Question

The solution set of the inequality \( |3x| \geq |6 - 3x| \) is: 

• A. (−∞, 1]
• B. [1, ∞)
• C. (−∞, 1) ∪ (1,∞)
• D. (−∞, −1) ∪ (−1,∞)

Hint:

When solving inequalities involving absolute values, it’s important to break them into different cases based on the possible signs of the expressions inside the absolute values. Each case leads to a different inequality that you can solve. If you end up with a contradiction (like \( 0 \leq -6 \)), that case provides no valid solutions. Once all cases are considered, the union of their solutions will give you the final answer.

Answer 1

The Correct Option is B

To resolve the inequality \( |3x| \geq |6 - 3x| \), we must utilize the definitions of absolute values. The absolute value of a number \( a \), denoted \( |a| \), is defined as:

• \(|a| = a\) when \( a \geq 0 \)
• \(|a| = -a\) when \( a<0 \)

Let's define the expressions \( |3x| \) and \( |6 - 3x| \) separately:

• \(|3x| = 3x\) if \( x \geq 0 \) and \(|3x| = -3x\) if \( x<0 \)
• \(|6 - 3x| = 6 - 3x\) if \( 6 - 3x \geq 0 \) and \(|6 - 3x| = -(6 - 3x)\) if \( 6 - 3x<0 \)

Next, we identify the critical points that delineate intervals for \( x \). Setting \( 6 - 3x = 0 \) yields \( x = 2 \).

We will now examine the inequality across three intervals defined by \( x = 0 \) and \( x = 2 \):

• Interval 1: \( x<0 \)
• Interval 2: \( 0 \leq x<2 \)
• Interval 3: \( x \geq 2 \)

The inequality is evaluated for each interval:

• For \( x<0 \): The inequality becomes \(-3x \geq -(6 - 3x)\). This simplifies to \(-3x \geq -6 + 3x\), which further simplifies to \(0 \geq -6 + 6x\), or \(6x \geq 6\). This implies \(x \geq 1\). This solution is incompatible with the condition \( x<0 \).
• For \( 0 \leq x<2 \): The inequality is \(3x \geq 6 - 3x\). Simplifying gives \(6x \geq 6\), or \(x \geq 1\). Within this interval, the solution is \(x \in [1,2)\).
• For \( x \geq 2 \): The inequality is \(3x \geq 3x - 6\). This simplifies to \(6 \geq 0\), which is universally true for all \( x \geq 2\).

Combining the solutions from each interval, the complete solution set is \( x \in [1, \infty) \).

Therefore, the solution set for the inequality is \([1, \infty)\).