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 any value in the interval

A
(0,2π)
B
(π,0)
C
(π2π2)
D
[0,π]

Solution

The correct option is D $$[0, \pi]$$For given quadratic to posses real roots, its discriminant should be non- negative$$\Rightarrow \cos^2p-4\sin p(\cos p-1)\ge 0$$We can observe that $$\cos^2\ge 0, \cos p -1\le 0\forall x\in R$$So for discriminant to be always non-negative $$\sin p\ge 0$$and we know $$\sin x$$ positive in 1st and 2nd quadrant hence interval of $$p$$ is $$[0,\pi]$$Mathematics

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