Q. If $f(x) = \sin{x + \cos{x,\ \ g(x) = x^{2} - 1}}$, then $g(f(x))$ is invertible in the domain
A) $\left\lbrack 0,\ \frac{\pi}{2} \right\rbrack$
B) $\left\lbrack - \frac{\pi}{4},\ \frac{\pi}{4} \right\rbrack$
C) $\left\lbrack - \frac{\pi}{2},\ \frac{\pi}{2} \right\rbrack\$
D) $\lbrack 0,\ \pi\rbrack$
Explanation
(b)
$\because g\left( f(x) \right) = \left( \sin{x + \cos x} \right)^{2} - 1$, is invertible ($ie$, bijective)
$\Rightarrow g\left( f(x) \right) = \sin{2x}$, is bijective
We know $\sin x$ is bijective only when $x \in \left\lbrack - \frac{\pi}{2},\ \frac{\pi}{2} \right\rbrack$
Thus, $g(f(x))$ is bijective if, $- \frac{\pi}{2} \leq 2x \leq \frac{\pi}{2}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - \frac{\pi}{4} \leq x \leq \frac{\pi}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.