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

$f^{\prime }\left(3\right)$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $\frac{2}{3}$

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

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

$\underset{h\to 0}{lim}\frac{f\left(\frac{x+h}{x}\right)}{h}=\underset{h\to 0}{lim}\frac{f(1+\frac{h}{x})}{h}$

$\underset{h\to 0}{lim}\frac{f(1+\frac{h}{x})}{\frac{h}{x}.x}=\frac{1}{x}{f}^{\prime }\left(1\right)$

$=\frac{1}{x}.2=\frac{2}{x}$

$\therefore f^{\prime }\left(3\right)=\frac{2}{3}$