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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)
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B
(π,0)
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C
(π2π2)
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D
[0,π]
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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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