| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Graph y = a|bx+c| + d: identify vertex and intercepts |
| Difficulty | Moderate -0.3 This is a straightforward modulus function question requiring standard techniques: finding intercepts by substituting x=0 or y=0, finding the vertex from the critical point where the expression inside the modulus equals zero, and solving a modulus equation by considering cases. All steps are routine applications of well-practiced methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02l Modulus function: notation, relations, equations and inequalities1.02s Modulus graphs: sketch graph of |ax+b| |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \(P(0,\ 3a)\) | B1 | For \((0, 3a)\). Condone just \(y = 3a\) as long as there isn't a value for \(x\) coordinate (apart from 0). Note \(P = 3a\) or \((3a, 0)\) is B0 unless \(y = 3a\) previously seen |
| (ii) \(Q(a,\ 0)\), \(R\left(\frac{7}{3}a,\ 0\right)\) | B1 B1 | For either coordinate \((a, 0)\) or \(\left(\frac{7}{3}a, 0\right)\). Condone just \(x = a\) or \(x = \frac{7}{3}a\) as long as no \(y\) coordinate (that isn't 0). Note \(Q = a\) or \((0, a)\) is B0 unless \(x = a\) previously seen. SC: \(Q = a\) AND \(R = \frac{7}{3}a\) or vice versa is B1 B0. Question states simplest form so don't accept e.g. \(\frac{10}{6}a\) for \(\frac{5}{3}a\) |
| (iii) \(S\left(\frac{5}{3}a,\ -2a\right)\) | B1 | Allow coordinates given as \(x = \ldots\), \(y = \ldots\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3x - 5a - 2a = x - 2a \Rightarrow x = \ldots\) or \(-(3x-5a) - 2a = -(x-2a) \Rightarrow x = \ldots\) | M1 | Attempts to solve one correct equation without moduli, found correctly. Condone \(\Rightarrow a = \ldots\). Either taking both positive aspects: \(3x - 5a - 2a = x - 2a\) (i.e. \(3x - 7a = x - 2a\)), or both negative aspects: \(-3x + 5a - 2a = -x + 2a\) |
| \(x = \frac{5}{2}a\) or \(x = \frac{1}{2}a\) | A1 | One correct value of \(x\) from a correctly produced equation |
| \(3x - 5a - 2a = x - 2a\) and \(-(3x-5a) - 2a = -(x-2a) \Rightarrow x = \ldots\) | dM1 | Attempts to solve both equations |
| \(x = \frac{5}{2}a\) and \(x = \frac{1}{2}a\) | A1 | Both correct values of \(x\) (from correct equations) with no other values given. Also allow if 4 answers found, followed by 2 correct answers chosen and other 2 incorrect discarded |
## Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| (i) $P(0,\ 3a)$ | B1 | For $(0, 3a)$. Condone just $y = 3a$ as long as there isn't a value for $x$ coordinate (apart from 0). Note $P = 3a$ or $(3a, 0)$ is B0 unless $y = 3a$ previously seen |
| (ii) $Q(a,\ 0)$, $R\left(\frac{7}{3}a,\ 0\right)$ | B1 B1 | For either coordinate $(a, 0)$ or $\left(\frac{7}{3}a, 0\right)$. Condone just $x = a$ or $x = \frac{7}{3}a$ as long as no $y$ coordinate (that isn't 0). Note $Q = a$ or $(0, a)$ is B0 unless $x = a$ previously seen. SC: $Q = a$ AND $R = \frac{7}{3}a$ or vice versa is B1 B0. Question states simplest form so don't accept e.g. $\frac{10}{6}a$ for $\frac{5}{3}a$ |
| (iii) $S\left(\frac{5}{3}a,\ -2a\right)$ | B1 | Allow coordinates given as $x = \ldots$, $y = \ldots$ |
**(4 marks)**
---
## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3x - 5a - 2a = x - 2a \Rightarrow x = \ldots$ **or** $-(3x-5a) - 2a = -(x-2a) \Rightarrow x = \ldots$ | M1 | Attempts to solve one **correct** equation without moduli, found correctly. Condone $\Rightarrow a = \ldots$. Either taking both positive aspects: $3x - 5a - 2a = x - 2a$ (i.e. $3x - 7a = x - 2a$), or both negative aspects: $-3x + 5a - 2a = -x + 2a$ |
| $x = \frac{5}{2}a$ **or** $x = \frac{1}{2}a$ | A1 | One correct value of $x$ from a correctly produced equation |
| $3x - 5a - 2a = x - 2a$ **and** $-(3x-5a) - 2a = -(x-2a) \Rightarrow x = \ldots$ | dM1 | Attempts to solve **both** equations |
| $x = \frac{5}{2}a$ **and** $x = \frac{1}{2}a$ | A1 | Both correct values of $x$ (from correct equations) with no other values given. Also allow if 4 answers found, followed by 2 correct answers chosen and other 2 incorrect discarded |
**(4 marks) Total 8**6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5abaa077-1da4-4023-b442-194f6972095b-16_652_835_292_616}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the graph with equation
$$y = | 3 x - 5 a | - 2 a$$
where $a$ is a positive constant.\\
The graph
\begin{itemize}
\item cuts the $y$-axis at the point $P$
\item cuts the $x$-axis at the points $Q$ and $R$
\item has a minimum point at $S$
\begin{enumerate}[label=(\alph*)]
\item Find, in simplest form in terms of $a$, the coordinates of
\begin{enumerate}[label=(\roman*)]
\item point $P$
\item points $Q$ and $R$
\item point $S$
\end{enumerate}\item Find, in simplest form in terms of $a$, the values of $x$ for which
\end{itemize}
$$| 3 x - 5 a | - 2 a = | x - 2 a |$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2023 Q6 [8]}}