A) \[p\in (-\pi ,0)\]
B) \[p\in \left( -\frac{\pi }{2},\frac{\pi }{2} \right)\]
C) \[p\in (0,\pi )\]
D) \[p\in (0,2\pi )\]
Correct Answer: C
Solution :
Given equation is \[(\cos \,p-1){{x}^{2}}+(\cos \,p)x+\sin \,p=0\] \[\because \] Roots are real. \[\therefore \] \[D\ge 0\] \[\Rightarrow \] \[{{\cos }^{2}}\,p-4(\cos \,p-1)\,\sin \,p\ge 0\] \[\Rightarrow \] \[{{\cos }^{2}}\,p-4\,\cos \,p\,\sin \,p+4\,\sin \,p\ge 0\] \[\Rightarrow \] \[{{(\cos \,p-2\,\sin \,p)}^{2}}-4\,{{\sin }^{2}}\,p+4\,\sin p\ge 0\] \[\Rightarrow \] \[{{(\cos \,p-2\,\sin \,p)}^{2}}\] \[+4\,\sin \,p\,(1-sin\,p)\ge 0\] ?. (i) Since, \[(1-\sin \,P)\ge \,0\] for all real p, and \[\sin \,p\,>0\] for \[0<p<\pi \] \[\therefore \] \[4\,\,\sin \,p\,(1-sin\,p)\ge 0\] when \[0<p<\pi \]
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