APPLICATION OF DERIVATIVES — MATHS STD 12 Science — Question

2. $\text{Decreasing in }\Big(\frac{\pi}{2},\pi\Big)$

Solution:

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

$\therefore\ \text{f(x)}=12\sin^2\text{x}\cdot\cos\text{x}-12\sin\text{x}\cdot\cos\text{x}+12\cos\text{x}$

$=12\cos\text{x}\big[\sin^2\text{x}-\sin\text{x}+1\big]$

$=12\cos\text{x}\big[\sin^2\text{x}+(1-\sin\text{x}\big]$

Now, $1-\sin\text{x}\geq0\text{ and }\sin^2\text{x}\geq0$

$\therefore\ \sin^2\text{x}+\text{f}-\sin\text{x}>0$

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

So, f(x) is increasing when $\text{x}\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big)$

and f'(x) < 0, when $\cos\text{x}<0\text{ i.e., x}\in\Big(\frac{\pi}{2},\frac{3\pi}{2}\Big)$

Hence, f(x) is decreasing when $\text{x}\in\Big(\frac{\pi}{2},\frac{3\pi}{2}\Big)$

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