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: Determine the discriminant requirement for real roots. A quadratic equation \( ax^2 + bx + c = 0 \) has real roots if its discriminant, \( b^2 - 4ac \), is non-negative. \[ b^2 - 4ac \geq 0. \]

Step 2: Apply the discriminant condition to the provided equation. For the equation \( (\cos p - 1)x^2 + (\cos p)x + \sin p = 0 \), we have \( a = (\cos p - 1) \), \( b = \cos p \), and \( c = \sin p \). The discriminant condition becomes: \[ (\cos p)^2 - 4(\cos p - 1)(\sin p) \geq 0. \] This simplifies to: \[ (\cos p)^2 \geq 4(\cos p - 1)(\sin p). \] 

Step 3: Evaluate the inequality. The inequality holds true for \( p \in (0, \pi) \) because both \( \cos p \) and \( \sin p \) are positive in this interval. In other regions, negative values of \( \cos p \) and/or \( \sin p \) may invalidate the inequality. 

Step 4: State the conclusion. Consequently, the applicable interval for \( p \) is \( (0, \pi) \).