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Solve (|x – 1| – 3) (|x + 2| – 5) < 0. Then

(a) –7 < x < – 2 

(b) 3 < x < 4 

(c) x < – 7 and x > 4 

(d) Both (a) and (b)

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(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.

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