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

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

Step 1: Find critical points

First, to find where the function \( f(x) = \frac{|x-1|}{x^2} \) changes its behavior, consider the derivative. To differentiate, first manage the absolute value function by defining it piecewise:- For \( x \geq 1 \), \( |x-1| = x-1 \)- For \( x < 1 \), \( |x-1| = -(x-1) = 1-x \)Calculate the derivative for both cases separately.

Step 2: Differentiate for \( x \geq 1 \)

For \( x \geq 1 \), the function becomes \( f(x) = \frac{x-1}{x^2} \). Use the quotient rule to differentiate:\[ f'(x) = \frac{(1 \cdot x^2 - (x-1) \cdot 2x)}{x^4} = \frac{x^2 - 2x(x-1)}{x^4} = \frac{x^2 - 2x^2 + 2x}{x^4} \]Simplify to:\[ f'(x) = \frac{-x^2 + 2x}{x^4} = \frac{2x - x^2}{x^4} \]

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

For \( x < 1 \), the function becomes \( f(x) = \frac{1-x}{x^2} \). Again, use the quotient rule:\[ f'(x) = \frac{0 \cdot x^2 - (1-x) \cdot 2x}{x^4} = \frac{-2x(1-x)}{x^4} = \frac{-2x + 2x^2}{x^4} \]Simplify to:\[ f'(x) = \frac{2x^2 - 2x}{x^4} \]

Step 4: Analyze derivative for critical points

Set the derivatives to zero to find critical points:1. For \( x \geq 1 \), solve \( 2x - x^2 = 0 \): - Factor as \( x(2-x) = 0 \) - Critical points are \( x = 0 \) and \( x = 2 \) (consider domain for validity)2. For \( x < 1 \), solve \( 2x^2 - 2x = 0 \): - Factor as \( 2x(x-1) = 0 \) - Critical points are \( x = 0 \) and \( x = 1 \) (consider domain for validity)

Step 5: Test intervals and Monotonicity

Now, test intervals between the critical points and the given intervals:- **For \( x \geq 1 \):** - Test point between 1 and 2 (say 1.5): If \( f'(1.5) > 0 \), the function increases, else decreases. - After 2, test a point (say 3): If \( f'(3) < 0 \), the function decreases, else increases.- **For \( x < 1 \):** - Test point between 0 and 1 (say 0.5): If \( f'(0.5) < 0 \), the function decreases, else increases. Through testing and evaluating these test points within the derivatives, determine where the function increases or decreases.

Key Concepts

Critical Points

In calculus, the concept of critical points is crucial in understanding where a function changes its behavior, such as increasing or decreasing. A critical point occurs where the derivative of a function is either zero or undefined. By finding these critical points, we can identify the potential places where the function might have a local maximum or minimum, or where the behavior of the function changes.
Consider the function \( f(x) = \frac{|x-1|}{x^2} \). To find the critical points, we first manage the absolute value by defining the function piecewise. This approach helps us to handle the, often complex, absolute values efficiently:

• For \( x \geq 1 \), the function simplifies to \( \frac{x-1}{x^2} \).
• For \( x < 1 \), it becomes \( \frac{1-x}{x^2} \).

To find critical points, we compute the derivative for these two cases. We then set these derivatives equal to zero. Once we solve these equations, we observe where critical points occur. Only by factoring and solving such equations can we pinpoint the specific \( x \) values leading to a critical point. Finally, take into consideration the domain restrictions in these solutions before concluding the valid critical points.

Derivative

The derivative of a function is a powerful tool in calculus. It provides crucial information about the function's rate of change, indicating how the function behaves as \( x \) changes. The derivative can tell us if a function is increasing, decreasing, or constant at specific intervals.
For our function \( f(x) = \frac{|x-1|}{x^2} \), we find the derivative by applying the quotient rule. Given its piecewise nature, we derive separately for the cases \( x \geq 1 \) and \( x < 1 \):

• For \( x \geq 1 \), the derivative is \( f'(x) = \frac{2x - x^2}{x^4} \).
• For \( x < 1 \), it becomes \( f'(x) = \frac{2x^2 - 2x}{x^4} \).

These derivatives inform us about the function's slopes in those ranges. Simplifying these derivatives, we can observe how they might reach zero or change signs. Using these calculations, we find where the function's critical points lie and analyze the intervals carefully.
Understanding how to apply the quotient rule in such contexts opens up paths to unravel complex behaviors of functions like \( f(x) \).

Function Behavior

Analyzing the behavior of a function involves understanding where it increases or decreases. This is determined by scrutinizing the signs of the derivative in specific intervals. We use critical points as reference points to separate these intervals and then test the derivative's sign within each one.
For the given function \( f(x) = \frac{|x-1|}{x^2} \), after identifying and computing the derivative, we analyze the behavior as follows:

• For \( x \geq 1 \), test between 1 and 2, like at \( x = 1.5 \). If \( f'(1.5) > 0 \), the function increases; if \( f'(1.5) < 0 \), it decreases. Similarly, test after 2, say at \( x = 3 \).
• For \( x < 1 \), test between 0 and 1 at a point like \( x = 0.5 \). Again, a positive derivative indicates increasing behavior, while a negative indicates decreasing.

This testing helps us sketch a clearer picture of the function's behavior across its domain. When the derivative is positive in an interval, the function is increasing; when negative, it decreases. Thus, evaluating these conditions around critical points allows us to map out the full behavior of \( f(x) \). This analysis empowers you to predict how the function acts across its domain.