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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the maximum and minimum values of \( x + \sin\: 2x\) on \([0, 2\pi].\)

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  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
  • $\large\frac{d}{dx}$$(\sin x)=\cos x$
Step 1:
Let $f(x)=x+\sin 2x$
$f'(x)=1+2\cos 2x$
For maxima and minima
$1+2\cos 2x=0$
$2\cos 2x=-1$
$\cos 2x=\large\frac{-1}{2}$
Step 2:
Now we find $f(x)$ at $x=0,\large\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3},$$2\pi$
We have $f'(x)=1+2\cos 2x$
$f(x)=x+\sin 2x$
$f(0)=0+\sin 0=0$
$\qquad=\large\frac{\pi}{3}+\frac{\sqrt 3}{2}$
$\qquad=\large\frac{2\pi}{3}+\frac{\sqrt 3}{2}$
$\qquad=\large\frac{4\pi}{3}+\frac{\sqrt 3}{2}$
$\qquad=\large\frac{5\pi}{3}+\frac{\sqrt 3}{2}$
$f(2\pi)=2\pi+\sin 2\pi$
Hence maximum $f(x)=2\pi$
Minimum $f(x)=0$
answered Aug 8, 2013 by sreemathi.v

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