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Q.

Theequation(cosp1)x2+(cosp)x+sinp=0hasrealroots,then'p'cantakeanyvalueintheinterval

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a

(0,  2π)

b

(π,  0)

c

(π2,π2)

d

(0,  π)

answer is D.

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Detailed Solution

GivenEquation(cosp1)x2+(cosp)x+sinp=0hasrealrootsthenD0b24ac0cos2p4(cosp1)sinp0cos2p4cosp.sinp+4sinp0(cosp2sinp)24sin2p+4sinp0(cosp2sinp)2+4sinp(1sinp)0  (cosp2sinp)20,1  valuesof  p  1sinp0  for  p(0,  π)

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