Q. 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 $(fog) (x)$ denote $f(g(x))$ and $(gof) (x)$ denote $g(f(x))$. Then which of the following is (are) true ?
Solution:
$f \left(x\right) = sin \left(\frac{\pi}{6}sin\left(\frac{\pi}{2}sin \,x\right)\right)$
$- 1 \le sin x \le 1 as x ∈ R$
$-\frac{\pi}{2} \le \frac{\pi}{2} sinx \le \frac{\pi}{2}$
$-1 \le sin\left(\frac{\pi}{2}sin\, x\right)\le 1$
$-\frac{\pi}{6} \le \frac{\pi}{6} sin \left(\frac{\pi}{2}sin\, x\right) \le \frac{\pi}{6}$
$-\frac{1}{2}\le sin\left(\frac{\pi}{6}sin\left(\frac{\pi}{2}sin\, x\right)\right) \le \frac{1}{2}$
Range $∈\left[-\frac{1}{2}, \frac{1}{2}\right]$
So option $\left(A\right)$ is correct.
f o g(x)
Range of g(x) $= \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] ∩$ 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
$\displaystyle \lim_{x\to0} \frac{sin\left(\frac{\pi}{6}sin\left(\frac{\pi}{2}sin\, x\right)\right)}{\frac{\pi}{2}sin\, x} = \displaystyle \lim_{x\to 0} \left(\frac{sin\left(\frac{\pi }{6}sin\left(\frac{\pi }{2}sin\, x\right)\right)}{\frac{\pi }{6}sin\left(\frac{\pi }{2}sin\, x\right)}\right)\left(\frac{\frac{\pi }{6}sin\left(\frac{\pi }{2}sin\, x\right)}{\left(\frac{\pi}{2}sin\, x\right)}\right)$
$= \displaystyle \lim_{x\to 0} \frac{\pi}{6}.\left(1\right) = \frac{\pi}{6}$
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) $= \frac{\pi}{2}sin f\left(x\right)$
$= \frac{\pi}{2}sin\frac{\left(sin \left(\frac{\pi}{6}sin \left(\frac{\pi}{2} sin \,x\right)\right)\right) = 1}{\frac{1}{2} to \frac{1}{2}}$
$sin\left(\frac{1}{2} to \frac{1}{2}\right) = \frac{2}{\pi}\approx0.6379$ (not possible)
$\frac{1}{2} \approx 28.5^{\circ}$
So option $\left(D\right)$ is not correct.
