# Mark (√) against the correct answer in the following:

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 $(f)=(b)(-\infty,-1]$