Question
The equation $$\left( {\cos p - 1} \right){x^2} + \left( {\cos p} \right)x + \sin p = 0$$ In the variable $$x,$$ has real roots. Then $$p$$ can take any value in the interval
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 :
The given equation is
$$\left( {\cos p - 1} \right){x^2} + \left( {\cos p} \right)x + \sin p = 0$$
For this equation to have real roots $$D \geqslant 0$$
$$\eqalign{
& \Rightarrow \,\,{\cos ^2}p - 4\sin p\left( {\cos p - 1} \right) \geqslant 0 \cr
& \Rightarrow \,\,{\cos ^2}p - 4\sin p\cos p + 4{\sin ^2}p + 4\sin p - 4{\sin ^2}p \geqslant 0 \cr
& \Rightarrow \,\,{\left( {\cos p - 2\sin p} \right)^2} + 4\sin p\left( {1 - \sin p} \right) \geqslant 0 \cr} $$
For every real value of $$p{\left( {\cos p - 2\sin p} \right)^2} \geqslant 0$$ and
$$1 - \sin p \geqslant 0\,\,\,\,\,\,\,\,\therefore \,\,D \geqslant 0,\,\,\forall \,p \in \left( {0,\pi } \right)$$