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Question
Find the difference between the greatest and least values of the function f(x) = sin2x – x, on `[- pi/2, pi/2]`
Solution
f(x) = sin2x – x
⇒ f′(x) = 2 cos2x – 1
Therefore, f′(x) = 0
⇒ cos2x = `1/2`
⇒ 2x is `(-pi)/3` or `pi/3`
⇒ x = `- pi/6` or `pi/6`
`"f"(- pi/2) = sin(- pi) + pi/2 = pi/2`
`"f"(- pi/6) = sin(-(2pi)/6) + pi/6 - - sqrt(3)/2 + pi/6`
`"f"(pi/6) = sin((2pi)/6) - pi/6 = sqrt(3)/2 - pi/6`
`"f"(pi/2) = sin(pi) - pi/2 = - pi/2`
Clearly, `pi/2` is the greatest value and `- pi/2` is the least.
Therefore, difference = `pi/2 + pi/2` = π
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