Question

The equation \( (\cos p - 1)x^2 + (\cos p)x + \sin p = 0 \), where \( x \) is a variable with real roots. Then the interval of \( p \) may be any one of the following:

• A. \( (0, 2\pi) \)
• B. \( (-\pi, 0) \)
• C. \( (-\frac{\pi}{2}, \frac{\pi}{2}) \)
• D. \( (0, \pi) \)

Hint:

For a quadratic equation to have real roots, the discriminant (i.e., \( b^2 - 4ac \)) must be non-negative.

Answer 1

The Correct Option is D

Step 1: Identify the discriminant condition for real roots. For a quadratic equation \( ax^2 + bx + c = 0 \) to have real roots, the discriminant must be greater than or equal to 0. \[ b^2 - 4ac \geq 0. \] 

Step 2: Apply this condition to the given equation. In the given equation \( (\cos p - 1)x^2 + (\cos p)x + \sin p = 0 \), - \( a = (\cos p - 1), b = \cos p, c = \sin p \). So, the discriminant condition is: \[ (\cos p)^2 - 4(\cos p - 1)(\sin p) \geq 0. \] Simplifying, we get: \[ (\cos p)^2 \geq 4(\cos p - 1)(\sin p). \] 

Step 3: Analyze the inequality. For \( p \in (0, \pi) \), \( \cos p \) is positive, and \( \sin p \) is also positive, making the inequality valid. However, for other intervals, \( \cos p \) and/or \( \sin p \) could be negative, invalidating the inequality. 

Step 4: Conclusion. Therefore, the interval for \( p \) is \( (0, \pi) \).