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 $(f\circ g)\left(x\right)$ denotes $f\left(g\right(x\left)\right)$ 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)

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\pi }{6}$

$f\left(x\right)=\sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)\right)$
We know that
$\Rightarrow \frac{\pi }{2}\sin x\in [-\frac{\pi }{2},\frac{\pi }{2}]\Rightarrow \frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)\in [-\frac{\pi }{6},\frac{\pi }{6}]\Rightarrow \sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)\right)\in [-\frac{1}{2},\frac{1}{2}]$
Therefore, the range of $f$ is $[\frac{-1}{2},\frac{1}{2}]$

$f\left(x\right)=\sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)\right)g\left(x\right)=\frac{\pi }{2}\sin x\Rightarrow f\circ g\left(x\right)=\sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin \left(\frac{\pi }{2}\sin x\right)\right)\right)$
Similarly, we can solve
Thereore, the range of $f\circ g$ is $[-\frac{1}{2},\frac{1}{2}]$

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

$g\circ f\left(x\right)=1\Rightarrow \frac{\pi }{2}\sin \left(\sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)\right)\right)=1\Rightarrow \sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)\right)={\sin }^{-1}\frac{2}{\pi }\Rightarrow \sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)\right)>\frac{\pi }{6}(\because {\sin }^{-1}\frac{2}{3.14}>{\sin }^{-1}\frac{1}{2})\Rightarrow f\left(x\right)>\frac{1}{2}(\because \frac{\pi }{6}=0.523)$
Which is beyond the range of the function $f\left(x\right)$, so its not possible.