(a) (– 5, – 1) ∪ (1, 5)
Let |x| = p, where p > 0 …(i)
So |p – 3| < 2 and |p – 2| < 3 …(ii)
⇒ 1 < p < 5 and – 1 < p < 5 …(iii)
(∴ |x – a| < r ⇒ a – r < x < a + r)
Therefore, the conditions (i), (ii) and (iii) are satisfied by 1 < p < 5, i.e. 1 < | x | < 5, i.e., | x | > 1 and | x | < 5
i.e., x < – 1 or x > 1 and – 5 < x < 5
⇒ \(x\)∈ (– 5, – 1) ∪ (1, 5).