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

Answer: 3

$f\left(xy\right)=f\left(x\right)\cdot f\left(y\right)$
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,
$f^{\prime }\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}$
$=\underset{h\to 0}{\lim }f\left(x\right)\left(\frac{f\left(1+\frac{h}{x}\right)-f\left(1\right)}{h}\right)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{f\left(x\right)}{x}{f}^{\prime }\left(1\right)$
$\Rightarrow \frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\frac{k}{x}$
$\Rightarrow 1nf\left(x\right)=k1nx+c$
$f\left(1\right)=1\Rightarrow 1n1=k1n1+c\Rightarrow c=0$
$\Rightarrow 1nf\left(x\right)=k1nx\Rightarrow f\left(x\right)=x^k$ but $f\left(0\right)=1$
$\Rightarrow k=0$
$\therefore f\left(x\right)=1$
$\frac{dy}{dx}=f\left(x\right)=1$
$\Rightarrow y=x+c,y\left(0\right)=1$
$1\Rightarrow c=1$
$\Rightarrow y=x+1$
$\therefore y\left(\frac{1}{4}\right)+y\left(\frac{3}{4}\right)=\frac{1}{4}+1+\frac{3}{4}+1=3$