A) \[\frac{x+\sqrt{{{x}^{2}}-4}}{2}\]
B) \[\frac{x}{1+{{x}^{2}}}\]
C) \[\frac{x-\sqrt{{{x}^{2}}-4}}{2}\]
D) \[1+\sqrt{x-4}\]
Correct Answer: A
Solution :
We have,\[f(x)=x\frac{1}{x}\] let \[f(x)=y=x+\frac{1}{x}\] \[\Rightarrow \] \[xy={{x}^{2}}+1\] \[\Rightarrow \] \[{{x}^{2}}-yx+1=0\] \[\Rightarrow \] \[x=\frac{y\pm \sqrt{{{y}^{2}}-4}}{2}={{f}^{-1}}(y)\] \[\Rightarrow \] \[{{f}^{-1}}(x)=\frac{x\pm \sqrt{{{x}^{2}}-4}}{2}\] \[\Rightarrow \]\[{{f}^{-1}}(x)=\frac{x+\sqrt{{{x}^{2}}-4}}{2}\] (neglecting\[-ve\]as \[x>1\])
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