Functions Question 630
Question: The natural domain of the real valued function defined by $ f(x)=\sqrt{x^{2}-1}+\sqrt{x^{2}+1} $ is
[SCRA 1996]
Options:
A) $ 1<x<\infty $
B) $ -\infty <x<\infty $
C) $ -\infty <x<-1 $
D) $ (-\infty ,\ \infty )-(-1,\ 1) $
Show Answer
Answer:
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) $ .
BETA
