A2 June 2025 Q3
3.
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
Use algebra to determine the values of \(x\) for which
\[\frac{x^2 - 4}{|x - 5|} \gt 4x + 8\](6)
| Scheme | Marks | AO |
|---|---|---|
| \(x = 5\) or \(x \neq 5\) | B1 | 2.2a |
| \(\{x \gt 5\}\) \(x^2 - 4 = (8 + 4x)(x - 5) \Rightarrow 3x^2 - 12x - 36 = 0 \Rightarrow x = \ldots\) or \(\{x \lt 5\}\) \(x^2 - 4 = (8 + 4x)(5 - x) \Rightarrow 5x^2 - 12x - 44 = 0 \Rightarrow x = \ldots\) | M1 | 1.1b |
| \(\{x \gt 5\}\) \(x^2 - 4 = (8 + 4x)(x - 5) \Rightarrow 3x^2 - 12x - 36 = 0 \Rightarrow x = \ldots\) And \(\{x \lt 5\}\) \(x^2 - 4 = (8 + 4x)(5 - x) \Rightarrow 5x^2 - 12x - 44 = 0 \Rightarrow x = \ldots\) | M1 | 3.1a |
| Any 2 of: \(x = -2, \dfrac{22}{5}, 6\) | A1 | 1.1b |
| \(x \lt -2,\ \dfrac{22}{5} \lt x \lt 5,\ \ 5 \lt x \lt 6\) o.e. | A1 A1 | 1.1b 3.2a |
| (6) | ||
| (6 marks) |
Notes
B1: Deduces that \(x = 5\) is a critical value. May be seen stated or as part of their final answer \(x \neq 5\)
M1: Attempts to find the critical values by solving a quadratic equation for \(x \gt 5\) or for \(x \lt 5\)
M1: A complete method to find all the critical values by solving two quadratic equations. E.g. considers both \(x \gt 5\) and \(x \lt 5\)
A1: Obtains any 2 correct critical values as long as one of the previous method marks has been scored
A1: Obtains any 2 correct regions. Note that \(\dfrac{22}{5} \lt x \lt 6,\ \ x \neq 5\) counts as 2 correct regions.
A1: Obtains the fully correct regions with no extras. Note that \(x \lt -2,\ \dfrac{22}{5} \lt x \lt 6,\ \ x \neq 5\) is also fully correct. If uses set notation it must be correct, or, \(\cup\) not \(\cap\) isw
Alternative 1 by squaring:
| Scheme | Marks | AO |
|---|---|---|
| \(x = 5\) or \(x \neq 5\) | B1 | 2.2a |
| \(\dfrac{x^2 - 4}{|x - 5|} = 4x + 8 \Rightarrow \dfrac{x^4 - 8x^2 + 16}{x^2 - 10x + 25} = 16x^2 + 64x + 64\) \(x^4 - 8x^2 + 16 = (16x^2 + 64x + 64)(x^2 - 10x + 25)\) | M1 | 3.1a |
| \(15x^4 - 96x^3 - 168x^2 + 960x + 1584 = 0 \Rightarrow x = \ldots\) | M1 | 1.1b |
| Any 2 of: \(x = -2, \dfrac{22}{5}, 6\) | A1 | 1.1b |
| \(x \lt -2,\ \dfrac{22}{5} \lt x \lt 5,\ \ 5 \lt x \lt 6\) o.e. | A1 A1 | 1.1b 3.2a |
B1: Deduces that \(x = 5\) is a critical value. May be seen stated or as part of their final answer \(x \neq 5\)
M1: Attempts to find square both sides and multiples up or vice versa in order to obtain a quartic equation that will give the critical values
M1: A complete method to find all four of the critical values by solving a quartic equation may use a calculator
A1: Obtains any 2 correct critical values, as long as they have scored at least one of the previous method marks. May be seen as an inequality.
A1: Obtains any 2 correct regions. Note that \(\dfrac{22}{5} \lt x \lt 6,\ \ x \neq 5\) counts as 2 correct regions.
A1: Obtains the fully correct regions with no extras. Note that \(x \lt -2,\ \dfrac{22}{5} \lt x \lt 6,\ \ x \neq 5\) is also fully correct. If uses set notation it must be correct, or, \(\cup\) not \(\cap\) isw
Alternative 2
| Scheme | Marks | AO |
|---|---|---|
| \(x = 5\) | B1 | 2.2a |
| \((x^2 - 4)(x - 5) = (8 + 4x)(x - 5)^2 \Rightarrow (x - 5)(3x^2 - 12x - 36) = 0 \Rightarrow x = \ldots\) or \(\dfrac{(x^2 - 4) - (4x + 8)(x - 5)}{x - 5} = \dfrac{3x^2 - 12x - 36}{x - 5} = 0 \Rightarrow x = \ldots\) OR \((x^2 - 4)(5 - x) = (8 + 4x)(5 - x)^2 \Rightarrow (5 - x)(5x^2 - 12x - 44) \Rightarrow x = \ldots\) or \(\dfrac{(x^2 - 4) - (4x + 8)(5 - x)}{5 - x} = \dfrac{5x^2 - 12x - 44}{5 - x} = 0 \Rightarrow x = \ldots\) | M1 | 3.1a |
| \((x^2 - 4)(x - 5) = (8 + 4x)(x - 5)^2 \Rightarrow (x - 5)(3x^2 - 12x - 36) = 0 \Rightarrow x = \ldots\) or \(\dfrac{(x^2 - 4) - (4x + 8)(x - 5)}{x - 5} = \dfrac{3x^2 - 12x - 36}{x - 5} = 0 \Rightarrow x = \ldots\) AND \((x^2 - 4)(5 - x) = (8 + 4x)(5 - x)^2 \Rightarrow (5 - x)(5x^2 - 12x - 44) = 0 \Rightarrow\) or \(\dfrac{(x^2 - 4) - (4x + 8)(5 - x)}{5 - x} = \dfrac{5x^2 - 12x - 44}{5 - x} = 0 \Rightarrow x = \ldots\) | M1 | 1.1b |
| Any 2 of: \(x = -2, \dfrac{22}{5}, 6\) | A1 | 1.1b |
| \(x \lt -2,\ \dfrac{22}{5} \lt x \lt 5,\ \ 5 \lt x \lt 6\) o.e. | A1 A1 | 1.1b 3.2a |
B1: Deduces that \(x = 5\) is a critical value. May be seen stated or as part of their final answer \(x \neq 5\)
M1: Either attempts to multiply both sides by \((x - 5)^2\) or \((5 - x)^2\) and solves to find the critical values. Or collects onto one side with a denominator of \(x - 5\) or \(5 - x\), uses a common denominator to combine and then finds the critical values may use a calculator
M1: Attempts to multiply both sides by \((x - 5)^2\) and \((5 - x)^2\) and solves to find the critical values. may use a calculator Or collects onto one side with a denominator of \(x - 5\) and \(5 - x\), uses a common denominator to combine and then finds the critical values
A1: Obtains any 2 correct critical values, as long as they have scored one of the previous method marks and using a correct equation
A1: Obtains any 2 correct regions. Note that \(\dfrac{22}{5} \lt x \lt 6,\ \ x \neq 5\) counts as 2 correct regions.
A1: Obtains the fully correct regions with no extras. Note that \(x \lt -2,\ \dfrac{22}{5} \lt x \lt 6,\ \ x \neq 5\) is also fully correct