If the roots of the given equation $(\cos p-1) x^2+(\cos p) x+\sin p=0$ are real, then

Question

If the roots of the given equation $(\cos p-1) x^2+(\cos p) x+\sin p=0$ are real, then, $p \in(-\pi, 0)$, $p \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$, $p \in(0, \pi)$, $p \in(0,2 \pi)$

MARKS App · Accepted Answer

$(\cos p-1) x^2+(\cos p) x+\sin p=0$ Since, roots are real $\begin{array}{rlrl} \Rightarrow & \Delta \geq 0 \Rightarrow b^2-4 a c \geq 0 \ & (\cos p)^2-4(\cos p-1) (\sin p) \geq 0 \ & \cos ^2 p \geq 0 \text { and }(\cos p-1) \leq 0 \forall P \in \mathbf{R} \ & \therefore \sin p \geq 0 \ & \therefore p \in(0, \pi) \end{array}$ Hence, option (c) is correct.

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