33also f''(x)<0
Let\ f:R^{+}\rightarrow R\ be\ a\ strictly\ increasing\ function\ such\ that\ f(x)>-\frac{1}{x}\forallx>0\\ And\ also\ f(x).f\left(f(x)+\frac{1}{x}\right)=1\\ \\ Match\ the\ following:\\ \\ f(1)=\\ f_{max}x\ in\ [1,2]=\\ \\f_{min}x\ in\ [1,2]\\ \\ \int_{e}^{e^2}f(x)dx=\\ \\ \\ \\ \frac{1+\sqrt{5}}{2}\\ \frac{1-\sqrt{5}}{2}\\ \frac{1-\sqrt{5}}{4}\\ \frac{1+\sqrt{5}}{4}\\
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13 Answers
62sparkle check this question once.. you have made some mistake somewhere! :(
according to you, puttig x=1 gives!
f(1)>-1>0
1sorry,iam new to latex
it should be f(x)>-1/x for all x>0
(FIITJEE AITS question)
33f(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]
341one more clarification:
You say f(x) is strictly increasing. So, fmin(x) in [1,2] should be f(1). So two of the options should be the same. They arent.
1i re-checked the options,no two options are same...but this is a multiple choice q...so it is possible.
33f(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}
____________
33\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}
33fmin(x)=f(1)=(1-√5)/2
fmax(x)=f(2)=(1-√5)/4
e∫e2f(x)=(1-√5)/2