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JEE Main 9 Apr 2024 Shift 2 Solved Paper
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© examsnet.com
Question : 12
Total: 90
The sum of the coefficient of
x
2
∕
3
and
x
−
2
∕
5
in the binomial expansion of
(
x
2
∕
3
+
1
2
x
−
2
∕
5
)
9
is :
[9 Apr 2024 Shift 2]
21
∕
4
69
∕
16
63
∕
16
19
∕
4
Validate
Solution:
👈: Video Solution
T
r
+
1
=
9
C
r
(
x
2
∕
3
)
9
−
r
(
x
−
2
∕
5
2
)
r
=
9
C
r
(
1
2
)
r
(
r
)
(
σ
−
2
π
3
⋅
2
r
5
)
for coefficient of
x
2
∕
3
, put
6
−
2
r
3
−
2
r
5
=
2
3
⇒
r
=
5
∴
Coefficient of
x
2
∕
3
is
=
9
C
5
(
1
5
)
5
For coefficient of
x
−
2
∕
5
, put
6
−
2
r
3
−
2
r
5
=
−
2
5
⇒
r
=
6
Coefficient of
x
−
2
∕
5
is
9
C
6
(
1
2
)
6
Sum
=
9
C
5
(
1
2
)
5
+
9
C
6
(
1
2
)
6
=
21
4
© examsnet.com
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