MCQ
If $\alpha $ is a root of $25{\cos ^2}\theta + 5\cos \theta - 12 = 0$, $\pi /2 < \alpha < \pi $, then $\sin 2\alpha $ is equal to
- A$24/25$
- ✓$ - 24/25$
- C$13/18$
- D$ - 13/18$
$\therefore$ $25{\cos ^2}\alpha + 5\cos \alpha - 12 = 0$
==> $\cos \alpha = \frac{{ - 5 \pm \sqrt {25 + 1200} }}{{50}}$ $ = \frac{{ - 5 \pm 35}}{{50}}$
==> $\cos \alpha = - 4/5$ $[ \because \pi /2 < \alpha < \pi \Rightarrow \cos \alpha < 0]$
$\therefore $$\sin 2\alpha = 2\sin \alpha \cos \alpha = - 24/25$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.