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 $f\left(1\right)=0;f^{\prime }\left(1\right)=2$

$f\left(x\right)-f\left(y\right)$ is equal to

• A. $f\left(\frac{y}{x}\right)$
• B. $f\left(\frac{x}{y}\right)$
• C. $f\left(2x\right)$
• D. $f\left(2y\right)$

Answer 1

Answer: (B)

$f\left(\frac{x}{y}\right)$

$f\left(x\right)=\frac{1}{2}\left[f\right(xy)+f\left(\frac{x}{y}\right)]$ ...(i)

Put $x=y$ and $y=x$

$f\left(y\right)=\frac{1}{2}\left[f\right(xy)+f\left(\frac{y}{x}\right)]$ ...(ii)

On subtracting (ii) from (i), we have

$f\left(x\right)-f\left(y\right)=\frac{1}{2}\{f\left(\frac{x}{y}\right)-f\left(\frac{y}{x}\right)\}$ ....(iii)

Put $x=1$ in eqn (i), we get

$f\left(1\right)=\frac{1}{2}\left[f\right(y)+f\left(\frac{1}{y}\right)]$

Given $f\left(1\right)=0$

$\Rightarrow f\left(y\right)=-f\left(\frac{1}{y}\right)$
Hence, $f\left(\frac{x}{y}\right)=-f\left(\frac{y}{x}\right)$

$\therefore$ Eq. (iii), becomes $f\left(x\right)-f\left(y\right)=f\left(\frac{x}{y}\right)$