Problem 47 - 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

The cubic has a minimum at \(\sqrt{\frac{p}{3}}\) and a maximum at \(-\sqrt{\frac{p}{3}}\).

Step 1: Differentiate the Cubic Function

Given the cubic function \( f(x) = x^3 - px + q \), find the derivative to determine the critical points. The derivative is \( f'(x) = 3x^2 - p \).

Step 2: Set the Derivative to Zero

Find the critical points by setting \( f'(x) = 0 \):\[ 3x^2 - p = 0 \]Solving for \( x \), we get:\[ 3x^2 = p \]\[ x^2 = \frac{p}{3} \]\[ x = \pm \sqrt{\frac{p}{3}} \]

Step 3: Determine the Nature of Critical Points

Using the second derivative test to determine if these critical points are maxima or minima.The second derivative is given by \( f''(x) = 6x \).- Evaluate \( f''(\sqrt{\frac{p}{3}}) = 6\sqrt{\frac{p}{3}} > 0 \), indicating \( x=\sqrt{\frac{p}{3}} \) is a minimum.- Evaluate \( f''(-\sqrt{\frac{p}{3}}) = -6\sqrt{\frac{p}{3}} < 0 \), indicating \( x=-\sqrt{\frac{p}{3}} \) is a maximum.

Key Concepts

Maxima and Minima

To understand the behavior of cubic functions at particular points, we use the concepts of maxima and minima. For a function, these points represent where the function reaches its highest or lowest values locally. In the context of the cubic function given in the exercise, maxima and minima are determined by examining the critical points. These points are where the slope of the function, given by its derivative, is zero.

Knowing the critical points is just the start. To identify whether these points are indeed maxima or minima, we apply tests such as the second derivative test. In this test, a positive value at a critical point suggests a minimum because the slope changes from negative to positive. Conversely, a negative value indicates a maximum as the slope transitions from positive to negative. Therefore, identifying these points is key in understanding the curve's shape and behavior.

Derivative

The derivative of a function reveals a lot about its dynamic features like slope and rate of change. When dealing with a cubic function such as our original exercise, the derivative helps us locate critical points.

The process begins by taking the derivative of the cubic function, which in this exercise is given by:

• Original function: \( f(x) = x^3 - px + q \)
• Derivative: \( f'(x) = 3x^2 - p \)

This function shows how the slope of the cubic plot changes. Setting the derivative to zero, \( f'(x) = 0 \), provides potential points where the slope might be a peak or trough. Solving this equation gives the critical points that are crucial for further analysis.

Understanding differentiation is not just about finding slopes; it's about understanding trends and predicting behaviors of various functions, which is essential in calculus and higher-level mathematics.

Critical Points

Critical points are essential to understanding the behavior of functions. For the cubic function in question, the critical points are found by setting the derivative equal to zero. These critical points signify potential maxima or minima of the function, depending on the concavity indicated by the second derivative.

Here’s how you can find them:

• First, differentiate the function to get the derivative: \( f'(x) = 3x^2 - p \).
• Set this derivative equal to zero to find the critical points: \( 3x^2 - p = 0 \).
• Solve the equation to find \( x = \pm \sqrt{\frac{p}{3}} \).

These values are potential maxima or minima. The importance of critical points lies in how they help us map and interpret the function's graph. By examining them using graphical or numerical methods, we can understand regions of increase or decrease within the function and predict behavior in analytical work.