June 2018 Paper 2 Q6
6 \(\mathrm{f}(x) = x^2 - 7 \qquad\) for all values of \(x\)
\(\mathrm{g}(x) = 1 - 3x \qquad\) for \(\ -4 \leqslant x \leqslant 4\)
(a) Work out the range of \(\mathrm{f}(x)\).
Give your answer as an inequality. [1 mark]
(b) Work out the range of \(\mathrm{g}(x)\).
Give your answer as an inequality. [2 marks]
(c) Solve \(\quad 2\mathrm{f}(x) = \mathrm{g}(x)\)
You must show your working.
Give your answers to 3 decimal places. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(\mathrm{f}(x) \geqslant -7\) or \(-7 \leqslant \mathrm{f}(x)\) | B1 |
Additional guidance
| \(\mathrm{f}(x)\) may be replaced by \(y\) or f or f\(x\) or g(\(x\)) or g or g\(x\) or \(x^2 - 7\) | |
| \(x \geqslant -7\) | B0 |
| \(\geqslant -7\) | B0 |
| Condone \(-7 \leqslant \mathrm{f}(x) \lt \infty\) or \(-7 \leqslant \mathrm{f}(x) \leqslant \infty\) or \(-7 \leqslant \mathrm{f}(x) \lt\) or \(-7 \leqslant \mathrm{f}(x) \leqslant\) | B1 |
| \([-7,\ \infty)\) or \([-7,\ \infty]\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| \(-11 \leqslant \mathrm{g}(x) \leqslant 13\) or \(13 \geqslant \mathrm{g}(x) \geqslant -11\) | B2 | B1 \(\mathrm{g}(x) \geqslant -11\) or \(\mathrm{g}(x) \leqslant 13\) on their own or embedded within an inequality or \(-11 \lt \mathrm{g}(x) \lt 13\) or \([-11,\ 13]\) or \(-11 \leqslant x \leqslant 13\) |
Additional guidance
| \(\mathrm{g}(x)\) may be replaced by \(y\) or g or g\(x\) or f(\(x\)) or f or f\(x\) or \(1 - 3x\) in B2 or B1 responses | |
| \(\mathrm{g}(x) \geqslant -11 \quad \mathrm{g}(x) \leqslant 13\) | B1 |
| \(-11\) to 13 inclusive (‘inclusive’ must be seen) Do not allow if 24 also seen | B1 |
| B1 may be seen with an incorrect inequality eg1 \(\ -11 \lt \mathrm{g}(x) \leqslant 13\) eg2 \(\ -11 \leqslant \mathrm{g}(x) \lt 13\) eg3 \(\ 0 \lt \mathrm{g}(x) \leqslant 13\) eg4 \(\ 13 \leqslant \mathrm{g}(x) \geqslant -11\) | B1 B1 B1 B1 |
| \([-11,\ 13)\) or \((-11,\ 13]\) or \((-11,\ 13)\) | B0 |
| \(-11 \lt x \leqslant 13\) or \(-11 \leqslant x \lt 13\) or \(-11 \lt x \lt 13\) | B0 |
| \(\{-11,\ -10,\ -9,\ \ldots\ldots\ 0,\ 1,\ 2,\ 3,\ \ldots\ldots,\ 12,\ 13\}\) | B0 |
| Answer | Mark | Comments |
|---|---|---|
| \(2x^2 - 14\) | M1 | |
| \(2x^2 + 3x - 15\ (= 0)\) or \(-2x^2 - 3x + 15\ (= 0)\) or \(2x^2 + 3x = 15\) or \(-2x^2 - 3x = -15\) | A1 | |
| \(\dfrac{-3 \pm \sqrt{3^2 - 4 \times 2 \times -15}}{2 \times 2}\) or \(\dfrac{-3 \pm \sqrt{9 + 120}}{4}\) or \(\dfrac{-3 \pm \sqrt{129}}{4}\) | M1 | oe eg \(-\dfrac{3}{4} \pm \sqrt{\dfrac{15}{2} + \left(\dfrac{3}{4}\right)^2}\) correct method to solve their 3-term quadratic implied by correct solutions to their 3-term quadratic to at least 2 dp |
| 2.089 \(\quad -3.589\) | A1ft | correct or ft M1A0M1 or M0A0M1 must both be rounded to 3 decimal places |
Additional guidance
| 2nd M1 Allow correct factorisation of their 3-term quadratic if it does factorise | |
| 2nd M1 Allow correct use of formula even if discriminant is negative | |
| Two ‘correct’ solutions to at least 2 decimal places implies M1A1M1 eg 2.09 and \(-3.59\) | M1A1M1A0 |
| 2.089 and \(-3.589\) in working but only one on answer line | M1A1M1A0 |
| Answers only 2.089 \(\quad -3.589\) | M1A1M1A1 |
| Answer only 2.089 | Zero |
| Answer only \(-3.589\) | Zero |
| \(2x^2 - 7\) from incorrect expansion leading to 1.386 \(\quad -2.886\) | M0A0M1A1ft |
| \(x^2 - 14\) from incorrect expansion leading to 2.653 \(\quad -5.653\) | M0A0M1A1ft |
| \(2x^2 - 14\) and \(2x^2 + 3x - 13\ (= 0)\) Answers 1.908 \(\quad -3.408\) | M1A0 M1A1ft |