Question

Let $f: \left[0, 1\right]\to R$ be such that $f\left(xy\right) =f\left(x\right)\cdot f\left(y\right)$, for all $x, y \in \left[0, 1\right]$, and $f \left(0\right)\ne0$.If $y=y\left(x\right)$satisfies the differential equation,$\frac{dy}{dx} = f\left(x\right)$ with $y\left(0\right) =1$, then $y\left(\frac{1}{4}\right) +y\left(\frac{3}{4}\right)$ is equal to

Answer 1

$f \left(xy\right) =f\left(x\right)\cdot f\left(y\right)$
Put $x =y =0$ in $\left(1\right)$ to get $f\left(0\right) =1$
Put $x =y=1$ in $\left(1\right)$ to get $f\left(1\right) =0$ or $f\left(1\right) =1$
$f\left(1\right) =0$ is rejected else $y=1$ in $\left(1\right)$ gives $f\left(x\right) =0$
imply $f \left(0\right) =0$
Hence $f\left(0\right) =1$ and $f\left(1\right) =1$
By first principle derivative formula,
$f' \left(x\right) =\lim\limits_{h\to0} \frac{f\left(x +h\right) - f\left(x\right)}{h}$
$=\displaystyle\lim_{h\to0} f\left(x\right)\left(\frac{f \left(1+\frac{h}{x}\right) -f\left(1\right)}{h}\right)$
$\Rightarrow f'\left(x\right) =\frac{f\left(x\right)}{x} f'\left(1\right)$
$\Rightarrow \frac{f'\left(x\right)}{f\left(x\right)} =\frac{k}{x}$
$\Rightarrow 1n f\left(x\right) =k \,1n\, x +c$
$f \left(1\right) = 1\Rightarrow 1n 1 = k \,1n\, 1 +c\Rightarrow c=0$
$\Rightarrow 1n f\left(x\right) = k\, 1n\, x \Rightarrow f\left(x\right) =x^{k}$ but $f\left(0\right) =1$
$\Rightarrow k =0$
$\therefore f \left(x\right) = 1$
$\frac{dy}{dx} = f\left(x\right) = 1$
$\Rightarrow y = x +c, y\left(0\right) = 1$
$1 \Rightarrow c =1$
$\Rightarrow y = x +1$
$\therefore y \left(\frac{1}{4}\right) + y\left(\frac{3}{4}\right) = \frac{1}{4} + 1+\frac{3}{4} +1 =3$