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Question

The equation (cosp1)x2+cospx+sinp=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,π]
For given quadratic to posses real roots, its discriminant should be non- negative
cos2p4sinp(cosp1)0
We can observe that cos20,cosp10xR
So for discriminant to be always non-negative sinp0
and we know sinx positive in 1st and 2nd quadrant
hence interval of p is [0,π]

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