Find : i) 4thterm of (8 - x)^1/3
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MATHS
If x>0 and the 4
th
term in the expansion of (2+
8
3
x)
10
has maximum value then find the range of x.
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ANSWER
(2+
8
3
x)
10
=2
10
(1+
16
3
x)
10
∣
∣
∣
∣
∣
T
3
T
4
∣
∣
∣
∣
∣
≥1 ∵T
4
has the maximum numerical value.
∣
∣
∣
∣
∣
T
5
T
4
∣
∣
∣
∣
∣
≥1⇒
∣
∣
∣
∣
∣
T
4
T
5
∣
∣
∣
∣
∣
≤1
(1+x)
n
has
∣
∣
∣
∣
∣
T
r
T
r+1
∣
∣
∣
∣
∣
=
r
n−r+1
×x
Consider the expansion
(1+
16
3
x)
10
which is of the form (1+x)
n
where x→
16
3x
and n=10
⇒
T
r
T
r+1
=
r
10−r+1
×
16
3x
=
16
8x
=
2
x
We have
∣
∣
∣
∣
∣
T
3
T
4
∣
∣
∣
∣
∣
≥1
⇒
∣
∣
∣
∣
2
x
∣
∣
∣
∣
≥1
⇒∣x∣≥2 ......(1)
⇒
∣
∣
∣
∣
∣
T
4
T
5
∣
∣
∣
∣
∣
≤1
⇒
∣
∣
∣
∣
∣
4
10−4+1
×
16
3x
∣
∣
∣
∣
∣
≤1
⇒
∣
∣
∣
∣
∣
4
7
×
16
3x
∣
∣
∣
∣
∣
≤1
⇒21∣x∣≤64
⇒∣x∣≤
21
64
.......(2)
From (1)⇒x
2
≥4
From (2)⇒x
2
≤(
21
64
)
2
If x
2
≥4⇒x≥2 or x≤−2
If x
2
≤(
21
64
)
2
⇒x≤
21
64
or
21
−64
≤x≤
21
64
∴ the range of x is 2≤x≤
21
64
or
21
−64
≤x≤−2