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Question:
Let $f:[0,1] \rightarrow \mathrm{R}$ be such that $f(\mathrm{xy})=f(\mathrm{x}) . f(\mathrm{y})$ for all $\mathrm{x}, \mathrm{y}, \varepsilon[0,1]$, and $f(0) \neq 0$. If $\mathrm{y}=\mathrm{y}(\mathrm{x})$ satisfies the
differential equation, $\frac{\mathrm{dy}}{\mathrm{dx}}=f(\mathrm{x})$ with
$y(0)=1$, then $y\left(\frac{1}{4}\right)+y\left(\frac{3}{4}\right)$ is equal to
Correct Option: , 2
Solution:
$f(x y)=f(x) . f(y)$
$f(0)=1$ as $f(0) \neq 0$
$\Rightarrow f(x)=1$
$\frac{\mathrm{dy}}{\mathrm{dx}}=f(\mathrm{x})=1$
$\Rightarrow y=x+c$
At, $x=0, y=1 \Rightarrow \mathrm{c}=1$
$y=x+1$
$\Rightarrow \mathrm{y}\left(\frac{1}{4}\right)+\mathrm{y}\left(\frac{3}{4}\right)=\frac{1}{4}+1+\frac{3}{4}+1=3$
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