Question

Domain of the function $f(x) = \sqrt{2 - 2x - x^2 } $ is

• A. $ - \sqrt{3} \le x \le + \sqrt{3}$
• B. $-1 - \sqrt{3} \le x \le -1 + \sqrt{3}$
• C. $-2 \le x \le 2 $
• D. $ - 2+ \sqrt{3} \le x \le - 2 - \sqrt{3}$

Answer 1

Answer: (B)

Given $f(x) = \sqrt{2 - 2x - x^2} \ge 0$
Now f(x) is well defined when :
$\sqrt{2 - 2x - x^2 } \ge 0$
or ,$ 2 - 2x - x^2 \ge 0$
or, $x^2 + 2x - 2 \le 0$
$(x + 1 - \sqrt{3})(x +1 + \sqrt{3}) \le 0$
or $-1 - \sqrt{3} \le x \le - 1 + \sqrt{3}$