Practicing Success
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 |
$(x-a)^{2 n-2}+b$ $b-1-(x+a)^{2 n+1}$ $b-(x-a)^{2 n+2}$ $(x-a)^{2 n+2}+b$ |
$b-(x-a)^{2 n+2}$ |
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. |