| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | October |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Write inequalities from graph |
| Difficulty | Moderate -0.8 This question requires finding equations of a parabola (using vertex form with given minimum) and a straight line (from two points), then writing two inequalities to define the shaded region. While it involves multiple steps, each is a standard A-level technique with all necessary information provided explicitly, making it easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \((-2,13)\) into \(y = mx + 25\) and finds \(m\) | M1 | Attempts equation of line |
| \(y = 6x + 25\) | A1 | Must be in form \(y = \ldots\) but can be implied from inequality |
| Attempts to use intercept \((0,25)\) in \(y = a(x \pm 2)^2 + 13\) to find \(a\) | M1 | Attempts equation of \(C\) |
| \(y = 3(x+2)^2 + 13\) or \(y = 3x^2 + 12x + 25\) | A1 | Do not accept just \(C = 3x^2+12x+25\) for A1 |
| Region \(R\): \(3(x+2)^2 + 13 < y < 6x + 25\) | B1ft | Fully defines region \(R\); allow strict or non-strict inequalities used consistently |
## Question 7:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $(-2,13)$ into $y = mx + 25$ and finds $m$ | M1 | Attempts equation of line |
| $y = 6x + 25$ | A1 | Must be in form $y = \ldots$ but can be implied from inequality |
| Attempts to use intercept $(0,25)$ in $y = a(x \pm 2)^2 + 13$ to find $a$ | M1 | Attempts equation of $C$ |
| $y = 3(x+2)^2 + 13$ or $y = 3x^2 + 12x + 25$ | A1 | Do not accept just $C = 3x^2+12x+25$ for A1 |
| Region $R$: $3(x+2)^2 + 13 < y < 6x + 25$ | B1ft | Fully defines region $R$; allow strict or non-strict inequalities used consistently |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{dc0ac5df-24a7-41b5-8410-f0e9b332ba64-16_868_805_242_632}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of a curve $C$ with equation $y = \mathrm { f } ( x )$ and a straight line $l$.\\
The curve $C$ meets $l$ at the points $( - 2,13 )$ and $( 0,25 )$ as shown.\\
The shaded region $R$ is bounded by $C$ and $l$ as shown in Figure 1.\\
Given that
\begin{itemize}
\item $\mathrm { f } ( x )$ is a quadratic function in $x$
\item ( $- 2,13$ ) is the minimum turning point of $y = \mathrm { f } ( x )$\\
use inequalities to define $R$.
\end{itemize}
\hfill \mbox{\textit{Edexcel Paper 1 2020 Q7 [5]}}