A) \[\frac{x+\sqrt{{{x}^{2}}-4}}{2}\]
B) \[\frac{x}{1+{{x}^{2}}}\]
C) \[\frac{x-\sqrt{{{x}^{2}}-4}}{2}\]
D) \[1+\sqrt{{{x}^{2}}-4}\]
Correct Answer: A
Solution :
Let\[y=f(x)\Rightarrow x={{f}^{-1}}(y)\]and let\[y=x+\frac{1}{x}\] \[\Rightarrow \]\[y=\frac{{{x}^{2}}+1}{x}\] \[\Rightarrow \]\[{{x}^{2}}-xy+1=0\] \[\Rightarrow \]\[x=\frac{y\pm \sqrt{{{y}^{2}}-4}}{2}\]\[\Rightarrow \]\[{{f}^{-1}}(y)=\frac{y\pm \sqrt{{{y}^{2}}-4}}{2}\]\[\Rightarrow \]\[{{f}^{-1}}(x)=\frac{x\pm \sqrt{{{x}^{2}}-4}}{2}\] Since, the range of the inverse function is \[[1,\infty )\]. \[\Rightarrow \]\[{{f}^{-1}}(x)=\frac{x+\sqrt{{{x}^{2}}-4}}{2}\] If we consider\[{{f}^{-1}}(x)=\frac{x-\sqrt{{{x}^{2}}-4}}{2}.\] Then, \[{{f}^{-1}}(x)>1.\]. This is possible only, if \[{{(x-2)}^{2}}>{{x}^{2}}-4\]\[\Rightarrow \]\[{{x}^{2}}+4-4x>{{x}^{2}}-4\] \[\Rightarrow \]\[8>4x\]\[\Rightarrow \]\[x<2,\]but it is given x > 2 Hence, option [a] is correct.
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