#### Explain solution rd sharma class 12 chapter 16 Increasing and decreasing function exercise multiple choice question, question 33

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

Hints: Find the derivative of given equation.

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

Solution: We have,

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

\begin{aligned} f^{\prime}(x) &=12 \sin ^{2} x \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}

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

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

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

So, $f(x)$ is increasing when $x\epsilon \left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$

$f'{x}<0$ , when $\cos x>0$ , i.e., $x\epsilon \left ( \frac{\pi}{2}, \frac{3\pi}{2} \right )$

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

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