Q. If the roots of the given equation $\left( \cos{p - 1} \right)x^{2} + \left( \cos p \right)x + \sin{p = 0}$ are real, then
A) $p \in ( - \pi,\ 0)$
B) $p \in \left( - \frac{\pi}{2},\ \frac{\pi}{2} \right)$
C) $p \in (0,\ \pi)$
D) $p \in (0,\ 2\pi)$
Explanation
(c)
Here, $D \geq 0$
$\Rightarrow \ \ \cos^{2}{p - 4\left( \cos{p - 1} \right)\sin{p \geq 0}}$
$\Rightarrow \ \ \cos^{2}{p - 4\cos{p\sin{p + 4\sin{p \geq 0}}}}$
$\Rightarrow \ \ \left( \cos{p - 2\sin p} \right)^{2} + 4\sin{p\left( 1 - \sin p \right) \geq 0}$ ...(i)
Since, $\left( 1 - \sin p \right) \geq 0$ for all real $p$ and $\sin{p > 0}$ for $0 < p < \pi$
$\therefore 4\sin{p(1 - \sin p) \geq 0}$ when $0 < p < \pi$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.