Problem 174 - 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) is correct.

Step 1: Understand the Derivative

The function we have is a cubic of the form \( f(x) = x^3 - px + q \). To find critical points, we need to first differentiate this function. The derivative is \( f'(x) = 3x^2 - p \).

Step 2: Solve for Critical Points

To find the critical points, we set the derivative equal to zero: \( 3x^2 - p = 0 \). Solving for \( x \) gives \( x^2 = \frac{p}{3} \), which implies \( x = \pm \sqrt{\frac{p}{3}} \).

Step 3: Determine Nature of Critical Points

To determine whether these critical points are minima or maxima, we need the second derivative: \( f''(x) = 6x \). Evaluate \( f''(x) \) at each critical point. For \( x = \sqrt{\frac{p}{3}} \), \( f''(x) = 6\sqrt{\frac{p}{3}} > 0 \), indicating a minimum. For \( x = -\sqrt{\frac{p}{3}} \), \( f''(x) = 6(-\sqrt{\frac{p}{3}}) < 0 \), indicating a maximum.

Key Concepts

Critical Points

Critical points in a function are crucial for understanding where the function might change direction, such as turning from increasing to decreasing, or vice versa. In the context of a cubic function like \( f(x) = x^3 - px + q \), critical points occur where the derivative of the function equals zero. Essentially, these points are candidates for local maxima, minima, or saddle points.
For the cubic function given, the derivative is \( f'(x) = 3x^2 - p \). Critical points are determined by solving the equation \( 3x^2 - p = 0 \), leading to the solutions \( x = \pm \sqrt{\frac{p}{3}} \). These solutions indicate the x-values where critical points are located. Thus, understanding how to find these points is the first step in analyzing the function's behavior.

Derivative Analysis

Analyzing the derivative of a function helps us understand the behavior of the original function. For cubic functions like \( f(x) = x^3 - px + q \), the first derivative, \( f'(x) = 3x^2 - p \), provides valuable information about the slope of the tangent to the curve at any given point. When \( f'(x) = 0 \), it indicates that the tangent is horizontal, and the function has a potential maximum or minimum at that point.
The derivative \( f'(x) \) also provides information about where the function is increasing or decreasing:

• If \( f'(x) > 0 \), the function is increasing.
• If \( f'(x) < 0 \), the function is decreasing.

This analysis becomes vital when determining where the cubic function reaches its critical points and how it behaves around those points.

Second Derivative Test

Once critical points are identified, the second derivative test helps to classify the nature of these points. In simple terms, the second derivative examines the concavity of the function at the critical points.
For the cubic function \( f(x) = x^3 - px + q \), the second derivative is \( f''(x) = 6x \). By evaluating \( f''(x) \) at the critical points \( x = \pm \sqrt{\frac{p}{3}} \), we can determine whether these points correspond to local minima or maxima.

• If \( f''(x) > 0 \), the function is concave up at that point, indicating a local minimum.
• If \( f''(x) < 0 \), the function is concave down, indicating a local maximum.

For \( x = \sqrt{\frac{p}{3}} \), \( f''(x) = 6\sqrt{\frac{p}{3}} \) is positive, indicating a local minimum. At \( x = -\sqrt{\frac{p}{3}} \), \( f''(x) = 6(-\sqrt{\frac{p}{3}}) \) is negative, indicating a local maximum. This powerful test allows us to precisely classify critical points and understand their roles in the overall shape of the function.