Problem 25 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

If \(x\) satisfies \(|x-1|+|x-2|+|x-3| \geq 6\) then (a) \(0 \leq x \leq 4\) (b) \(x \leq-2\) or \(x \geq 4\) (c) \(x \leq 0\) or \(x \geq 4\) (d) None of these

Step-by-Step Solution

Answer

Option (c): \(x \leq 0\) or \(x \geq 4\).

Step 1: Consider Subcases for Absolute Values

To solve the inequality \(|x-1|+|x-2|+|x-3| \geq 6\), we need to consider different cases based on the values of \(x\). These cases revolve around where each expression inside the absolute value changes sign, occur at \(x = 1\), \(x = 2\), and \(x = 3\). Thus, we will explore values in the following intervals: \(x < 1\), \(1 \leq x < 2\), \(2 \leq x < 3\), and \(x \geq 3\).

Step 2: Solve for Case 1: \(x < 1\)

In this case, \(|x-1| = 1-x\), \(|x-2| = 2-x\), and \(|x-3| = 3-x\). Thus, the inequality becomes:\[(1-x) + (2-x) + (3-x) \geq 6\]\[6 - 3x \geq 6\]\[-3x \geq 0\]\[x \leq 0\].So in this interval \(x < 1\), the solution is \(x \leq 0\).

Step 3: Solve for Case 2: \(1 \leq x < 2\)

Here, \(|x-1| = x-1\), \(|x-2| = 2-x\), and \(|x-3| = 3-x\). The inequality becomes:\[(x-1) + (2-x) + (3-x) \geq 6\]\[4 - x \geq 6\]\[-x \geq 2\]\[x \leq -2\].This is a contradiction since this condition can't be satisfied in the interval \([1, 2)\).

Step 4: Solve for Case 3: \(2 \leq x < 3\)

For this range, \(|x-1| = x-1\), \(|x-2| = x-2\), and \(|x-3| = 3-x\). The inequality is:\[(x-1) + (x-2) + (3-x) \geq 6\]\[x \geq 6\].Since for \(2 \leq x < 3\), \(x\) can never be \( \geq 6\), there are no solutions in this interval.

Step 5: Solve for Case 4: \(x \geq 3\)

Here \(|x-1| = x-1\), \(|x-2| = x-2\), and \(|x-3| = x-3\). The equation becomes:\[(x-1) + (x-2) + (x-3) \geq 6\]\[3x - 6 \geq 6\]\[3x \geq 12\]\[x \geq 4\].Thus, the solution in this interval is \(x \geq 4\).

Step 6: Combine All Solutions

From steps in individual cases, we have solutions \(x \leq 0\) from Case 1 and \(x \geq 4\) from Case 4. Combining, the solution for \(x\) that fits the original inequality is \(x \leq 0\) or \(x \geq 4\).

Step 7: Determine the Correct Option

From the combined solutions, the condition matches option (c): \(x \leq 0\) or \(x \geq 4\).

Key Concepts

Absolute Value

Absolute value is a concept in mathematics that denotes the distance of a number from zero on the number line, regardless of direction. It is always non-negative. For any real number \(a\), the absolute value is represented as \(|a|\). The absolute value removes any negative signs, essentially measuring the magnitude of a number.

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

In the context of this exercise, we are dealing with expressions like \(|x-1|\), where the formula changes depending on if \(x\) is less than, equal to, or greater than 1. This requires examining separate cases, as the value inside the absolute value impacts the inequality directly. Understanding absolute value enables you to break down and simplify complex inequality problems, making them easier to solve.

Piecewise Functions

Piecewise functions are defined by different expressions depending on the input's interval. They are a powerful tool to describe situations where a function exhibits distinct behaviors across different domains. In this problem, the expression \(|x-1| + |x-2| + |x-3|\) can be thought of as a piecewise function.Each segment of the piecewise function corresponds to a different arrangement of the inequality based on the critical points \(x = 1\), \(x = 2\), and \(x = 3\). For example:

• For \(x < 1\), the expressions are \(|x-1| = 1-x\), \(|x-2| = 2-x\), and \(|x-3| = 3-x\).
• For \(1 \leq x < 2\), the expressions change to \(|x-1| = x-1\), \(|x-2| = 2-x\), and \(|x-3| = 3-x\).

This segmentation allows us to simplify and calculate each interval separately and logically determine the contributing solutions to the inequality.

Interval Notation

Interval notation is a convenient way to represent sets of numbers, particularly solutions to inequalities. It denotes intervals on a number line using brackets and parentheses to describe whether an endpoint is included.For instance, consider the following notations:

• \([a, b]\): Both \(a\) and \(b\) are included in the set.
• \((a, b]\): \(a\) is not included, but \(b\) is.
• \([a, b)\): \(a\) is included, but \(b\) isn't.
• \((a, b)\): Neither \(a\) nor \(b\) is included.

In this exercise, we use interval notation to express the solutions found in each case. For example, solutions such as \(x \leq 0\) and \(x \geq 4\) are expressed in interval notation as \((-67, 0]\) and \([4, 67)\), respectively. Interval notation is essential for clearly and concisely expressing solution sets without ambiguity.

Case Analysis

Case analysis is a problem-solving technique crucial for handling complex problems with multiple conditions, such as those involving absolute values. By breaking down an inequality into several cases based on critical points where expressions change behavior, you can simplify and solve each part separately.In the problem \(|x-1| + |x-2| + |x-3| \geq 6\), we perform case analysis by considering the intervals formed by \(x = 1\), \(x = 2\), and \(x = 3\). This means considering:

• The interval \(x < 1\), where each absolute expression results in subtraction from fixed numbers.
• The interval \(1 \leq x < 2\), which demands a different form owing to changed signs inside the absolute values.
• The interval \(x \geq 3\), where all expressions are simply positive differences.

This systematic approach allows us to find separate solutions for each interval, which we then combine to find the overall possible solutions. Effective case analysis ensures thorough exploration of all possibilities, guarding against overlooked solutions.