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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
Find the maximum and minimum value of $\text{x}+\sin 2\text{x }\text{ on }[0,\ 2\pi]$
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Answer

Let $\text{f}\text{(x)}=\text{x}+\sin2\text{x}\ \Rightarrow\ \text{f}'\text{(x)}=1+2\cos2\text{x}$ Now $\text{f}'\text{(x)} =0\ \Rightarrow\ 1+2\cos2\text{x}=0\ \Rightarrow\ 2\cos2\text{x}=-1$ $\Rightarrow\ \cos2\text{x}=\frac{- 1}{2}=-\cos\frac{\pi}{3}=\cos\Big(\pi-\frac{\pi}{3}\Big)=\cos\frac{2\pi}{3}$ $\Rightarrow\ 2\text{x}=2\text{n}\pi\pm \frac{2\pi}{3}\text{ when }\text{n}\in\text{z}$ $\Rightarrow \ \text{x}=\text{n}\pi\pm\frac{\pi}{3}$ $\text{For } \text{n}=0,\ \text{x}=\pm\frac {\pi}{3}$ $\text{ But}\ \text{x}=-\frac{\pi}{3}\in[0,\ 2\pi],\text{ therefore }\text{ x}=\frac{\pi}{3}$ $\text{For }\text{n}=1,\ \text{x}=\pi\pm \frac{\pi}{3}=\pi+\frac{\pi}{3}\text{ and }\pi-\frac{\pi}{3}$ $\text{For }\text{n}=2,\ \text{x}=2\pi\pm\frac {\pi}{3}$$\text{But }\text{ x}=2\pi+\frac{\pi}{3}>2\pi\in[0,\ 2\pi],\text{therefore }\text{ x}=2\pi- \frac{\pi}{3}=\frac{5\pi}{3}$ Therefore, it is clear that the only turning point of f (x) given by x + sin 2x which belong to given closed interval $[0,\ 2\pi]$ are, $\text{x}=\frac {\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}$
$\text {At x }=\frac{\pi}{3}$ $\text{f}\Big (\frac{\pi}{3}\Big)=\frac{\pi}{3}+\sin\frac{2\pi}{3}=\frac{\pi}{3}+\frac{\sqrt{3}}{2}=1.05+0.87=1.92\text{ nearly}$
$\text {At x }=\frac{2\pi}{3}$ $\text{f}\Big (\frac{2\pi}{3}\Big)=\frac{2\pi}{3}+\sin\frac{4\pi}{3}=2\pi-\frac{\sqrt{3}}{2}=2.10-0.87=1.23\text{ nearly}$
$\text {At x} = \frac{4\pi}{3}$ $\text{f}\Big (\frac{4\pi}{3}\Big)=\frac{4\pi}{3}+\sin\frac{8\pi}{3}=\frac{4\pi}{3}+\frac{\sqrt{3}}{2}=4\times1.05+0.87=5.07\text{ nearly}$
$\text {At x} = \frac{5\pi}{3}$ $\text{f}\Big (\frac{5\pi}{3}\Big)=\frac{5\pi}{3}+\sin\frac{10\pi}{3}=\frac{5\pi}{3}-\frac{\sqrt{3}}{2}=5\times1.05-0.87=4.38\text{ nearly}$
$\text {At x} = 0$ $\text{f}(0) =0+\sin0=0$
$\text {At x }= 2\pi$ $\text{f}(2\pi)=2\pi+\sin4\pi=2\pi+0=2\pi=2\times3.14=6.28\text{ nearly}$
Therefore, Maximum value $=2\pi$ and minimum value = 0

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