Q. Domain of the function $f(x) = \sqrt{2 - 2x - x^{2}}$ is
A) $- \sqrt{3} \leq x \leq + \sqrt{3}$
B) $- 1 - \sqrt{3} \leq x \leq - 1 + \sqrt{3}$
C) $- 2 \leq x \leq 2$
D) $- 2 + \sqrt{3} \leq x \leq - 2 - \sqrt{3}$
Explanation
(b)
For domain of $f(x),\ 2 - 2x - x^{2} \geq 0$
$\Rightarrow \ \ \ \ \ \ \ \ \ x^{2} + 2x - 2 \leq 0$
$\Rightarrow \ \ \ \ \ \ - 1 - \sqrt{3} \leq x < - 1 + \sqrt{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.