A2 June 2019 Paper 2 Q3
3 The set \(\mathcal{A}\) is defined by \(\mathcal{A} = \{x : -\sqrt{2} \lt x \lt 0\} \cup \{x : 0 \lt x \lt \sqrt{2}\}\)
Which of the inequalities given below has \(\mathcal{A}\) as its solution?
Circle your answer. [1 mark]
- \(|x^2 - 1| \gt 1\)
- \(|x^2 - 1| \geqslant 1\)
- \(|x^2 - 1| \lt 1\)
- \(|x^2 - 1| \leqslant 1\)
| Scheme | Marks | AO |
|---|---|---|
| Circles correct answer. | B1 | 2.2a |
| (1 mark) |