Problem 117 - A Complete Resource Book in Mathematics for JEE Main

Question

The values of \(x\) for which the functions \(f(x)=x-3\) and \(\phi(x)=4-x\) satisfy the inequality \(|f(x)+\phi(x)|<\) \(|f(x)|+|\phi(x)|\) are (A) \([3,4]\) (B) \((-\infty, \infty)\) (C) \((-\infty, \infty)-[3,4]\) (D) None of these

Step-by-Step Solution

Answer

The solution is (C) \((- \infty, \infty) - [3,4]\).

Step 1: Simplify the Left-Hand Side

We start by calculating \(|f(x)+\phi(x)|\). Given \(f(x) = x - 3\) and \(\phi(x) = 4 - x\), we have:\[f(x) + \phi(x) = (x - 3) + (4 - x) = 1.\]Thus, \(|f(x) + \phi(x)| = |1| = 1.\) This simplifies the left-hand side of the inequality to 1.

Step 2: Simplify the Right-Hand Side

Next, calculate \(|f(x)| + |\phi(x)|\). We know:\[f(x) = x - 3 \quad \text{and} \quad \phi(x) = 4 - x.\]Hence, \(|f(x)| = |x - 3|\) and \(|\phi(x)| = |4 - x|.\)The right-hand side becomes:\[|x - 3| + |4 - x|.\]

Step 3: Understanding the Absolute Values

Note that \(|x - 3| = |3 - x|\). Thus,\[|f(x)| + |\phi(x)| = |x - 3| + |4 - x| = |3 - x| + |4 - x|.\]By examining the expression, realize that \(3 - x\) and \(4 - x\) have different signs between 3 and 4, implying the expression reduces differently across these intervals. Consider the value in the interval to see how the expressions change.

Step 4: Evaluate the Inequality in Different Cases

Determine how the inequality \(|1| < |x-3| + |4-x|\) behaves in different ranges of \(x\).1. **When \(x < 3\):** - \(|x - 3| = 3 - x\) - \(|4 - x| = 4 - x\) - Thus, \(|x - 3| + |4 - x| = 7 - 2x\). Since \(7 - 2x > 1\) for any \(x < 3\), the inequality holds.2. **When \(3 \leq x \leq 4\):** - As \(x\) moves from 3 to 4, \(|x-3| = x-3\) but \(|4-x| = 4-x\). - Hence, \(|x - 3| + |4 - x| = x-3 + 4-x = 1.\) Since \(|1| = 1\), the inequality becomes \(1 < 1\) which does not hold.3. **When \(x > 4\):** - \(|x - 3| = x - 3\) - \(|4 - x| = x - 4\) - Thus, \(|x - 3| + |4 - x| = 2x - 7\). Given \(2x - 7 > 1\), the inequality holds for \(x > 4\).

Step 5: Identify Solution Set

The inequality \(|f(x) + \phi(x)| < |f(x)| + |\phi(x)|\) holds for the intervals where \(x < 3\) or \(x > 4\). This means that the inequality is not satisfied for \(x\) in \([3,4]\). In the interval notation, these values are \((-\infty, 3) \cup (4, \infty)\). Thus, the solution set is all real numbers except those in the interval \([3,4]\).

Key Concepts

Absolute Value

Absolute value refers to the distance of a number from zero on the number line, irrespective of its direction. For any real number \(x\), the absolute value is represented as \(|x|\). This concept plays a key role when dealing with inequalities involving relevant expressions. In the exercise we've looked at, absolute value is used to ensure expressions are handled correctly by considering both positive and negative contributions.

The absolute value function is defined as follows:

• If \(x \geq 0\), then \(|x| = x\).
• If \(x < 0\), then \(|x| = -x\).

When solving the inequality from the exercise, understanding how to split the analysis based on the positive or negative values of \(f(x)\) and \(\phi(x)\) is essential. The breakdown between different intervals of \(x\) is vital as it allows for different expressions arising from the absolute value conditions.

These conditions guide us in managing and simplifying terms like \(|x - 3|\) and \(|4 - x|\). This results in analyzing separately over different intervals of \(x\) and observing how each expression behaves.

Interval Notation

Interval notation is a convenient way of representing subsets of real numbers, which often come up in solutions to inequalities. It uses intervals to describe a range of values that satisfy certain conditions. In this exercise, the solutions are presented using interval notation, highlighting the ranges where the given inequality is valid.

Intervals can be either open or closed:

• Open intervals, written as \((a, b)\), do not include the endpoints \(a\) and \(b\).
• Closed intervals, indicated by square brackets like \([a, b]\), include both endpoints.
• Combinations of open and closed intervals, such as \((a, b]\) or \([a, b)\), include one endpoint but not the other.

For the given inequality, the solution is \((-\infty, 3) \cup (4, \infty)\), denoting values of \(x\) that are not between 3 and 4, since within this latter closed interval the inequality is not satisfied. Understanding interval notation is crucial for clearly representing and interpreting ranges of variable validity.

Function Analysis

Function analysis involves understanding a function's behavior over different intervals. This generally includes determining values or ranges where inequalities involving functions hold true, as seen in this exercise.

Analyzing the functions and their combinations helps identify critical points and intervals where expressions change behavior. Given the two functions \(f(x) = x - 3\) and \(\phi(x) = 4 - x\), it is essential to investigate how their sum and independent absolute values interact across different ranges.

Different strategies are applied:

• When considering points where \(x < 3\), \(|x-3| = 3-x\) and \(|4-x| = 4-x\), simplifying into \(7 - 2x\).
• Similarly, for \(x > 4\), \(|x-3| = x-3\) and \(|4-x| = x-4\) results in \(2x-7\).
• In the range \([3, 4]\), we observe that simplification results in \(1\) implying no solvable inequality within this interval.

These analyses allow us to establish the exact intervals where the inequality holds and where it breaks down, which in this example leads to our solution being those values not in \([3, 4]\). Recognizing such different outcomes over various segments is a key facet of function analysis.