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(C) Given: $f(x) = x^3 - x$ and $g(x) = \sin(2x)$.We need to solve $f(g(x)) > 0$.$f(g(x)) = (\sin 2x)^3 - \sin 2x > 0$.Factorizing the expression: $\s

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$\left(\frac{\pi}{2}, \frac{3\pi}{4}\right) \cup \left(\frac{3\pi}{4}, \pi\right)$

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(C) Given: $f(x) = x^3 - x$ and $g(x) = \sin(2x)$.We need to solve $f(g(x)) > 0$.$f(g(x)) = (\sin 2x)^3 - \sin 2x > 0$.Factorizing the expression: $\sin 2x (\sin^2 2x - 1) > 0$.Since $\sin^2 2x - 1 = -\cos^2 2x$,we have: $-\sin 2x \cos^2 2x > 0$.This implies $\sin 2x \cos^2 2x For this to hold,we must have $\sin 2x In the interval $x \in (0, 2\pi)$,$2x \in (0, 4\pi)$.$\sin 2x Also,$\cos 2x 
eq 0$ implies $2x 
eq \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$,so $x 
eq \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.Excluding these points from the interval,we get $x \in (\frac{\pi}{2}, \frac{3\pi}{4}) \cup (\frac{3\pi}{4}, \pi) \cup (\frac{3\pi}{2}, \frac{7\pi}{4}) \cup (\frac{7\pi}{4}, 2\pi)$.Comparing with the given options,the interval $(\frac{\pi}{2}, \frac{3\pi}{4}) \cup (\frac{3\pi}{4}, \pi)$ is a subset of the solution set.

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

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