June 2017 Paper 2 Q9
9 \(-11 \lt 5x \leqslant 5 \qquad\) and \(\qquad 6x + 7 \leqslant 4x + 4\)
Show that there is exactly one integer that \(x\) can be. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(-\dfrac{11}{5} \lt x \leqslant \dfrac{5}{5}\) or \(-2.2 \lt x \leqslant \dfrac{5}{5}\) | M1 | oe eg \(x \leqslant \dfrac{5}{5}\) and \(x \gt -\dfrac{11}{5}\) |
| \(-\dfrac{11}{5} \lt x \leqslant 1\) or \(-2.2 \lt x \leqslant 1\) or \(-2 \leqslant x \leqslant 1\) or \(-2, -1, 0, 1\) | A1 | oe eg \(x \leqslant 1\) and \(x \gt -\dfrac{11}{5}\) |
| \(6x - 4x \leqslant 4 - 7\) or \(2x \leqslant -3\) | M1 | oe Collects terms |
| \(x \leqslant -\dfrac{3}{2}\) or \(x \leqslant -1.5\) or \(x \lt -\dfrac{3}{2}\) or \(x \lt -1.5\) or \(x \leqslant -2\) or \(-2, -3\) (, \(-4\), ....) | A1 | \(-2.2 \lt x \leqslant -1.5\) or \(-2 \leqslant x \leqslant -1.5\) implies M1A1M1A1 |
| \(-2\) with no other values given | A1 | Must have gained M1A1M1A1 |
| Alternative method 2 | ||
| Shows that \(-2\) satisfies either \(-11 \lt 5x \leqslant 5\) or \(6x + 7 \leqslant 4x + 4\) | M1 | eg \(-11 \lt -10 \leqslant 5\) or \(5x = -10\) and yes |
| Shows that \(-2\) satisfies both \(-11 \lt 5x \leqslant 5\) and \(6x + 7 \leqslant 4x + 4\) | A1 | |
| Shows that \(-1\) does not satisfy \(6x + 7 \leqslant 4x + 4\) or shows that \(-3\) does not satisfy \(-11 \lt 5x \leqslant 5\) | M1 | eg \(-6 + 7 \gt -4 + 4\) |
| Shows that \(-1\) does not satisfy \(6x + 7 \leqslant 4x + 4\) and shows that \(-3\) does not satisfy \(-11 \lt 5x \leqslant 5\) | A1 | |
| \(-2\) with no other values given | A1 | Must have gained M1A1M1A1 |
| Alternative method 3 | ||
| \(-\dfrac{11}{5} \lt x \leqslant \dfrac{5}{5}\) or \(-2.2 \lt x \leqslant \dfrac{5}{5}\) | M1 | oe eg \(x \leqslant \dfrac{5}{5}\) and \(x \gt -\dfrac{11}{5}\) |
| \(-\dfrac{11}{5} \lt x \leqslant 1\) or \(-2.2 \lt x \leqslant 1\) or \(-2 \leqslant x \leqslant 1\) or \(-2, -1, 0, 1\) | A1 | oe eg \(x \leqslant 1\) and \(x \gt -\dfrac{11}{5}\) |
| Shows that \(-2\) satisfies \(6x + 7 \leqslant 4x + 4\) or shows that \(-1\) does not satisfy \(6x + 7 \leqslant 4x + 4\) | M1 | eg \(6 \times -2 + 7 = -5\) and \(4 \times -2 + 4 = -4\) ✓ |
| Shows that \(-2\) satisfies \(6x + 7 \leqslant 4x + 4\) and shows that \(-1\) does not satisfy \(6x + 7 \leqslant 4x + 4\) | A1 | |
| \(-2\) with no other values given | A1 | Must have gained M1A1M1A1 |
| Alternative method 4 | ||
| \(6x - 4x \leqslant 4 - 7\) or \(2x \leqslant -3\) | M1 | oe Collects terms |
| \(x \leqslant -\dfrac{3}{2}\) or \(x \leqslant -1.5\) or \(x \lt -\dfrac{3}{2}\) or \(x \lt -1.5\) or \(x \leqslant -2\) or \(-2, -3\) (, \(-4\), ....) | A1 | |
| Shows that \(-2\) satisfies \(-11 \lt 5x \leqslant 5\) or shows that \(-3\) does not satisfy \(-11 \lt 5x \leqslant 5\) | M1 | eg \(-11 \lt -10 \leqslant 5\) or \(5x = -10\) and yes |
| Shows that \(-2\) satisfies \(-11 \lt 5x \leqslant 5\) and shows that \(-3\) does not satisfy \(-11 \lt 5x \leqslant 5\) | A1 | |
| \(-2\) with no other values given | A1 | Must have gained M1A1M1A1 |
Additional guidance
| Allow eg max 1 and min \(-2.2\) for \(-2.2 \lt x \leqslant 1\), unless contradicted by a list of values | |
| Condone omission of non-critical values from lists eg \(-2, -1, 1\) | |
| Using = signs when solving inequalities can score M marks only unless recovered | |
| Incorrect notation eg \(\leqslant\) for \(\lt\) can score M marks only | |
| If answers to trials evaluated they must be correct | |
| Choose the scheme that favours the student | |
| \(-2\) identified as the only integer with no valid working | Zero |