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Difficulty: Very hardLinear Inequalities and Absolute Value

How many integer values of xx satisfy the compound absolute value inequality 1≤∣∣x−4∣−3∣≤51 \le ||x - 4| - 3| \le 5?

  1. A
    10
  2. B
    12
  3. 15Answer
  4. D
    17
  5. E
    18

Answer

There are 15 integer values of xx that satisfy the given compound inequality.
Solving 1≤∣∣x−4∣−3∣≤51 \le ||x - 4| - 3| \le 5 requires breaking the nested absolute value into its boundary constraints. The upper bound ∥x−4∣−3∣≤5\|x - 4| - 3| \le 5 restricts xx to [−4,12][-4, 12]. The lower bound ∣∣x−4∣−3∣≥1||x - 4| - 3| \ge 1 requires either ∣x−4∣≥4|x - 4| \ge 4 (giving x≤0x \le 0 or x≥8x \ge 8) or ∣x−4∣≤2|x - 4| \le 2 (giving 2≤x≤62 \le x \le 6). Taking the intersection produces three distinct inclusive integer intervals: [−4,0][-4, 0], [2,6][2, 6], and [8,12][8, 12]. Each interval contains 5 integers, yielding a total of 15 integer solutions.

Step-by-Step Solution

1
Decompose the double inequality into two separate absolute value inequalities: ∣∣x−4∣−3∣≤5||x - 4| - 3| \le 5 and ∣∣x−4∣−3∣≥1||x - 4| - 3| \ge 1.
Two simultaneous inequalities to solve for xx.
A double inequality a≤∣u∣≤ba \le |u| \le b requires satisfying both ∣u∣≤b|u| \le b and ∣u∣≥a|u| \ge a.
2
Solve the upper bound inequality ∣∣x−4∣−3∣≤5||x - 4| - 3| \le 5.
−5≤∣x−4∣−3≤5  ⟹  −2≤∣x−4∣≤8-5 \le |x - 4| - 3 \le 5 \implies -2 \le |x - 4| \le 8. Since ∣x−4∣≥0≥−2|x - 4| \ge 0 \ge -2 is always true, this simplifies to ∣x−4∣≤8  ⟹  −4≤x≤12|x - 4| \le 8 \implies -4 \le x \le 12.
Absolute value is non-negative, so the lower bound of −2-2 imposes no extra constraint.
3
Solve the lower bound inequality ∣∣x−4∣−3∣≥1||x - 4| - 3| \ge 1.
This splits into two cases: ∣x−4∣−3≥1|x - 4| - 3 \ge 1 OR ∣x−4∣−3≤−1|x - 4| - 3 \le -1.
Case A: ∣x−4∣≥4  ⟹  x−4≥4|x - 4| \ge 4 \implies x - 4 \ge 4 or x−4≤−4  ⟹  x≥8x - 4 \le -4 \implies x \ge 8 or x≤0x \le 0.
Case B: ∣x−4∣≤2  ⟹  −2≤x−4≤2  ⟹  2≤x≤6|x - 4| \le 2 \implies -2 \le x - 4 \le 2 \implies 2 \le x \le 6.
The absolute value inequality ∣u∣≥1|u| \ge 1 holds when u≥1u \ge 1 or u≤−1u \le -1.
4
Intersect the solution set from the upper bound [−4,12][-4, 12] with the solution set from the lower bound (−∞,0]∪[2,6]∪[8,∞)(-\infty, 0] \cup [2, 6] \cup [8, \infty).
The valid solution set is x∈[−4,0]∪[2,6]∪[8,12]x \in [-4, 0] \cup [2, 6] \cup [8, 12].
Both conditions must hold simultaneously.
5
Count the integer values in each of the three valid intervals.
Interval [−4,0][-4, 0] has 5 integers: {−4,−3,−2,−1,0}\{-4, -3, -2, -1, 0\}.
Interval [2,6][2, 6] has 5 integers: {2,3,4,5,6}\{2, 3, 4, 5, 6\}.
Interval [8,12][8, 12] has 5 integers: {8,9,10,11,12}\{8, 9, 10, 11, 12\}.
Total integer solutions = 5+5+5=155 + 5 + 5 = 15.
The number of integers in an inclusive integer range [a,b][a, b] is b−a+1b - a + 1.

Key Concept

Linear Inequalities and Absolute Value
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Topics

Linear Inequalities and Absolute Value

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