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

Step 1: Understanding the Concept:
For the quadratic equation to have real roots, its discriminant must be non-negative (\( D \ge 0 \)).
: Key Formula or Approach:
For \( Ax^2 + Bx + C = 0 \), \( D = B^2 - 4AC \ge 0 \).
Step 2: Detailed Explanation:
Given equation: \( (\cos p - 1)x^{2} + (\cos p)x + \sin p = 0 \).
Discriminant \( D = (\cos p)^{2} - 4(\cos p - 1)\sin p \).
For real roots:
\[ \cos^2 p - 4\sin p(\cos p - 1) \ge 0 \]
We can analyze the interval \( (0, \pi) \). In this interval, \( \sin p>0 \).
Also, \( \cos p - 1 \) is always \( \le 0 \) because the maximum value of \( \cos p \) is 1.
Thus, \( -4\sin p(\cos p - 1) \) becomes \( (\text{negative}) \times (\text{positive}) \times (\text{negative or zero}) \), which is \( \ge 0 \).
Since \( \cos^2 p \) is always \( \ge 0 \), the sum \( \cos^2 p + [-4\sin p(\cos p - 1)] \) is always \( \ge 0 \) for \( p \in (0, \pi) \).
Step 3: Final Answer:
The equation has real roots if \( p \) lies in the interval \( (0, \pi) \).