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Let $f(x)=\sin 2 x+\cos 2 x$ and $g(x)=x^2-1$ then $g(f(x))$ is invertible in the domain
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Verified Answer
The correct answer is:
$x \in\left[\frac{-\pi}{8}, \frac{\pi}{8}\right]$
$f(x)=\sin 2 x+\cos 2 x$ and $g(x)=x^2-1$
$\begin{aligned} & g[f(x)]=\left[(\sin 2 x+\cos 2 x)^2-1\right] \\ & g[f(x)]=\sin ^2 2 x+\cos ^2 2 x+\sin 4 x-1 \\ & g[f(x)]=\sin 4 x \\ & \text { Let } g f(x)=y\end{aligned}$
$y=\sin 4 x \quad\left[\because\right.$ Domain of $\sin x$ is $\left.-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\therefore$ Domain of $\sin 4 x=\left[-\frac{\pi}{8}, \frac{\pi}{8}\right]$
$\begin{aligned} & g[f(x)]=\left[(\sin 2 x+\cos 2 x)^2-1\right] \\ & g[f(x)]=\sin ^2 2 x+\cos ^2 2 x+\sin 4 x-1 \\ & g[f(x)]=\sin 4 x \\ & \text { Let } g f(x)=y\end{aligned}$
$y=\sin 4 x \quad\left[\because\right.$ Domain of $\sin x$ is $\left.-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\therefore$ Domain of $\sin 4 x=\left[-\frac{\pi}{8}, \frac{\pi}{8}\right]$
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