Functions

First line is

f(x)>-1/x>0 ??

f(f(x)+1/x)=1/f(x)

Now as x increases... 1/f(x) decreases.. (as f(x) increases..)
so f(f(x)+1/x) decreases..
so f(x)+1/x decreases..

or f'(x)-1/x²<0
0<f'(x)<1/x²

Now... [12]

also f''(x)<0

one more clarification:

You say f(x) is strictly increasing. So, f_min(x) in [1,2] should be f(1). So two of the options should be the same. They arent.

f(x)*f(f(x)+\frac{1}{x})=1
putting x=f^-1(x)

f(f^{-1}(x))*f(f(f^{-1}(x))+\frac{1}{f^{-1}(x)})=1

x*f((x)+\frac{1}{f^{-1}(x)})=1

f((x)+\frac{1}{f^{-1}(x)})=\frac{1}{x}

f((x)+\frac{1}{f^{-1}(x)})=f(f^{-1}\frac{1}{x})

So..
(x)+\frac{1}{f^{-1}(x)}=f^{-1}\frac{1}{x})

(as f is strictly increasing... so one one)

let f^-1(x)=g(x)

(x)+\frac{1}{g(x)}=g(\frac{1}{x}) ..(i)
put x=1/x

\frac{1}{x}+\frac{1}{g(\frac{1}{x})}=g(x)
this gives g(\frac{1}{x})=\frac{x}{xg(x)-1}
putting this g(1/x) in (i)

\frac{xg(x)+1}{g(x)}=\frac{x}{xg(x)-1}
Putting xg(x)=y
y²-y-1=0

xg(x)=(-1±√5)/2

xg(x)=K

xf^-1(x)=K

f^-1(x)=K/x

f(K/x)=x
f(x)=k/x

but k<0 as f'(x)>0

therefore

f(x)=\frac{1-\sqrt{5}}{2x}

____________

\int_{e}^{e^2}{f(x)}=\int_{e}^{e^2}{\frac{1-\sqrt{5}}{2x}}=\frac{1-\sqrt{5}}{2}(ln(e^2)-ln(e))=\frac{1-\sqrt{5}}{2}

f_min(x)=f(1)=(1-√5)/2
f_max(x)=f(2)=(1-√5)/4
_e∫^e2f(x)=(1-√5)/2

par iska koi chota tarika nahi hai kya?? [12]