satisfies the differential equation,
Question:

Let $f:[0,1] \rightarrow \mathrm{R}$ be such that $f(x y)=f(x) f(y)$, for all $x, y \in[0,1]$, and $f(0) \neq 0$. If $y=y(x)$ satisfies the

differential equation, $\frac{d y}{d x}=f(x)$ with $y(0)=1$, then

$y\left(\frac{1}{4}\right)+y\left(\frac{3}{4}\right)$ is equal to:

1. (1) 3

2. (2) 4

3. (3) 2

4. (4) 5

Correct Option: 1

Solution:

$f(x y)=f(x) \cdot f(y)$….(1)

Put $x=y=0$ in (1) to get $f(0)=1$

Put $x=y=1$ in (1) to get $f(1)=0$ or $f(1)=1$

$f(1)=0$ is rejected else $y=1$ in (1) gives $f(x)=0$

$\operatorname{imply} f(0)=0$

Hence, $f(0)=1$ and $f(1)=1$

By first principle derivative formula,

$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} f(x)\left(\frac{f\left(1+\frac{h}{x}\right)-f(1)}{h}\right)$

$\Rightarrow \quad f^{\prime}(x)=\frac{f(x)}{x} f^{\prime}(1)$

$\Rightarrow \quad \frac{f^{\prime}(x)}{f(x)}=\frac{k}{x} \Rightarrow \ln f(x)=k \ln x+c$

$f(1)=1 \quad \Rightarrow \quad \ln 1=k \ln 1+\mathrm{c} \Rightarrow c=0$

$\Rightarrow \quad \ln f(x)=k \ln x \Rightarrow f(x)=x^{k}$ but $f(0)=1$

$\Rightarrow \quad k=0$

$\therefore \quad f(x)=1$

$\frac{d y}{d x}=f(x)=1 \Rightarrow y=x+c, y(0)=1 \Rightarrow c=1$

$\Rightarrow \quad y=x+1$

$\therefore \quad y\left(\frac{1}{4}\right)+y\left(\frac{3}{4}\right)=\frac{1}{4}+1+\frac{3}{4}+1=3$