Multiple choice

Suppose cubic equation $x^{3}-px+q=0$ has three distinct & real roots $(p, q> 0)$ then which of the following is true for $f\left ( x \right )=x^{3}-px+q=0$

  1. maximum at $-\sqrt{p/3}$ & minimum at $\sqrt{p/3}$
  2. maximum at both $\sqrt{p/3}$ and $-\sqrt{p/3}$
  3. minimum at both $\sqrt{p/3}$ and $-\sqrt{p/3}$

  4. None of the above
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A Correct answer
Explanation

f'(x) = 3x^2 - p. Setting f'(x) = 0 gives x = +/- sqrt(p/3). Using the second derivative test, f''(x) = 6x. f''(sqrt(p/3)) > 0 (minimum), f''(-sqrt(p/3)) < 0 (maximum).

AI explanation

To find extrema, take the first derivative to get f'(x) = 3x^2 - p and set it to zero, yielding x = sqrt(p/3) and x = -sqrt(p/3). The second derivative is f''(x) = 6x; evaluating at -sqrt(p/3) gives a negative value indicating a maximum, and evaluating at sqrt(p/3) gives a positive value indicating a minimum. Therefore, the function has a maximum at -sqrt(p/3) and a minimum at sqrt(p/3).

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