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Question

Let f be a function satisfying the condition $$\lambda f(xy)=\dfrac{f(x)}{y}+\dfrac{f(y)}{x}$$ $$\forall$$ x, y $$> 0$$. If $$f(x)$$ is differentiable and $$f(1)=1$$, then the value of $$\displaystyle\lim_{x\rightarrow\infty}x$$ $$f(x)$$ is?


A
1
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B
2
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C
3
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D
4
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Solution

The correct option is A $$1$$
$$\lambda f(xy)=\dfrac{f(x)}{y}+\dfrac{f(y)}{x}$$

Let $$y=1$$ $$f(1)=1$$ or $$f(y)=1$$

$$\lambda f(x)=\dfrac{f(x)}{1}+\dfrac{f(y)}{x}$$

$$\lambda f(x)-f(x)=\dfrac{1}{x}$$

$$(\lambda -1)f(x)=\dfrac{1}{x}$$

Only $$f(x)=\dfrac{1}{x};(\lambda =2)$$

satisfies equation

hence , $$f(x)=\dfrac{1}{x}$$

$$\underset{x\rightarrow \infty }{\lim} x f(x)$$

$$\underset{x\rightarrow \infty }{\lim} x\times \dfrac{1}{x}=1$$

Maths

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