Question

# Let $$f:[0,1]\rightarrow R$$ be such that $$f(xy)=f(x).f(y)$$ for all x,y,$$\in$$[0,1], and $$f(0)\neq 0$$. If $$y=y(x)$$ satisfies the differential equation , $$\dfrac{dy}{dx}=f(x)$$ with $$y(0)=1$$, then $$y\left(\dfrac{1}{4}\right)+y\left(\dfrac{3}{4}\right)$$ is equal to

A
4
B
3
C
5
D
2

Solution

## The correct option is B $$3$$$$f(xy) =f(x).f(y)$$$$f(0)=1$$ as $$f(0)\neq 0$$$$\Rightarrow f(x)=1$$$$\dfrac{dy}{dx}=f(x)=1$$$$\Rightarrow y=x+c$$At, $$x=0,y=1 \Rightarrow c=1$$$$y=x+1$$$$\Rightarrow y\left(\dfrac{1}{4}\right)+y\left(\dfrac{3}{4}\right)=\dfrac{1}{4}+1+\dfrac{3}{4}+1=3$$Mathematics

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