| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2006 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area between curve and line |
| Difficulty | Moderate -0.3 This is a structured, multi-part C1 question that guides students through standard techniques: finding areas of rectangles, basic integration, area between curves, differentiation, tangent equations, and solving quadratic inequalities. While it requires multiple steps and careful setup, each individual part uses routine methods with clear scaffolding, making it slightly easier than the average A-level question. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| \(y_D = 3 + 1 = 4\) or \(y_C = 12 - 8 = 4\) | M1 | Attempt at either y coordinate |
| Area \(ABCD = 3 \times 4 = 12\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^3 - \frac{x^4}{4}(+C)\) | M1 | Increase one power by 1 |
| A1 | — | One term correct unsimplified |
| A1 | 3 | All correct unsimplified (condone no +C) |
| Answer | Marks | Guidance |
|---|---|---|
| Sub limits –1 and 2 into their (b)(i) ans | M1 | May use both –1, 0 and 0, 2 instead |
| \([8-4]-\left[-1-\frac{1}{4}\right] = 5\frac{1}{4}\) | A1 | — |
| Shaded area = "their" (rectangle – integral) | M1 | Alt method: difference of two integrals |
| \(= 12 - 5\frac{1}{4} = 6\frac{3}{4}\) | A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 6x - 3x^2\) | M1 | One term correct |
| A1 | 2 | All correct (no +C etc) |
| Answer | Marks | Guidance |
|---|---|---|
| When \(x = 1, y = 2\) when \(x = 1\), | B1 | May be implied by correct tgt equation |
| \(\frac{dy}{dx} = 3\) as 'their' grad of tgt | M1√ | Ft their derivative when x = 1 |
| Tangent is \(y - 2 = 3(x-1)\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Decreasing when \(\frac{dy}{dx} = 6x - 3x^2 < 0\) | M1 | Watch no fudging here! May work backwards in proof. |
| \(3(2x - x^2) < 0 \Rightarrow x^2 - 2x > 0\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Two critical points 0 and 2 | M1 | Marked on diagram or in solution |
| \(x > 2, x < 0\) ONLY | A1 | 2 |
## 8(a)
$y_D = 3 + 1 = 4$ or $y_C = 12 - 8 = 4$ | M1 | Attempt at either y coordinate
Area $ABCD = 3 \times 4 = 12$ | A1 | 2 | —
## 8(b)(i)
$x^3 - \frac{x^4}{4}(+C)$ | M1 | Increase one power by 1
| A1 | — | One term correct unsimplified
| A1 | 3 | All correct unsimplified (condone no +C)
## 8(b)(ii)
Sub limits –1 and 2 into their (b)(i) ans | M1 | May use both –1, 0 and 0, 2 instead
$[8-4]-\left[-1-\frac{1}{4}\right] = 5\frac{1}{4}$ | A1 | —
Shaded area = "their" (rectangle – integral) | M1 | Alt method: difference of two integrals
$= 12 - 5\frac{1}{4} = 6\frac{3}{4}$ | A1 | 4 | CSO. Attempted M2, A2
## 8(c)(i)
$\frac{dy}{dx} = 6x - 3x^2$ | M1 | One term correct
| A1 | 2 | All correct (no +C etc)
## 8(c)(ii)
When $x = 1, y = 2$ when $x = 1$, | B1 | May be implied by correct tgt equation
$\frac{dy}{dx} = 3$ as 'their' grad of tgt | M1√ | Ft their derivative when x = 1
Tangent is $y - 2 = 3(x-1)$ | A1 | 3 | Any correct form; $y = 3x - 1$ etc
## 8(c)(iii)
Decreasing when $\frac{dy}{dx} = 6x - 3x^2 < 0$ | M1 | Watch no fudging here! May work backwards in proof.
$3(2x - x^2) < 0 \Rightarrow x^2 - 2x > 0$ | A1 | 2 | AG (be convinced no step incorrect)
## 8(d)
Two critical points 0 and 2 | M1 | Marked on diagram or in solution
$x > 2, x < 0$ ONLY | A1 | 2 | or M1 A0 for $0 < x < 2$ or $0 > x > 2$; **SC B1** for $x > 2$ (or $x < 0$)
---
**TOTAL: 75 marks**8 The diagram shows the curve with equation $y = 3 x ^ { 2 } - x ^ { 3 }$ and the line $L$.\\
\includegraphics[max width=\textwidth, alt={}, center]{81f6fc30-982b-47b5-bab3-076cc0cc6563-5_479_816_406_596}
The points $A$ and $B$ have coordinates $( - 1,0 )$ and $( 2,0 )$ respectively. The curve touches the $x$-axis at the origin $O$ and crosses the $x$-axis at the point $( 3,0 )$. The line $L$ cuts the curve at the point $D$ where $x = - 1$ and touches the curve at $C$ where $x = 2$.
\begin{enumerate}[label=(\alph*)]
\item Find the area of the rectangle $A B C D$.
\item \begin{enumerate}[label=(\roman*)]
\item Find $\int \left( 3 x ^ { 2 } - x ^ { 3 } \right) \mathrm { d } x$.
\item Hence find the area of the shaded region bounded by the curve and the line $L$.
\end{enumerate}\item For the curve above with equation $y = 3 x ^ { 2 } - x ^ { 3 }$ :
\begin{enumerate}[label=(\roman*)]
\item find $\frac { \mathrm { d } y } { \mathrm {~d} x }$;
\item hence find an equation of the tangent at the point on the curve where $x = 1$;
\item show that $y$ is decreasing when $x ^ { 2 } - 2 x > 0$.
\end{enumerate}\item Solve the inequality $x ^ { 2 } - 2 x > 0$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2006 Q8 [18]}}