Question

If$f\left(x\right)=\sin x,-\pi /2\le x\le \pi /2,$ then

• A. $f\left(x\right)$ is increasing in the interval $[-\pi /2,\pi /2]$
• B. $f\left\{f\left(x\right)\right\}$ is increasing in the interval $[-\pi /2,\pi /2]$
• C. $f\left\{f\left(x\right)\right\}$ is decreasing in $[-\pi /2,0]$ and increasing in $[0,\pi /2]$
• D. $f\left\{f\left(x\right)\right\}$ s invertible in $[-\pi /2,\pi /2]$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $f\left(x\right)$ is increasing in the interval $[-\pi /2,\pi /2]$
B $f\left\{f\left(x\right)\right\}$ is increasing in the interval $[-\pi /2,\pi /2]$
D $f\left\{f\left(x\right)\right\}$ s invertible in $[-\pi /2,\pi /2]$

Given, $f\left(x\right)=\sin x$
$f^{\prime }\left(x\right)=\cos x>0$ in interval $[\frac{-\pi }{2},\frac{\pi }{2}]$, so f(x) is increasing in interval $[\frac{-\pi }{2},\frac{\pi }{2}]$
$f\left(f\right(x\left)\right)=\sin \left(\sin x\right)$
$f^{\prime }\left(f\right(x\left)\right)=\cos \left(\sin x\right)\cos x>0$ in interval $[\frac{-\pi }{2},\frac{\pi }{2}]$, so $f\left(x\right)$ is increasing in interval $[\frac{-\pi }{2},\frac{\pi }{2}]$
$y=\sin \left(\sin x\right)$
${\sin }^{-1}y=\sin x$
Thus $f\left(f\right(x\left)\right)$ is invertible.