If the roots of the given equation:
\( \Large \left(\cos p-1\right)x^{2}+ \left(\cos p\right)x+\sin p = 0 \) are real, then:


A) \( \Large P \epsilon \left(- \pi ,0\right) \)

B) \( \Large P \epsilon \left(- \frac{ \pi }{2}, \frac{ \pi }{2} \right) \)

C) \( \Large P \epsilon \left(0, \pi \right) \)

D) \( \Large P \epsilon \left(0, 2 \pi \right) \)

Correct answer:
C) \( \Large P \epsilon \left(0, \pi \right) \)

Description for Correct answer:

Given equation

\( \Large \left(\cos P-1\right)x^{2}+ \left(\cos P\right)x+\sin P=0 \)

since, roots are real, its discriminant  \( \Large \phi \ge 0 \)

=> \( \Large \cos^{2}P-4\cos P \sin P+4 \sin P \ge 0 \)

=> \( \Large \left(\cos P - 2 \sin P\right)^{2}-4 \sin^{2} P+4 \sin P \ge 0 \)...(i)

Now, \( \Large \left(1 - \sin P\right)\ge 0 \) for all real P and \( \Large \sin P > 0 \) for \( \Large 0 < P < \pi \). Therefore, \( \Large 4 \sin P \left(1-\sin P\right)\ge 0 \) when \( \Large 0 < P < \pi \) or  \( \Large P \epsilon (0, \pi) \)



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