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Difficulty: HardReal Numbers, Number Line, and Absolute Value

On the real number line, pp and qq are real numbers such that ∣3−2p∣≤5|3 - 2p| \le 5 and ∣q+4∣=2|q + 4| = 2. What is the maximum possible value of ∣p2−q∣|p^2 - q|?

  1. 22Answer
  2. B
    18
  3. C
    14
  4. D
    10
  5. E
    7

Answer

22
Solving ∣3−2p∣≤5|3 - 2p| \le 5 gives −5≤3−2p≤5  ⟹  −1≤p≤4-5 \le 3 - 2p \le 5 \implies -1 \le p \le 4, so p2p^2 can range from 00 up to 1616. Solving ∣q+4∣=2|q + 4| = 2 yields q=−2q = -2 or q=−6q = -6. To maximize ∣p2−q∣|p^2 - q|, we combine the maximum possible value of p2p^2 (1616) with q=−6q = -6, obtaining ∣16−(−6)∣=22|16 - (-6)| = 22.

Step-by-Step Solution

1
Solve the absolute value inequality ∣3−2p∣≤5|3 - 2p| \le 5 for pp.
−1≤p≤4-1 \le p \le 4
−5≤3−2p≤5  ⟹  −8≤−2p≤2-5 \le 3 - 2p \le 5 \implies -8 \le -2p \le 2. Dividing by −2-2 and reversing the inequality direction gives −1≤p≤4-1 \le p \le 4.
2
Determine the range of possible values for p2p^2.
0≤p2≤160 \le p^2 \le 16
Since pp spans from −1-1 to 44 (which includes 00), the minimum square is 02=00^2 = 0 and the maximum square is 42=164^2 = 16.
3
Solve the absolute value equation ∣q+4∣=2|q + 4| = 2 for qq.
q=−2q = -2 or q=−6q = -6
q+4=2  ⟹  q=−2q + 4 = 2 \implies q = -2, and q+4=−2  ⟹  q=−6q + 4 = -2 \implies q = -6.
4
Find the combination of p2p^2 and qq that maximizes ∣p2−q∣|p^2 - q|.
Maximum value is 22
Pairing p2=16p^2 = 16 with q=−6q = -6 yields ∣16−(−6)∣=22|16 - (-6)| = 22, which is greater than ∣16−(−2)∣=18|16 - (-2)| = 18.

Key Concept

Absolute value inequalities, real number bounds, and distance optimization.
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Topics

Real Numbers, Number Line, and Absolute Value

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