CAT 2020 Quant Question [Slot 2] with Solution 08

Question

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

  1. 4
  2. 5
  3. 6
  4. 7
Option: 4
Solution:

\({x^2} - 2|x| + |a - 2| = 0\)

\(|x| = \frac{{2 \pm \sqrt {4 - 4(|a - 2|)} }}{2}\)

\(|x| = 1 \pm \sqrt {1 - |a - 2|} \)

If \(a > 2;|a - 2| = a - 2\)

\(|x| = 1 \pm \sqrt {1 - (a - 2)} \)

\( = 1 \pm \sqrt {3 - a} \)

since \(x\) is integer \(3 - a \ge 0\)

\(a \le 3\)

The possible values of a is \( = 3\)

Then \(x =  \pm 1;\)

If \(a = 2,|x| = |1 \pm 1|, \Rightarrow x =  \pm 2,0\)

If \(a < 2,|a - 2| = 2 - a\)

\(|x| = 1 \pm \sqrt {1 - (2 - a)} \)

\(|x| = 1 \pm \sqrt {a - 1} \)

Since x is integer \(a - 1 \ge 0 \Rightarrow a \ge 1\)

\(\therefore \) The possible values of \(a\) is 1

If \(a = 1,|x| = 1 \Rightarrow x =  \pm 1\)

\(\therefore \) The possible pairs =(-1,3), (1,3), (1,1), (-1,1), (2,2),(-2,2), (0,2) i.e., 7

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