Question

The domain of the function ${\cos }^{-1}\left(\frac{1}{1-x}\right)$ is equal to

• A. $[0,2]$
• B. $R-[0,2]$
• C. $R-(0,2)$
• D. $[-1,1]$

Answer 1

The correct option is C $R-(0,2)$

${\cos }^{-1}\left(\frac{1}{1-x}\right)$ exists if $-1\le \frac{1}{1-x}\le 1$

Case $1:$
$\frac{1}{1-x}\ge -1$
$\Rightarrow \frac{1}{1-x}+1\ge 0$
$\Rightarrow \frac{x-2}{x-1}\ge 0$
$\Rightarrow x\notin [1,2)\cdots \left(1\right)$

Case $2:$
$\frac{1}{1-x}\le 1$
$\Rightarrow \frac{1}{1-x}-1\le 0$
$\Rightarrow \frac{x}{x-1}\ge 0$
$\Rightarrow x\notin (0,1]\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$ we get
$x\notin (0,2)$
or, $x\in R-(0,2)$