The function f(x)=4 sin ^3 x-6 sin ^2 x+12 si | vedclass

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(D) Given,$f(x)=4 \sin ^3 x-6 \sin ^2 x+12 \sin x+100$. First,we find the derivative $f'(x)$: $f'(x) = 12 \sin^2 x \cos x - 12 \sin x \cos x + 12 \cos

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decreasing in $\left(\frac{\pi}{2}, \pi\right)$

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(D) Given,$f(x)=4 \sin ^3 x-6 \sin ^2 x+12 \sin x+100$. First,we find the derivative $f'(x)$: $f'(x) = 12 \sin^2 x \cos x - 12 \sin x \cos x + 12 \cos x$. Factoring out $12 \cos x$: $f'(x) = 12 \cos x (\sin^2 x - \sin x + 1)$. Consider the quadratic expression $g(t) = t^2 - t + 1$ where $t = \sin x$. The discriminant $D = (-1)^2 - 4(1)(1) = 1 - 4 = -3 Since the coefficient of $t^2$ is positive and $D Thus,the sign of $f'(x)$ depends only on $\cos x$. $f'(x) > 0$ when $\cos x > 0$,which occurs in $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$. $f'(x) Therefore,$f(x)$ is strictly decreasing in the interval $(\frac{\pi}{2}, \pi)$.

The function $f(x)=4 \sin ^3 x-6 \sin ^2 x+12 \sin x+100$ is strictly

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