Question 46

Mathematics Trigonometry Simple Identities Medium

The equation \((\cos p-1) x^{2}+(\cos p) x+\sin p=0\) where \(x\) is a variable has real roots. Then the interval of \(p\) is

(A) \((0,2 \pi)\)
(B) \((-\pi, 0)\)
(C) \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)
(D) \((0, \pi)\)
View Dynamic Solution & Explanation
Correct Solution: Option D

Step-by-step Solution:

To obtain real roots, \[\] \((\cos p)^{2} \geq 4(\cos p-1) \sin p\). \[\] \((\cos p)^{2}-4 \cos p \sin p+4 \sin p \geq 0\) \[\] \(\Rightarrow \cos ^{2} p-4 \cos p \sin p+4 \sin ^{2} p\) \( +4 \sin p-4 \sin ^{2} p \geq 0 \Rightarrow(\cos p-2 \sin p)^{2}+4\left(\sin p-\sin ^{2} p\right) \geq 0\) \[\] \((\cos p-2 \sin p)^{2}\) is always + ve. \(\left(\sin p-\sin ^{2} p\right)\) is also positive, where \( < p < \pi\), it can be shown that when \(p\) lies in III or IV quadrants, \(\left(\sin p-\sin ^{2} p\right)\) becomes negative.\[\]Choice (D)\[\]