If f: R rightarrow [-1, 1] and g: R rightarro | vedclass

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(A) Let $f(x) = y$. Since $f: R \rightarrow [-1, 1]$ is surjective,$y$ takes all values in $[-1, 1]$.Given $\sin \left(g(x) - \frac{\pi}{3}\right) = \

Vedclass · Accepted Answer

$\left[0, \frac{2 \pi}{3}\right]$

Vedclass · Answer

(A) Let $f(x) = y$. Since $f: R ightarrow [-1, 1]$ is surjective,$y$ takes all values in $[-1, 1]$.Given $\sin \left(g(x) - \frac{\pi}{3}ight) = \frac{y}{2} \sqrt{4 - y^2}$.Let $y = 2 \sin 	heta$. Since $y \in [-1, 1]$,$\sin 	heta \in \left[-\frac{1}{2}, \frac{1}{2}ight]$,so $	heta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}ight]$.Then $\frac{y}{2} \sqrt{4 - y^2} = \sin 	heta \sqrt{4 - 4 \sin^2 	heta} = \sin 	heta \sqrt{4 \cos^2 	heta} = 2 \sin 	heta \cos 	heta = \sin(2 	heta)$.Thus,$\sin \left(g(x) - \frac{\pi}{3}ight) = \sin(2 	heta)$.Since $	heta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}ight]$,$2 	heta \in \left[-\frac{\pi}{3}, \frac{\pi}{3}ight]$.Therefore,$g(x) - \frac{\pi}{3} = 2 	heta \in \left[-\frac{\pi}{3}, \frac{\pi}{3}ight]$.$g(x) \in \left[-\frac{\pi}{3} + \frac{\pi}{3}, \frac{\pi}{3} + \frac{\pi}{3}ight] = \left[0, \frac{2 \pi}{3}ight]$.Since $g$ is surjective,the codomain $A = \left[0, \frac{2 \pi}{3}ight]$.

If $f: R \rightarrow [-1, 1]$ and $g: R \rightarrow A$ are two surjective mappings and $\sin \left(g(x) - \frac{\pi}{3}\right) = \frac{f(x)}{2} \sqrt{4 - f^2(x)}$,then $A =$

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