Within (0, π), on which interval is f(x) = sin 2x strictly decreasing?

Increasing & Decreasing Functions · Question 54 of 74

Question 54

MediumOriginal question

Within $(0, \pi)$ , on which interval is $f(x) = \sin 2x$ strictly decreasing?

(0, \frac{\pi}{2})

Correct answerYour answer

(\frac{\pi}{2}, \pi)

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(0, \frac{\pi}{4}) \cup (\frac{3 \pi}{4}, \pi)

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(\frac{\pi}{4}, \frac{3 \pi}{4})

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Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

$f'(x) = 2\cos 2x$ , which is negative when $2x$ is between $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ .
So $x$ is between $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ .

Worth noting

$\left(\frac{\pi}{2}, \pi\right)$ is where $\sin x$ falls; doubling the angle halves the interval's ends.

Ten questions from Increasing & Decreasing Functions, at your own pace.

Ten questions from Increasing & Decreasing Functions, at your own pace.