# 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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