Question

The equation \((\cos p - 1)x^2 + \cos p \, x + \sin p = 0\) has real roots. Then \(p\) lies in

• A. \((0,\pi)\)
• B. \((-\pi,0)\)
• C. \((-\pi/2,\pi/2)\)
• D. \((-\pi,\pi)\)

Hint:

If coefficient gives negative sign, check sign of sine/cosine instead of expanding fully.

Answer 1

The Correct Option is A

Concept: For real roots: \[ D \ge 0 \]

Step 1: Identify coefficients.
\[ a = \cos p -1,\quad b = \cos p,\quad c = \sin p \]

Step 2: Discriminant.
\[ D = b^2 - 4ac = \cos^2 p - 4(\cos p -1)\sin p \]

Step 3: Analyze condition.
For real roots, expression must be non-negative. Observe: \[ \cos p -1 \le 0 \text{ for all } p \] So, \[ -4(\cos p -1)\sin p \ge 0 \quad \text{when } \sin p>0 \] Thus: \[ D \ge 0 \quad \text{when } \sin p>0 \]

Step 4: Final interval.
\[ \sin p>0 \Rightarrow p \in (0,\pi) \]