Let $f(x)=x^{n+1}+a x^n$, where a > 0. Then, x = 0, is a point of
Answer & explanation
Correct answer: option 3
We have,
$f(x)=x^{n+1}+a x^n$
$\Rightarrow f^{\prime}(x)=(n+1) x^n+n a x^{n-1}$
$\Rightarrow f^{\prime}(x)=[(n+1) x+a] x^{n-1}$
If n is even, then
(LHD at x = 0) > 0 and (RHD at x = 0) < 0.
Thus, f(x) has a local maximum at x = 0.