Question:
Mark (√) against the correct answer in the following:
Let $f(x)=e^{\sqrt{x^{2}-1}} \cdot \log (x-1) .$ Then, dom $(f)=?$
A. $(-\infty, 1]$
B. $[-1, \infty)$
C. $(1, \infty)$
D. $(-\infty,-1] \cup(1, \infty)$
Solution:
$\mathrm{f}(\mathrm{x})=\mathrm{e}^{\sqrt{\mathrm{x}^{2}-1}} \log (\mathrm{x}-1)$
$x-1>0$
$\Rightarrow x>1$
And
$\Rightarrow \mathrm{x}^{2}-1 \geq 0$
$\Rightarrow \mathrm{x}^{2} \geq 1$
$\Rightarrow-1 \leq \mathrm{x} \geq 1$
Taking the intersection we get
$\operatorname{Dom}(f)=(1, \infty)$
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