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 $-1\le \frac{1}{1-x}\le 1$
$\Rightarrow \frac{1}{|1-x|}\le 1$
$\Rightarrow |1-x|\ge 1$

Case $1:$
If $1-x\ge 0$ or $x\le 1,$
then $1-x\ge 1$
$\Rightarrow x\le 0\cdots \left(1\right)$

Case $2:$
If $1-x<0$ or $x>1,$
then $x-1\ge 1$
$\Rightarrow x\ge 2\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$ we get
$x\in (-\infty ,0]\cup [2,\infty )$
or, $x\in R-(0,2)$