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Application of Derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplication of Derivatives1 Mark
MCQ
Choose the correct answer from the given four options : The function $\text{f(x)}=4\sin^3\text{x}-6\sin^2\text{x}+12\sin\text{x}+100$ is strictly :
  • A
    $\text{Increasing in }\Big(\pi,\frac{3\pi}{2}\Big)$
  • ✓
    $\text{Decreasing in }\Big(\frac{\pi}{2},\pi\Big)$
  • C
    $\text{Decreasing in }\Big[\frac{-\pi}{2},\frac{\pi}{2}\Big]$
  • D
    $\text{Decreasing in }\Big[0,\frac{\pi}{2}\Big]$
✓

Answer

Correct option: B.
$\text{Decreasing in }\Big(\frac{\pi}{2},\pi\Big)$
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, $1-\sin\text{x}\geq0$ 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 $\Big(\frac{\pi}{2},\pi\Big)$

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