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Difficulty: Very hardReal Numbers, Number Line, and Absolute Value

Let xx and yy be real numbers such that ∣x2−4∣=2y|x^2 - 4| = 2y and ∣y−3∣≤1|y - 3| \le 1. Which of the following could be the value of xx? Select all such values.

  1. −3-3Answer
  2. 00Answer
  3. C
    22
  4. 33Answer
  5. E
    44

Answer

The values that xx could take are −3-3, 00, and 33.
The inequality ∣y−3∣≤1|y - 3| \le 1 restricts yy to the closed interval [2,4][2, 4]. Consequently, 2y2y lies in [4,8][4, 8], which means 4≤∣x2−4∣≤84 \le |x^2 - 4| \le 8. Breaking this compound inequality into cases yields x=0x = 0 (from x2=0x^2 = 0) or x∈[−12,−8]∪[8,12]x \in [-\sqrt{12}, -\sqrt{8}] \cup [\sqrt{8}, \sqrt{12}]. Since 8≈2.83\sqrt{8} \approx 2.83 and 12≈3.46\sqrt{12} \approx 3.46, the integer values −3-3 and 33 fall in these intervals, alongside 00. Thus, −3-3, 00, and 33 are all valid choices.

Step-by-Step Solution

1
Determine the acceptable range for yy from the given absolute value inequality.
Solving ∣y−3∣≤1|y - 3| \le 1 gives −1≤y−3≤1-1 \le y - 3 \le 1, which simplifies to 2≤y≤42 \le y \le 4.
The inequality ∣y−c∣≤r|y - c| \le r represents all values of yy within distance rr from cc.
2
Relate the range of yy to the absolute value expression ∣x2−4∣|x^2 - 4|.
Since 2≤y≤42 \le y \le 4, multiplying by 22 yields 4≤2y≤84 \le 2y \le 8. Therefore, 4≤∣x2−4∣≤84 \le |x^2 - 4| \le 8.
Substitute 2y=∣x2−4∣2y = |x^2 - 4| into the inequality derived for yy.
3
Analyze the two cases for the absolute value equation ∣x2−4∣|x^2 - 4|.
Case 1: 4≤x2−4≤8  ⟹  8≤x2≤124 \le x^2 - 4 \le 8 \implies 8 \le x^2 \le 12, which gives x∈[−12,−8]∪[8,12]x \in [-\sqrt{12}, -\sqrt{8}] \cup [\sqrt{8}, \sqrt{12}]. Case 2: 4≤−(x2−4)≤8  ⟹  4≤4−x2≤8  ⟹  −4≤−x2≤4  ⟹  −4≤x2≤04 \le -(x^2 - 4) \le 8 \implies 4 \le 4 - x^2 \le 8 \implies -4 \le -x^2 \le 4 \implies -4 \le x^2 \le 0. Since x2≥0x^2 \ge 0 for all real numbers, x2=0  ⟹  x=0x^2 = 0 \implies x = 0.
Absolute value ∣A∣|A| splits into AA when A≥0A \ge 0 and −A-A when A<0A < 0.
4
Evaluate the test values against the valid domains for xx.
For x=−3x = -3, x2=9x^2 = 9, which satisfies 8≤9≤128 \le 9 \le 12. For x=0x = 0, x2=0x^2 = 0, which satisfies x2=0x^2 = 0. For x=3x = 3, x2=9x^2 = 9, which satisfies 8≤9≤128 \le 9 \le 12. The values x=2x = 2 and x=4x = 4 give x2=4x^2 = 4 and x2=16x^2 = 16, neither of which fall into the valid intervals.
Checking test options confirms which values fall within the solution intervals.

Key Concept

Absolute Value Equations and System Constraints on Real Numbers
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Topics

Real Numbers, Number Line, and Absolute Value

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