C3 June 2016 Q1
1. The functions f and g are defined by\[\mathrm{f} : x \to 7x - 1, \qquad x \in \mathbb{R}\]\[\mathrm{g} : x \to \frac{4}{x - 2}, \qquad x \neq 2, x \in \mathbb{R}\]
| Scheme | Marks |
|---|---|
| \(\mathrm{fg}(x) = \dfrac{28}{x - 2} - 1 \qquad \left(= \dfrac{30 - x}{x - 2}\right)\) | M1 |
| Sets \(\mathrm{fg}(x) = x \Rightarrow \dfrac{28}{x - 2} - 1 = x\) \(\Rightarrow 28 = (x + 1)(x - 2)\) | |
| \(\Rightarrow x^2 - x - 30 = 0\) | M1 |
| \(\Rightarrow (x - 6)(x + 5) = 0\) \(\Rightarrow x = 6, x = -5\) | dM1 A1 |
| (4) |
Alt 1(a)
| Scheme | Marks |
|---|---|
| \(\mathrm{fg}(x) = x \Rightarrow \mathrm{g}(x) = \mathrm{f}^{-1}(x)\) \(\dfrac{4}{x - 2} = \dfrac{x + 1}{7}\) | M1 |
| \(\Rightarrow x^2 - x - 30 = 0\) | M1 |
| \(\Rightarrow (x - 6)(x + 5) = 0\) \(\Rightarrow x = 6, x = -5\) | dM1 A1 |
| 4 marks |
S. Case
| Uses \(\mathrm{gf}(x)\) instead \(\mathrm{fg}(x)\) | Makes an error on \(\mathrm{fg}(x)\) | Marks |
|---|---|---|
| \(\dfrac{4}{7x - 1 - 2} = x\) | Sets \(\mathrm{fg}(x) = x \Rightarrow \dfrac{7 \times 4}{7 \times (x - 2)} - 1 = x\) | M0 |
| \(\Rightarrow 7x^2 - 3x - 4 = 0\) \(\Rightarrow (7x + 4)(x - 1) = 0\) | \(\Rightarrow x^2 - x - 6 = 0\) \(\Rightarrow (x + 2)(x - 3) = 0\) | M1 |
| \(\Rightarrow x = -\dfrac{4}{7}, \quad x = 1\) | \(\Rightarrow x = -2, \quad x = 3\) | dM1 A0 |
| 2 out of 4 marks |
Notes
M1: Sets or implies that \(\mathrm{fg}(x) = \dfrac{28}{x - 2} - 1\) Eg accept \(\mathrm{fg}(x) = 7\left(\dfrac{4}{x - 2}\right) - 1\) followed by \(\mathrm{fg}(x) = \dfrac{7 \times 4}{x - 2} - 1\)
Alternatively sets \(\mathrm{g}(x) = \mathrm{f}^{-1}(x)\) where \(\mathrm{f}^{-1}(x) = \dfrac{x \pm 1}{7}\)
Note that \(\mathrm{fg}(x) = 7\left(\dfrac{4}{x - 2}\right) - 1 = \dfrac{28}{7(x - 2)} - 1\) is M0
M1: Sets up a 3TQ (= 0) from an attempt at \(\mathrm{fg}(x) = x\) or \(\mathrm{g}(x) = \mathrm{f}^{-1}(x)\)
dM1: Method of solving 3TQ (= 0) to find at least one value for \(x\). See "General Priciples for Core Mathematics" on page 3 for the award of the mark for solving quadratic equations
This is dependent upon the previous M. You may just see the answers following the 3TQ.
A1: Both \(x = 6\) and \(x = -5\)
| Scheme | Marks |
|---|---|
| \(a = 6\) | B1 ft |
| (1) | |
| (5 marks) |
Notes
B1ft: For \(a = 6\) but you may follow through on the largest solution from part (a) provided more than one answer was found in (a). Accept 6, \(a = 6\) and even \(x = 6\)
Do not award marks for part (a) for work in part (b).