KCET · Maths · Functions
Let \(f(x)=\sin 2 x+\cos 2 x\) and \(g(x)=x^2-1\) then \(g(f(x))\) is invertible in the domain
- A \(x \in\left[\frac{-\pi}{8}, \frac{\pi}{8}\right]\)
- B \(x \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
- C \(x \in\left[0, \frac{\pi}{4}\right]\)
- D \(x \in\left[\frac{-\pi}{4}, \frac{\pi}{4}\right]\)
Answer & Solution
Correct Answer
(A) \(x \in\left[\frac{-\pi}{8}, \frac{\pi}{8}\right]\)
Step-by-step Solution
Detailed explanation
\(f(x)=\sin 2 x+\cos 2 x\) 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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