Question

If $f(x)=\sin \,x+\cos \,x, g(x)=x^{2}-1$, then $g(f(x))$ is invertible in the domain

• A. $\left[0, \frac{\pi}{2}\right]$
• B. $\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$
• C. $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
• D. $[0, \pi]$

Answer 1

Answer: (B)

$f(x)=\sin\, x+\cos\, x, g(x)=x^{2}-1$
$\Rightarrow g(f(x))=(\sin\, x+\cos\, x)^{2}-1=\sin \,2 x$
Clearly $g(f(x))$ is invertible in
$-\frac{\pi}{2} \leq 2 x \leq \frac{\pi}{2}$
$[\because \sin \,\theta$ is invertible when $-\pi / 2 \leq \theta \leq \pi / 2]$
$\Rightarrow -\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$