Problem 51 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(a, b \in R\) be such that the fucntion \(f\) given by \(f(x)=\) \(\ln |x|+b x^{2}+a x, x \neq 0\) has extreme values at \(x=-1\) and \(x=2\). Statement \(1: f\) has local maximum at \(x=-1\) and at \(x=2\). Statement 2: \(a=\frac{1}{2}\) and \(b=\frac{-1}{4}\). (A) Statement- 1 is false, Statement- 2 is true. (B) Statement \(-1\) is true, statement- 2 is true, statement-2 is a correct explanation for Statement-1. (C) Statement \(-1\) is true, statement \(-2\) is true; statement-2 is not a correct explanation for Statement-1. (D) Statement- 1 is true, statement- 2 is false.

Step-by-Step Solution

Answer

(C) Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1.

Step 1: Formulate the Problem

We are given the function \( f(x) = \ln |x| + bx^2 + ax \). We need to find the values of \( a \) and \( b \) such that this function has extreme values (either minima or maxima) at \( x = -1 \) and \( x = 2 \).

Step 2: Differentiate the Function

To find the extrema, we take the derivative of \( f(x) \). Hence, \( f'(x) = \frac{d}{dx}(\ln |x|) + \frac{d}{dx}(bx^2) + \frac{d}{dx}(ax) \). This gives \( f'(x) = \frac{1}{x} + 2bx + a \).

Step 3: Set Derivative to Zero

Extreme values occur where \( f'(x) = 0 \). Set \( f'(x) = \frac{1}{x} + 2bx + a = 0 \) and find the solutions for \( x = -1 \) and \( x = 2 \).

Step 4: Solve for Parameters at \(x = -1\)

Substituting \( x = -1 \) into \( f'(x) = 0 \), we get \( \frac{1}{-1} + 2b(-1) + a = 0 \), simplifying to \( -1 - 2b + a = 0 \), so \( a = 1 + 2b \).

Step 5: Solve for Parameters at \(x = 2\)

Substituting \( x = 2 \) into \( f'(x) = 0 \), we get \( \frac{1}{2} + 2b(2) + a = 0 \), simplifying to \( \frac{1}{2} + 4b + a = 0 \).

Step 6: Solve the System of Equations

We now have the system of equations: 1) \( a = 1 + 2b \) and 2) \( \frac{1}{2} + 4b + a = 0 \). Substitute equation 1 into equation 2: \( \frac{1}{2} + 4b + (1 + 2b) = 0 \). This simplifies to \( \frac{3}{2} + 6b = 0 \).

Step 7: Calculate \(b\) and \(a\)

Solving \( \frac{3}{2} + 6b = 0 \), we find \( b = -\frac{1}{4} \). Substitute back into \( a = 1 + 2b \) to get \( a = 1 - \frac{1}{2} = \frac{1}{2} \).

Step 8: Check Second Derivative Test

To confirm the nature of the critical points, compute the second derivative \( f''(x) \). \( f''(x) = -\frac{1}{x^2} + 2b \). Check \( f''(-1) \) and \( f''(2) \) to determine if the points are maxima or minima. Since \( b = -\frac{1}{4} \), both calculations confirm maxima since both \( f''(-1) \) and \( f''(2) \) are negative.

Step 9: Conclude

Statement 1 is true because \( f(x) \) has local maxima at \( x = -1 \) and \( x = 2 \). Statement 2 is also true, as \( a = \frac{1}{2} \) and \( b = -\frac{1}{4} \). However, Statement 2 is not a correct mathematical explanation for Statement 1; while the parameters correspond to the extreme values, they do not explain why these points are specifically maxima.

Key Concepts

Extreme Values

Extreme values in calculus refer to the maximum or minimum points of a function. These points are critical because they reveal the highest or lowest values that the function can achieve within a specified interval. When we talk about finding extreme values, we are often interested in identifying and classifying these points.

To determine extreme values, the process usually involves differentiating the function and finding where the derivative equals zero, which gives us the critical points. Then, we use further tests such as the second derivative test to confirm if these critical points are maxima or minima. For example, if both the first and second derivatives of a function can be calculated and analyzed, we can accurately determine whether a function reaches a peak or a trough at a given point.

Understanding extreme values is crucial not only for solving mathematical problems but also because of its applications in various fields like economics and engineering, where optimization is key.

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative represents the rate of change of a function concerning one of its variables. In calculus, differentiation is a fundamental tool used to understand the behavior of functions.

The notation for the first derivative of a function \( f(x) \) is often written as \( f'(x) \) or \( \frac{df}{dx} \). It gives the slope of the tangent line to the graph of the function at any point \( x \). If the function represents a real-world scenario, the derivative can depict how one quantity changes in response to changes in another.

Differentiation is essential for finding critical points where the derivative equals zero. In the exercise above, we differentiated the function \( f(x) = \ln |x| + bx^2 + ax \) to obtain \( f'(x) = \frac{1}{x} + 2bx + a \). This step is critical as it sets up the conditions needed to determine the location of extreme values within the function.

Critical Points

Critical points of a function are where the derivative is either zero or undefined. These points are of interest because they can potentially be locations of extreme values, or notable changes in the graph of the function.

To find the critical points, we set the derivative equal to zero and solve for the variable. In the given problem, setting \( f'(x) = \frac{1}{x} + 2bx + a = 0 \) allowed us to solve for the critical points of \( x = -1 \) and \( x = 2 \). These represent the x-values where potential maximum or minimum values occur.

Once the critical points are located, further tests can determine whether these points are relative minima, maxima, or neither. For students, understanding critical points is essential for analyzing the behavior of functions and for further pursuits in calculus and beyond.

Second Derivative Test

The second derivative test helps us classify the critical points as local maxima, minima, or points of inflection. It involves taking the second derivative of the function and evaluating it at critical points.

The test operates on the principle that:

• If \( f''(x) > 0 \) at the critical point, the function has a local minimum there.
• If \( f''(x) < 0 \) at the critical point, the function has a local maximum there.
• If \( f''(x) = 0 \), the test is inconclusive, and other methods must be used.

In our exercise, we derived \( f''(x) = -\frac{1}{x^2} + 2b \), and found at both \( x = -1 \) and \( x = 2 \), \( f''(x) \) is negative, confirming local maxima at these points.

Understanding this test allows students to not just identify critical points, but effectively determine their nature, thus gaining deeper insight into the function’s shape and behavior.