| Exam Board | CAIE |
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| Module | P2 (Pure Mathematics 2) |
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| Year | 2022 |
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| Session | November |
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| Topic | Modulus function |
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2 The solutions of the equation \(| 4 x - 1 | = | x + 3 |\) are \(x = p\) and \(x = q\), where \(p < q\).
Find the exact values of \(p\) and \(q\), and hence determine the exact value of \(| p - 2 | - | q - 1 |\).