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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
At what points on the interval $[0,\ 2\pi]$ does the function sin 2x attain its maximum value?
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Answer

Let $\text{f}\text{(x)} =\sin2\text{x}\ \Rightarrow\ \text{f}'\text{(x)}=2\cos2\text{x}$
Now $\text{f}'\text{(x)}=0\ \Rightarrow\ 2\cos2\text{x}=0$$\Rightarrow\ \ 2\text{x}=(2\text{n}+1)\frac{\pi}{2}$
$\Rightarrow\ \text{x}=(2\text{n}+1)\frac{\pi}{4}$
$\text{Putting n }=0,1,2,3;\ \text{x}=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\in[0,\ 2\pi]$
$\text{Now }\ \text{f}\text{(x)}=\sin2\text{x}$
$ \therefore \ \text{f}\Big[(2\text{n}+1)\frac{\pi}{4}\Big]=\sin(2\text{n}+1)\frac{\pi}{2}$ $=\sin\bigg(\text{n}\pi+\frac{\pi}{2}\bigg)$$=(-1)^\pi\sin\frac{\pi}{2}=(-1)^\pi$
Putting n = 0, 1, 2, 3;
$\text{f}\bigg(\frac{\pi}{4}\bigg)=(-1)^0=1\ \ \text{f}\bigg(\frac{3\pi}{4}\bigg)=(-1)^1=-1 $
$\text{f}\bigg(\frac{5\pi}{4}\bigg)=(-1)^2=1\ \ \text{f}\bigg(\frac{7\pi}{4}\bigg)=(-1)^3=-1 $
$\text{Also}\ \text{f}(0)=\sin0=0\text{ and }\ \text{f}(2\pi)=\sin4\pi=0$
Since f(x) attains its maximum value 1 at $\text{x}=\frac{\pi}{4} \text{ and }\text{ x}=\frac{5\pi}{4}.$
Therefore, the required points are $\Big(\frac{\pi}{4},1\Big)\text{ and }\Big(\frac{5\pi}{4},1\Big).$

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