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Let f(x)=x^(n+1)+ax^n, "where " a gt 0. ...

Let f(x)`=x^(n+1)+ax^n, "where " a gt 0`. Then, x=0 is point of

A

local minimum for any integer n

B

local minimum if n is an even integer

C

local maximum if n is an even integer

D

local minimum if n is am odd interger

Text Solution

Verified by Experts

The correct Answer is:
C

We have
`f(x)=x^(n+1)+ax^n`
`rArr f(x) =(n+1)x^n +nax^(n-1)`
`rArr f(x)=[(n+1)x+a]x^(n-1)`
If n is even , then
`(LHD at x =0 ) gt 0 and (RHD at x=0)lt 0 `
Thus f(x) has a local maximum at x=0
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