(d) Both (a) and (b)
(|x – 1| – 3) (|x + 2| – 5) < 0
The product of the two factors is < 0, i.e., negative when one of the factors is positive and one negative.
Case I : (|x – 1| – 3) > 0 and (|x + 2| – 5) < 0
⇒ |x – 1| > 3 and |x + 2| < 5
⇒ – (x – 1) > 3, (x – 1) > 3 and –(x + 2) < 5, x + 2 < 5
⇒ – x > 2, x > 4 and – x < 7, x < 3
⇒ \(x\) < – 2, x > 4 and x > –7, x < 3
Combining we get – 7 < x < – 2
Case II : (|x – 1| – 3) < 0 and (|x + 2| – 5) > 0
⇒ |x – 1| < 3 and |x + 2| > 5
⇒ – (x – 1) < 3, (x – 1) < 3 and (x + 2) > 5, – (x + 2) > 5
⇒ – \(x\) < 2, \(x\) < 4 and \(x\) > 3, – \(x\) > 7
⇒ \(x\) > – 2, \(x\) < 4 and \(x\) > 3, \(x\) < – 7
⇒ 3 < \(x\) < 4
∴ –7 < \(x\) < – 2 and 3 < \(x\) < 4 is the solution.