Q79 / 170 · Manipal (MET) 2018 · Mathematics
The equation $(\cos p-1) x^2+\cos p x+\sin p=0$ in the variable $x$, has real roots. Then, $p$ can take any value in the interval
- Correct answerA$(0, \pi)$
- B$(-\pi, 0)$
- C$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
- D$(-\pi, \pi)$
Solution
Since, the given equation has real roots.
$\begin{aligned}
& \therefore \text { Discriminant, } \\
& D=\cos ^2 p-4(\cos p-1) \sin p \geq 0 \\
& \Rightarrow \quad D=(\cos p-2 \sin p)^2+4 \sin p(1-\sin p) \geq 0 \\
& \Rightarrow \quad 0 \lt p \lt \pi
\end{aligned}$
Hence, the correct answer is option: 'A'.