Problem 47 - A Complete Resource Book in Mathematics for JEE Main

Question

Suppose the cube \(x^{3}-p x+q\) has three distinct real roots where \(p>0\) and \(q>0\). Then which one of the following holds? (A) The cubic has minima at \(\sqrt{\frac{p}{3}}\) and maxima at \(-\sqrt{\frac{p}{3}}\) (B) The cubic has minima at \(-\sqrt{\frac{p}{3}}\) and maxima at \(\sqrt{\frac{p}{3}}\) (C) The cubic has minima at both \(\sqrt{\frac{p}{3}}\) and \(-\sqrt{\frac{p}{3}}\) (D) The cubic has maxima at both \(\sqrt{\frac{p}{3}}\) and \(-\sqrt{\frac{p}{3}}\)

Step-by-Step Solution

Answer

The cubic has minima at \(\sqrt{\frac{p}{3}}\) and maxima at \(-\sqrt{\frac{p}{3}}\). (Option A)

Step 1: Calculate the Derivative

To find the minima and maxima, we need to calculate the derivative of the given cubic function, which is \( f(x) = x^3 - px + q \). The derivative \( f'(x) \) is\[ f'(x) = 3x^2 - p. \]

Step 2: Set the Derivative Equal to Zero

To find the critical points where the maxima and minima occur, set the derivative equal to zero:\[ 3x^2 - p = 0. \]

Step 3: Solve for Critical Points

Solve for \( x \) to find the critical points:\[ 3x^2 = p \]\[ x^2 = \frac{p}{3} \]\[ x = \pm \sqrt{\frac{p}{3}}. \]

Step 4: Determine Nature of Critical Points

To determine whether each critical point is a maxima or a minima, apply the second derivative test.Compute the second derivative:\[ f''(x) = 6x. \]

Step 5: Apply Second Derivative Test

Evaluate the second derivative at \( x = \sqrt{\frac{p}{3}} \):\[ f''(\sqrt{\frac{p}{3}}) = 6\sqrt{\frac{p}{3}} > 0, \] so \( x = \sqrt{\frac{p}{3}} \) is a minima.Evaluate the second derivative at \( x = -\sqrt{\frac{p}{3}} \):\[ f''(-\sqrt{\frac{p}{3}}) = 6(-\sqrt{\frac{p}{3}}) < 0, \] so \( x = -\sqrt{\frac{p}{3}} \) is a maxima.

Step 6: Final Conclusion

Based on the second derivative tests, we conclude that:- The cubic has a minima at \( x = \sqrt{\frac{p}{3}} \).- The cubic has a maxima at \( x = -\sqrt{\frac{p}{3}} \).Therefore, the correct option is (A).

Key Concepts

Real Roots

Cubic functions can have up to three real roots. A real root of a cubic function is a value of the variable that satisfies the equation, meaning the graph of the cubic function will cross the x-axis at this point.
For a cubic function given as \(f(x) = x^3 - px + q\), having three distinct real roots suggests the graph intersects the x-axis at three separate points.
These roots represent the inputs for which the value of the function is zero, and the presence of distinct roots often leads to interesting characteristics regarding turning points.

• Three distinct real roots indicate the cubic function's graph shifts the x-axis three times.
• The polynomial's nature ensures that it will have one real root and possibly two others which can be real or complex depending on the discriminant.
• Given \(p > 0\) and \(q > 0\), these conditions can help dictate the nature and positions of the roots.

Cubic Derivative

The first derivative of a cubic function helps in finding critical points which can later be analyzed to understand the behavior of the function.
For the cubic function \(f(x) = x^3 - px + q\), the derivative is given by \(f'(x) = 3x^2 - p\).
This derivative function is quadratic and doesn't directly tell us the positions of the maxima or minima, but indicates where the slope of the function equals zero, signaling critical points.

• Calculate the derivative to find how the slope of the cubic function changes.
• This step is crucial for identifying critical points where the function changes direction.
• The derivative also provides insight into the function's increasing or decreasing nature.

Second Derivative Test

The second derivative test is used to determine the concavity of a function at its critical points, and to decide whether each critical point is a maxima or minima.
For our given function, the second derivative is \(f''(x) = 6x\).
The sign of the second derivative at a given critical point determines the nature of that point.

• At \(x = \sqrt{\frac{p}{3}}\), \(f''(x) = 6\sqrt{\frac{p}{3}} > 0\), indicating the function is concave up, so it has a local minima.
• At \(x = -\sqrt{\frac{p}{3}}\), \(f''(x) = 6(-\sqrt{\frac{p}{3}}) < 0\), indicating the function is concave down, so it has a local maxima.
• The second derivative test effectively distinguishes between maxima and minima by analyzing concavity.

Critical Points

Critical points of a function occur where the first derivative is zero or undefined. These points are potential candidates for local maxima or minima.
In the case of our cubic function, \(3x^2 - p = 0\) leads to critical points at \(x = \pm \sqrt{\frac{p}{3}}\).
These are the points where the cubic function could switch from increasing to decreasing or vice versa.

• Recognizing critical points involves solving the derivative equation for zero.
• Critical points alone do not reveal their nature, which requires additional tests like the second derivative test.
• In practical applications, these points could represent equilibrium states or points of stability or instability.