Question

If $4f\left(x\right)-f\left(\frac{1}{x}\right)={\log }_e{x}^5$, then

• A. $f\left(x\right)={\log }_{e}x$
• B. $f\left(e^{2x}\right)$ is discontinuous for at least one $x\in [1,2]$
• C. $f\left(e^{2x}\right)$ is continuous on $[-1,1]$
• D. $f^{\prime }\left(e^x\right)=1$

Answer 1

Answer: (A), (C)

The correct options are
A $f\left(x\right)={\log }_{e}x$
C $f\left(e^{2x}\right)$ is continuous on $[-1,1]$

$4f\left(x\right)-f\left(\frac{1}{x}\right)=5{\log }_{e}x\cdots \left(1\right)$
Replace $x$ by $\frac{1}{x}$
$4f\left(\frac{1}{x}\right)-f\left(x\right)=5{\log }_e\left(\frac{1}{x}\right)$
$\Rightarrow 4f\left(\frac{1}{x}\right)-f\left(x\right)=-5{\log }_{e}x\cdots \left(2\right)$

$4\times \left(1\right)+\left(2\right)$
$\Rightarrow 16f\left(x\right)-f\left(x\right)=15{\log }_{e}x$
$\Rightarrow f\left(x\right)={\log }_{e}x$

$f\left(e^x\right)=x\Rightarrow f^{\prime }\left(e^x\right)=e^{-x}$
$f\left(e^{2x}\right)=2x$ which is a polynomial, so continuous everywhere.