Problem 80 - A Complete Resource Book in Mathematics for JEE Main

Question

The solution set of \(\left|\frac{x+1}{x}\right|+|x+1|=\frac{(x+1)^{2}}{|x|}\) is (A) \(\\{x \mid x \geq 0\\}\) (B) \(\\{x \mid x>0\\} \cup\\{-1\\}\) (C) \(\\{-1,1\\}\) (D) \(\\{x \mid x \geq 1\) or \(x \leq-1\\}\)

Step-by-Step Solution

Answer

The solution set is (D) \( \{x | x \geq 1 \text{ or } x \leq -1\} \).

Step 1: Simplify the Expression

We start by examining the equation. The expression \( \left|\frac{x+1}{x}\right| + |x+1| \) can be rewritten as two separate cases, depending on the sign of \( x \):1. If \( x > 0 \), both \( \frac{x+1}{x} \) and \( x+1 \) are positive, so the expression becomes \( \frac{x+1}{x} + (x+1) = \frac{(x+1)^2}{x} \).2. If \( x < 0 \), the expression becomes \( -\frac{x+1}{x} + |x+1| = \frac{(x+1)^2}{|x|} = \frac{(x+1)^2}{-x} \).3. If \( x = 0 \), the original equation is undefined due to division by zero.

Step 2: Analyze the Case for x > 0

Consider the scenario where \( x > 0 \):The equation simplifies to \( \frac{x+1}{x} + x+1 = \frac{(x+1)^2}{x} \).Both sides are equal, confirming that \( x > 0 \) satisfies the equation. However, we want to check the boundary if \( x = 1 \), plugging into the original form matches both sides as equal. So \( x \geq 1 \) works.

Step 3: Analyze the Case for x < 0

Consider the scenario where \( x < 0 \):The equation becomes \( -\frac{x+1}{x} + |x+1| = \frac{(x+1)^2}{-x} \). Both expressions expand to the same result only when \( x = -1 \). Plugging \( x = -1 \) verifies that it satisfies the original equation, making \( x = -1 \) a part of the solution set.

Step 4: Conclusion on Solution Set

From both cases:- For \( x > 0 \), \( x \ge 1 \) satisfies the equation.- For \( x < 0 \), only \( x = -1 \) satisfies the equation.Thus the solution set is \( \{ x | x \geq 1 \text{ or } x = -1 \} \). This matches option (D).

Key Concepts

Absolute Value

Absolute value is like measuring how far a number is from zero on a number line. It doesn't worry about which direction—left or right—the number is, it only counts the distance. This means

• The absolute value of positive numbers is the number itself.
• The absolute value of negative numbers is also positive.
• The absolute value of zero is zero.

To compute the absolute value of a number like \(x\), we use the notation \(|x|\). For instance, \(|-3| = 3\) and \(|5| = 5\). In our exercise, the goal was to manage expressions like \(|\frac{x+1}{x}|\) and \(|x+1|\). Both were transformed depending on whether \(x\) was positive, negative, or zero. Keep in mind special cases, like division by zero, results in the expression being undefined.

Inequalities

In mathematics, inequalities are like scales measuring different amounts. They show the relation between two expressions that are not always equal. We use signs like

• '\(<\)' for less than,
• '\(>\)' for greater than,
• '\(\leq\)' for less than or equal to, and
• '\(\geq\)' for greater than or equal to.

The step-by-step problem solution factored inequalities into understanding solutions for \(x\) where \(x \geq 1\) or special values like \(x = -1\). These areas tell us where our original equation holds true for specific intervals of \(x\). If the inequality represents \(x \geq 1\), it tells us that all numbers greater than or equal to 1 satisfy the condition. Keep these concepts in mind when solving complex equations as they guide the values that make an equation true.

Quadratic Expressions

Quadratic expressions are like the powerhouse of algebra, often taking the form \(ax^2 + bx + c\). These are polynomials of degree 2. In our exercise, we encounter quadratic expressions when terms like \((x+1)^2\) appear. Simply put, a quadratic expression can be expanded, factored, or solved.Expanding \((x+1)^2\) gives us \(x^2 + 2x + 1\), showing how multiplication affects the expression. When solving or simplifying equations, identifying these expressions helps in setting factorable equations or understanding their roots.When you solve the original exercise, it involves simplifying these expressions to find where they equal each other or not. Recognizing patterns like perfect squares and knowing how to manipulate them mathematically is essential. Practice solving various quadratic forms to build confidence in this key area of algebra.