Question

If $2f\left(x\right)=f\left(xy\right)+f\left(\frac{x}{y}\right)$ for all positive values of $x$ and $y,f\left(1\right)=0$ and $f^{\prime }\left(1\right)=1$,then Find $f\left(e\right)$

Answer 1

Substitute $x=1$ in $2f\left(x\right)=f\left(xy\right)+f\left(\frac{x}{y}\right)$ ...(i)
$2f\left(1\right)=f\left(y\right)+f\left(\frac{1}{y}\right)$ ...(ii)
$\Rightarrow f\left(y\right)=-f\left(\frac{1}{y}\right)$
Replacing $x$ by $y$ and $y$ by $x$ in Eq. (i), we get
$2f\left(y\right)=f\left(yx\right)+f\left(\frac{y}{x}\right)$ .....(iii)
Eqs. (i) and (iii),
$2\left\{f\right(x)-f(y\left)\right\}=f\left(\frac{x}{y}\right)-\{-f\left(\frac{x}{y}\right)\}=2f\left(\frac{x}{y}\right)$ ...(iv)
$f\left(x\right)-f\left(y\right)=f\left(\frac{x}{y}\right)$ ...(v)
$\underset{h\to 0}{lim}\frac{f(1+h)-f\left(1\right)}{h}=f^{\prime }\left(1\right)=1$
$\underset{h\to 0}{lim}\frac{f(1+h)}{h}=1$, as $f\left(1\right)=0$
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f(1+\frac{h}{x})}{h}=\frac{1}{x}$
$f\left(x\right)=\log \left|x\right|+c$
$f\left(1\right)=0\Rightarrow c=0$
$f\left(e\right)=1$