Question

Given $f^{\prime }\left(1\right)=1$ and $\frac{d}{dx}\left(f\right(2x\left)\right)=f^{\prime }\left(x\right)$ for all $x>0$. If $f^{\prime }\left(x\right)$ is differentiable then there exists a number $cϵ(2,4)$ such that $f"\left(c\right)$ equals

• A. $-1/4$
• B. $-1/8$
• C. $1/4$
• D. $1/8$

Answer 1

The correct option is B $-1/8$

$\frac{d}{dx}\left(f\left(2x\right)\right)=f^{{}^{\prime }}\left(x\right)\Rightarrow 2f^{{}^{\prime }}\left(2x\right)=f^{{}^{\prime }}\left(x\right)$
As $f^{{}^{\prime }}\left(1\right)=1\Rightarrow f^{{}^{\prime }}\left(2\right)=\frac{1}{2}\Rightarrow f^{{}^{\prime }}\left(x\right)=\frac{1}{x}$
Now using mean value theorem
$f^{\prime \prime }\left(x\right)=\frac{f^{{}^{\prime }}\left(4\right)-f^{{}^{\prime }}\left(2\right)}{4-2}=-\frac{1}{8}$
Hence, option 'B' is correct.