JAMB 2000 · UME · Q23

Find the minimum value of the function $f(\theta) = \dfrac{2}{3 - \cos\theta}$ for $0 \le \theta \le 2\pi$ .

Worked solution (try it first)

The top is fixed at 2, so the fraction is smallest when the bottom, $3 - \cos\theta$ , is largest.
$\cos\theta$ is at least $-1$ , so the bottom is at most $3 - (-1) = 4$ , at $\theta = \pi$ .
So the minimum is $\frac24 = \frac12$ , option A.