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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
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B
3
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C
5
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D
2
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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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