Absolute Value Equation Pairs

CAT 2020 Slot 2 · Quantitative Ability · Hard · Algebra

In how many ways can a pair of integers (x,a)(x, a) be chosen such that x2−2∣x∣+∣a−2∣=0x^2 - 2|x| + |a - 2| = 0?

  1. A.

    7

  2. B.

    6

  3. C.

    4

  4. D.

    5

Answer

A

Explanation

Rewrite as ∣a−2∣=2∣x∣−x2=2∣x∣−∣x∣2=∣x∣(2−∣x∣)|a - 2| = 2|x| - x^2 = 2|x| - |x|^2 = |x|(2 - |x|). Since ∣a−2∣≥0|a - 2| \ge 0, we must have ∣x∣(2−∣x∣)≥0|x|(2 - |x|) \ge 0. Since ∣x∣≥0|x| \ge 0, we need 2−∣x∣≥0  ⟹  ∣x∣≤22 - |x| \ge 0 \implies |x| \le 2. Since xx is an integer, ∣x∣|x| can be 0,1,20, 1, 2.

Case 1: ∣x∣=0  ⟹  x=0|x| = 0 \implies x = 0. ∣a−2∣=0  ⟹  a=2|a - 2| = 0 \implies a = 2. Pair: (0,2)(0, 2) — 1 solution.

Case 2: ∣x∣=1  ⟹  x=1|x| = 1 \implies x = 1 or x=−1x = -1. ∣a−2∣=1(2−1)=1  ⟹  a−2=1|a - 2| = 1(2 - 1) = 1 \implies a - 2 = 1 or −1  ⟹  a=3-1 \implies a = 3 or a=1a = 1. Pairs: (1,3),(1,1),(−1,3),(−1,1)(1, 3), (1, 1), (-1, 3), (-1, 1) — 4 solutions.

Case 3: ∣x∣=2  ⟹  x=2|x| = 2 \implies x = 2 or x=−2x = -2. ∣a−2∣=2(2−2)=0  ⟹  a=2|a - 2| = 2(2 - 2) = 0 \implies a = 2. Pairs: (2,2),(−2,2)(2, 2), (-2, 2) — 2 solutions.

Total pairs =1+4+2=7= 1 + 4 + 2 = 7.

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