Grading Note: Full credit (6 pts) requires identifying No Solution and explaining that absolute value cannot be negative. If a student blindly splits into \(-4 > 5x + 15 > 4\), award max 2 pts.
#7 \(|2x - 8| \ge -3\) All Real Numbers • \((-\infty, \infty)\) or \(\mathbb{R}\)
Mathematical Justification: The absolute value of any real expression is non-negative (\(|2x - 8| \ge 0\)). Since any non-negative number is greater than or equal to \(-3\), the inequality is universally true for every real value of \(x\).
Rubric: 4 pts for correct interval \((-\infty, \infty)\) + 2 pts for conceptual explanation of absolute value non-negativity.
#8 \(|3x + 6| \le 0\) Single Point: \(\{-2\}\) or \(x = -2\)
Mathematical Justification: An absolute value can never be strictly negative (\(|3x + 6| < 0\) has no solution). Thus, the only admissible solutions occur when \(|3x + 6| = 0 \implies 3x + 6 = 0 \implies x = -2\).
Geometric Meaning: The solution set consists of exactly one point on the number line rather than an interval or segment.
#9 Manufacturing Tolerance Model (Titanium Piston Pins) Interval: \([18.46\text{ mm}, 18.54\text{ mm}]\)
(a) Model Formulation (3 pts):
\(|d - \text{target}| \le \text{tolerance}\)
\(|d - 18.50| \le 0.04\)
(b) Analytical Solution (3 pts):
\(-0.04 \le d - 18.50 \le 0.04\)
\(18.46 \le d \le 18.54\) (or \([18.46, 18.54]\))
#10 \(4|7 - 2x| - 8 > 16\) Answer: \(\left(-\infty, \frac{1}{2}\right) \cup \left(\frac{13}{2}, \infty\right)\) or \((-\infty, 0.5) \cup (6.5, \infty)\)
Isolate: \(4|7 - 2x| > 24 \implies |7 - 2x| > 6\)
Branch 1: \(7 - 2x > 6 \implies -2x > -1 \implies x < \frac{1}{2}\)
Branch 2: \(7 - 2x < -6 \implies -2x < -13 \implies x > \frac{13}{2}\)
Graph Key: Open circles at 0.5 and 6.5, shaded outward
0.5 6.5
Teacher Key • Part 2 Solutions (Problems 6–10) Page 2 of 3
Honors Solutions Part 3: Advanced Analytical Extensions (Problems 11–15)
Comprehensive rubrics for variable bounds, critical regions, and proof critiques.
8 Points Each
#11 \(|2x - 3| \le x + 6\) (Variable Right-Hand Side) Answer: \([-1, 9]\) or \(-1 \le x \le 9\)
Domain Restriction: Variable RHS must be non-negative: \(x + 6 \ge 0 \implies x \ge -6\).
Compound Setup: \(-(x + 6) \le 2x - 3 \le x + 6\)
• Left: \(-x - 6 \le 2x - 3 \implies -3 \le 3x \implies x \ge -1\)
• Right: \(2x - 3 \le x + 6 \implies x \le 9\)
Intersection with domain: \([-1, 9] \cap [-6, \infty) = [-1, 9]\).
#12 \(|x + 2| + |x - 4| \ge 8\) (Sum of Absolute Values) Answer: \((-\infty, -3] \cup [5, \infty)\)
Critical Points: \(x = -2\) and \(x = 4\). Partition the number line into 3 intervals:
Region 1 (\(x < -2\)):
\(-(x+2) - (x-4) \ge 8\)
\(-2x + 2 \ge 8 \implies x \le -3\)
⇒ \((-\infty, -3]\)
Region 2 (\(-2 \le x \le 4\)):
\((x+2) - (x-4) \ge 8\)
\(6 \ge 8\) (False)
⇒ No Solution
Region 3 (\(x > 4\)):
\((x+2) + (x-4) \ge 8\)
\(2x - 2 \ge 8 \implies x \ge 5\)
⇒ \([5, \infty)\)
#13 \(2|3x - 1| - 7 \ge 4x + 3\) Answer: \(\left(-\infty, -\frac{4}{5}\right] \cup [6, \infty)\) or \((-\infty, -0.8] \cup [6, \infty)\)
Isolate: \(2|3x - 1| \ge 4x + 10 \implies |3x - 1| \ge 2x + 5\)
• Case 1: \(3x - 1 \ge 2x + 5 \implies x \ge 6\)
• Case 2: \(3x - 1 \le -(2x + 5) \implies 5x \le -4 \implies x \le -\frac{4}{5}\)
Union of disjoint solution sets: \(\left(-\infty, -\frac{4}{5}\right] \cup [6, \infty)\).
#14 Aerospace Sensor Thermal Envelope Answer: \([15^\circ\text{C}, 45^\circ\text{C}]\) or \(15 \le T \le 45\)
Constraint 1 (Safe Operation): \(|0.5T - 15| \le 10 \implies -10 \le 0.5T - 15 \le 10 \implies 5 \le 0.5T \le 25 \implies 10 \le T \le 50\)
Constraint 2 (Heater Stays OFF): \(\text{NOT }(|T - 30| > 15) \iff |T - 30| \le 15 \implies -15 \le T - 30 \le 15 \implies 15 \le T \le 45\)
Compound Intersection: \([10, 50] \cap [15, 45] = [15^\circ\text{C}, 45^\circ\text{C}]\).
#15 Conceptual Error Analysis & Critique Correct Solution: \((-\infty, \infty)\) or \(\mathbb{R}\)
Flaw 1 (Fundamental Definition): The student mechanically applied the positive-number splitting rule to a negative bound. By definition, \(|4x - 12| \ge 0\) for every real number \(x\). Since any non-negative number is strictly greater than \(-8\), all real numbers satisfy the inequality without needing algebraic steps.
Flaw 2 (Compound Logic Failure): The student formulated an "OR" compound disjunction (\(x > 1\) OR \(x < 5\)), whose mathematical union is the entire real line \((-\infty, \infty)\). However, they incorrectly recorded the intersection \((1, 5)\) (an "AND" condition), demonstrating a fundamental failure of set-theory logic.
Teacher Key • Part 3 Solutions (Problems 11–15) • Form A Page 3 of 3
