Question

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

• A. Range of f is $[-\frac{1}{2},\frac{1}{2}]$
• B. Range of $f\cdot g$ 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 $xϵR$ such that $(g\cdot f)\left(x\right)=1$

Answer 1

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

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

$f\left(x\right)=\sin \left(\frac{\pi }{6}\sin \left(\frac{\pi }{2}\sin x\right)\right)$

$\frac{\pi \sin x}{2}$ varies from $-\frac{\pi }{2}$ to $\frac{\pi }{2}$.

Hence, $f\left(x\right)$ varies from $\sin \left(\frac{-\pi }{6}\right)$ to $\sin \left(\frac{-\pi }{6}\right)$

Hence, the range of $f\left(x\right)$ is $[\frac{-1}{2},\frac{1}{2}]$

The range of $f.g\left(x\right)$ is also $[\frac{-1}{2},\frac{1}{2}]$ with the extreme values attained at $\sin x=\pm 1$.
$\underset{x\to 0}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{t\to 0}{lim}\frac{\sin \left(\frac{\pi t}{6}\right)}{t}=\frac{\pi }{6}$
Hence, options A , B and C are correct.