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JEE Main Math Class 11 Binomial Theorem Part 2 Questions
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© examsnet.com
Question : 12
Total: 115
n
−
1
C
r
=
(
k
2
−
8
)
n
C
r
+
1
if and only if :
[27-Jan-2024 Shift 1]
2
√
2
<
k
≤
3
2
√
3
<
k
≤
3
√
2
2
√
3
<
k
<
3
√
3
2
√
2
<
k
<
2
√
3
Validate
Solution:
👈: Video Solution
n
−
1
C
r
=
(
k
2
−
8
)
n
C
r
+
1
r
+
1
≥
0
,
r
≥
0
╰
─
─
▾
─
─
╯
r
≥
0
n
−
1
C
r
n
C
r
+
1
=
k
2
−
8
r
+
1
n
=
k
2
−
8
⇒
k
2
−
8
>
0
(
k
−
2
√
2
)
(
k
+
2
√
2
)
>
0
k
∈
(
−
∞
,
−
2
√
2
)
∪
(
2
√
2
,
∞
)
⋅
⋅
⋅
⋅
⋅
⋅
⋅
(
i
)
∴
n
≥
r
+
1
,
r
+
1
n
≤
1
⇒
k
2
−
8
≤
1
k
2
−
9
≤
0
−
3
≤
k
≤
3
⋅
⋅
⋅
⋅
⋅
⋅
⋅
(
i
i
)
From equation (I) and (II) we get
k
∈
[
−
3
,
−
2
√
2
)
∪
(
2
√
2
,
3
]
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