Question

Let $f:[0,1]\to R$ be such that $f\left(xy\right)=f\left(x\right).f\left(y\right)$ for all x,y,$\in$[0,1], 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

• A. $4$
• B. $3$
• C. $5$
• D. $2$

Answer 1

The correct option is B $3$

$f\left(xy\right)=f\left(x\right).f\left(y\right)$

$f\left(0\right)=1$ as $f\left(0\right)\ne 0$
$\Rightarrow f\left(x\right)=1$

$\frac{dy}{dx}=f\left(x\right)=1$
$\Rightarrow y=x+c$

At, $x=0,y=1\Rightarrow c=1$

$y=x+1$

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