Q99 / 180 · KCET 2022 · Mathematics
The function $f(x)=4 \sin ^3 x-6 \sin ^2 x$
$+12 \sin x+100$ is strictly
- Adecreasing in $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
- Bdecreasing in $\left[0, \frac{\pi}{2}\right]$
- Cincreasing in $\left(\pi, \frac{3 \pi}{2}\right)$
- Correct answerDdecreasing in $\left(\frac{\pi}{2}, \pi\right)$
Solution
Given, $f(x)=4 \sin ^3 x-6 \sin ^2 x+12 \sin x+100$
$$
\begin{aligned}
f^{\prime}(x) & =12 \sin ^2 x \cdot \cos x-12 \sin x \cos x+12 \cos x \\
& =12 \cos x\left[\sin ^2 x-\sin x+1\right] \\
& =12 \cos x\left[\sin ^2 x+(1-\sin x)\right]
\end{aligned}
$$
We know, $1-\sin x \geq 0$ and $\sin ^2 x \geq 0$
$$
\therefore \sin ^2 x+1-\sin x>0
$$
Therefore, $f^{\prime}(x)>0$, when $\cos x>0 \Rightarrow x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$
So, $f(x)$ is increasing when $x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$
and $f^{\prime}(x) < 0$, when $\cos x < 0 \Rightarrow x \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$
So, $f(x)$ is decreasing when $x \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$
Hence, $f(x)$ is decreasing in $\left(\frac{\pi}{2}, \pi\right)$.
Hence, the correct answer is option: 'D'.