A) \[1<x<\infty \]
B) \[-\infty <x<\infty \]
C) \[-\infty <x<-1\]
D) \[(-\infty ,\ \infty )-(-1,\ 1)\]
Correct Answer: D
Solution :
\[f(x)=\sqrt{{{x}^{2}}-1}+\sqrt{{{x}^{2}}+1}\,\,\Rightarrow \,\,f(x)={{y}_{1}}+{{y}_{2}}\] Domain of \[{{y}_{1}}=\sqrt{{{x}^{2}}-1}\,\Rightarrow \,\,{{x}^{2}}-1\ge 0\,\,\Rightarrow \,\,{{x}^{2}}\ge 1\] \[x\in (-\,\infty ,\,\,\infty )-(-1,\,\,1)\] and Domain of \[{{y}_{2}}\] is real number, \[\therefore \] Domain of \[f(x)=(-\infty ,\,\,\infty )-(-1,\,\,1)\].
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