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Question

If the equation (cosp1)x2+(cosp)x+sinp=0 in x has real roots, then the set of values of p is

A
[0,2π]
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B
[π,0]
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C
[π2,π2]
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D
[0,π]
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Solution

The correct option is D [0,π]
Discriminant of given equation is,
D=cos2p4sinp(cosp1)
D=cos2p+4sinp(1cosp)
For equation to posses real solution, D0
now 1cosp11cosp0
thus for D0 only we need is, sinp0
p[0,π]

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