Complex Numbers And Quadratic Equations question 774
Question: If the roots of the given equation $ (\cos p-1)x^{2}+(\cos p)x+\sin p=0 $ are real, then [IIT 1990; RPET 1995]
Options:
A) $ p\in (-\pi ,0) $
B) $ p\in ( -\frac{\pi }{2},\frac{\pi }{2} ) $
C) $ p\in (0,\pi ) $
D) $ p\in (0,2\pi ) $
Show Answer
Answer:
Correct Answer: C
Solution:
Given equation $ (\cos p-1)x^{2}+(\cos p)x+\sin p=0 $ Its discriminant $ D\ge 0 $ since roots are real
Þ $ {{\cos }^{2}}p-4(\cos p-1)\sin p\ge 0 $
Þ $ {{\cos }^{2}}p-4\cos p\sin p+4\sin p\ge 0 $
Þ $ {{(\cos p-2\sin p)}^{2}}-4{{\sin }^{2}}p+4\sin p\ge 0 $
Þ $ {{(\cos p-2\sin p)}^{2}}+4\sin p(1-\sin p)\ge 0 $ …….(i) Now $ (1-\sin p)\ge 0 $ for all real p, $ \sin p>0 $ for $ 0<p<\pi . $ Therefore $ 4\sin p(1-\sin p)\ge 0 $ when $ 0<p<\pi $ or $ p\in (0,\pi ) $
BETA
