Question

Let $f:[0,1]\to R$ be such that $f\left(xy\right)=f\left(x\right)\cdot f\left(y\right)$, 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 $y\left(0\right)=1,\text{then}y\left(\frac{1}{4}\right)+y\left(\frac{3}{4}\right)$ is equal to :

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is B $3$

$f\left(xy\right)=f\left(x\right)\cdot f\left(y\right)$
Put $x=0,y=0$
$f\left(0\right)=\left(f\right(0){)}^2$
$\Rightarrow f\left(0\right)=1\text{as}f\left(0\right)\ne 0$

Now, put $y=0$
$f\left(0\right)=f\left(x\right)\cdot f\left(0\right)$
$\Rightarrow f\left(x\right)=1$

As given, $\frac{dy}{dx}=f\left(x\right)$
$\Rightarrow y=\int f\left(x\right)dx=\int 1dx$
$\Rightarrow y=x+c$
At $x=0,y\left(0\right)=1$ $\Rightarrow$ $c=1$
$\therefore y\left(x\right)=x+1$

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