| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Completing the square and sketching |
| Type | Complete square then find vertex/turning point |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring completing the square (routine AS technique), finding a turning point from completed square form (direct reading), and integration to find area. The integration setup is clear from the diagram, requiring only standard polynomial integration between obvious limits. Slightly easier than average due to the scaffolded structure and standard techniques throughout. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x) = -3x^2 + 12x + 8 = -3(x \pm 2)^2 + \ldots\) | M1 | Attempts to take out common factor and complete the square; or compares \(-3x^2+12x+8\) to \(ax^2+2abx+ab^2+c\) |
| \(= -3(x-2)^2 + \ldots\) | A1 | Proceeds to \(-3(x-2)^2 + \ldots\) or finds \(a=-3, b=-2\) |
| \(= -3(x-2)^2 + 20\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Coordinates of \(M = (2, 20)\) | B1ft | One correct coordinate |
| B1ft | Correct coordinates; follow through on their \((-b, c)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int -3x^2 + 12x + 8 \, dx = -x^3 + 6x^2 + 8x\) | M1 A1 | M1: any correct index; A1: fully correct (may be unsimplified) |
| Method to find \(R =\) their \(2 \times 20 - \int_0^{2}(-3x^2+12x+8)\,dx\) | M1 | Look for their \(2 \times \text{"20"} - \int_0^{"2"} f(x)\,dx\) |
| \(R = 40 - \left[-2^3 + 24 + 16\right]\) | dM1 | Correct application of limits on their integrated function; their 2 must be used |
| \(= 8\) | A1 | Shows area of \(R = 8\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int 3x^2 - 12x + 12 \, dx = x^3 - 6x^2 + 12x\) | M1 A1 | — |
| \(R = \int_0^{2} 3x^2 - 12x + 12 \, dx\) | M1 | — |
| \(R = 2^3 - 6\times2^2 + 12\times2 = 8\) | dM1 A1 | — |
## Question 14:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = -3x^2 + 12x + 8 = -3(x \pm 2)^2 + \ldots$ | M1 | Attempts to take out common factor and complete the square; or compares $-3x^2+12x+8$ to $ax^2+2abx+ab^2+c$ |
| $= -3(x-2)^2 + \ldots$ | A1 | Proceeds to $-3(x-2)^2 + \ldots$ or finds $a=-3, b=-2$ |
| $= -3(x-2)^2 + 20$ | A1 | — |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Coordinates of $M = (2, 20)$ | B1ft | One correct coordinate |
| | B1ft | Correct coordinates; follow through on their $(-b, c)$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int -3x^2 + 12x + 8 \, dx = -x^3 + 6x^2 + 8x$ | M1 A1 | M1: any correct index; A1: fully correct (may be unsimplified) |
| Method to find $R =$ their $2 \times 20 - \int_0^{2}(-3x^2+12x+8)\,dx$ | M1 | Look for their $2 \times \text{"20"} - \int_0^{"2"} f(x)\,dx$ |
| $R = 40 - \left[-2^3 + 24 + 16\right]$ | dM1 | Correct application of limits on their integrated function; their 2 must be used |
| $= 8$ | A1 | Shows area of $R = 8$ |
**Alt(c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int 3x^2 - 12x + 12 \, dx = x^3 - 6x^2 + 12x$ | M1 A1 | — |
| $R = \int_0^{2} 3x^2 - 12x + 12 \, dx$ | M1 | — |
| $R = 2^3 - 6\times2^2 + 12\times2 = 8$ | dM1 A1 | — |
---\begin{enumerate}
\item A curve $C$ has equation $y = \mathrm { f } ( x )$ where
\end{enumerate}
$$f ( x ) = - 3 x ^ { 2 } + 12 x + 8$$
(a) Write $\mathrm { f } ( x )$ in the form
$$a ( x + b ) ^ { 2 } + c$$
where $a$, $b$ and $c$ are constants to be found.
The curve $C$ has a maximum turning point at $M$.\\
(b) Find the coordinates of $M$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{235cd1dc-a3ab-473a-bf77-3e41b274dfd8-34_735_841_913_612}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of the curve $C$.\\
The line $l$ passes through $M$ and is parallel to the $x$-axis.\\
The region $R$, shown shaded in Figure 3, is bounded by $C , l$ and the $y$-axis.\\
(c) Using algebraic integration, find the area of $R$.
\hfill \mbox{\textit{Edexcel AS Paper 1 2021 Q14 [10]}}