At which point in (0, π) does f(x) = sin 2x - x have a local maximum?

Question 44

HardOriginal question

At which point in $(0, \pi)$ does $f(x) = \sin 2x - x$ have a local maximum?

\frac{\pi}{6}

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\frac{\pi}{4}

Correct answerYour answer

\frac{\pi}{3}

Correct answerYour answer

\frac{5 \pi}{6}

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Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

$f'(x) = 2\cos 2x - 1 = 0$ when $\cos 2x = \frac{1}{2}$ : $2x = \frac{\pi}{3}$ or $\frac{5\pi}{3}$ , so $x = \frac{\pi}{6}$ or $\frac{5\pi}{6}$ .
$f''(x) = -4\sin 2x$ : at $\frac{\pi}{6}$ it is $-2\sqrt{3} < 0$ , a local maximum; at $\frac{5\pi}{6}$ it is $2\sqrt{3} > 0$ , a local minimum.

Worth noting

$\frac{5\pi}{6}$ is the other critical point, where $f'' > 0$ : the local minimum. $\frac{\pi}{4}$ is where $\sin 2x$ alone peaks.

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