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Question
Find the maximum and minimum values of x + sin 2x on [0, 2π].
Solution
Let f (x) = x + sin2x, 0 ≤ x ≤ 2π
⇒ f' (x) = 1 + 2cos 2x
⇒ For critical points, let f' (x) = 0
⇒ 1 + cos 2x = 0
⇒ `cos 2x = -1/2`
⇒ `cos 2x = -cos pi/3`
(If 0< x < 2π, then 0< 2x < 4π)
`⇒ cos 2x = cos (pi- pi/3), cos (pi + pi/2), cos (3pi - pi/3), cos (3pi + pi/3)`
⇒ `2x = (2pi)/3 , (4pi)/3, (8pi)/3, (10pi)/3`
⇒ `x = pi/3, (2pi)/3, (4pi)/3, (5pi)/3`
So, for finding maximum and minimum, we evaluate f (x) at `0, 2pi , pi/3, (2pi)/3, (4pi)/3, (5pi)/3`
Now f(0) = 0 + sin 0 =
f (2π) = 2π + sin 4π = 2π + 0 = 2π
`f (pi/3) = pi/3 + sin (2pi)/3 = pi/3 + sin (pi - pi/3)`
= `pi/3 + sin pi/3 = pi/3 + sqrt3/2`
`f ((2pi)/3) = (2pi)/3 + sin (4pi)/3 = (2pi)/3 + sin (pi + pi/3)`
= `(2pi)/3 -sin pi/3 = (2pi)/3 - sqrt3/2`
`f((4pi)/3) = (4pi)/3 + sin (8pi)/3 = (4pi)/3 + sin (2pi + (2pi)/3)`
= `(4pi)/3 + sin (2pi)/3 = (4pi)/3 + sqrt3/2`
and `f ((5pi)/3) = (5pi)/3 + sin (10pi)/3 = (5pi)/3 + sin (3pi + pi/3)`
= `(5pi)/3 -sin pi/3 = (5pi)/3 - sqrt3/2`
Thus, maximum value of f (x) = 2π at x = 2π and minimum value of f (x) = 0 at x = 0.
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