Question

The function $f(x)=4 \sin ^3 x-6 \sin ^2 x+12 \sin x+100$ 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 $\left[0, \frac{\pi}{2}\right]$

Answer 1

Answer: (B)

$\because f(x)=4 \sin ^3 x-6 \sin ^2 x+12 \sin x+100$
$\Rightarrow f^{\prime}(x) =12 \sin ^2 x \cdot \cos x-12 \sin x \cdot \cos x + 12 \cos x + 0$
$ =12 \cos x\left(\sin ^2 x-\sin x+1\right) $
Since, in second quadrant $\sin x$ is $+v e$ and the cos $x$ is $-v e$.
So, $f^{\prime}(x) < 0$ for all $x \in\left(\frac{\pi}{2}, \pi\right)$