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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
Find the intervals in which the function f given by $\text{f}\text{(x)}=\frac{4\sin\text{x}-2\text{x}-\text{x}\cos\text{x}}{2+\cos\text{x}}$ is (i) increasing (ii) decreasing.
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Answer

$\text{f}\text{(x)}=\frac{4\sin\text{x}-2\text{x}-\text{x}\cos\text{x}}{2+\cos\text{x}}$
$\therefore\ \text{f}'\text{x}=\frac{(2+\cos\text{x})(4\cos\text{x}-2-\cos\text{x}+\text{x}\sin\text{x})-(4\sin\text{x}-2\text{x}-\text{x}\cos\text{x})(-\sin\text{x})}{(2+\cos\text{x})^2}$
$=\frac{(2+\cos\text{x})(3\cos\text{x}-2+\text{x}\sin\text{x})+\sin\text{x}(4\sin\text{x}-2\text{x}-\text{x}\cos\text{x})}{(2+\cos\text{x})^2}$
$=\frac{6\cos\text{x}-4+2\text{x}\sin\text{x}+3\cos^2\text{x}-2\cos\text{x}+\text{x}\sin\text{x}\cos\text{x}+4\sin^2\text{x}-2\text{x}\sin\text{x}-\text{x}\sin\text{x}\cos\text{x}}{(2+\cos\text{x})^2}$ Now, f'(x) = 0
$=\frac{4\cos\text{x}-4+3\cos^2\text{x}+4\sin^2\text{x}}{(2+\cos\text{x})^2}$
$=\frac{4\cos\text{x}-4+3\cos^2\text{x}+4-4\cos^2\text{x}}{(2+\cos\text{x})^2}$
$=\frac{4\cos\text{x}-\cos^2\text{x}}{(2+\cos\text{x})^2}$ $=\frac{\cos\text{x}(4-\cos\text{x})}{(2+\cos\text{x})^2}$
$\Rightarrow\ \cos\text{x}=0\text{ or }\cos\text{x}=4$
$\text{But}, \cos\text{x}\neq4$
$\therefore\ \cos\text{x}=0$
$\Rightarrow\text{x}=\frac{\pi}{2}.\frac{3\pi}{2}$
Now, $\text{x}=\frac{\pi}{2}\text{ and x }=\frac{3\pi}{2}$ divides $(0,\ 2\pi)$ into three disjoint intervals i.e.,
$\Big(0,\frac{\pi}{2}\Big),\Big(\frac{\pi}{2},\frac{3\pi}{2}\Big),\text{and}\Big(\frac{3\pi}{2},2\pi\Big).$
In intervals $\Big(0,\frac{\pi}{2}\Big)\text{ and }\Big(\frac{3\pi}{2},2\pi\Big),\text{f}'\text{x}>0.$
Thus, f(x) is increasing for $0<\text{x}<\frac{\text{x}}{2}\text{ and }\frac{3\pi}{2}<\text{x}<2\pi.$
In the intervals $\Big(\frac{\pi}{2},\frac{3\pi}{2}\Big),\text{f}'\text{(x)}<0.$
Thus, f(x) is decreasing for $\frac{\pi}{2}<\text{x}<\frac{3\pi}{2}.$

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