If the roots of the given equation (cos p-1) | vedclass

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Question

(C) Given equation: $(\cos p-1) x^2+(\cos p) x+\sin p=0$. Since the roots are real,the discriminant $\Delta \geq 0$. $\Delta = b^2 - 4ac = (\cos p)^2

Vedclass · Accepted Answer

$p \in(0, \pi)$

Vedclass · Answer

(C) Given equation: $(\cos p-1) x^2+(\cos p) x+\sin p=0$. Since the roots are real,the discriminant $\Delta \geq 0$. $\Delta = b^2 - 4ac = (\cos p)^2 - 4(\cos p - 1)(\sin p) \geq 0$. $\cos^2 p - 4\sin p \cos p + 4\sin p \geq 0$. For the quadratic equation to exist,the coefficient of $x^2$ must be non-zero: $\cos p - 1 
eq 0 \Rightarrow \cos p 
eq 1$. Since $\cos^2 p \geq 0$ and $(\cos p - 1)  0$. Thus,$p \in (0, \pi)$. Hence,option $C$ is correct.

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

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