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

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

The correct answer is Option (2) → decreasing in $\left( \frac{\pi}{2}, \pi \right)$ ##

We have,

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

$∴f'(x) = 12 \sin^2 x \cdot \cos x - 12 \sin x \cdot \cos x + 12 \cos x$

$= 12 [\sin^2 x \cdot \cos x - \sin x \cdot \cos x + \cos x]$

$= 12 \cos x [\sin^2 x - \sin x + 1]$

$\Rightarrow f'(x) = 12 \cos x [\sin^2 x + (1 - \sin x)] \quad \dots(i)$

$∵1 - \sin x \geq 0 \text{ and } \sin^2 x \geq 0$

$∴\sin^2 x + 1 - \sin x \geq 0$

Hence, $f'(x) > 0$, when $\cos x > 0$ i.e., $x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$.

So, $f(x)$ is increasing when $x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ and $f'(x) < 0$, when $\cos x < 0$ i.e., $x \in \left( \frac{\pi}{2}, \frac{3\pi}{2} \right)$.

Hence, $f(x)$ is decreasing when $x \in \left( \frac{\pi}{2}, \frac{3\pi}{2} \right)$.

Since, $\left( \frac{\pi}{2}, \pi \right) \in \left( \frac{\pi}{2}, \frac{3\pi}{2} \right)$

Hence, $f(x)$ is decreasing in $\left( \frac{\pi}{2}, \pi \right)$.