10 </span>'/{ \html '<span class="right_angle">2 </span>' \html '<span class="right_angle">8</span>'} k^2=405 \Rightarrow10.9/2k^2=405 \Rightarrowk^2=405/45=9 \Rightarrowk^2=9 \Rightarrowk=\sqrt9 \Rightarrowk=\pm3" >
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NDA Math Binomial Theorem
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© examsnet.com
Question : 24
Total: 50
If the constant term in the expansion of
(
√
x
−
k
x
2
)
10
is 405, then what can be the values of k?
[NDA II 2019]
±2
±3
±5
±9
Validate
Solution:
Let
(
r
+
1
)
th term in the expansion of
(
√
x
−
k
x
2
)
10
be independent of x
Now,
T
r
+
1
=
10
C
r
(
√
x
)
10
−
r
⋅
(
−
k
x
2
)
r
=
10
C
r
x
5
−
r
2
⋅
k
r
⋅
x
−
2
r
=
10
C
r
x
5
−
r
2
−
2
r
⋅
k
r
⇒
T
r
+
1
=
10
C
r
x
5
−
5
r
2
⋅
k
r
.....(i)
This will be independent of x
if
5
−
5
r
2
=
0
⇒
10
−
5
r
2
=
0
⇒
10
−
5
r
=
0
⇒
5
r
=
10
⇒
r
=
10
5
=
2
Putting
r
=
2
in (i), we get
T
3
=
10
C
2
k
2
It is given that the term independent of x is equal to 405
∴
10
C
2
k
2
=
405
⇒
10
2
8
k
2
=
405
⇒
10.9
2
k
2
=
405
⇒
k
2
=
405
45
=
9
⇒
k
2
=
9
⇒
k
=
√
9
⇒
k
=
±
3
© examsnet.com
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