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
The roots are real when \(D = \cos^2 p - 4(\cos p - 1)\sin p \ge 0\).
\(\cos p - 1 \le 0\) always. When \(\sin p > 0\), that is, \(p \in (0, \pi)\), the term \(-4(\cos p - 1)\sin p \ge 0\). Then \(D \ge \cos^2 p \ge 0\), so the roots are real.
So the required interval is \((0, \pi)\).
Answer: (D)
Reinforce your conceptual grasp with related questions from NIMCET 2011:
Take timed full-length mock tests with authentic question distributions, negative marking (+12/-3 in Math), and instant All India Rank calculation.
Start Free Mock Test Series →