Question

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

• A. 4
• B. 5
• C. 6
• D. 7

Answer 1

The Correct Option is D

The correct answer is (D): \(7\)

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

\(|x|=\frac{2±\sqrt{4-4(|a-2|})}{2}\)

\(|x|=1±\sqrt{1-|a-2|}\)

If \(a>2;|a-2|=a-2\)

\(|x|=1±\sqrt{1-(a-2)}\)

= \(1±\sqrt{3-a}\)

since \(x\) is integer \(3-a≥0\)

\(a≤3\)

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

Then \(x = ±1\);

If \(a=2,|x|=|1±1|,⇒x=±2,0\)

If \(a<2,|a-2|=2-a\)

\(|x|=1±\sqrt{1-(2-a)}\)

\(|x|=1±\sqrt{a-1}\)

Since \(x\) is integer \(a-1≥0⇒a≥1\)

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

If \(a=1,|x|=1⇒x=±1\)

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

Answer 2

\(|x|^2 – 2|x| + |a – 2| = 0\)
\(|x|^2 – 2|x| + 1 = 0\) is the square of a quadratic number
The value of constant cannot be more than 1
So \(|a – 2|\) = 0 or = 1
\(|x|^2 – 2|x| = 0\)
\(|x|^2 = 2|x|\)
\(x = 0 \space or\space  2 \space or -2\)
For all these possibilities value of a = 2
\(|x|^2 – 2|x| + 1 = 0\)
\((|x| - 1)^2 = 0\)
\(|x| = 1\)
So, \(x = 1 \space or \space x = -1\)
Then \(|a – 2| = 1, a = 3 \space or \space a = 1\)
4 combinations of \((x,a)\) are possible already we have 3

∴ Total 7 pairs