If $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ and $\mathrm{g}: \mathbb{R} \rightarrow \mathbb{R}$ are defined by $\mathrm{f}(\mathrm{x})=\mathrm{x}^3-\mathrm{x}$ and $g(x)=\sin 2 x$, then the value of $x \in(0,2 \pi)$ that satisfy $\mathrm{f}(\mathrm{g}(\mathrm{x}))0$, lie in the interval

Question

If $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ and $\mathrm{g}: \mathbb{R} \rightarrow \mathbb{R}$ are defined by $\mathrm{f}(\mathrm{x})=\mathrm{x}^3-\mathrm{x}$ and $g(x)=\sin 2 x$, then the value of $x \in(0,2 \pi)$ that satisfy $\mathrm{f}(\mathrm{g}(\mathrm{x}))>0$, lie in the interval

Aicharya · Accepted Answer

Given : $f(x)=x^3-x$ and $g(x)=\sin (2 x)$ $\begin{aligned} & \because \quad f(g(x))>0 \ & \Rightarrow \quad f(\sin 2 x)>0 \ & \Rightarrow \sin ^3 2 x-\sin 2 x>0 \ & \Rightarrow \quad\left(\sin ^2 2 x-1 ight) \sin 2 x>0 \ & \Rightarrow \quad-\cos ^2 2 x \sin 2 x>0 \Rightarrow \cos ^2 2 x \sin 2 x < 1 \ & \Rightarrow \quad \sin 2 x < 0 ext { and } \cos ^2 2 x eq 0 \ & \Rightarrow x \in\left(\frac{\pi}{2}, \pi ight) \cup\left(\frac{3 \pi}{2}, 2 \pi ight) \ & \Rightarrow \cos 2 x eq 0 \Rightarrow 2 x eq \frac{\pi}{2}, \frac{3 \pi}{2} \ & \Rightarrow x eq \frac{\pi}{4}, \frac{3 \pi}{4} \ & herefore x \in\left(\frac{\pi}{2}, \frac{3 \pi}{4} ight) \cup\left(\frac{3 \pi}{4}, \pi ight) \cup\left(\frac{3 \pi}{2}, 2 \pi ight) \ & ext { Also } x \in\left(\frac{\pi}{2}, \frac{3 \pi}{4} ight) \cup\left(\frac{3 \pi}{4}, \pi ight)\end{aligned}$

If $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ and $\mathrm{g}: \mathbb{R} \rightarrow \mathbb{R}$ are…

Solution