Question

Let $f:R^{+}\to R$ be a function which satifies :
$f\left(x\right)f\left(y\right)=f\left(xy\right)+2(\frac{1}{x}+\frac{1}{y}+1)$ for $x,y>0$, then $f\left(x\right)$ can be :

• A. $\frac{2x+1}{x}$
• B. $-\left(\frac{x+1}{x}\right)$
• C. $\frac{5x+2}{2x}$
• D. $\frac{2(2x+1)}{x}$

Answer 1

The correct option is A $\frac{2x+1}{x}$

Given function is $f:R^{+}\to R$

Also given $f\left(x\right)f\left(y\right)=f\left(xy\right)+2(\frac{1}{x}+\frac{1}{y}+1)$ for $x,y>0$

Put $x=y=1$ in the above function, we get

${\left[f\right(1\left)\right]}^2=f\left(1\right)+6$

$\Rightarrow {\left[f\right(1\left)\right]}^2-f\left(1\right)-6=0$

$\Rightarrow \left[f\right(1)+2)\left]\right[f\left(1\right)-3]=0$

The root of the equation is $f\left(1\right)=-2$ or $f\left(1\right)=3$

Since, the function domain is $R^{+}$, so the $f\left(1\right)=3$ is accepted.

Now, put the $y=1$ in the given function, we get

$f\left(x\right)f\left(1\right)=f\left(x\right)+2(\frac{1}{x}+1+1)$

$\Rightarrow 3f\left(x\right)=f\left(x\right)+2(\frac{1}{x}+2)$

$\Rightarrow 2f\left(x\right)=2(\frac{1}{x}+2)$

$\Rightarrow f\left(x\right)=2+\frac{1}{x}=\frac{2x+1}{x}$