Problem 174 - A Complete Resource Book in Mathematics for JEE Main 2017

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

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

Step 1: Identify the Function

The given cubic function is \( f(x) = x^3 - px + q \). We need to analyze this function to determine where it achieves its minima and maxima.

Step 2: Find the Derivative

To find the critical points, we need the derivative of the function. The derivative is \( f'(x) = 3x^2 - p \).

Step 3: Set Derivative to Zero

For minima and maxima, set the derivative equal to zero: \( 3x^2 - p = 0 \). Solving this gives \( x^2 = \frac{p}{3} \), or \( x = \sqrt{\frac{p}{3}} \) and \( x = -\sqrt{\frac{p}{3}} \).

Step 4: Determine Minima and Maxima Using Second Derivative

Calculate the second derivative \( f''(x) = 6x \). Substitute the critical points:1. \( f''(\sqrt{\frac{p}{3}}) = 6\sqrt{\frac{p}{3}} \), which is positive since \( p > 0 \). Thus, \( \sqrt{\frac{p}{3}} \) is a point of minima.2. \( f''(-\sqrt{\frac{p}{3}}) = 6(-\sqrt{\frac{p}{3}}) \), which is negative. Thus, \( -\sqrt{\frac{p}{3}} \) is a point of maxima.

Step 5: Choose the Correct Option

Based on the analysis, the function has a minima at \( \sqrt{\frac{p}{3}} \) and a maxima at \( -\sqrt{\frac{p}{3}} \). This corresponds to option (A).

Key Concepts

Real Roots

Cubic equations of the form \(x^3 - px + q\) are fascinating because they can have three distinct real roots. Understanding the concept of real roots is essential because it helps us determine where a function might cross the x-axis. For cubic functions like this, having three real roots implies that the curve crosses the x-axis at three distinct points. This behavior typically indicates a more complex shape with turns and bounces across the axis. The real roots are crucial because they tell us not just where the equation equals zero, but also provide insights into the intervals where the function might be increasing or decreasing. It adds layers of depth in analyzing the function and predicting its behavior.

Critical Points

To understand critical points, remember they are where a function's derivative equals zero. In other words, these are places on the graph where the slope is flat, meaning the function might be changing direction. For the cubic equation \(f(x) = x^3 - px + q\), we find the derivative, \(f'(x) = 3x^2 - p\). By setting this derivative to zero, we can solve for \(x\), which gives us the critical points. In this exercise, the critical points were identified as \(x = \sqrt{\frac{p}{3}}\) and \(x = -\sqrt{\frac{p}{3}}\). These points are essential because they help us explore potential maxima and minima, the turning points of our cubic function.

Derivative Analysis

Derivative analysis is a tool for finding where a function increases or decreases and locating critical points. It involves taking the first and second derivatives of the function, each offering crucial insights. First, we took the derivative \(f'(x) = 3x^2 - p\) which revealed where the slope of the function is zero by setting it to zero. This step helps find potential maxima or minima locations. For each solution \(x = \sqrt{\frac{p}{3}}\) and \(x = -\sqrt{\frac{p}{3}}\), we took the second derivative \(f''(x) = 6x\) to test the nature of these critical points. Positive second derivative values confirm a minima, while negative ones indicate maxima. Thus, derivative analysis shows not only where the critical points are but also their roles in the function's shape.

Minima and Maxima

Minima and maxima in a function are the lowest and highest points, respectively, on its graph. Identifying these points is important not just for understanding the function's shape, but also for practical applications like optimization. For the function \(f(x) = x^3 - px + q\), to determine which critical points are minima or maxima, we used the second derivative test. At \(x = \sqrt{\frac{p}{3}}\), the second derivative \(f''(x) = 6\sqrt{\frac{p}{3}}\) is positive, meaning the slope is increasing, indicating a minimum point. At \(x = -\sqrt{\frac{p}{3}}\), the second derivative \(f''(x) = 6(-\sqrt{\frac{p}{3}})\) is negative, showing a decreasing slope, which identifies it as a maximum point. This analysis confirms option (A) where the function has a minima at \(\sqrt{\frac{p}{3}}\) and a maxima at \(-\sqrt{\frac{p}{3}}\).