WAEC 2022 · Paper 1 · Q24

If $g(x) = \sqrt{1 - x^2}$ , find the domain of $g(x)$ .

Worked solution (try it first)

A square root is only real when the number under it is not negative, so you need $1 - x^2 \ge 0$ .
That is $x^2 \le 1$ , which holds for $-1 \le x \le 1$ .
So the domain is $-1 \le x \le 1$ , option C.