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Question

If f is a differentiable function satisfying f(xy)=f(x)+f(y)+x+y1xy for all x,y>0 and f(1)=2, then the value of [f(e100)] is
(where [.] represents the greatest integer function)

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Solution

Given : f(xy)=f(x)+f(y)+x+y1xy
Putting y=x=1, we get
f(1)=2f(1)+1f(1)=1
Also, f(1)=2

Now, we know that
f(x)=limh0f(x+h)f(x)hf(x)=limh0f(x(1+hx))f(x)hf(x)=limh0f(x)+f(1+hx)+x+1+hx1x(1+hx)f(x)hf(x)=limh0f(1+hx)+x2+hx(x+h)hf(x)=limh0x(x+h)f(1+hx)+x2+hh(x2+hx)f(x)=1x2limh0x2[f(1+hx)+1]+h[xf(1+hx)+1]hx2f(x)=limh0x[f(1+hx)f(1)]hx+(1x) (f(1)=1)x2f(x)=xf(1)+(1x)f(x)=1+xx2f(x)=1x+ln|x|+c
Putting x=1, we get
1=1+0+cc=0
Therefore, f(e100)=1e100+100
[f(e100)]=99 (1e100(0,1))

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