Find the maximum and minimum values of $x + sin 2 x$ on $[0, 2 π]$ .

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Solution

Let
$f (x) = x + sin 2 x$

$f^{'} (x) = 1 + 2 cos 2 x$

For critical points,

$1 + 2 cos 2 x = 0$
Or
$2 cos 2 x = - 1$

$cos 2 x = \frac{- 1}{2}$

$2 x = \frac{2 π}{3}, \frac{4 π}{3}$

$x = \frac{π}{3}, \frac{2 π}{3}$

Now
$f (\frac{π}{3}) = \frac{π}{3} + sin (\frac{2 π}{3})$

$= \frac{π}{3} + \frac{\sqrt{3}}{2}$ ...(i)

$f (\frac{2 π}{3}) = \frac{2 π}{3} + sin (\frac{4 π}{3})$

$= \frac{2 π}{3} - \frac{\sqrt{3}}{2}$ ...(ii)
Hence
Maximum value is
$\frac{π}{3} + \frac{\sqrt{3}}{2}$ and Minimum Value is

$\frac{2 π}{3} - \frac{\sqrt{3}}{2}$

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