Let f(x) = sin left(frac{pi}{6} sin left(frac | vedclass

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(A) Given $g(x) = \frac{\pi}{2} \sin x$. Then $f(x) = \sin \left(\frac{1}{3} g(g(x))\right)$.Since the range of $\sin x$ is $[-1, 1]$,the range of $g(

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$(A, B, C)$

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(A) Given $g(x) = \frac{\pi}{2} \sin x$. Then $f(x) = \sin \left(\frac{1}{3} g(g(x))ight)$.Since the range of $\sin x$ is $[-1, 1]$,the range of $g(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.For $f(x)$,the inner argument is $\frac{\pi}{6} \sin(	heta)$ where $	heta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Since $\sin(	heta) \in [-1, 1]$,the argument of the outer sine is in $[-\frac{\pi}{6}, \frac{\pi}{6}]$. Thus,the range of $f$ is $[\sin(-\frac{\pi}{6}), \sin(\frac{\pi}{6})] = [-\frac{1}{2}, \frac{1}{2}]$. So $(A)$ is true.For $f(g(x))$,the range of $g(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$,so the range of $f(g(x))$ is also $[-\frac{1}{2}, \frac{1}{2}]$. So $(B)$ is true.For $(C)$,$\lim_{x ightarrow 0} \frac{f(x)}{g(x)} = \lim_{x ightarrow 0} \frac{\sin(\frac{\pi}{6} \sin(\frac{\pi}{2} \sin x))}{\frac{\pi}{2} \sin x}$. Using $\sin 	heta \approx 	heta$,we get $\frac{\frac{\pi}{6} \cdot \frac{\pi}{2} \sin x}{\frac{\pi}{2} \sin x} = \frac{\pi}{6}$. So $(C)$ is true.For $(D)$,$(g \circ f)(x) = \frac{\pi}{2} \sin(f(x))$. Since the range of $f(x)$ is $[-\frac{1}{2}, \frac{1}{2}]$,the range of $\sin(f(x))$ is $[\sin(-1/2), \sin(1/2)]$. Since $\sin(1/2) < 1$,$\frac{\pi}{2} \sin(f(x)) < \frac{\pi}{2} \cdot 1 = \frac{\pi}{2} \approx 1.57$. However,$\frac{\pi}{2} \sin(1/2) \approx 1.57 \cdot 0.479 \approx 0.75$. Since $1$ is in the range $[-\frac{\pi}{2} \sin(1/2), \frac{\pi}{2} \sin(1/2)]$,there exists $x$ such that $(g \circ f)(x) = 1$. So $(D)$ is true.

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