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 :
Clearly, for\[f:[1,\,\infty )\to [2,\,\,\infty ),\,\,f(x)=x+\frac{1}{x}\] is a bijection. Let\[f(x)\]. Then,\[x+\frac{1}{x}=y\] \[\Rightarrow \] \[{{x}^{2}}-xy+1=0\] \[\Rightarrow \] \[x=\frac{y\pm \sqrt{{{y}^{2}}-4}}{2}\] \[\Rightarrow \] \[x=\frac{y+\sqrt{{{y}^{2}}-4}}{2}\] \[[\because \,\,x\ge 1]\] \[\Rightarrow \] \[{{f}^{-1}}(y)=\frac{y+\sqrt{{{y}^{2}}-4}}{2}\] Hence, \[{{f}^{-1}}(x)=\frac{x+\sqrt{{{x}^{2}}-4}}{2},\,\,\forall x\in [1,\,\,\infty )\]
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