Problem 138 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(x=-1\) and \(x=2\) are extreme points of \(f(x)=\alpha \log |x|+\beta x^{2}+x\), then (A) \(\alpha=-6, \beta=\frac{1}{2}\) (B) \(\alpha=-6, \beta=-\frac{1}{2}\) (C) \(\alpha=2, \beta=-\frac{1}{2}\) (D) \(\alpha=2, \beta=\frac{1}{2}\)

Step-by-Step Solution

Answer

The correct option is (C) \( \alpha=2, \beta=-\frac{1}{2} \).

Step 1: Find the first derivative of the function

The given function is \( f(x) = \alpha \log |x| + \beta x^2 + x \). To find critical points, we need the first derivative, \( f'(x) \). We apply differentiation: \( f'(x) = \frac{\alpha}{x} + 2\beta x + 1 \).

Step 2: Set the first derivative equal to zero at extreme points

Because \(x = -1\) and \(x = 2\) are given as extreme points, they are critical points where \( f'(x) = 0 \). Thus, we will evaluate and set the derivative equal to zero at each point: - At \(x = -1\): \( f'(-1) = \frac{\alpha}{-1} + 2\beta(-1) + 1 = -\alpha - 2\beta + 1 = 0 \).- At \(x = 2\): \( f'(2) = \frac{\alpha}{2} + 2\beta(2) + 1 = \frac{\alpha}{2} + 4\beta + 1 = 0 \).

Step 3: Solve the system of equations

Now solve the system:1. \( -\alpha - 2\beta + 1 = 0 \)2. \( \frac{\alpha}{2} + 4\beta + 1 = 0 \)Let's solve this system: From equation 1:\( -\alpha = 2\beta - 1 \) implies \( \alpha = 1 - 2\beta \).Substitute \( \alpha = 1 - 2\beta \) into equation 2:\[ \frac{1 - 2\beta}{2} + 4\beta + 1 = 0 \]Multiply through by 2 to eliminate the fraction:\[ 1 - 2\beta + 8\beta + 2 = 0 \]Combine terms: \( 6\beta + 3 = 0 \).Solve for \( \beta \): \( \beta = -\frac{1}{2} \).Substitute \( \beta \) back into \( \alpha = 1 - 2\beta \) to get \( \alpha = 1 - 2(-\frac{1}{2}) = 1 + 1 = 2 \).

Step 4: Confirm solution with choices

We found that \( \alpha = 2 \) and \( \beta = -\frac{1}{2} \). Matching these to the given options, we see that option (C) \( \alpha=2, \beta=-\frac{1}{2} \) is correct.

Key Concepts

Critical Points

Critical points in calculus are where the first derivative of a function is zero or undefined. This makes them key candidates for locating extreme points, which are important in determining maximum and minimum values of functions. For the function \( f(x) = \alpha \log |x| + \beta x^2 + x \), critical points are found by solving \( f'(x) = 0 \).

Critical points coincide with potential extreme points, like peaks or troughs on a curve. In the exercise, \( x = -1 \) and \( x = 2 \) were given as extreme points, indicating \( f'(x) \) equaled zero at these values. By substituting these x-values into the derivative, we formulated equations that helped solve for the unknown variables \( \alpha \) and \( \beta \).

• Recognize that if a function has critical points at given x-values, the derivative should evaluate to zero at these points.
• Knowing how to identify critical points is integral for optimization problems in calculus.

Differentiation

Differentiation is a mathematical process used to find the derivative of a function. The derivative measures how the function's output value changes as its input value changes. In the provided solution, the derivative of the function \( f(x) = \alpha \log |x| + \beta x^2 + x \) was obtained as \( f'(x) = \frac{\alpha}{x} + 2\beta x + 1 \).

Differentiation uses rules such as the product rule, quotient rule, and chain rule. In this exercise, the chain rule is implicitly used for the absolute logarithm term. Differentiation provided the equations needed to solve for the variables \( \alpha \) and \( \beta \).

• The derivative found through differentiation is central to predicting where a function increases or decreases.
• Using differentiation allows one to construct equations that can be solved for unknown variables in the function.

System of Equations

A system of equations is a set of simultaneous equations that are solved together due to mutual dependencies. In our exercise, two equations arose out of setting the derivative \( f'(x) \) equal to zero at the critical points \( x = -1 \) and \( x = 2 \).

The system was:\(-\alpha - 2\beta + 1 = 0\) and \(\frac{\alpha}{2} + 4\beta + 1 = 0\). Solving it involved substitution and simplification: expressing one variable in terms of another using one equation and then substituting back.

Systems of equations can be solved by different methods, including substitution, elimination, or using matrices. For this problem, substitution was chosen:

• Identifying dependent equations allows for solving the system step-by-step.
• Succeeding in these steps helps to determine complex variable values systematically.

Finally, finding that \( \alpha = 2 \) and \( \beta = -\frac{1}{2} \) showed the method's effectiveness in confirming the correct option.