Question
The equation $$\left( {\cos p - 1} \right){x^2} + \left( {\cos p} \right)x + \sin p = 0$$ in $$x$$ has real roots. Then the set of values of $$p$$ is
A.
$$\left[ {0,2\pi } \right]$$
B.
$$\left[ { - \pi ,0} \right]$$
C.
$$\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]$$
D.
$$\left[ {0,\pi } \right]$$
Answer :
$$\left[ {0,\pi } \right]$$
Solution :
$$\eqalign{
& D \geqslant 0 \cr
& \Rightarrow \,\,{\cos ^2}p - 4\sin p \cdot \left( {\cos p - 1} \right) \geqslant 0 \cr
& {\text{or, }}{\cos ^2}p + 4\sin p\left( {1 - \cos p} \right) \geqslant 0\,\,\,\,\,\,\,.....\left( 1 \right) \cr} $$
$${\text{As }}{\cos ^2}p \geqslant 0\,\,{\text{and }}1 - \cos p \geqslant 0,\left( 1 \right)$$ will hold for all $$\sin p \geqslant 0.\,{\text{So, }}p \in \left[ {0,\pi } \right].$$