Problem 120 - 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

Option (A) is correct: the function increases in \((-\infty, 0) \cup (1,2)\).

Step 1: Understand the Problem

We are tasked with determining in which intervals the function \( f(x) = \frac{|x-1|}{x^2} \) is increasing or decreasing. Analyzing functions for increasing or decreasing behavior typically involves finding the derivative and checking its sign in different intervals.

Step 2: Find the Derivative

Calculate the derivative of \( f(x) \). First, express \( |x - 1| \) as two cases: \( x - 1 \) for \( x \geq 1 \) and \( 1 - x \) for \( x < 1 \).For \( x \geq 1 \), \( f(x) = \frac{x - 1}{x^2} \).For \( x < 1 \), \( f(x) = \frac{1 - x}{x^2} \).Now, differentiate for each case.

Step 3: Differentiate for \(x \geq 1\)

For \( x \geq 1 \), \( f(x) = \frac{x - 1}{x^2} \).Using the quotient rule: \( f'(x) = \frac{(x^2)(1) - (x - 1)(2x)}{(x^2)^2} \).This simplifies to \( f'(x) = \frac{x^2 - 2x^2 + 2x}{x^4} = \frac{2x - x^2}{x^4} \).Thus, \( f'(x) = \frac{x(2 - x)}{x^4} \).

Step 4: Differentiate for \(x < 1\)

For \( x < 1 \), \( f(x) = \frac{1 - x}{x^2} \).Using the quotient rule: \( f'(x) = \frac{-(x^2) - (1 - x)(2x)}{(x^2)^2} \).This simplifies to \( f'(x) = \frac{-x^2 - 2x + 2x^2}{x^4} = \frac{x^2 - 2x}{x^4} \).Thus, \( f'(x) = \frac{x(x - 2)}{x^4} \).

Step 5: Analyze the Derivative for Increasing or Decreasing Intervals

The function is increasing where \( f'(x) > 0 \) and decreasing where \( f'(x) < 0 \). Analyze the signs for different intervals:- For \( x > 1 \): \( f'(x) = \frac{x(2 - x)}{x^4} \). This is positive when \( 2 - x > 0 \) or \( x < 2 \), making \( 1 < x < 2 \) an interval of increase.- For \( x < 1 \): \( f'(x) = \frac{x(x - 2)}{x^4} \). This is negative for \( 0 < x < 1 \), hence the function is decreasing.Additionally, for \( x < 0 \), \( f'(x) > 0 \) implies an increasing interval because the numerator and denominator will both be negative.

Step 6: Evaluate Options

Compare intervals of increase and decrease against the provided options:- Option (A) suggests increasing in \((-\infty, 0) \cup (1,2)\).- Option (B) suggests increasing in \((0,1) \cup (2, \infty)\).- Option (C) suggests decreasing in \((0,1) \cup (2, \infty)\).- Option (D) suggests decreasing in \((-\infty, \infty) \cup (1,2)\).Our analysis shows the function is increasing in \((-\infty, 0) \cup (1,2)\) and decreasing in other intervals.

Key Concepts

Derivative

In calculus, the derivative of a function describes how the function's output value changes as its input values change. It's like the slope of the function at any given point, telling us if the function is going up, down, or staying constant. For our given function, \( f(x) = \frac{|x-1|}{x^2} \), finding the derivative helps us figure out in which intervals the function is increasing or decreasing.
The process starts by understanding how to differentiate \( |x - 1| \), which must be broken into two cases due to the absolute value. For values \( x \geq 1 \), the expression becomes \( x - 1 \), while for \( x < 1 \), it's \( 1 - x \). Next, we use the differentiation rules to derive each piece separately, employing the Quotient Rule since the function is a ratio of polynomials.
By calculating the derivatives for both cases, we effectively determine how the function behaves over different parts of the domain. This derivative tells us where changes occur, enabling us to accurately predict the function’s behavior, such as when it will switch from increasing to decreasing.

Increasing and Decreasing Functions

To determine where a function is increasing or decreasing, we inspect the sign of the derivative. A positive derivative indicates the function is increasing, while a negative derivative points to a decreasing function.
In the exercise, after deriving \( f'(x) \) for different cases, the signs of these derivatives are evaluated over specific intervals.
For example:

• For \( x > 1 \), the derivative is \( f'(x) = \frac{x(2-x)}{x^4} \). It tells us the function increases in the interval \( 1 < x < 2 \) because the derivative is positive.
• For \( x < 1 \), the derivative changes to \( f'(x) = \frac{x(x-2)}{x^4} \), decreasing in the interval \( 0 < x < 1 \) since the derivative is negative.

Additionally, when \( x < 0 \), both the numerator and denominator of \( f'(x) \) are negative, ensuring the function increases there. Understanding these signs is vital, as it allows us to sketch the function’s graph accurately, predicting how its slope changes over different parts of its domain.

Quotient Rule

The Quotient Rule is essential when you're dealing with the division of two functions, like in our scenario with \( f(x) = \frac{|x-1|}{x^2} \). It is expressed as \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \), where \( u \) and \( v \) are both differentiable functions.
This rule helps us find the derivative of complicated fractions like our case, by systematically applying differentiation rules to each component function. Start by identifying the numerator (\( u \)) and the denominator (\( v \)).

• For \( x \geq 1 \), where \( u = x - 1 \) and \( v = x^2 \), the derivative \( f'(x) \) is calculated using the Quotient Rule to be \( \frac{x(2-x)}{x^4} \).
• For \( x < 1 \), with \( u = 1 - x \) and \( v = x^2 \), similarly, \( f'(x) \) becomes \( \frac{x(x-2)}{x^4} \).

Mastering the Quotient Rule offers a clear pathway through complex functions by breaking them down into digestible parts. This enables us to deal efficiently with any potential changes in the function’s rate over its domain.