Find the general term in the expansion of (1 + px)^-1/p.
Answers
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Step-by-step explanation:
x
1+px
−
1−px
x−2
2x+1
,0≤x≤1
,−1≤x<0
Since, f(x) is continuous in [−1,1], so f(x) is continuous at x=0
Now, f(0)=
2
−1
(by def of f(x))
LHL=
x→0
−
lim
f(x)
=
h→0
lim
f(0−h)
=
h→0
lim
−h
1−ph
−
1+ph
=
h→0
lim
−h
1−ph
−
1+ph
×
1−ph
+
1+ph
1−ph
+
1+ph
=
h→0
lim
−h(
1−ph
+
1+ph
)
1−ph−1−ph
=
h→0
lim
−h(
1−ph
+
1+ph
)
−2ph
LHL=
2
2p
=p
Since, f(x) is continuous at x=0
LHL=f(0)
⇒p=−
2
1
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