Question

At what points in the interval [0, 2π], does the function sin 2x attain its maximum value?

Answer 1

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`.