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The natural domain of the real valued function defined by $f(x)=\sqrt{x^2-1}+\sqrt{x^2+1}$ is
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$(-\infty, \infty)-(-1,1)$
$f(x)=\sqrt{x^2-1}+\sqrt{x^2+1} \Rightarrow f(x)=y_1+y_2$
Domain of $y_1=\sqrt{x^2-1} \Rightarrow x^2-1 \geq 0 \Rightarrow x^2 \geq 1$
$x \in(-\infty, \infty)-(-1,1)$ and Domain of $y_2$ is real number, $\therefore$ Domain of $f(x)=(-\infty, \infty)-(-1,1)$.
Domain of $y_1=\sqrt{x^2-1} \Rightarrow x^2-1 \geq 0 \Rightarrow x^2 \geq 1$
$x \in(-\infty, \infty)-(-1,1)$ and Domain of $y_2$ is real number, $\therefore$ Domain of $f(x)=(-\infty, \infty)-(-1,1)$.
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