Question

Let $f\left(x\right)=\sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)\right)$ for all $x\in R$ and $g\left(x\right)=\frac{\pi }{2}\sin x$ for all $x\in R.$ Let $(f\circ g)\left(x\right)$ denote $f\left(g\right(x\left)\right)$ and $(g\circ f)\left(x\right)$ denotes $g\left(f\right(x\left)\right).$ Then which of the following is (are) true?

• A. Range of $f$ is $[\frac{-1}{2},\frac{1}{2}]$
• B. Range of $f\circ g$ is $[\frac{-1}{2},\frac{1}{2}]$
• C. $\underset{x\to 0}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\pi }{6}$
• D. There is an $x\in R$ such that $(g\circ f)\left(x\right)=1$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Range of $f$ is $[\frac{-1}{2},\frac{1}{2}]$
B Range of $f\circ g$ is $[\frac{-1}{2},\frac{1}{2}]$
C $\underset{x\to 0}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\pi }{6}$

For option $\left(1\right)$:

$f\left(x\right)=\sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)\right)$
Let $\frac{\pi }{2}\sin x=\theta \Rightarrow \theta \in [-\frac{\pi }{2},\frac{\pi }{2}]$
Then $f\left(x\right)=\sin \left(\frac{\pi }{6}\sin \theta \right)$
Let $\frac{\pi }{6}\sin \theta =\gamma \Rightarrow \gamma \in [-\frac{\pi }{6},\frac{\pi }{6}]$
$\Rightarrow f\left(x\right)=\sin \gamma$
Therefore, Range of $f$ is $[\frac{-1}{2},\frac{1}{2}]$

For option $\left(2\right)$:

$f\left(x\right)=\sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)\right)$,
$g\left(x\right)=\frac{\pi }{2}\sin x$
$\Rightarrow f\circ g\left(x\right)=\sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin \left(\frac{\pi }{2}\sin x\right)\right)\right)$
From the above procedure, we can say that the range of $f\circ g$ is $[\frac{-1}{2},\frac{1}{2}]$

For option $\left(3\right)$:
$\underset{x\to 0}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to 0}{lim}\frac{\sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)\right)}{\frac{\pi }{2}\sin x}$
$\Rightarrow \underset{x\to 0}{lim}\frac{\sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)\right)}{\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)}\times \frac{\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)}{\frac{\pi }{2}\sin x}=\frac{\pi }{6}$

For option $\left(4\right)$:

$g\circ f\left(x\right)=\frac{\pi }{2}\sin \left(\sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)\right)\right)=1\Rightarrow \sin (\sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right))\right)=\frac{2}{\pi }=\frac{2}{3.14}>\frac{1}{2}$
which is beyond the range of the function.
So, option $\left(4\right)$ is not correct.