JEE (Main)-2023 : Phase-2 (06-04-2023)-Morning
MATHEMATICS
Sol.
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.
Tn = 5 +
The number of ways to distribute 20 chocolates
among three students such that each student will
get atleast one chocolate is
(1)
22
C2
(2)
19
(3)
19
C3
(4)
22
Tn = 5 + 6 ( n − 1) + ( n − 1)( n − 2 )
Tn = n 2 + 3n + 1
C2
C3
Tn =
Answer (2)
x  1, y  1, z  1
 where x1 = x + 1

y1 = y + 1


z1 = z + 1

Number of ways =
=
4.
17 +3 −1
C3 −1
19
C2
+n
Mean of first 15 numbers is 12 and variance is 14.
Mean of next 15 numbers is 14 and variance is a. If
variance of all 30 numbers is 13, then a is equal to
(1) 12
(2) 14
(3) 10
(4) 3
Sol. 2 =
(1)
14
(2)
15
(3)
15
(4)
14
C7
C6
C8
13 =
C8
xi2
2
– (x )
n
(14 + 144)  15 + (a + 196)  15
– (13)2
30
 a = 10
Answer (3)
 1 
Cr ( x 4 )15 −r  − 3 
 x 
5.
r
15
If the image of point P(1, 2, 3) about the plane 2x –
y + 3y = 2 is Q, then the area of triangle PQR, where
coordinates of R is (4, 10, 12)
Cr x 60 − 4r − 3r ( − 1)r
15
Now, 60 – 7r = 18
(1)
1531
2
(2)
1675
2
(3)
2443
2
(4)
1784
2
r=6
 coefficient of x18 is
15
C6
Sum of first 20 turns of the series 5, 11, 19, 29, 41,
…… is
(1) 3130
(2) 3520
(3) 2790
(4) 1880
Answer (2)
2
Answer (3)
15
 4 1 
x − 3 
x 

3.
3 ( n )( n + 1)
= 2870 + 630 + 20 = 3520
The coefficient of x18 in the expansion of
=
6
+
20  21 41 3  20  21
+
+ 20
6
2
S20 =
x1 + y1 + z1 = 17
Sol. Tr +1 =
( n )( n + 1)( 2n + 1)
n = 20
Sol. x + y + z = 20
2.
n −1
(12 + ( n − 2) 2)
2
Answer (1)
- 12 -
JEE (Main)-2023 : Phase-2 (06-04-2023)-Morning
Sol. Image formula
x −1 y − 2 z − 3
2−2+9−2
=
=
= −2 

2
−1
3
14


 x = –1
=
1
5log2 − 6 + 3
9
=
1
5 ( log2 ) − 3 
9
2
y=3
18  f ( x ) dx = 10log2 − 6
z=0
1
7.
The sum of roots of x 2 − 8 x + 15 − 2 x + 7 = 0 is
(1) 11 + 3
(2) 11 − 3
(3) 9 + 3
PQ = −2i + j − 3k
(4) 9 − 3
Answer (3)
PR = 3i + 8 j + 9k
Sol. x 2 − 8 x + 15 − 2 x + 7 = 0
1
Area = PQ  PR
2
6.

( x − 3)( x − 5) − 2 x + 7 = 0
1
= 33i + 9 j − 19k
2
Let x  3 or x  5, then
1531
=
2
x2 – 10x + 22 = 0
x2 – 8x + 15 – 2x + 7 = 0
2
x=
5  3
but x  (–, 3]  [5, )
3
1

x =5 + 3
 1 1
If 5f ( x ) + 4f   = + 3 , then 18  f ( x ) dx is
x x
(1) 10log2 + 6
(2) 10log2 − 6
Now,
(3) 5log2 + 6
(4) 5log2 − 6
x  (3, 5), then
Answer (2)
–x2 + 8x – 15 – 2x + 7 = 0


 1 1
5  5f ( x ) + 4f   = + 3 
x
x
 


1
Sol.
x→
x
  1

4  5f   + 4f ( x ) = x + 3 
 x

x2 – 6x + 8 = 0
...(i)
x = 2, 4
 x=4
 Sum of roots = 9 + 3
...(ii)
8.
(1) (P  R)  Q
5
9f ( x ) = + 15 − 4 x − 12
x
5
9f ( x ) = − 4 x + 3
x
2
 f ( x ) dx =
1
1
9
(P  Q)  (R  Q) is equivalent to
(2) (P  R)  Q
(3) (Q  R)  (P  R) (4) (R  P)  (Q  R)
Answer (1)
Sol. (P  Q)  (R  Q)
= (~P  Q)  (~R  Q)
2
5

  x − 4 x + 3  dx
1
= (~P  ~R)  Q
= ~ (PR)  Q
2
1
= 5log x − 2 x 2 + 3 x 
1
9
= (PR)  Q
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JEE (Main)-2023 : Phase-2 (06-04-2023)-Morning
9.
Let a = 2i + 3 ˆj + 4kˆ, b = iˆ − 2 jˆ − 2kˆ, c = −iˆ + 4 jˆ + 3kˆ


x 2 ( x sec 2 x + tan x )
( x tan x + 1)2
and d is a vector perpendicular to both b and c
(1) 720
(2) 640
(3) 680
(4) 760
11. From the top of 30 m tower AB the angle of
depression to another tower’s QP base and top is
60º and 30º respectively. Another point C lies on
tower AB such that CQ is parallel to BP (where B
and P are the base of towers). Then the area of
BCQP is
Answer (1)
Sol. b  c = 2iˆ − jˆ + 2kˆ

d = (2iˆ − jˆ + 2kˆ )
(1) 600
a · d = 18
(3) 600
(
)
3 −1
(
300 (
(2) 600
(4)
)
3 − 1)
3 +1
Answer (1)
 =2
Sol.
| a  d |2 = | a |2 · | d |2 −(a · d )2

−x2
+ 2log | x sin x + cos x | + c
x tan x + 1
=
and a · d = 18 , then | a  d |2 is
dx
= 720
10. The integration

x 2 ( x sec 2 x + tan x )
( x tan x + 1)2
dx is
(1)
x
+ log | x sin x + cos x | + c
x tan x + 1
(2)
x
− log | x sin x + cos x | + c
x tan x + 1
(3)
−x
+ 2log | x sin x + cos x | + c
x tan x + 1
(4)
x2
+ 2log | x sin x + cos x | + c
x tan x + 1
tan60º =
2
BP = 10 3
tan15º =
x
2
·
I
(
( x sec 2 x + tan x )
( x tan x + 1)
2
(
Now area = BP × QP
(
= 10 3 60 − 20 3
I
Let xsinx + cosx = t
= 600
(
)
3 −1
 dy 
12. If 2y x + 3 x y = 20 , then 
is equal to

 dx at ( 2,2 )
(xcosx + sinx – sinx) dx = dt
(1)
dt
= 2log t + c
t
= 2log | x sin x + cos x | + c
)
= 600 3 − 600
x
dx
x tan x + 1
x cos x
= 2
dx
x sin x + cos x
= 2
)
= 60 − 20 3
− x2
2x
+
dx
( x tan x + 1)  x tan x + 1
I = 2
)
 QP = 30 − 10 3 2 − 3
dx
II
=
AC
10 3
 AC = 10 3 2 − 3
Answer (3)
Sol.
30
BP
− ( 8 + 12ln2 )
(12 + 8ln2 )
(3) 8ln + 12
Answer (2)
- 14 -
(2)
− ( 8ln2 + 12 )
( 8 + 12ln2 )
(4) 8 + 12ln2
JEE (Main)-2023 : Phase-2 (06-04-2023)-Morning
Sol. 2y x + 3 x y = 20
Sol.
…(i)
Let u = yx
lnu = xlny
1 dy x dy
=
+ ln y
u dx y dx
Let v = xy
lnv = y lnx
1 dy y dy
= +
ln x
v dx x dx
Now (i) differentiate w.r.t. x
Equation of OG :
dy
dv
2
+3
=0
dx
dx
Equation of BE :
  y dy
 x dy


2y x 
+ ln y  + 3  x y  +
ln x   = 0

 y dx

  x dx
Put x = 2, y = 2
 dy

 dy

8
+ ln 2  + 12 1 +
ln 2  = 0
 dx

 dx

13. Number of words with (or) without meaning using
all the letters of the word ASSASSINATION such
that all the vowels come together is
(1) 38004
(2) 38042
(3) 50400
(4) 60200
15.
16.
17.
Answer (3)
Sol. SSSSNTN
18.
A AIAIO
19.
7
20.
Number of required words

SECTION - B
8!
6!

= 50400
4! 2! 3! 2!
14. If a cuboid has its sides along axes with lengths 3,
4 and 5, find the shortest distance between body
diagonal and the side not containing the vertices of
body diagonal.
(1)
(2)
(3)
(4)
20
41
12
5
Numerical Value Type Questions: This section
contains 10 questions. In Section B, attempt any five
questions out of 10. The answer to each question is a
NUMERICAL VALUE. For each question, enter the
correct numerical value (in decimal notation,
truncated/rounded-off to the second decimal place;
e.g., 06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using
the mouse and the on-screen virtual numeric keypad in
the place designated to enter the answer.
21. Let a1, a2, a3,… an are in arithmetic progression
having common difference as ‘d’. The value of
15
d
n → n
lim
34
18
5
Answer (2)


1
1
1


+
+ ...
 a + a

a
+
a
a
+
a
1
2
2
3
n
−
1
n


is _______
Answer (1)
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JEE (Main)-2023 : Phase-2 (06-04-2023)-Morning
d
n → n
Sol. lim
d
n → n
= lim
= lim
n →
n −4


1
1
1


+
+ ...
 a + a
a2 + a3
an –1 + an 
2
 1
 a 2 – a1
a – a2
a – an –1 


+ 3
+ ... n
 a2 – a1
a3 – a2
an – an –1 

d 1
–
n d
(
an – a1
4
 1
 −1 
3 4 
C4  2 4 
 


T
 

 = 6
Sol. 5 =
n −4
4
T5 '
 −1 
 1
n
4
4




C4 3
2


 


 
n
)
 1
=  24 
 
 
1  a1 + ( n – 1) d – a1 

n → d 
n


= lim
 1  a
a 
d
= lim 
 1 + d – – 1  
n
d  
n →  d  n

n− 8
4
n −4−4

( 6)

n −8 1
=
4
2
 1
. 3 4 
 
 
n −4−4
1
= ( 6) 2
 n–8=2
=1
 n = 10
22. Matrix A is 2 × 2 matrix and A2 = I, no elements of
24. If 2nC3 : nC3 = 10, then
the matrix is zero, let sum of diagonal elements is a
and det(A) = b, then the value of
3a2
+
b2
is
n 2 + 3n
n 2 − 3n + 4
is equal to
Answer (02)
Answer (1)
u v  u
Sol. 

w x  w
1
= (6)2
Sol.
v
x 
u2 + vw = 1
uv + vx = 0  u = –v
wu + wx = 0  u = –x  u + x = 0 = a
3! ( n − 3 ) !
2n !

= 10
3! ( 2n − 3 )!
n!

( 2n )( 2n − 1)( 2n − 2 )
= 10
n ( n − 1)( n − 2 )

4 ( 2n − 1) = 10n − 20
 2n = 16  n = 8
wv + x2 = 1

|A2| = |I|
det(A) = 1
n 2 + 3n
2
n − 3n + 4
=
64 + 24
88
=
= 02
64 − 24 + 4 44
 b = 1
25. The number of points of non-differentiability of the
function f(x) = [4 + 13sinx] in (0, 2) is ____.
 3a2 + b2 = 3 × 0 + 1 = 1
Answer (50)
Sol. Number of points of non-differentiability
4 + [13 sinx] is 4 × 12 + 2 = 50 (by graph)
23. Ratio of terms of 5th term from beginning and 5th
n
term from end is
 1

 4
1 
. The value
6 : 1 in  2 +
1 


34 

26.
27.
28.
of n is _______.
29.
Answer (10)
30.
❑
❑
- 16 -
❑
for