Answer :

It is given that f (x) = sin2x, x [0, 2π]

f’(x) = 2cos2x

Now, f’(x) = 0

cos2x = 0

2x = 0

x =

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

f’ =

f’ =

f’ =

f’ =

f(0) = sin0, f(2π) = sin2π = 0

Therefore, we have the absolute maximum value of f on [0, ] is 1 occurring at

x = and x =.

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