Which of the following inequalities has a solution set that, when graphed on the number line, is a single line...

1. Translate the problem requirements: We need to find which inequality creates a solution set that forms exactly one finite line segment when graphed. This means we want solutions that are bounded (have both upper and lower limits) and form a continuous interval.
2. Analyze each inequality's geometric representation: Visualize what each inequality looks like on the number line to identify patterns of finite vs infinite solution sets.
3. Identify bounded vs unbounded solutions: Distinguish between inequalities that create infinite rays (unbounded) versus finite intervals (bounded on both sides).
4. Verify the finite solution candidate: Solve the inequality that appears to create a bounded interval to confirm it produces exactly one finite line segment.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're looking for in plain English. We want an inequality whose solution set forms exactly one finite line segment when we draw it on a number line.

Think of it this way: imagine you're highlighting portions of a number line with a marker. We want the inequality that creates exactly one highlighted segment that has a clear starting point and a clear ending point - not something that goes on forever in one or both directions.

For a solution set to be a single finite line segment, it must be:

• Continuous (no gaps or breaks)
• Bounded on both sides (has both a minimum and maximum value)
• Connected (forms one piece, not multiple separate pieces)

Process Skill: TRANSLATE - Converting the geometric requirement into mathematical properties

2. Analyze each inequality's geometric representation

Let's think through what each inequality looks like on a number line by using concrete examples and everyday reasoning.

Choice A: \(\mathrm{x}^4 \geq 1\)
Let's think: when is \(\mathrm{x}^4\) at least 1? Well, \(\mathrm{x}^4\) is always positive, and it equals 1 when \(\mathrm{x} = 1\) or \(\mathrm{x} = -1\). For values between -1 and 1 (like \(\mathrm{x} = 0.5\)), we get \(\mathrm{x}^4 = 0.0625\), which is less than 1. So this inequality is satisfied when \(\mathrm{x} \leq -1\) or \(\mathrm{x} \geq 1\). This creates two separate rays going in opposite directions - not one finite segment.

Choice B: \(\mathrm{x}^3 \leq 27\)
Since \(\mathrm{x}^3 = 27\) when \(\mathrm{x} = 3\), this means \(\mathrm{x} \leq 3\). This creates one ray starting from 3 and going left forever - it's bounded on the right but unbounded on the left, so not finite.

Choice C: \(\mathrm{x}^2 \geq 16\)
Since \(\mathrm{x}^2 = 16\) when \(\mathrm{x} = 4\) or \(\mathrm{x} = -4\), this means \(\mathrm{x} \leq -4\) or \(\mathrm{x} \geq 4\). Like choice A, this creates two separate infinite rays - not one finite segment.

Choice D: \(2 \leq |\mathrm{x}| \leq 5\)
The absolute value \(|\mathrm{x}|\) is at least 2 and at most 5. This means x is either between -5 and -2, or between 2 and 5. That's two separate finite segments, not one.

Choice E: \(2 \leq 3\mathrm{x} + 4 \leq 6\)
This looks promising! It's a compound inequality that bounds the expression \(3\mathrm{x} + 4\) between two finite values.

Process Skill: VISUALIZE - Creating mental pictures of solution sets on the number line

3. Identify bounded vs unbounded solutions

From our analysis above, let's categorize what we found:

Unbounded solutions (infinite length):

• Choice A: Two infinite rays
• Choice B: One infinite ray
• Choice C: Two infinite rays

Bounded but multiple segments:

• Choice D: Two separate finite segments

Potentially one finite segment:

• Choice E: Appears to create one bounded interval

Only Choice E has the potential to create exactly one finite line segment.

4. Verify the finite solution candidate

Let's solve Choice E step by step: \(2 \leq 3\mathrm{x} + 4 \leq 6\)

We can break this compound inequality into two parts:

• \(2 \leq 3\mathrm{x} + 4\) AND \(3\mathrm{x} + 4 \leq 6\)

Solving the left inequality: \(2 \leq 3\mathrm{x} + 4\)
Subtract 4 from both sides: \(-2 \leq 3\mathrm{x}\)
Divide by 3: \(-\frac{2}{3} \leq \mathrm{x}\)

Solving the right inequality: \(3\mathrm{x} + 4 \leq 6\)
Subtract 4 from both sides: \(3\mathrm{x} \leq 2\)
Divide by 3: \(\mathrm{x} \leq \frac{2}{3}\)

Combining both conditions: \(-\frac{2}{3} \leq \mathrm{x} \leq \frac{2}{3}\)

This creates exactly one finite line segment on the number line, from \(-\frac{2}{3}\) to \(\frac{2}{3}\).

Process Skill: MANIPULATE - Solving compound inequalities systematically

5. Final Answer

Choice E is correct. The inequality \(2 \leq 3\mathrm{x} + 4 \leq 6\) has the solution set \(-\frac{2}{3} \leq \mathrm{x} \leq \frac{2}{3}\), which graphs as exactly one finite line segment on the number line.

All other choices either create infinite rays or multiple separate segments, but only Choice E creates the single finite segment we were looking for.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misinterpreting "single line segment of finite length"
Students often confuse this requirement and think any bounded solution qualifies. They might select Choice D (\(2 \leq |\mathrm{x}| \leq 5\)) because it involves finite values, not recognizing that this creates TWO separate segments: one from -5 to -2 and another from 2 to 5. The key word "single" means exactly one connected piece, not just finite boundaries.

Faltering Point 2: Failing to systematically categorize solution types
Students may jump directly to solving equations without first understanding what each inequality type typically produces. They miss that \(\mathrm{x}^2 \geq 16\) and \(\mathrm{x}^4 \geq 1\) will always create two separate rays (due to the even powers), while absolute value inequalities like \(|\mathrm{x}| \geq 2\) also split into two regions. Without this systematic thinking, they waste time on detailed calculations for obviously wrong choices.

Faltering Point 3: Not recognizing compound linear inequalities as the target pattern
Students may not immediately identify that only a compound linear inequality of the form "a ≤ linear expression ≤ b" can produce exactly one finite segment. They might get distracted by the more complex-looking choices with powers and absolute values, missing that the "simplest-looking" Choice E is actually the only one that can satisfy the geometric requirement.

Errors while executing the approach

Faltering Point 1: Sign errors when solving compound inequalities
When solving \(2 \leq 3\mathrm{x} + 4 \leq 6\), students often make arithmetic mistakes, particularly when subtracting 4 from all parts or dividing by 3. A common error is getting the direction wrong and writing \(\frac{2}{3} \leq \mathrm{x} \leq -\frac{2}{3}\), which makes no mathematical sense since \(\frac{2}{3}\) cannot be less than \(-\frac{2}{3}\).

Faltering Point 2: Incorrectly handling absolute value inequalities
For Choice D (\(2 \leq |\mathrm{x}| \leq 5\)), students often forget that \(|\mathrm{x}| \geq 2\) means \(\mathrm{x} \leq -2\) OR \(\mathrm{x} \geq 2\), not \(-2 \leq \mathrm{x} \leq 2\). This leads them to incorrectly conclude that Choice D gives one segment from -5 to 5, when it actually gives two separate segments.

Faltering Point 3: Computational errors with even powers
When analyzing \(\mathrm{x}^4 \geq 1\) or \(\mathrm{x}^2 \geq 16\), students sometimes forget that these equations have both positive and negative solutions. They might correctly find that \(\mathrm{x}^2 \geq 16\) gives \(|\mathrm{x}| \geq 4\), but then incorrectly interpret this as \(-4 \leq \mathrm{x} \leq 4\) instead of the correct \(\mathrm{x} \leq -4\) or \(\mathrm{x} \geq 4\).

Faltering Point 3: Second-guessing the "simple" answer
Students often assume that GMAT questions must have complex solutions and become suspicious of Choice E because it "looks too straightforward." They might change their answer to something that seems more mathematically sophisticated, not realizing that the complexity lies in recognizing the geometric requirements, not in the algebraic manipulation.