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