# The function $f (x) = 4 sin^3x - 6 sin^2x + 12 sinx + 100$ is strictly A. increasing in$\left ( \pi,\frac{3\pi}{2} \right )$ B. decreasing in $\left ( \frac{\pi}{2},\pi \right )$ C. decreasing in $\left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$ D. decreasing in $\left ( 0,\frac{\pi}{2} \right )$

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

Apply the first derivative and get

$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(4 \sin ^{3} \mathrm{x}-6 \sin ^{2} \mathrm{x}+12 \sin \mathrm{x}+100\right)}{\mathrm{dx}}$
Apply sum rule and get
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(4 \sin ^{3} \mathrm{x}\right)}{\mathrm{dx}}-\frac{\mathrm{d}\left(6 \sin ^{2} \mathrm{x}\right)}{\mathrm{dx}}+\frac{\mathrm{d}(12 \sin \mathrm{x})}{\mathrm{dx}}+0$
Then apply power rule and get
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=12 \sin ^{2} \mathrm{x} \frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}-12 \sin \mathrm{x} \frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}+12 \frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}$

Now apply the derivative,
$\\ { \Rightarrow f' (x)=12 sin\textsuperscript{2}x (cos x)-12sin x (cos x)+12(cos x)}\\ { \Rightarrow f' (x)=12 sin\textsuperscript{2}x cos x-12sin x cos x +12 \cos x}\\ { \Rightarrow f' (x)=12 cos x(sin\textsuperscript{2}x -sin x +1)}\\$

Now,$1-sin x \geq 0$ and $sin\textsuperscript{2}x \geq 0$

Hence $sin\textsuperscript{2}x -sin x +1 \geq 0$

Therefore, $f'(x) > 0, \text{ when } \cos x > 0, \text{ i.e., } x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
Hence f(x) is increasing when $\mathrm{x} \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$
and $f^{\prime}(x)<0$, when $\cos x<0, i.e. \mathrm{x} \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$
Hence f(x) is decreasing when $\mathrm{X} \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$
Now $\left(\frac{\pi}{2}, \pi\right) \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$
Hence, f(x) is decreasing in $\left(\frac{\pi}{2}, \pi\right)$
So the correct answer is option B