Multiple choice

Consider the equation $az^2 + z + 1 = 0$ having purely imaginary root where $a = \cos \theta + i \sin \theta, i = \sqrt{-1}$ and function $f(x) = x^3 - 3x^2 + 3(1 + \cos \theta)x + 5,$ then answer the following questions. Which of the following is true?

  1. $f(x) = 0$ has three real distinct roots
  2. $f(x) = 0$ has one positive real root
  3. $f(x) = 0$ has one negative real root
  4. $f(x) = 0$ has three but not distinct roots
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C Correct answer
Explanation

The root of az^2 + z + 1 = 0 is purely imaginary (let it be ki). a(k^2 i^2) + ki + 1 = 0 => -ak^2 + ki + 1 = 0. Since a = cos(theta) + i sin(theta), this leads to cos(theta) = 0, so theta = pi/2. f(x) = x^3 - 3x^2 + 3(1+0)x + 5 = x^3 - 3x^2 + 3x + 5. f(-1) = -1 - 3 - 3 + 5 = -2. f(0) = 5. By IVT, there is a root between -1 and 0.

AI explanation

To ensure the quadratic equation az^2 + z + 1 = 0 has purely imaginary roots, the real and imaginary parts of a must satisfy specific conditions from the quadratic formula. The discriminant is 1 - 4a, and for the root to be purely imaginary, we set 1 - 4a = -t^2 for some real t, leading to the real part of a being 1/4. Since a = cos(theta) + i sin(theta), this means cos(theta) = 1/4, and it follows that sin^2(theta) = 15/16. We substitute this into the function f(x) = x^3 - 3x^2 + 3(1 + cos theta)x + 5, and analyzing its derivative f'(x) = 3x^2 - 6x + 3(1 + 1/4) = 3(x - 1)^2 + 3/4 shows the function is always increasing. Evaluating the limits as x approaches negative and positive infinity shows the graph crosses the x-axis exactly once on the left side, meaning it has one negative real root.

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