Question

Let $f\left(x\right)=\frac{1}{2}\left[f\right(xy)+f\left(\frac{x}{y}\right)]$ for $x,y\in R^{+}$ such that $f\left(1\right)=0;f^{\prime }\left(1\right)=2$

$f\left(e\right)$ is equal to

• A. $2$
• B. $1$
• C. $3$
• D. $4$

Answer 1

Answer: (A)

$2$

$f\left(x\right)$ is of the form $a\log \left(x\right)$, where '$a$' is a constant.
Hence, $\frac{1}{2}\left[f\right(xy)+f(\frac{x}{y}\left)\right]$
$=\frac{1}{2}\left[a\log \right(xy)+a\log (\frac{x}{y}\left)\right]$
$=\frac{1}{2}[a\log x+a\log y+a\log x-a\log y]$
$=a\log \left(x\right)$.
Now $f\left(1\right)=0$
And $f^{\prime }\left(1\right)=2$
$\Rightarrow {\frac{a}{x}}_{x=1}=2$
$\Rightarrow a=2$
Hence, $f\left(x\right)=2\log \left(x\right)$
Hence, $f\left(e\right)=2\log \left(e\right)$ $=2$