June 2022 Paper 2 Q1
1 In this question you must show detailed reasoning.
Solve the following equations.
| Scheme | Marks | AO |
|---|---|---|
| DR \(\dfrac{x(x + 2) - (x - 1)(x + 1)}{(x + 1)(x + 2)}\) or \(\dfrac{x^2 + 2x - x^2 + 1}{x^2 + 3x + 2}\) oe \((= 0)\) | M1 M1 | 1.1 1.1 |
| \(x = -\dfrac{1}{2}\) | A1 | 1.1 |
| [3] |
Notes
M1: M1 for \(x(x + 2) - (x + 1)(x - 1)\) oe
M1: Multiply out brackets. Allow one error
Ignore denominator even if “= 0”
A1: NB correct with no working: SC B1
Alternative method
| Scheme | Marks |
|---|---|
| \(x(x + 2) = (x + 1)(x - 1)\) | M1 |
| \(x^2 + 2x = x^2 - 1\) or \(2x = -1\) oe | M1 |
| \(x = -\dfrac{1}{2}\) | A1 |
M1: M1 for attempt “cross-multiply”.
M1: Multiply out brackets. Allow one error
| Scheme | Marks | AO |
|---|---|---|
| DR Solve quadratic in \(\frac{1}{x^3}\) or \(x^3\) or \(u\) \(\left(= x^3 \text{ or } \frac{1}{x^3}\right)\) using any correct method. | M1 | 3.1a |
| \(\dfrac{1}{x^3}\) (or \(u\)) \(= 1\ \& -\frac{1}{8}\) or \(x^3\) (or \(u\)) \(= 1\ \& -8\) or correct factorisation of quadratic | B1 | 1.1 |
| \(x = 1\ \&\ x = -2\) with no extras | B1f | 1.1 |
| [3] |
Notes
M1: or cubic in \(x\). Condone quadratic in \(x\) with \(x = \frac{1}{x^3}\) or \(x = x^3\)
Must see attempt at correct method for this mark
Allow arithmetical errors
B1: Can be scored without M1. Condone \(x = 1, -\frac{1}{8}\) or \(x = 1, -8\)
Ignore \(x^3 = 0\), if seen, for this mark
B1f: ft their \(x^3\) or \(\dfrac{1}{x^3}\). If also \(x = 0\), B0
NB correct with no working: M0B0B1
| Scheme | Marks | AO |
|---|---|---|
| DR eg \((x^2 - 7)\ln 3 = \ln\dfrac{1}{243}\) or \(x^2 - 7 = \log_3\left(\dfrac{1}{243}\right)\) or \(3^{x^2 - 7} = 3^{-5}\) or \(x^2 - 7 = -5\) or \(3^{x^2} = 3^2\) | M1 | 3.1a |
| \(x = \pm\sqrt{2}\) or \(\pm 1.41\) (3 sf) | A1 | 1.1 |
| [2] |
Notes
M1: Condone incorrect or omitted brackets
Any correct step after log(both sides)
or ANY correct step using indices
A1: NB correct with no working or T & I: SC B1