| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Finding quadratic constants from algebraic conditions |
| Difficulty | Moderate -0.3 This is a structured multi-part question covering standard P1 content: finding a line equation from two points, using symmetry of quadratics about the vertex, and writing the quadratic in completed square form. Part (b) requires recognizing that the vertex lies midway between roots (a key insight but commonly taught), while parts (a), (c), and (d) are routine applications of formulas. Slightly easier than average due to the scaffolded structure and straightforward calculations. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points1.02i Represent inequalities: graphically on coordinate plane1.03a Straight lines: equation forms y=mx+c, ax+by+c=0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = 3x +\) "c" or \(y =\) "m"\(x - 12\) | M1 | Attempts form \(y = mx+c\) with \(m\) or \(c\) correct. May be implied by e.g. \(m=3\) or \(\frac{12}{4}\) or \(c=-12\) |
| \(y = 3x - 12\) | A1 | Full correct equation required including "\(y=\)". Condone \(\frac{12}{4}\) for 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(k = 10\) | B1 | Condone \(x=10\) and condone \((10,0)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| E.g. \(y = A(x-4)(x-10)\) or \(y = C(x-7)^2 - 18\) | M1 | Condone \(A\), \(C=1\) or any other constant. Possible to try with all three coordinates |
| E.g. \(-18 = A(7-4)(7-10) \Rightarrow A = \ldots\) or \(0 = C(4-7)^2 - 18 \Rightarrow C = \ldots\) | dM1 | Full attempt at equation with attempt at finding \(A\) or \(C\). Depends on first mark |
| \(y = 2(x-4)(x-10)\), \(y = 2(x-7)^2 - 18\) o.e. | A1 | "\(y=\)" is not required here, just a correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Two of \(y > 3x-12\), \(y < 2(x-4)(x-10)\), \(x > 0\), \(x < 4\) | M1 | Accept \(\leqslant \leftrightarrow <\) and \(\geqslant \leftrightarrow >\). Follow through on (a) and (c) provided (a) is linear and (c) is quadratic |
| E.g. \(3x-12 < y < 2(x-4)(x-10)\), \(0 < x < 4\) | A1 | Fully defines region correctly (not ft here). Accept \(0 < x < p\) as long as \(4 \leqslant p \leqslant \frac{23}{2}\) |
# Question 8(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = 3x +$ "c" or $y =$ "m"$x - 12$ | M1 | Attempts form $y = mx+c$ with $m$ or $c$ correct. May be implied by e.g. $m=3$ or $\frac{12}{4}$ or $c=-12$ |
| $y = 3x - 12$ | A1 | Full correct equation required including "$y=$". Condone $\frac{12}{4}$ for 3 |
# Question 8(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $k = 10$ | B1 | Condone $x=10$ and condone $(10,0)$ |
# Question 8(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| E.g. $y = A(x-4)(x-10)$ or $y = C(x-7)^2 - 18$ | M1 | Condone $A$, $C=1$ or any other constant. Possible to try with all three coordinates |
| E.g. $-18 = A(7-4)(7-10) \Rightarrow A = \ldots$ or $0 = C(4-7)^2 - 18 \Rightarrow C = \ldots$ | dM1 | Full attempt at equation with attempt at finding $A$ or $C$. Depends on first mark |
| $y = 2(x-4)(x-10)$, $y = 2(x-7)^2 - 18$ o.e. | A1 | "$y=$" is not required here, just a correct expression |
# Question 8(d):
| Answer | Mark | Guidance |
|--------|------|----------|
| Two of $y > 3x-12$, $y < 2(x-4)(x-10)$, $x > 0$, $x < 4$ | M1 | Accept $\leqslant \leftrightarrow <$ and $\geqslant \leftrightarrow >$. Follow through on (a) and (c) provided (a) is linear and (c) is quadratic |
| E.g. $3x-12 < y < 2(x-4)(x-10)$, $0 < x < 4$ | A1 | Fully defines region correctly (not ft here). Accept $0 < x < p$ as long as $4 \leqslant p \leqslant \frac{23}{2}$ |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bb21001f-fe68-4776-992d-ede1aae233d7-20_728_885_248_584}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the straight line $l$ and the curve $C$.\\
Given that $l$ cuts the $y$-axis at - 12 and cuts the $x$-axis at 4 , as shown in Figure 2,
\begin{enumerate}[label=(\alph*)]
\item find an equation for $l$, writing your answer in the form $y = m x + c$, where $m$ and $c$ are constants to be found.
Given that $C$
\begin{itemize}
\item has equation $y = \mathrm { f } ( x )$ where $\mathrm { f } ( x )$ is a quadratic expression
\item has a minimum point at $( 7 , - 18 )$
\item cuts the $x$-axis at 4 and at $k$, where $k$ is a constant
\item deduce the value of $k$,
\item find $\mathrm { f } ( x )$.
\end{itemize}
The region $R$ is shown shaded in Figure 2.
\item Use inequalities to define $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2023 Q8 [8]}}