| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Graph y = a|bx+c| + d: identify vertex and intercepts |
| Difficulty | Standard +0.3 This is a straightforward modulus function question requiring standard techniques: finding the vertex by setting the expression inside the modulus to zero, substituting x=0 and y=0 for intercepts, and solving simultaneous equations. All steps are routine applications of modulus function properties with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02s Modulus graphs: sketch graph of |ax+b|1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{b}{2}, a\right)\) | B1B1 | B1 for one correct coordinate \(x=\frac{1}{2}b\) or \(y=a\); second B1 for both correct. May be seen on sketch. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((0, a-b)\) | B1 | Allow \(x=0\) and \(y=a-b\) or just \(y=a-b\) without \(x=0\). If more than 1 point offered with no clear decision, score B0. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{b-a}{2}, 0\right)\) and \(\left(\frac{a+b}{2}, 0\right)\) | B1B1 | Allow just \(x=\frac{b-a}{2}\) or \(x=\frac{a+b}{2}\) without \(y=0\). Allow equivalent expressions e.g. \(\frac{-a-b}{-2}\) for \(\frac{a+b}{2}\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct V-shaped graph sketch | B1B1 | B1: A V shaped graph anywhere on axes. B1: Correct shape with minimum on \(y\)-axis below \(x\)-axis, symmetrical about \(y\)-axis, with left branch of \(y=\ |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-x-1 = 2x+a-b,\ x=-3 \Rightarrow 2=-6+a-b\) or \(x-1=a+b-2x,\ x=5 \Rightarrow 5-1=a+b-10\) | M1 | Forms one correct equation in \(a\) and \(b\) with modulus signs removed |
| Both equations formed correctly | dM1 | A1 on ePEN. Forms 2 correct equations in \(a\) and \(b\) |
| \(a-b=8\) and \(a+b=14 \Rightarrow a=\ldots\) or \(b=\ldots\) | ddM1 | Solves correct equations to find a value for \(a\) or \(b\) |
| \(a=11,\ b=3\) | A1 | Correct answers with no working scores no marks |
## Question 8:
**Part (a)(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{b}{2}, a\right)$ | B1B1 | B1 for one correct coordinate $x=\frac{1}{2}b$ or $y=a$; second B1 for both correct. May be seen on sketch. |
**Part (a)(ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(0, a-b)$ | B1 | Allow $x=0$ and $y=a-b$ or just $y=a-b$ without $x=0$. If more than 1 point offered with no clear decision, score B0. |
**Part (a)(iii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{b-a}{2}, 0\right)$ and $\left(\frac{a+b}{2}, 0\right)$ | B1B1 | Allow just $x=\frac{b-a}{2}$ or $x=\frac{a+b}{2}$ without $y=0$. Allow equivalent expressions e.g. $\frac{-a-b}{-2}$ for $\frac{a+b}{2}$. |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct V-shaped graph sketch | B1B1 | B1: A V shaped graph anywhere on axes. B1: Correct shape with minimum on $y$-axis below $x$-axis, symmetrical about $y$-axis, with left branch of $y=\|x\|-1$ intersecting left branch of $y=a-\|2x-b\|$ and right branch intersecting right branch. Intersections must not both be on $x$-axis. |
**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-x-1 = 2x+a-b,\ x=-3 \Rightarrow 2=-6+a-b$ **or** $x-1=a+b-2x,\ x=5 \Rightarrow 5-1=a+b-10$ | M1 | Forms one correct equation in $a$ and $b$ with modulus signs removed |
| Both equations formed correctly | dM1 | A1 on ePEN. Forms 2 correct equations in $a$ and $b$ |
| $a-b=8$ and $a+b=14 \Rightarrow a=\ldots$ or $b=\ldots$ | ddM1 | Solves correct equations to find a value for $a$ or $b$ |
| $a=11,\ b=3$ | A1 | Correct answers with no working scores no marks |
---8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{76989f19-2624-4e86-a8ee-4978dd1014c2-22_652_634_255_717}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
\section*{In this question you must show all stages of your working.}
\section*{Solutions relying on calculator technology are not acceptable.}
The graph shown in Figure 2 has equation
$$y = a - | 2 x - b |$$
where $a$ and $b$ are positive constants, $a > b$
\begin{enumerate}[label=(\alph*)]
\item Find, giving your answer in terms of $a$ and $b$,
\begin{enumerate}[label=(\roman*)]
\item the coordinates of the maximum point of the graph,
\item the coordinates of the point of intersection of the graph with the $y$-axis,
\item the coordinates of the points of intersection of the graph with the $x$-axis.
On page 24 there is a copy of Figure 2 called Diagram 1.
\end{enumerate}\item On Diagram 1, sketch the graph with equation
$$y = | x | - 1$$
Given that the graphs $y = | x | - 1$ and $y = a - | 2 x - b |$ intersect at $x = - 3$ and $x = 5$
\item find the value of $a$ and the value of $b$
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{76989f19-2624-4e86-a8ee-4978dd1014c2-24_675_652_1959_712}
\captionsetup{labelformat=empty}
\caption{Diagram 1}
\end{center}
\end{figure}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2024 Q8 [11]}}