JAMB 2017 · UTME · Q6✱✱

If $\frac{\pi}{2} \le \theta < 2\pi$ , find the maximum value of $f(\theta) = \dfrac{4}{6 + 2\cos\theta}$ .

Worked solution (try it first)

The top is fixed at 4, so $f(\theta)$ is largest when the bottom, $6 + 2\cos\theta$ , is smallest.
$\cos\theta$ is smallest, $-1$ , at $\theta = \pi$ , which is inside the range.
So the maximum is $\frac{4}{6 - 2} = \frac44 = 1$ , option B.