| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Area with Turning Points |
| Difficulty | Standard +0.8 This question requires finding a turning point by differentiation (using product rule), then setting up and evaluating a definite integral with polynomial terms. While the individual techniques are standard AS-level content, the multi-step nature (find turning point, determine correct bounds, integrate x³ and x² terms, evaluate exactly) and the need to coordinate these steps makes it moderately challenging—above average but not exceptionally difficult. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(y=(x-2)^2(x+3)=(x^2-4x+4)(x+3)=x^3-1x^2-8x+12\) | B1 | Correct expansion seen at some point |
| Attempt to find \(x\)-coordinate of maximum: expand and differentiate then solve \(\frac{dy}{dx}=0\), OR use product rule then solve \(\frac{dy}{dx}=0\) | M1 | Problem-solving mark for method |
| \(y=x^3-x^2-8x+12 \Rightarrow \frac{dy}{dx}=3x^2-2x-8\) | M1 | At least two terms correctly differentiated |
| \(\frac{dy}{dx}=0 \Rightarrow (x-2)(3x+4)=0\) | M1 | Sets \(\frac{dy}{dx}=0\), solves by correct method |
| \(x = -\frac{4}{3}\) | A1 | Correct value |
| Attempt area under \(y=(x-2)^2(x+3)\) between two values: expand then apply two limits | M1 | Problem-solving mark for integration method |
| \(\displaystyle\int\!\left(x^3-x^2-8x+12\right)dx = \left[\frac{x^4}{4}-\frac{x^3}{3}-4x^2+12x\right]\) | M1 | At least two terms correctly integrated |
| Uses top limit \(2\) and bottom limit \(x=-\frac{4}{3}\): \(\left[\frac{x^4}{4}-\frac{x^3}{3}-4x^2+12x\right]_{-\frac{4}{3}}^{2}\) | M1 | Correct limits applied |
| \(\frac{28}{3} - \left(-\frac{1744}{81}\right) = \frac{2500}{81}\) | A1 | Correct final answer \(R=\frac{2500}{81}\) |
## Question 14:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $y=(x-2)^2(x+3)=(x^2-4x+4)(x+3)=x^3-1x^2-8x+12$ | B1 | Correct expansion seen at some point |
| Attempt to find $x$-coordinate of maximum: expand and differentiate then solve $\frac{dy}{dx}=0$, OR use product rule then solve $\frac{dy}{dx}=0$ | M1 | Problem-solving mark for method |
| $y=x^3-x^2-8x+12 \Rightarrow \frac{dy}{dx}=3x^2-2x-8$ | M1 | At least two terms correctly differentiated |
| $\frac{dy}{dx}=0 \Rightarrow (x-2)(3x+4)=0$ | M1 | Sets $\frac{dy}{dx}=0$, solves by correct method |
| $x = -\frac{4}{3}$ | A1 | Correct value |
| Attempt area under $y=(x-2)^2(x+3)$ between two values: expand then apply two limits | M1 | Problem-solving mark for integration method |
| $\displaystyle\int\!\left(x^3-x^2-8x+12\right)dx = \left[\frac{x^4}{4}-\frac{x^3}{3}-4x^2+12x\right]$ | M1 | At least two terms correctly integrated |
| Uses top limit $2$ and bottom limit $x=-\frac{4}{3}$: $\left[\frac{x^4}{4}-\frac{x^3}{3}-4x^2+12x\right]_{-\frac{4}{3}}^{2}$ | M1 | Correct limits applied |
| $\frac{28}{3} - \left(-\frac{1744}{81}\right) = \frac{2500}{81}$ | A1 | Correct final answer $R=\frac{2500}{81}$ |14.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fa7abe9f-f5c0-4578-afd1-73176c717536-30_673_819_246_623}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Figure 5 shows a sketch of the curve $C$ with equation $y = ( x - 2 ) ^ { 2 } ( x + 3 )$
The region $R$, shown shaded in Figure 5, is bounded by $C$, the vertical line passing through the maximum turning point of $C$ and the $x$-axis.
Find the exact area of $R$.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\hfill \mbox{\textit{Edexcel AS Paper 1 Q14 [9]}}