Multiple choice

The equation $(cos p -1)x^2 + (cos p) x + sin p = 0$ in the variable x, has real roots. Then p can take any value in the interval

  1. $(0, 2 \pi)$
  2. $(-\pi, 0)$
  3. $(-\pi/2, \pi/2)$
  4. $(0, \pi)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For real roots, the discriminant D = b^2 - 4ac must be >= 0. Here, (cos p)^2 - 4(cos p - 1)(sin p) >= 0. Testing p in (0, pi) shows that for values like pi/2, the expression simplifies to 0, and the inequality holds for the specified range.

AI explanation

For the quadratic equation to have real roots, its discriminant must be greater than or equal to zero. The discriminant is cos^2 p - 4(cos p - 1) sin p, which must be non-negative. At the boundaries p = 0 and p = pi, sin p equals 0 and the discriminant evaluates to 0. For any value of p strictly between 0 and pi, sin p is positive while (cos p - 1) is negative, ensuring the entire discriminant expression is positive. Therefore, the complete set of values p can take is the interval (0, pi).

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