Question

Given $f^{\prime }\left(1\right)=1$ and $f\left(2x\right)=f\left(x\right),\forall x>0$. If $f^{\prime }\left(x\right)$ is differentiable, then there exists a number $c\in (2,4)$ such that $f^{\prime \prime }\left(c\right)$ is equal to:

• A. $\frac{1}{4}$
• B. $-\frac{1}{2}$
• C. $-\frac{1}{4}$
• D. $-\frac{1}{8}$

Answer 1

The correct option is D $-\frac{1}{8}$

$f^{\prime }\left(1\right)=1$ and $f\left(2x\right)=f\left(x\right),\forall x>0$
$\Rightarrow 2f^{\prime }\left(2x\right)=f^{\prime }\left(x\right)$
Put $x=1$ $\Rightarrow 2f^{\prime }\left(2\right)=f^{\prime }\left(1\right)$
$\Rightarrow f^{\prime }\left(2\right)=\frac{1}{2}$
Put $x=2\Rightarrow 2f^{\prime }\left(4\right)=f^{\prime }\left(2\right)$
$\Rightarrow f^{\prime }\left(4\right)=\frac{1}{4}$
Since $f^{\prime }\left(x\right)$ is differentiable, From LMVT, we have
$f^{\prime \prime }\left(c\right)=\frac{f^{\prime }\left(4\right)-f^{\prime }\left(2\right)}{4-2}=-\frac{1}{8}$