Question

The equation $(\cos p - 1)x^{2}+ (\cos p)\,x + \sin p = 0$ in the variable $x$ has real roots. Then $p$ can take values in the interval

• A. $(0, \pi)$
• B. $(-\pi, 0)$
• C. $\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$
• D. $(0, 2\pi)$

Hint:

For quadratic $ax^2+bx+c=0$, discriminant $D = b^2 - 4ac \geq 0$ for real roots. Check the sign of each trigonometric term in the given interval.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a quadratic to have real roots, the discriminant $D = b^2 - 4ac \ge 0$.
Step 2: Detailed Explanation:
$D = \cos^2 p - 4(\cos p - 1)\sin p \ge 0$.
Analysis of this inequality shows it is satisfied for $p \in (0,\pi)$, where $\sin p>0$ and the constraint on $\cos p$ is met.
Step 3: Final Answer:
$p \in (0, \pi)$.