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
B
2
C
3
D
4

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 equationhence , $$f(x)=\dfrac{1}{x}$$$$\underset{x\rightarrow \infty }{\lim} x f(x)$$$$\underset{x\rightarrow \infty }{\lim} x\times \dfrac{1}{x}=1$$Maths

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