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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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At what points in the interval \([0, 2\pi]\), does the function \(\sin\: 2x\) attain its maximum value?

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Toolbox:
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
  • $\large\frac{d}{dx}$$\sin x=\cos x$
Step 1:
$f(x)=\sin 2x$
$f'(x)=2\cos 2x$
For maxima and minima $f'(x)=0$
$2\cos 2x=0$
$\cos 2x=0$
$2x=\cos^{-1}0$
$2x=\large\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2}$
$x=\large\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$
Step 2:
Now,we find $f(x)$ at $x=0,\large\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$$,2\pi$
$f(0)=0$
$f(\large\frac{\pi}{4})$$=\sin\large\frac{\pi}{2}$$=1$
$f(\large\frac{3\pi}{4})$$=\sin\large\frac{3\pi}{2}$$=-1$
$f(\large\frac{5\pi}{4})$$=\sin\large\frac{5\pi}{2}$$=1$
$f(\large\frac{7\pi}{4})$$=\sin\large\frac{7\pi}{2}$$=-1$
$f(2\pi)=\sin 2\pi=0$
Hence maximum value of $f(x)=1$ at $x=\large\frac{\pi}{4},\large\frac{5\pi}{4}$
answered Aug 7, 2013 by sreemathi.v
 

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