Question

If $f$ is a differentiable function satisfying $f\left(xy\right)=f\left(x\right)+f\left(y\right)+\frac{x+y-1}{xy}$ for all $x,y>0$ and $f^{\prime }\left(1\right)=2,$ then the value of $\left[f\right(e^{100}\left)\right]$ is
(where $[.]$ represents the greatest integer function)

Answer 1

Given : $f\left(xy\right)=f\left(x\right)+f\left(y\right)+\frac{x+y-1}{xy}$
Putting $y=x=1$, we get
$f\left(1\right)=2f\left(1\right)+1\Rightarrow f\left(1\right)=-1$
Also, $f^{\prime }\left(1\right)=2$

Now, we know that
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(x(1+\frac{h}{x})\right)-f\left(x\right)}{h}\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f\left(x\right)+f(1+\frac{h}{x})+\frac{x+1+\frac{h}{x}-1}{x(1+\frac{h}{x})}-f\left(x\right)}{h}\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(1+\frac{h}{x})+\frac{x^2+h}{x(x+h)}}{h}\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{x(x+h)f(1+\frac{h}{x})+x^2+h}{h(x^2+hx)}\Rightarrow f^{\prime }\left(x\right)=\frac{1}{x^2}\underset{h\to 0}{lim}\frac{x^2[f(1+\frac{h}{x})+1]+h[xf(1+\frac{h}{x})+1]}{h}\Rightarrow x^2{f}^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{x[f(1+\frac{h}{x})-f(1\left)\right]}{\frac{h}{x}}+(1-x)(\because f(1)=-1)\Rightarrow x^2{f}^{\prime }\left(x\right)=x{f}^{\prime }\left(1\right)+(1-x)\Rightarrow f^{\prime }\left(x\right)=\frac{1+x}{x^2}\Rightarrow f\left(x\right)=-\frac{1}{x}+\ln |x|+c$
Putting $x=1$, we get
$-1=-1+0+c\Rightarrow c=0$
Therefore, $f\left(e^{100}\right)=-\frac{1}{e^{100}}+100$
$\Rightarrow \left[f\right(e^{100}\left)\right]=99(\because \frac{1}{e^{100}}\in (0,1\left)\right)$