| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Inequalities |
| Type | Graph feasible region from inequalities |
| Difficulty | Moderate -0.3 This is a standard linear programming question requiring students to find constants from a given area and then write inequalities for a region. Part (a) involves setting up equations from the intersection conditions and using the area formula for a region bounded by lines (likely a trapezium or triangle), which is routine P1 content. Part (b) is straightforward identification of inequalities from a diagram. While multi-step, it requires only standard techniques with no novel insight, making it slightly easier than average. |
| Spec | 1.02i Represent inequalities: graphically on coordinate plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0 = 10 - 2x \Rightarrow x = 5\) or \(y = 2, y = 10 - 2x \Rightarrow x = 4\) | B1 | Correct \(x\) intercept for \(y = 10 - 2x\) or correct \(x\) coordinate for intersection of \(y = 2\) with \(y = 10 - 2x\). May be seen on diagram, sketch or implied by working. |
| e.g. \(\frac{1}{2} \times 2(5 + 4 - a) = \frac{27}{4}\) or \(\frac{1}{2} \times 2\left(5 + 4 - \frac{2}{k}\right) = \frac{27}{4}\) (Trapezium) | M1 | Correct method for area set equal to \(\frac{27}{4}\) to form equation in \(a\) or \(k\). May use trapezium, triangle + trapezium, triangle + rectangle + triangle, or 2 triangles. Note with correct "5" and "4" the area is \(9 - a\) or \(9 - \frac{2}{k}\) |
| \(\Rightarrow k = \frac{8}{9}\), \(a = \frac{9}{4}\) | A1 | Correct value for \(a\) or \(k\) as single value not a calculation |
| A1ft | Correct values for both \(a\) and \(k\). Follow through using \(ak = 2\) from first value provided M1 scored. Allow \(k = 0.\dot{8}\), \(a = 2.25\) or \(2\frac{1}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Two of \(y \geq \frac{8}{9}x\), \(y \leq 10 - 2x\), \(x > \frac{9}{4}\) | M1 | Any two of the three correct inequalities with their \(k\) and/or \(a\) or letters \(a\) and \(k\). Inequalities in \(y\) may be combined e.g. \(\frac{8}{9}x < y < 10 - 2x\) counts as two. Accept \(<\) or \(\leq\) for M. Do not allow incorrect notation such as \(R_2 \geq \frac{8}{9}x\), \(R_2 \leq 10-2x\) |
| All three: \(y \geq \frac{8}{9}x\), \(y \leq 10 - 2x\), \(x > \frac{9}{4}\) | A1 | All three correct with correct values of \(a\) and \(k\) — no follow through. Note \(y = \frac{8}{9}x\) and \(y = 10-2x\) intersect when \(x = \frac{45}{13}\), so upper limit on \(x\) may be this value or greater. Also accept limits on \(y\) provided values are in correct range, e.g. \(y > 2\), \(y \geq 2\), \(2 < y < \frac{11}{2}\), \(2 \leq y \leq \frac{11}{2}\) |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = 10 - 2x \Rightarrow x = 5$ or $y = 2, y = 10 - 2x \Rightarrow x = 4$ | B1 | Correct $x$ intercept for $y = 10 - 2x$ or correct $x$ coordinate for intersection of $y = 2$ with $y = 10 - 2x$. May be seen on diagram, sketch or implied by working. |
| e.g. $\frac{1}{2} \times 2(5 + 4 - a) = \frac{27}{4}$ or $\frac{1}{2} \times 2\left(5 + 4 - \frac{2}{k}\right) = \frac{27}{4}$ (Trapezium) | M1 | Correct method for area set equal to $\frac{27}{4}$ to form equation in $a$ or $k$. May use trapezium, triangle + trapezium, triangle + rectangle + triangle, or 2 triangles. Note with correct "5" and "4" the area is $9 - a$ or $9 - \frac{2}{k}$ |
| $\Rightarrow k = \frac{8}{9}$, $a = \frac{9}{4}$ | A1 | Correct value for $a$ or $k$ as single value not a calculation |
| | A1ft | Correct values for both $a$ and $k$. Follow through using $ak = 2$ from first value provided M1 scored. Allow $k = 0.\dot{8}$, $a = 2.25$ or $2\frac{1}{4}$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Two of $y \geq \frac{8}{9}x$, $y \leq 10 - 2x$, $x > \frac{9}{4}$ | M1 | Any two of the three correct inequalities with their $k$ and/or $a$ or letters $a$ and $k$. Inequalities in $y$ may be combined e.g. $\frac{8}{9}x < y < 10 - 2x$ counts as two. Accept $<$ or $\leq$ for M. Do not allow incorrect notation such as $R_2 \geq \frac{8}{9}x$, $R_2 \leq 10-2x$ |
| All three: $y \geq \frac{8}{9}x$, $y \leq 10 - 2x$, $x > \frac{9}{4}$ | A1 | All three correct with correct values of $a$ and $k$ — no follow through. Note $y = \frac{8}{9}x$ and $y = 10-2x$ intersect when $x = \frac{45}{13}$, so upper limit on $x$ may be this value or greater. Also accept limits on $y$ provided values are in correct range, e.g. $y > 2$, $y \geq 2$, $2 < y < \frac{11}{2}$, $2 \leq y \leq \frac{11}{2}$ |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a5a5dd8b-1438-4698-929a-c5e3d5ed0694-18_737_951_301_587}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The region $R _ { 1 }$, shown shaded in Figure 2, is defined by the inequalities
$$0 \leqslant y \leqslant 2 \quad y \leqslant 10 - 2 x \quad y \leqslant k x$$
where $k$ is a constant.\\
The line $x = a$, where $a$ is a constant, passes through the intersection of the lines $y = 2$ and $y = k x$\\
Given that the area of $R _ { 1 }$ is $\frac { 27 } { 4 }$ square units,
\begin{enumerate}[label=(\alph*)]
\item find
\begin{enumerate}[label=(\roman*)]
\item the value of $a$
\item the value of $k$
\end{enumerate}\item Define the region $R _ { 2 }$, also shown shaded in Figure 2, using inequalities.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2023 Q7 [6]}}