| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Solve quadratic inequality |
| Difficulty | Standard +0.8 This Further Maths question requires students to work backwards from a given solution set to determine unknown parameters in a product of quadratics. It demands understanding of sign analysis, factorization of the first quadratic to find roots at x=-3 and x=8, then deducing that the second quadratic must have roots at x=-9 and x=b, requiring both algebraic manipulation and careful reasoning about inequality solution sets. |
| Spec | 5.02c Linear coding: effects on mean and variance |
| Question number | Additional page, if required. Write the question numbers in the left-hand margin. |
| \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\) | |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Factorise \(x^2-5x-24 = (x-8)(x+3)\), so critical values include \(-3\) and \(8\), therefore \(b=8\) | 3.1a M1 | Factorise \(x^2-5x-24\) to \((x+m)(x+n)\) where \(m+n=-5\) or \(mn=-24\) and identifies \(m,n>2\) as \(b\). Or uses coefficients of quartic to form equation in \(a\) and/or \(b\). Or multiplies two or more of \((x+9),(x+3),(x-2),(x-b)\). |
| \(a=-18\) or \(b=8\) | 1.1b A1 | Expands \((x+9)(x-2)\) and identifies constant term as \(a\). Or correctly forms two equations in \(a\) and \(b\). |
| \((x+9)(x-2) = x^2+7x-18\), therefore \(a=-18\) | 3.1a M1 | Divides expanded quartic by quadratic/cubic formed from two or three of the factors. Or compares coefficients in expansions of \((x^2-5x-24)(x^2+7x+a)\) and \((x+9)(x+3)(x-2)(x-b)\). |
| \(a=-18\) and \(b=8\) | 1.1b A1 | Both values correct. |
## Question 14:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Factorise $x^2-5x-24 = (x-8)(x+3)$, so critical values include $-3$ and $8$, therefore $b=8$ | 3.1a M1 | Factorise $x^2-5x-24$ to $(x+m)(x+n)$ where $m+n=-5$ or $mn=-24$ and identifies $m,n>2$ as $b$. Or uses coefficients of quartic to form equation in $a$ and/or $b$. Or multiplies two or more of $(x+9),(x+3),(x-2),(x-b)$. |
| $a=-18$ **or** $b=8$ | 1.1b A1 | Expands $(x+9)(x-2)$ and identifies constant term as $a$. Or correctly forms two equations in $a$ and $b$. |
| $(x+9)(x-2) = x^2+7x-18$, therefore $a=-18$ | 3.1a M1 | Divides expanded quartic by quadratic/cubic formed from two or three of the factors. Or compares coefficients in expansions of $(x^2-5x-24)(x^2+7x+a)$ and $(x+9)(x+3)(x-2)(x-b)$. |
| $a=-18$ **and** $b=8$ | 1.1b A1 | Both values correct. |14 The inequality
$$\left( x ^ { 2 } - 5 x - 24 \right) \left( x ^ { 2 } + 7 x + a \right) < 0$$
has the solution set
$$\{ x : - 9 < x < - 3 \} \cup \{ x : 2 < x < b \}$$
Find the values of integers $a$ and $b$\\
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\hfill \mbox{\textit{AQA Further AS Paper 1 2023 Q14 [4]}}