Answer :

It is given that f (x) = x + sin 2x, x ∈ [0, 2π]

f’(x) = 1+ 2cos2x

Now, f’(x) = 0

2x = 2π � , n ϵ Z

⇒ x = nπ � , n ϵ Z

Now, we evaluate the value of f at critical point and at end points of the interval [0, 2π]

f’(0) = 0 + sin 0 = 0

f’(2π) = 2π + sin 4π = 2π + 0 = 2π

Therefore, we have the absolute maximum value of f on [0, 2π] is 2π occurring at

x = 2π and absolute minimum value of f(x) in the interval [0, 2π] is 0 occuring at x = 0.

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