Question

Let $2f\left(x\right)=f\left(xy\right)+f\left(\frac{x}{y}\right),x,y\in R^{+}.$ If $f\left(1\right)=0,$ then

• A. $f\left(x^n\right)=(n-1)f\left(x\right)$
• B. $f\left(x^n\right)=nf\left(x\right)$
• C. $f\left(x^n\right)=(n+1)f\left(x\right)$
• D. $f\left(x^n\right)=(n+2)f\left(x\right)$

Answer 1

The correct option is B $f\left(x^n\right)=nf\left(x\right)$

$2f\left(x\right)=f\left(xy\right)+f\left(\frac{x}{y}\right)$
$\Rightarrow f\left(xy\right)=2f\left(x\right)-f\left(\frac{x}{y}\right)...\left(i\right)$
Put $x=1$
$f\left(y\right)=2f\left(1\right)-f\left(\frac{1}{y}\right)\Rightarrow f\left(y\right)=-f\left(\frac{1}{y}\right)...\left(ii\right)(\because f(1)=0)$
Put $y=x$
$f\left(x^2\right)=2f\left(x\right)-f\left(1\right)\Rightarrow f\left(x^2\right)=2f\left(x\right)...\left(iii\right)$
Put $y=x^2$
$f\left(x^3\right)=2f\left(x^2\right)-f\left(\frac{1}{x}\right)\Rightarrow f\left(x^3\right)=2f\left(x\right)+f\left(x\right)(\because \text{from}(ii\left)\right)\Rightarrow f\left(x^3\right)=3f\left(x\right)$
$\Rightarrow f\left(x^n\right)=nf\left(x\right)$

Alternative Solution:
Take $f\left(x\right)=k\ln x$