Problem 138 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(x=-1\) and \(x=2\) are extreme points of \(f(x)=\alpha \log |x|+\beta x^{2}+x\), then [2013] (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

Option (C): \( \alpha = 2, \beta = -\frac{1}{2} \).

Step 1: Find Derivative

We start by finding the derivative of the function \( f(x) = \alpha \log |x| + \beta x^2 + x \). The derivative is given by \( f'(x) = \frac{\alpha}{x} + 2\beta x + 1 \).

Step 2: Set Derivative to Zero at Extreme Points

To find the extreme points, set \( f'(x) = 0 \) at \( x = -1 \) and \( x = 2 \).

Step 3: Solve Equations for \( x=-1 \)

Plug \( x = -1 \) into the derivative: \( f'(-1) = \frac{\alpha}{-1} + 2\beta(-1) + 1 = 0 \). This simplifies to \( -\alpha - 2\beta + 1 = 0 \), or \( \alpha + 2\beta = 1 \).

Step 4: Solve Equations for \( x=2 \)

Plug \( x = 2 \) into the derivative: \( f'(2) = \frac{\alpha}{2} + 2\beta(2) + 1 = 0 \). This simplifies to \( \frac{\alpha}{2} + 4\beta + 1 = 0 \), or \( \alpha + 8\beta = -2 \).

Step 5: Solve System of Equations

We have two equations: \( \alpha + 2\beta = 1 \) and \( \alpha + 8\beta = -2 \). Subtract the first equation from the second to find \( 6\beta = -3 \), giving \( \beta = -\frac{1}{2} \). Substitute \( \beta = -\frac{1}{2} \) back into the first equation to find \( \alpha + 2(-\frac{1}{2}) = 1 \), which simplifies to \( \alpha - 1 = 1 \) or \( \alpha = 2 \).

Step 6: Verify Solutions with Options

Our solutions \( \alpha = 2 \) and \( \beta = -\frac{1}{2} \) match option (C): \( \alpha = 2, \beta = -\frac{1}{2} \).

Key Concepts

Derivative

The derivative is a powerful tool in calculus that allows us to measure how a function changes as its input changes. Imagine you're driving a car on a road; the derivative tells you the speed at which you're passing each car. It gives us the rate of change for a function at any given point. For the function \( f(x) = \alpha \log |x| + \beta x^2 + x \), the derivative is found by applying the rules of differentiation to each term:

• The derivative of \( \log |x| \) is \( \frac{1}{x} \).
• The derivative of \( x^2 \) is \( 2x \).
• The derivative of \( x \) is simply \( 1 \).

Combine these, using the constants \( \alpha \) and \( \beta \), and you get \( f'(x) = \frac{\alpha}{x} + 2\beta x + 1 \). Newcomers to calculus might find derivatives tricky at first. But with practice, recognizing patterns and applying basic rules of differentiation becomes second nature.
When we set the derivative of a function equal to zero, \( f'(x) = 0 \), we are finding points where the function does not change, i.e., the function's extreme points like peaks (maximums) or troughs (minimums). For our problem, this is done at specific points \( x = -1 \) and \( x = 2 \).

Logarithmic Function

Logarithmic functions, such as \( \log |x| \), are vital in various scientific fields because they can model growth processes, the intensity of sound, and even the strength of earthquakes. A logarithmic function takes the form \( \log_b(x) \), meaning how many times we need to multiply the base \( b \) to get \( x \). In this context, \( \log |x| \) implies the natural (base \( e \)) logarithm of the absolute value of \( x \), written as \( \ln |x| \).
The absolute value ensures the argument of the logarithm remains positive, which is essential since logarithms are undefined for zero and negative values. This function's key property in the exercise is its derivative, \( \frac{1}{x} \), which measures how steeply the log function rises or falls as \( x \) changes.
In practical terms, knowing how to handle the derivative of \( \log |x| \) is useful for understanding problems where input changes rapidly and has a logarithmic growth or decay pattern. Consider how hackers scale up their level in a strategy game, the more they hack, incrementally, smaller and smaller their gains may grow, mirroring a logarithmic scale.

Solving Equations

Solving equations is the process of finding values for variables that make the equations true. In calculus, this often involves setting derivatives equal to zero to find extreme points. Think of it as solving a puzzle - finding the value of the unknown that fits the equation perfectly.
In our exercise, once you have the derivative \( f'(x) = \frac{\alpha}{x} + 2\beta x + 1 \), you set it to zero at the extreme points \( x = -1 \) and \( x = 2 \). This gives you two separate equations:

• For \( x = -1 \), the equation is \( -\alpha - 2\beta + 1 = 0 \).
• For \( x = 2 \), the equation is \( \frac{\alpha}{2} + 4\beta + 1 = 0 \).

These equations form a system of equations, which can be solved simultaneously. By manipulating these equations, you're able to find the values of \( \alpha \) and \( \beta \), which are \( \alpha = 2 \) and \( \beta = -\frac{1}{2} \) in this case.
Solving equations might seem daunting, but with systematic application of algebraic techniques, it becomes manageable. Understanding the underlying principles will enable you to tackle more complex problems with ease.