Question

Find the domain of $f\left(x\right)={\sin }^{-1}\left(\frac{3-2x}{5}\right)+\sqrt{3-x}$

• A. $x\in [-1,3]$
• B. $x\in [1,3]$
• C. $x\in [-3,-1]$
• D. $x\in [-1,0]$

Answer 1

The correct option is A $x\in [-1,3]$

For given function to be defined,
(i) $-1\le \frac{3-2x}{5}\le 1\Rightarrow -5\le 3-2x\le 5$
$\Rightarrow -8\le -2x\le 2\Rightarrow -1\le x\le 4$
(ii) $3-x\ge 0\Rightarrow x\le 3$
Thus considering both the cases required domain is $x\in [-1,3]$