Let $f ( x ) = \sin \left( \frac { \pi } { 6 } \sin \left( \frac { \pi } { 2 } \sin x \right) \right)$ for all $x \in \mathbb { R }$ and $g ( x ) = \frac { \pi } { 2 } \sin x$ for all $x \in \mathbb { R }$. Let $( f \circ g ) ( x )$ denote $f ( g ( x ) )$ and $( g \circ f ) ( x )$ 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 $f \circ g$ is $\left[ - \frac { 1 } { 2 } , \frac { 1 } { 2 } \right]$ (C) $\lim _ { x \rightarrow 0 } \frac { f ( x ) } { g ( x ) } = \frac { \pi } { 6 }$ (D) There is an $x \in \mathbb { R }$ such that $( g \circ f ) ( x ) = 1$
Let $f ( x ) = \sin \left( \frac { \pi } { 6 } \sin \left( \frac { \pi } { 2 } \sin x \right) \right)$ for all $x \in \mathbb { R }$ and $g ( x ) = \frac { \pi } { 2 } \sin x$ for all $x \in \mathbb { R }$. Let $( f \circ g ) ( x )$ denote $f ( g ( x ) )$ and $( g \circ f ) ( x )$ 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 $f \circ g$ is $\left[ - \frac { 1 } { 2 } , \frac { 1 } { 2 } \right]$\\
(C) $\lim _ { x \rightarrow 0 } \frac { f ( x ) } { g ( x ) } = \frac { \pi } { 6 }$\\
(D) There is an $x \in \mathbb { R }$ such that $( g \circ f ) ( x ) = 1$