Let f: [0,1] → R be such that f(xy) = f(x) . f(y), for all x, y ∈...

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 $f\left(0\right) = {f}^{2}\left(0\right) ⇒ f\left(0\right) = 0, 1$ but

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

$⇒ f\left(0\right) = 1,$

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

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

Hence $\dfrac{dy}{dx} = 1 ⇒ y = x + c$

$⇒ y = x + 1$ \left(sincey\left(0\right) = 1\right)$<br /><br />$⇒ y\left(\dfrac{1}{4}\right) + y\left(\dfrac{3}{4}\right)$</p><p>$= \left(\dfrac{1}{4} + 1\right) + \left(\dfrac{3}{4} + 1\right) = 3 .$