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Question

Let for all x>0,f(x)=limn⎜ ⎜ ⎜x1n11n⎟ ⎟ ⎟, then

A
f(x)+f(1x)=1
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B
f(xy)=f(x)+f(y)
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C
f(xy)=xf(y)+yf(x)
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D
f(xy)=xf(x)+yf(y)
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Solution

The correct option is D f(xy)=f(x)+f(y)
We know that, limn0xn1n=1
Now, as n,1n0

f(x)=limnx1n11n=logx
Consider, f(xy)=log(xy)
=logx+logy
=f(x)+f(y)
f(xy)=f(x)+f(y)

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