Mathematics - Functions Question with Solution | TestHub

Put $\mathrm{x}=\mathrm{y}=1,\left(\mathrm{f}(1{)}^2=3\mathrm{f}(1)-2\right)\Rightarrow \mathrm{f}(1)=1$ or 2

Let $\mathrm{f}(1)=1$, then put $\mathrm{y}=1$ $f(x)\cdot f(1)=f(x)+f(1)+f(x)-2$

$\Rightarrow \mathrm{f}(\mathrm{x})=1$ constant function