| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Write inequalities from graph |
| Difficulty | Standard +0.3 This is a multi-part question requiring students to write inequalities from a graph, find intersection points by solving a quadratic equation, and calculate an area using integration. While it involves several techniques (solving quadratics, integration, interpreting graphs), each component is standard AS-level material with straightforward execution. The area calculation requires subtracting one integral from another, which is routine. Slightly easier than average due to the visual scaffolding and standard procedures. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02i Represent inequalities: graphically on coordinate plane1.02q Use intersection points: of graphs to solve equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y \leq x + 2\) | B1 | AO2.2a – Deduces one correct inequality related to sloping line or curve; condone strict inequalities |
| \(y \geq x^2 - 4x - 12\) | (included above) | |
| \(y \geq 0\) | B1 | AO2.2a – Deduces the other two correct inequalities; condone strict inequalities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(A\) is \((-2, 0)\) | B1 | AO1.1b – States \(x\) coordinate of \(A\) is \(-2\) |
| \(B\) is \((6, 0)\) | B1 | AO1.1b – States \(x\) coordinate of \(B\) is \(6\) |
| \(x + 2 = x^2 - 4x - 12\), giving \(x^2 - 5x - 14 = 0\), \((x+2)(x-7)=0\) | M1 | AO1.1a – Eliminates \(y\) to obtain \(x\) coordinate of \(C = 7\) |
| \(C\) is point \((7, 9)\) | A1 | AO1.1b – Obtains correct \(y\) coordinates of \(A\), \(B\) and \(C\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Area of triangle under \(AC = 0.5 \times 9 \times 9 = 40.5\) | B1 | AO1.1b – Obtains correct value for area under \(AC\) |
| \(\int_6^7 (x^2 - 4x - 12)\, dx\) | M1 | AO1.1a – Integrates a quadratic expression with \(\frac{x^3}{3}\) term correct; PI by \(\frac{13}{3}\) ACF |
| \(= \left[\frac{x^3}{3} - 2x^2 - 12x\right]_6^7\) | A1 | AO1.1b – Integrates \(x^2 - 4x - 12\) completely correctly; condone \(+c\); PI by \(\frac{13}{3}\) ACF; condone integration of \(x^2 - 5x - 14\) correctly |
| \(= \frac{343}{3} - 98 - 84 - 72 + 72 + 72\) | M1 | AO1.1a – Substitutes a pair of limits into their integrated quadratic; must be three terms including subtraction; PI by \(\frac{13}{3}\) ACF |
| Shaded area \(= 40.5 - 4\frac{1}{3}\) | M1 | AO3.1a – Uses a correct method to combine areas that lead to the exact area of the shaded region |
| \(= 36\frac{1}{6}\) or \(\frac{217}{6}\) | R1 | AO2.1 – ISW |
## Question 9(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y \leq x + 2$ | B1 | AO2.2a – Deduces one correct inequality related to sloping line or curve; condone strict inequalities |
| $y \geq x^2 - 4x - 12$ | (included above) | |
| $y \geq 0$ | B1 | AO2.2a – Deduces the other two correct inequalities; condone strict inequalities |
---
## Question 9(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $A$ is $(-2, 0)$ | B1 | AO1.1b – States $x$ coordinate of $A$ is $-2$ |
| $B$ is $(6, 0)$ | B1 | AO1.1b – States $x$ coordinate of $B$ is $6$ |
| $x + 2 = x^2 - 4x - 12$, giving $x^2 - 5x - 14 = 0$, $(x+2)(x-7)=0$ | M1 | AO1.1a – Eliminates $y$ to obtain $x$ coordinate of $C = 7$ |
| $C$ is point $(7, 9)$ | A1 | AO1.1b – Obtains correct $y$ coordinates of $A$, $B$ and $C$ |
---
## Question 9(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Area of triangle under $AC = 0.5 \times 9 \times 9 = 40.5$ | B1 | AO1.1b – Obtains correct value for area under $AC$ |
| $\int_6^7 (x^2 - 4x - 12)\, dx$ | M1 | AO1.1a – Integrates a quadratic expression with $\frac{x^3}{3}$ term correct; PI by $\frac{13}{3}$ ACF |
| $= \left[\frac{x^3}{3} - 2x^2 - 12x\right]_6^7$ | A1 | AO1.1b – Integrates $x^2 - 4x - 12$ completely correctly; condone $+c$; PI by $\frac{13}{3}$ ACF; condone integration of $x^2 - 5x - 14$ correctly |
| $= \frac{343}{3} - 98 - 84 - 72 + 72 + 72$ | M1 | AO1.1a – Substitutes a pair of limits into their integrated quadratic; must be three terms including subtraction; PI by $\frac{13}{3}$ ACF |
| Shaded area $= 40.5 - 4\frac{1}{3}$ | M1 | AO3.1a – Uses a correct method to combine areas that lead to the exact area of the shaded region |
| $= 36\frac{1}{6}$ or $\frac{217}{6}$ | R1 | AO2.1 – ISW |
---9 The diagram below shows the graphs of $y = x ^ { 2 } - 4 x - 12$ and $y = x + 2$\\
\includegraphics[max width=\textwidth, alt={}, center]{11168e8f-5ba5-4d27-83ab-0327cc23d08c-10_933_912_358_566}
9
\begin{enumerate}[label=(\alph*)]
\item Write down three inequalities which together describe the shaded region.\\
9
\item Find the coordinates of the points $A , B$ and $C$.\\
9
\item Find the exact area of the shaded region.\\
Fully justify your answer.\\[0pt]
[6 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 2022 Q9 [12]}}