June 2018 Paper 2 Q1
1.
\[\mathrm{g}(x) = \frac{2x + 5}{x - 3} \qquad x \geqslant 5\]| Scheme | Marks | AO |
|---|---|---|
| Way 1: \(\mathrm{g}(5) = \dfrac{2(5) + 5}{5 - 3} = 7.5 \Rightarrow \mathrm{gg}(5) = \dfrac{2(\text{``}7.5\text{''}) + 5}{\text{``}7.5\text{''} - 3}\) | M1 | 1.1b |
| \(\mathrm{gg}(5) = \dfrac{40}{9}\ \left(\text{or } 4\dfrac{4}{9} \text{ or } 4.\dot{4}\right)\) | A1 | 1.1b |
| (2) |
Notes
(a) Way 2
| Scheme | Marks | AO |
|---|---|---|
| \(\mathrm{gg}(x) = \dfrac{2\left(\dfrac{2x + 5}{x - 3}\right) + 5}{\left(\dfrac{2x + 5}{x - 3}\right) - 3} \Rightarrow \mathrm{gg}(5) = \dfrac{2\left(\dfrac{2(5) + 5}{(5) - 3}\right) + 5}{\left(\dfrac{2(5) + 5}{(5) - 3}\right) - 3}\) | M1 | 1.1b |
| \(\mathrm{gg}(5) = \dfrac{40}{9}\ \left(\text{or } 4\dfrac{4}{9} \text{ or } 4.\dot{4}\right)\) | A1 | 1.1b |
| (2) |
M1: Full method of attempting \(\mathrm{g}(5)\) and substituting the result into g
Note: Way 2: Attempts to substitute \(x = 5\) into \(\dfrac{2\left(\dfrac{2x + 5}{x - 3}\right) + 5}{\left(\dfrac{2x + 5}{x - 3}\right) - 3}\), o.e. Note that \(\mathrm{gg}(x) = \dfrac{9x - 5}{14 - x}\)
A1: Obtains \(\dfrac{40}{9}\) or \(4\dfrac{4}{9}\) or \(4.\dot{4}\) or an exact equivalent
Note: Give A0 for 4.4 or 4.444... without reference to \(\dfrac{40}{9}\) or \(4\dfrac{4}{9}\) or \(4.\dot{4}\)
| Scheme | Marks | AO |
|---|---|---|
| {Range:} \(2 \lt y \leqslant \dfrac{15}{2}\) | B1 | 1.1b |
| (1) |
Notes
B1: States \(2 \lt y \leqslant \dfrac{15}{2}\) Accept any of \(2 \lt \mathrm{g} \leqslant \dfrac{15}{2},\ 2 \lt \mathrm{g}(x) \leqslant \dfrac{15}{2},\ \left(2, \dfrac{15}{2}\right]\)
Note: Accept \(\mathrm{g}(x) \gt 2\) and \(\mathrm{g}(x) \leqslant \dfrac{15}{2}\) o.e.
| Scheme | Marks | AO |
|---|---|---|
| Way 1: \(y = \dfrac{2x + 5}{x - 3} \Rightarrow yx - 3y = 2x + 5 \Rightarrow yx - 2x = 3y + 5\) | M1 | 1.1b |
| \(x(y - 2) = 3y + 5 \Rightarrow x = \dfrac{3y + 5}{y - 2}\ \left\{\text{or } y = \dfrac{3x + 5}{x - 2}\right\}\) | M1 | 2.1 |
| \(\mathrm{g}^{-1}(x) = \dfrac{3x + 5}{x - 2},\quad 2 \lt x \leqslant \dfrac{15}{2}\) | A1ft | 2.5 |
| (3) | ||
| (6 marks) |
Notes
(c) Way 2
| Scheme | Marks | AO |
|---|---|---|
| \(y = \dfrac{2x - 6 + 11}{x - 3} \Rightarrow y = 2 + \dfrac{11}{x - 3} \Rightarrow y - 2 = \dfrac{11}{x - 3}\) | M1 | 1.1b |
| \(x - 3 = \dfrac{11}{y - 2} \Rightarrow x = \dfrac{11}{y - 2} + 3\ \left\{\text{or } y = \dfrac{11}{x - 2} + 3\right\}\) | M1 | 2.1 |
| \(\mathrm{g}^{-1}(x) = \dfrac{11}{x - 2} + 3,\quad 2 \lt x \leqslant \dfrac{15}{2}\) | A1ft | 2.5 |
| (3) |
(c) Way 1
M1: Correct method of cross multiplication followed by an attempt to collect terms in \(x\) or terms in a swapped \(y\)
M1: A complete method (i.e. as above and also factorising and dividing) to find the inverse
A1ft: Uses correct notation to correctly define the inverse function \(\mathrm{g}^{-1}\), where the domain of \(\mathrm{g}^{-1}\) stated correctly or correctly followed through (using correct notation) on the values shown in their range in part (b). Allow \(\mathrm{g}^{-1} : x \rightarrow\). Condone \(\mathrm{g}^{-1} = \ldots\) Do not accept \(y = \ldots\)
Note: Correct notation is required when stating the domain of \(\mathrm{g}^{-1}(x)\). Allow \(2 \lt x \leqslant \dfrac{15}{2}\) or \(\left(2, \dfrac{15}{2}\right]\)
Do not allow any of e.g. \(2 \lt \mathrm{g} \leqslant \dfrac{15}{2},\ 2 \lt \mathrm{g}^{-1}(x) \leqslant \dfrac{15}{2}\)
Note: Do not allow A1ft for following through their range in (b) to give a domain for \(\mathrm{g}^{-1}\) as \(x \in \mathbb{R}\)
(c) Way 2
M1: Writes \(y = \dfrac{2x + 5}{x - 3}\) in the form \(y = 2 \pm \dfrac{k}{x - 3},\ k \neq 0\) and rearranges to isolate \(y\) and 2 on one side of their equation. Note: Allow the equivalent method with \(x\) swapped with \(y\)
M1: A complete method to find the inverse
A1ft: As in Way 1
Note: If a candidate scores no marks in part (c), but
- states the domain of \(\mathrm{g}^{-1}\) correctly, or
- states a domain of \(\mathrm{g}^{-1}\) which is correctly followed through on the values shown in their range in part (b)
then give special case (SC) M1 M0 A0