Mathematics - Functions Question with Solution | TestHub

Given relation is $f\left(xy\right)=f\left(x\right). f\left(y\right) \forall x, y\in R$ .....(i)

On putting $x= y=0,$ we get $f\left(0\right)=f^2\left(0\right)\Rightarrow f\left(0\right)=0, 1$ but

$f\left(0\right)\ne 0$

$\Rightarrow f\left(0\right)=1,$

Now if we put $y=0$ in (i), we get

$f\left(x\right)=1$

Hence $\frac{dy}{dx}=1\Rightarrow y=x+c$

$\Rightarrow y=x+1$ $\left(\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{e} y\left(0\right)=1\right)$

$\Rightarrow y\left(\frac{1}{4}\right)+y\left(\frac{3}{4}\right)$

$=\left(\frac{1}{4}+1\right)+\left(\frac{3}{4}+1\right)=3.$