Question

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 $A=$

• A. $\left[ 0 , \frac{2\pi}{3}\right]$
• B. [-1 , 1]
• C. $\left( \frac{-\pi}{2} , \frac{\pi}{2}\right)$
• D. $(0 , \pi )$

Answer 1

Answer: (A)

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

Answer 2

Let $f(x)=y,$ then $\sin \left(g(x)-\frac{\pi}{3}\right)=\frac{f(x)}{2} \sqrt{4-f^{2}(x)}$ $=\frac{y}{2} \sqrt{4-y^{2}}=t$ (let) $\Rightarrow y^{2}-\frac{y^{4}}{4}=t^{2}$ $\Rightarrow 4 y^{2}-y^{4}=4 t^{2}$ $\Rightarrow \left(y^{2}-2\right)^{2}=-4 t^{2}+4$ $\Rightarrow t^{2}=1-\frac{1}{4}\left(y^{2}-2\right)^{2}$ $\because f(x)=y \in[-1,1]$ $\Rightarrow y^{2} \in[0,1]$ $\therefore t^{2} \in\left[0, \frac{3}{4}\right]$ $\Rightarrow t \in\left[-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}\right]$ So, $\sin \left(g(x)-\frac{\pi}{3}\right) \in\left[-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}\right]$ $\Rightarrow g(x)-\frac{\pi}{3} \in\left[-\frac{\pi}{3}, \frac{\pi}{3}\right]$ $\Rightarrow g(x) \in\left[0, \frac{2 \pi}{3}\right]$