Problem 129 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The function \(f(x)=\frac{|x-1|}{x^{2}}\) (A) increases in \((-\infty, 0) \cup(1,2)\) (B) increases in \((0,1) \cup(2, \infty)\) (C) decreases in \((0,1) \cup(2, \infty)\) (D) decreases in \((-\infty, \infty) \cup(1,2)\)

Step-by-Step Solution

Answer

The function decreases in \((0,1) \cup (2,\infty)\), so the correct answer is (C).

Step 1: Simplify the Function

The function given is \( f(x) = \frac{|x-1|}{x^2} \). To analyze its behavior, we need to consider the absoluting value expression \(|x-1|\). This will split our function into two cases: \( x \geq 1 \) and \( x < 1 \). For \( x \geq 1 \), \(|x-1| = x - 1\), so \( f(x) = \frac{x-1}{x^2} \). For \( x < 1 \), \(|x-1| = -(x-1) = 1-x\), so \( f(x) = \frac{1-x}{x^2} \).

Step 2: Find Critical Points

To identify the function's increasing or decreasing regions, calculate its derivative \( f'(x) \). For \( x \geq 1 \), \( f(x) = \frac{x-1}{x^2} \). Use the quotient rule: \( f'(x) = \frac{(1)(x^2) - (x-1)(2x)}{(x^2)^2} = \frac{x^2 - 2x^2 + 2x}{x^4} = \frac{-x^2 + 2x}{x^4} = \frac{x(2-x)}{x^4} \). For \( x < 1 \), \( f(x) = \frac{1-x}{x^2} \). Again using the quotient rule: \( f'(x) = \frac{(-1)(x^2) - (1-x)(2x)}{(x^2)^2} = \frac{-x^2 + 2x - x^2}{x^4} = \frac{2x-x^2}{x^4} = \frac{x(2-x)}{x^4} \).

Step 3: Analyze Increasing/Decreasing Intervals

For both \( x < 1 \) and \( x \geq 1 \), the derivative \( f'(x) = \frac{x(2-x)}{x^4} \) is the same. Determine where the derivative is positive (increasing) or negative (decreasing). Solve \( x(2-x) > 0 \): it yields the intervals \( x \in (0, 1) \cup (2, \infty) \) for when \( f(x) \) is increasing. Solve \( x(2-x) < 0 \): it yields \( x \in (-\infty, 0) \cup (1, 2) \) for when \( f(x) \) is decreasing.

Key Concepts

Function Analysis

Function analysis is a crucial part of understanding calculus, as it involves breaking down a function to comprehend its behavior. This is particularly important when dealing with functions that include absolute values or require domain considerations. In the given problem, the function is defined as \( f(x) = \frac{|x-1|}{x^2} \). The presence of the absolute value necessitates a split of the analysis into different cases based on the value of \( x \).

• For \( x \geq 1 \), the absolute value resolves to \( x - 1 \), which simplifies the function to \( \frac{x-1}{x^2} \).
• For \( x < 1 \), the absolute value becomes \( 1 - x \), simplifying the function to \( \frac{1-x}{x^2} \).

Understanding how to handle and simplify such expressions is vital to correctly analyzing the function's behavior across its entire domain. Function analysis ultimately aids in determining intervals where the function is increasing or decreasing.

Increasing and Decreasing Functions

A function's growth can be identified by determining where it is increasing or decreasing. An increasing function is one where, as \( x \) gets larger, \( f(x) \) gets larger too. Conversely, a decreasing function is one where \( f(x) \) gets smaller as \( x \) increases.

To find these intervals for the function \( f(x) = \frac{|x-1|}{x^2} \), we first need its derivative. The derivative helps indicate where the function changes from increasing to decreasing, or vice versa. By solving the inequality \( x(2-x) \) relative to zero:

• \( x(2-x) > 0 \), shows the intervals where \( f(x) \) is increasing, identified as \((0, 1) \cup (2, \infty)\).
• \( x(2-x) < 0 \), reveals where \( f(x) \) is decreasing, given by \((-\infty, 0) \cup (1, 2)\).

Understanding these intervals is key to sketching graphs and studying the overall behavior of the function.

Derivative

In calculus, the derivative is an essential tool used to measure the rate of change of a function. It's akin to finding the slope of a curve at a specific point, providing insight into the function's behavior. For the provided function \( f(x) = \frac{|x-1|}{x^2} \), the derivative \( f'(x) \) plays a crucial role in determining the function's increasing and decreasing intervals.

To derive \( f'(x) \), the quotient rule is applied. The rule is utilized when taking the derivative of a function that is the ratio of two other functions. For our function, regardless of whether \( x \) is less than 1 or greater or equal to 1, the derivative simplifies to \( f'(x) = \frac{x(2-x)}{x^4} \).

The derivative tells us where the function may have turning points or where it changes direction. By identifying where \( f'(x) > 0 \) or \( f'(x) < 0 \), we determine the regions where \( f(x) \) is either increasing or decreasing.