JEE Advance 2024 Question Paper - 2 with Solutions | JEE Main & Advanced Previous Year Papers PDF Download

 Page 1


1
Answers & Solutions 
for
JEE (Advanced)-2024 (Paper-2) 
 
SECTION 1 (Maximum Marks : 12) 
• This section contains FOUR (04) questions.
• Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct answer.
• For each question, choose the option corresponding to the correct answer.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 If ONLY the correct option is chosen;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered); 
Negative Marks : -1 In all other cases.
1. Considering only the principal values of the inverse trigonometric functions, the value of
–1 –1
32
tan sin – 2cos
5
5
?? ? ? ? ?
? ? ?? ??
? ?
?? ??
 is 
(A)
7
24
(B)
–7
24
(C)
–5
24
(D)
5
24
Answer (B) 
Time : 3 hrs. Max. Marks: 180 
PART-I : MATHEMATICS 
JEE Adavanced 2024 Paper-2
Page 2


1
Answers & Solutions 
for
JEE (Advanced)-2024 (Paper-2) 
 
SECTION 1 (Maximum Marks : 12) 
• This section contains FOUR (04) questions.
• Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct answer.
• For each question, choose the option corresponding to the correct answer.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 If ONLY the correct option is chosen;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered); 
Negative Marks : -1 In all other cases.
1. Considering only the principal values of the inverse trigonometric functions, the value of
–1 –1
32
tan sin – 2cos
5
5
?? ? ? ? ?
? ? ?? ??
? ?
?? ??
 is 
(A)
7
24
(B)
–7
24
(C)
–5
24
(D)
5
24
Answer (B) 
Time : 3 hrs. Max. Marks: 180 
PART-I : MATHEMATICS 
JEE Adavanced 2024 Paper-2 2
Sol. 
–1 –1
32
tan sin – 2cos
5
5
?? ? ? ? ?
? ? ?? ??
? ?
?? ??
Let 
–1 –1
32 2
sin , 2cos cos
52
55
ß
= a =ß? =
3
sin
5
a= ?
3
tan
4
? a=
2
1
2tan 2
4
22
tan
1
3
1 – tan 1
24
ß
×
ß= = =
ß
-
? ( )
34
tan – tan 7
43
tan
1 tan tan 1 1 24
-
aß
a-ß = = = -
+ aß +
2. Let 
{ }
22
( , ) : 0, 0, 4 , 12 2 and 3 8 5 8 . = ?× = = = =- + = ?? S xy x y y xy x y x If the area of the region S is
2, a then a is equal to
(A)
17
2
(B)
17
3
(C)
17
4
(D)
17
5
Answer (B) 
Sol.  
22
4 , 12 2 2, 8 y xy x x y = = - ?= =
2
0
1
2 38
2
A xdx = +××
?
2
3
2
0
2 4 17
2 32 2 2 32 2
33 3
x
??
?? = × + = × + =
??
??
17
2
3
A =a ? a = ?
Option (B) is correct. 
3. Let . ? ? k If ( ) ( )
2
6
0
lim sin sin cos ,
?+
+ +=
x
x
kx x x e then the value of k is 
(A) 1 (B) 2
(C) 3 (D) 4
Answer (B) 
JEE Adavanced 2024 Paper-2
Page 3


1
Answers & Solutions 
for
JEE (Advanced)-2024 (Paper-2) 
 
SECTION 1 (Maximum Marks : 12) 
• This section contains FOUR (04) questions.
• Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct answer.
• For each question, choose the option corresponding to the correct answer.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 If ONLY the correct option is chosen;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered); 
Negative Marks : -1 In all other cases.
1. Considering only the principal values of the inverse trigonometric functions, the value of
–1 –1
32
tan sin – 2cos
5
5
?? ? ? ? ?
? ? ?? ??
? ?
?? ??
 is 
(A)
7
24
(B)
–7
24
(C)
–5
24
(D)
5
24
Answer (B) 
Time : 3 hrs. Max. Marks: 180 
PART-I : MATHEMATICS 
JEE Adavanced 2024 Paper-2 2
Sol. 
–1 –1
32
tan sin – 2cos
5
5
?? ? ? ? ?
? ? ?? ??
? ?
?? ??
Let 
–1 –1
32 2
sin , 2cos cos
52
55
ß
= a =ß? =
3
sin
5
a= ?
3
tan
4
? a=
2
1
2tan 2
4
22
tan
1
3
1 – tan 1
24
ß
×
ß= = =
ß
-
? ( )
34
tan – tan 7
43
tan
1 tan tan 1 1 24
-
aß
a-ß = = = -
+ aß +
2. Let 
{ }
22
( , ) : 0, 0, 4 , 12 2 and 3 8 5 8 . = ?× = = = =- + = ?? S xy x y y xy x y x If the area of the region S is
2, a then a is equal to
(A)
17
2
(B)
17
3
(C)
17
4
(D)
17
5
Answer (B) 
Sol.  
22
4 , 12 2 2, 8 y xy x x y = = - ?= =
2
0
1
2 38
2
A xdx = +××
?
2
3
2
0
2 4 17
2 32 2 2 32 2
33 3
x
??
?? = × + = × + =
??
??
17
2
3
A =a ? a = ?
Option (B) is correct. 
3. Let . ? ? k If ( ) ( )
2
6
0
lim sin sin cos ,
?+
+ +=
x
x
kx x x e then the value of k is 
(A) 1 (B) 2
(C) 3 (D) 4
Answer (B) 
JEE Adavanced 2024 Paper-2 3
Sol. ( ) ( )
2
6
0
lim sin sin cos
+
?
= ++=
x
x
l kx x x e
? ( ) ( )
0
2
ln lim sin sin cos 1
+
?
= ++-
x
l kx x x
x
?
( ) ( )
2
0
sin sin 1 cos
sin
ln lim 2 1
sin
+
?
? ? -
= ··+- ·
? ?
? ?
? ?
x
kx x
kx kx
lx
kx kx x
x
? lnl = 2(k + 1) ? l = e
2(k +1)
 = e
6 
k + 1 = 3 ? k = 2
4. Let : ? ?? f be a function defined by
2
2
sin , if 0,
()
0, if 0.
xx
fx
x
x
? p ? ?
?
? ? ?
=
? ? ?
?
=
?
Then which of the following statements is TRUE? 
(A) f(x) = 0 has infinitely many solutions in the interval
10
1
,
10
? ?
8
? ?
? ?
. 
(B) f(x) = 0 has no solutions in the interval 
1
,
? ?
8
? ?
p
? ?
. 
(C) The set of solutions of f(x) = 0 in the interval 
10
1
0,
10
??
??
??
 is finite. 
(D) f(x) = 0 has more than 25 solutions in the interval 
2
11
,
??
??
p
p ??
. 
Answer (D) 
Sol. 
2
2
sin , if 0,
()
0, if 0.
xx
fx
x
x
? p ? ?
?
? ? ?
=
? ? ?
?
=
?
f(x) = 0 ? sin
2
0
x
p ? ?
=
? ?
? ?
 
? 
2
x
p
= np 
?
2
1
x
n
=
? 
1
x
n
=
JEE Adavanced 2024 Paper-2
Page 4


1
Answers & Solutions 
for
JEE (Advanced)-2024 (Paper-2) 
 
SECTION 1 (Maximum Marks : 12) 
• This section contains FOUR (04) questions.
• Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct answer.
• For each question, choose the option corresponding to the correct answer.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 If ONLY the correct option is chosen;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered); 
Negative Marks : -1 In all other cases.
1. Considering only the principal values of the inverse trigonometric functions, the value of
–1 –1
32
tan sin – 2cos
5
5
?? ? ? ? ?
? ? ?? ??
? ?
?? ??
 is 
(A)
7
24
(B)
–7
24
(C)
–5
24
(D)
5
24
Answer (B) 
Time : 3 hrs. Max. Marks: 180 
PART-I : MATHEMATICS 
JEE Adavanced 2024 Paper-2 2
Sol. 
–1 –1
32
tan sin – 2cos
5
5
?? ? ? ? ?
? ? ?? ??
? ?
?? ??
Let 
–1 –1
32 2
sin , 2cos cos
52
55
ß
= a =ß? =
3
sin
5
a= ?
3
tan
4
? a=
2
1
2tan 2
4
22
tan
1
3
1 – tan 1
24
ß
×
ß= = =
ß
-
? ( )
34
tan – tan 7
43
tan
1 tan tan 1 1 24
-
aß
a-ß = = = -
+ aß +
2. Let 
{ }
22
( , ) : 0, 0, 4 , 12 2 and 3 8 5 8 . = ?× = = = =- + = ?? S xy x y y xy x y x If the area of the region S is
2, a then a is equal to
(A)
17
2
(B)
17
3
(C)
17
4
(D)
17
5
Answer (B) 
Sol.  
22
4 , 12 2 2, 8 y xy x x y = = - ?= =
2
0
1
2 38
2
A xdx = +××
?
2
3
2
0
2 4 17
2 32 2 2 32 2
33 3
x
??
?? = × + = × + =
??
??
17
2
3
A =a ? a = ?
Option (B) is correct. 
3. Let . ? ? k If ( ) ( )
2
6
0
lim sin sin cos ,
?+
+ +=
x
x
kx x x e then the value of k is 
(A) 1 (B) 2
(C) 3 (D) 4
Answer (B) 
JEE Adavanced 2024 Paper-2 3
Sol. ( ) ( )
2
6
0
lim sin sin cos
+
?
= ++=
x
x
l kx x x e
? ( ) ( )
0
2
ln lim sin sin cos 1
+
?
= ++-
x
l kx x x
x
?
( ) ( )
2
0
sin sin 1 cos
sin
ln lim 2 1
sin
+
?
? ? -
= ··+- ·
? ?
? ?
? ?
x
kx x
kx kx
lx
kx kx x
x
? lnl = 2(k + 1) ? l = e
2(k +1)
 = e
6 
k + 1 = 3 ? k = 2
4. Let : ? ?? f be a function defined by
2
2
sin , if 0,
()
0, if 0.
xx
fx
x
x
? p ? ?
?
? ? ?
=
? ? ?
?
=
?
Then which of the following statements is TRUE? 
(A) f(x) = 0 has infinitely many solutions in the interval
10
1
,
10
? ?
8
? ?
? ?
. 
(B) f(x) = 0 has no solutions in the interval 
1
,
? ?
8
? ?
p
? ?
. 
(C) The set of solutions of f(x) = 0 in the interval 
10
1
0,
10
??
??
??
 is finite. 
(D) f(x) = 0 has more than 25 solutions in the interval 
2
11
,
??
??
p
p ??
. 
Answer (D) 
Sol. 
2
2
sin , if 0,
()
0, if 0.
xx
fx
x
x
? p ? ?
?
? ? ?
=
? ? ?
?
=
?
f(x) = 0 ? sin
2
0
x
p ? ?
=
? ?
? ?
 
? 
2
x
p
= np 
?
2
1
x
n
=
? 
1
x
n
=
JEE Adavanced 2024 Paper-2 4
If x ?
10
1
,
10
? ?
8
? ?
? ?
 If 
1
, x
? ?
? 8
? ?
p
? ?
 If x ?
10
1
0,
10
??
??
??
10
11
,
10 n
? ?
? 8
? ?
? ?
 
11
,
n
? ?
? 8
? ?
p
? ?
 
10
(10 , ) n ?8 
(
10
0,10 n
?
?
?
( ] 0, n ? p n infinite 
(
10 2
0, (10 ) n
?
?
?
(
2
0, n
?
? p
?
 If x ?
2
11
,
??
??
p
p ??
Finite values of n n = 1, 2, 3 …9 
2
(, ) n ? pp 
24
(, ) n ?p p 
(9.8, 97.2 ...) n ? 
More than 25 solutions 
SECTION 2 (Maximum Marks : 12) 
• This section contains THREE (03) questions.
• Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is(are)
correct answer(s).
• For each question, choose the option(s) corresponding to (all) the correct answer(s).
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4  ONLY if (all) the correct option(s) is(are) chosen;
Partial Marks : +3  If all the four options are correct but ONLY three options are chosen;
Partial Marks : + 2  If three or more options are correct but ONLY two options are chosen, both of which
are correct; 
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a correct
option; 
Zero Marks : 0 If unanswered; 
Negative Marks : –2 In all other cases.
5. Let S be the set of all (a, ß) ? ? × ? such that
2
2
1
sin( )(log ) sin
lim 0
(log (1 ))
a
aß ß
?8
??
??
??
=
+
e
x
e
xx
x
xx
 
Then which of the following is (are) correct? 
(A) (–1, 3) ? S (B) (–1, 1) ? S
(C) (1, –1) ? S (D) (1, –2) ? S
Answer (B, C) 
JEE Adavanced 2024 Paper-2
Page 5


1
Answers & Solutions 
for
JEE (Advanced)-2024 (Paper-2) 
 
SECTION 1 (Maximum Marks : 12) 
• This section contains FOUR (04) questions.
• Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is the correct answer.
• For each question, choose the option corresponding to the correct answer.
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 If ONLY the correct option is chosen;
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered); 
Negative Marks : -1 In all other cases.
1. Considering only the principal values of the inverse trigonometric functions, the value of
–1 –1
32
tan sin – 2cos
5
5
?? ? ? ? ?
? ? ?? ??
? ?
?? ??
 is 
(A)
7
24
(B)
–7
24
(C)
–5
24
(D)
5
24
Answer (B) 
Time : 3 hrs. Max. Marks: 180 
PART-I : MATHEMATICS 
JEE Adavanced 2024 Paper-2 2
Sol. 
–1 –1
32
tan sin – 2cos
5
5
?? ? ? ? ?
? ? ?? ??
? ?
?? ??
Let 
–1 –1
32 2
sin , 2cos cos
52
55
ß
= a =ß? =
3
sin
5
a= ?
3
tan
4
? a=
2
1
2tan 2
4
22
tan
1
3
1 – tan 1
24
ß
×
ß= = =
ß
-
? ( )
34
tan – tan 7
43
tan
1 tan tan 1 1 24
-
aß
a-ß = = = -
+ aß +
2. Let 
{ }
22
( , ) : 0, 0, 4 , 12 2 and 3 8 5 8 . = ?× = = = =- + = ?? S xy x y y xy x y x If the area of the region S is
2, a then a is equal to
(A)
17
2
(B)
17
3
(C)
17
4
(D)
17
5
Answer (B) 
Sol.  
22
4 , 12 2 2, 8 y xy x x y = = - ?= =
2
0
1
2 38
2
A xdx = +××
?
2
3
2
0
2 4 17
2 32 2 2 32 2
33 3
x
??
?? = × + = × + =
??
??
17
2
3
A =a ? a = ?
Option (B) is correct. 
3. Let . ? ? k If ( ) ( )
2
6
0
lim sin sin cos ,
?+
+ +=
x
x
kx x x e then the value of k is 
(A) 1 (B) 2
(C) 3 (D) 4
Answer (B) 
JEE Adavanced 2024 Paper-2 3
Sol. ( ) ( )
2
6
0
lim sin sin cos
+
?
= ++=
x
x
l kx x x e
? ( ) ( )
0
2
ln lim sin sin cos 1
+
?
= ++-
x
l kx x x
x
?
( ) ( )
2
0
sin sin 1 cos
sin
ln lim 2 1
sin
+
?
? ? -
= ··+- ·
? ?
? ?
? ?
x
kx x
kx kx
lx
kx kx x
x
? lnl = 2(k + 1) ? l = e
2(k +1)
 = e
6 
k + 1 = 3 ? k = 2
4. Let : ? ?? f be a function defined by
2
2
sin , if 0,
()
0, if 0.
xx
fx
x
x
? p ? ?
?
? ? ?
=
? ? ?
?
=
?
Then which of the following statements is TRUE? 
(A) f(x) = 0 has infinitely many solutions in the interval
10
1
,
10
? ?
8
? ?
? ?
. 
(B) f(x) = 0 has no solutions in the interval 
1
,
? ?
8
? ?
p
? ?
. 
(C) The set of solutions of f(x) = 0 in the interval 
10
1
0,
10
??
??
??
 is finite. 
(D) f(x) = 0 has more than 25 solutions in the interval 
2
11
,
??
??
p
p ??
. 
Answer (D) 
Sol. 
2
2
sin , if 0,
()
0, if 0.
xx
fx
x
x
? p ? ?
?
? ? ?
=
? ? ?
?
=
?
f(x) = 0 ? sin
2
0
x
p ? ?
=
? ?
? ?
 
? 
2
x
p
= np 
?
2
1
x
n
=
? 
1
x
n
=
JEE Adavanced 2024 Paper-2 4
If x ?
10
1
,
10
? ?
8
? ?
? ?
 If 
1
, x
? ?
? 8
? ?
p
? ?
 If x ?
10
1
0,
10
??
??
??
10
11
,
10 n
? ?
? 8
? ?
? ?
 
11
,
n
? ?
? 8
? ?
p
? ?
 
10
(10 , ) n ?8 
(
10
0,10 n
?
?
?
( ] 0, n ? p n infinite 
(
10 2
0, (10 ) n
?
?
?
(
2
0, n
?
? p
?
 If x ?
2
11
,
??
??
p
p ??
Finite values of n n = 1, 2, 3 …9 
2
(, ) n ? pp 
24
(, ) n ?p p 
(9.8, 97.2 ...) n ? 
More than 25 solutions 
SECTION 2 (Maximum Marks : 12) 
• This section contains THREE (03) questions.
• Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is(are)
correct answer(s).
• For each question, choose the option(s) corresponding to (all) the correct answer(s).
• Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4  ONLY if (all) the correct option(s) is(are) chosen;
Partial Marks : +3  If all the four options are correct but ONLY three options are chosen;
Partial Marks : + 2  If three or more options are correct but ONLY two options are chosen, both of which
are correct; 
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a correct
option; 
Zero Marks : 0 If unanswered; 
Negative Marks : –2 In all other cases.
5. Let S be the set of all (a, ß) ? ? × ? such that
2
2
1
sin( )(log ) sin
lim 0
(log (1 ))
a
aß ß
?8
??
??
??
=
+
e
x
e
xx
x
xx
 
Then which of the following is (are) correct? 
(A) (–1, 3) ? S (B) (–1, 1) ? S
(C) (1, –1) ? S (D) (1, –2) ? S
Answer (B, C) 
JEE Adavanced 2024 Paper-2 5
Sol. 
2
2
1
sin( )sin (ln )
lim 0
(ln(1 ))
x
x x
x
xx
a
aß ß
?8
??
??
??
=
+
 
2
22
2
11
(sin )sin
lim 0
1
(ln(1 ))
x
x
xx
xx
x
?8
aß ß
??
??
??
= =
??
+
??
??
 
It is possible if aß + 2 > 0 
aß > – 2 
(A) aß = –3
(B) aß = –1
(C) aß = –1
(D) aß = –2
6. A straight line drawn from the point P(1,3,2), parallel to the line
2 46
12 1
xy z - --
= = , intersects the plane 
L1 : x – y + 3z = 6 at the point Q. Another straight line which passes through Q and is perpendicular to the plane 
L1 intersects the plane L2 : 2x – y + z = –4 at the point R. Then which of the following statements is(are) TRUE? 
(A) The length of the line segment PQ is 6
(B) The coordinates of R are (1,6,3)
(C) The centroid of the triangle PQR is 
4 14 5
,,
3 33
??
??
??
(D) The perimeter of the triangle PQR is 2 6 11 ++ 
Answer (A, C) 
Sol. Equation of line parallel to 
2 46
12 1
xy z - --
= = through P(1,3,2) is 
13 2
12 1
x y z - - -
= = = ? (let) 
Now, putting any point (? + 1, 2? + 3, ? + 2) in L1 
1 ?=
? Point Q(2,5,3)
Equation of line through Q(2,5,3) perpendicular to L1 is
2 53
1 13
xy z - --
= = = µ
-
 (Let) 
 Putting any point (µ + 2, –µ + 5, 3µ + 3) in L2 
1 µ= -
 ? Point R (1, 6, 0)
(A) 14 1 6 PQ= + + =
(B) R(1, 6, 0)
(C) Centroid 
4 14 5
,,
3 33
??
??
??
(D) PQ + QR + PR 6 11 13 = ++
JEE Adavanced 2024 Paper-2