A function f such that
$f'(a)=f''(a)=f'''(a)=...=f^{(2 n)}(a)=0$
and f has a local maximum value b at x = a, if f(x) is
Answer & explanation
Correct answer: option 3
It is given that
$f'(a)=f''(a)=...=f^{(2 n)}(a)=0$
$\Rightarrow x=a$ is root of $f(x)$ of order $(2 n+1)$ or more.
Also, it is given that $f(a)=b$. Therefore,
$f(x)=b \pm(x-a)^{2 n+2}$
If $f(x)=b-(x-a)^{2 n+2}$, then
$f^{\prime}(x)=-(2 n+2)(x-a)^{2 n+1}$
Clearly, $f^{\prime}(x)$ changes its sign from positive to negative in the neighbourhood of $x=a$.
Therefore, $f(x)$ attains a local maximum at $x=a$. Hence, option (c) is correct.