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JEE Main 28 June 2022 Shift 2 Solved Paper
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© examsnet.com
Question : 4
Total: 90
The term independent of
x
in the expansion of
(
1
−
x
2
+
3
x
3
)
(
5
2
x
3
−
1
5
x
2
)
11
,
x
≠
0
is :
[28-Jun-2022-Shift-2]
7
40
33
200
39
200
11
50
Validate
Solution:
General term of Binomial expansion
(
5
2
x
3
−
1
5
x
2
)
11
is
T
r
+
1
=
11
C
r
⋅
(
5
2
x
3
)
11
−
r
⋅
(
−
1
5
x
2
)
r
=
11
C
r
⋅
(
5
2
)
11
−
r
⋅
(
−
1
5
)
r
⋅
x
33
−
5
r
In the term,
(
1
−
x
2
+
3
x
3
)
(
5
2
x
3
−
1
5
x
2
)
11
Term independent of
x
is when
(1)
33
−
5
r
=
0
⇒
r
=
33
5
∉
integer
(2)
33
−
5
r
=
−
2
⇒
5
r
=
35
⇒
r
=
7
∈
integer
(3)
33
−
5
r
=
−
3
⇒
5
r
=
36
⇒
r
=
36
5
∉
integer
∴
Only for
r
=
7
independent of
x
term possible.
∴
Independent of
x
term
=
−
(
11
C
7
(
5
2
)
4
⋅
(
−
1
5
)
7
)
=
−
(
11
⋅
10
⋅
9
⋅
8
4
⋅
3
⋅
2
⋅
1
⋅
5
4
2
4
⋅
−
1
5
7
)
=
11
⋅
10
⋅
3
2
4
⋅
5
3
=
11
⋅
3
2
3
⋅
5
2
=
33
200
© examsnet.com
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