In how many ways can a pair of integers (x , a) be chosen such that \({x^2} - 2|x| + |a - 2| = 0\)?
- 4
- 5
- 6
- 7
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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