2ab) = ab(a + b)
2
 
  ( ) ( )
2 2
1 3 1 3
A A G G a b ab + ? = + ? 
 ? Option (1) is correct. 
12. Let [x] denote the greatest integer function and  
( ) ? ? ? ? ? ?
max 1 , 2 , 2 f x x x x x x = + + + + , 0 ? x ? 2. 
Let m be the number of points in [0, 2], where f is 
not continuous and n be the number of points in  
(0, 2), where f is not differentiable. Then  
(m + n)
2
 + 2 is equal to 
 (1) 2 
 (2) 11 
 (3) 6 
 (4) 3 
Answer (4) 
Sol.  ( )
? ?
? ?
? ?
max 1 , 2, 0 1
max 2, 2, 2 1 2
max 5, 4, 6 2
x x x x
f x x x x x
x
? + + ? ?
?
= + + + ? ?
?
?
=
?
 
 ? ( )
2 0 1
2 1 2
62
xx
f x x x
x
+ ? ? ?
?
= + ? ?
?
?
=
?
 
 ? ( )
2 0 2
62
xx
fx
x
+ ? ? ?
=
?
=
?
 
 f is not continuous at x = 2 
 f is differentiable in (0, 2) 
 ? m = 1, n = 0 
  (m + n)
2
 + 2 = 1 + 2 = 3. 
 Option (4) is correct