Multiple choice

The range of values of $a$ so that all the roots of the equation $2{ x }^{ 3 }-3{ x }^{ 2 }-12x+a=0$ are real and distinct, belongs to

  1. $\left( 7,20 \right) $
  2. $\left( -7,20 \right) $
  3. $\left( -20,7 \right) $
  4. $\left( -7,7 \right) $
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B Correct answer
Explanation

Let f(x) = 2x^3 - 3x^2 - 12x + a. For roots to be real and distinct, the local maximum and minimum must have opposite signs. f'(x) = 6x^2 - 6x - 12 = 6(x-2)(x+1). Critical points at x = 2, -1. f(2) = 16 - 12 - 24 + a = a - 20. f(-1) = -2 - 3 + 12 + a = a + 7. For opposite signs, (a-20)(a+7) < 0, so -7 < a < 20.

AI explanation

To ensure all three roots of the cubic 2x^3 - 3x^2 - 12x + a = 0 are real and distinct, the local maximum must be greater than zero and the local minimum must be less than zero. Taking the derivative gives 6x^2 - 6x - 12, which yields critical points at x = 2 and x = -1. Evaluating the original function at these points gives f(2) = a - 20 and f(-1) = a + 7, so we must have a - 20 < 0 and a + 7 > 0. Solving these inequalities places the value of a in the open interval (-7, 20).

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