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Question
At what points in the interval [0, 2π], does the function sin 2x attain its maximum value?
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Solution
Let f(x) = sin 2x, interval [0, 2π]
f‘(x) = 2 cos 2x
यदि f'(x) = 0 ⇒ 2 cos 2x = 0
⇒ 2x `= pi/2, (3pi)/2, (5pi)/2, (7 pi)/2 => x = pi/4, (3pi)/4, (5pi)/4, (7 pi)/4`
Hence we find `x = pi/4, (3pi)/4, (5pi)/4, (7 pi)/4` and the value of f at the endpoints of the interval [0, 2 `pi`].
At x = 0, f (0) = sin 0 = 0
x `= 2 pi at, f(2 pi) = sin 2 xx 2 pi = sin 4 pi = 0`
x`= pi/4 at, f(pi/4) = sin 2 xxpi/4 = sin pi/2 = 1`
x `= (3pi)/4 at, f((3 pi)/4) = sin (3 pi)/2 = - 1`
x `= (5pi)/4 at, f((5pi)/4) = sin (5 pi)/2 = 1`
x `= (7pi)/4 at, f((7pi)/4) = sin (7 pi)/2 = -1`
Thus, the function f(x) attains maximum value 1 at `= pi/4` and x`= (5 pi)/4`.
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