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Question

If (cosp1)x2+(cosp)x+sinp=0, xR has real roots for x, then the range of p is

A
[0,π]{2π}
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B
(0,π]
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C
[0,2π]
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D
{0}[π,2π]
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Solution

The correct option is A [0,π]{2π}
Given: (cosp1)x2+(cosp)x+sinp=0

When cosp1=0cosp=1, we get
sinp=0, so
0x2+x+0=0x=0p=0,2π

When cosp10cosp1, we get
(cosp1)x2+(cosp)x+sinp=0
For real roots
D0cos2p4(cosp1)sinp0cos2p4cospsinp+4sinp0(cosp2sinp)24sin2p+4sinp0(cosp2sinp)2+4sinp(1sinp)0
As (cosp2sinp)20, 1sinp0, so the inequality holds when
sinp0p(0,π] (cosp1)

Hence, the complete solution of p is [0,π]{2π}

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