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Question:

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

Explanation:

Given, x² − 2|x| + |a - 2| = 0

This can be written as

|x|² − 2|x| + 1 – 1 + |a - 2| = 0

⇒ (|x| - 1)² + |a - 2| - 1 = 0

Now, ((|x| - 1)² ≥ 0 and |a - 2| - 1 will be an integer since a is an integer.

∴ |a - 2| - 1 can take only take non-positive values i.e., 0 or -1.

Case 1: |a - 2| - 1 = 0 and (|x| - 1) = 0

⇒ |a – 2| = 1 and |x| = 1

⇒ a = 1 or 3 and x = ± 1

∴ 4 possible combinations of (x, a)

Case 2: |a - 2| - 1 = -1 and (|x| - 1) = ±1

|a - 2| = 0 and |x| = 0 or 2

⇒ a = 2 and x = 0 or ± 2

∴ 3 possible combinations of (x, a)

∴ Total 7 possible combination of (x, a) are there.

Hence, option (d).