Question

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

• A. Range of $f$ is $\left[-\frac{1}{2},\frac{1}{2}\right]$
• B. Range of $fog$ is $\left[-\frac{1}{2},\frac{1}{2}\right]$
• 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 $(gof)(x)=1$

Answer 1

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

$f\left(x\right)=sin\left(\frac{\pi }{6}sin\left(\frac{\pi }{2}sinx\right)\right)$
$-1\le sinx\le 1asx\in R$
$-\frac{\pi }{2}\le \frac{\pi }{2}sinx\le \frac{\pi }{2}$
$-1\le sin\left(\frac{\pi }{2}sinx\right)\le 1$
$-\frac{\pi }{6}\le \frac{\pi }{6}sin\left(\frac{\pi }{2}sinx\right)\le \frac{\pi }{6}$
$-\frac{1}{2}\le sin\left(\frac{\pi }{6}sin\left(\frac{\pi }{2}sinx\right)\right)\le \frac{1}{2}$
Range $\in \left[-\frac{1}{2},\frac{1}{2}\right]$
So option $\left(A\right)$ is correct.
f o g(x)
Range of g(x) $=\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\cap$ Domain of f(x) = R
Common $=\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$
Range of fog(x) = Range of f(x) when input $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$
$=\left[-\frac{1}{2},\frac{1}{2}\right]$
So option (B) is correct
$\underset{x\to 0}{\lim }\frac{sin\left(\frac{\pi }{6}sin\left(\frac{\pi }{2}sinx\right)\right)}{\frac{\pi }{2}sinx}=\underset{x\to 0}{\lim }\left(\frac{sin\left(\frac{\pi }{6}sin\left(\frac{\pi }{2}sinx\right)\right)}{\frac{\pi }{6}sin\left(\frac{\pi }{2}sinx\right)}\right)\left(\frac{\frac{\pi }{6}sin\left(\frac{\pi }{2}sinx\right)}{\left(\frac{\pi }{2}sinx\right)}\right)$
$=\underset{x\to 0}{\lim }\frac{\pi }{6}.\left(1\right)=\frac{\pi }{6}$
So option $\left(B\right)$ is correct
g o f(x)
Range of f(x) $=\left[-\frac{1}{2},\frac{1}{2}\right]$
g o f(x) $=\frac{\pi }{2}sinf\left(x\right)$
$=\frac{\pi }{2}sin\frac{\left(sin\left(\frac{\pi }{6}sin\left(\frac{\pi }{2}sinx\right)\right)\right)=1}{\frac{1}{2}to\frac{1}{2}}$
$sin\left(\frac{1}{2}to\frac{1}{2}\right)=\frac{2}{\pi }\approx 0.6379$ (not possible)
$\frac{1}{2}\approx 28.5^{\circ }$
So option $\left(D\right)$ is not correct.