| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2021 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve |f(x)| compared to |g(x)| with parameters: sketch then solve |
| Difficulty | Moderate -0.3 This is a straightforward modulus function question requiring standard techniques: finding the vertex by setting the inner expression to zero, sketching a V-shaped graph, and solving linear equations in cases. While it involves multiple parts and algebraic manipulation with a parameter, all steps follow routine procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02q Use intersection points: of graphs to solve equations1.02s Modulus graphs: sketch graph of |ax+b| |
| VIXV SIHIANI III IM IONOO | VIAV SIHI NI JYHAM ION OO | VI4V SIHI NI JLIYM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Either \(x=-\frac{a}{3}\) or \(y=a\) | B1 | |
| Correct coordinates \(\left(-\frac{a}{3}, a\right)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| V shape in quadrants 1 and 2 with vertex on negative \(x\)-axis | B1 | |
| Intersects/meets axes at \((0, 5a)\) and \((-5a, 0)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance Notes |
| Attempts to solve a correct equation e.g. \(x+5a=-3x-a+a\) or \(x+5a=3x+a+a\) | M1 | |
| \(x=-\frac{5}{4}a\) or \(x=\frac{3}{2}a\) | A1 | |
| \(x=-\frac{5}{4}a\) and \(x=\frac{3}{2}a\) | A1 | |
| \(x=\ldots \Rightarrow y=\ldots\) using either equation | dM1 | |
| \(\left(-\frac{5}{4}a,\frac{15}{4}a\right)\) and \(\left(\frac{3}{2}a,\frac{13}{2}a\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| One of \(x = -\frac{a}{3}\) or \(y = a\) | B1 | |
| Both coordinates correct: \(x = -\frac{a}{3}\) and \(y = a\) | B1 | May be written separately |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Correct shape in quadrants 1 and 2, vertex on negative \(x\)-axis | B1 | Condone asymmetric graph; ignore "dotted" lines |
| Correct intercepts as coordinates or marked on axes | B1 | Condone \((5a, 0)\) for \((0, 5a)\) if on correct axis; if graph only in Q2, can score for meeting axes at \((-5a, 0)\) and \((0, 5a)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempts to solve either correct equation (no modulus involved) | M1 | Allow \(x+5a=-3x\); \(x+5a=-3x-a+a\); \(x+5a=3x+2a\); \(x+5a=3x+a+a\). Do not condone incorrect combining of modulus terms |
| One correct value: \(x = -\frac{5}{4}a\) or \(x = \frac{3}{2}a\) | A1 | Following correct non-modulus equation; allow even if solution subsequently rejected |
| Both values correct \(x = -\frac{5}{4}a\) and \(x = \frac{3}{2}a\), no additional values | A1 | Cannot score if either solution rejected |
| Correct method to obtain at least one \(y\) value | dM1 | Evidence of embedded values leading to \(y=\) |
| Both coordinates \(\left(-\frac{5}{4}a, \frac{15}{4}a\right)\) and \(\left(\frac{3}{2}a, \frac{13}{2}a\right)\) | A1 | May be given as \(x=\ldots, y=\ldots\); no extras |
# Question 4(a):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Either $x=-\frac{a}{3}$ or $y=a$ | B1 | |
| Correct coordinates $\left(-\frac{a}{3}, a\right)$ | B1 | |
---
# Question 4(b):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| V shape in quadrants 1 and 2 with vertex on negative $x$-axis | B1 | |
| Intersects/meets axes at $(0, 5a)$ and $(-5a, 0)$ | B1 | |
---
# Question 4(c):
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Attempts to solve a correct equation e.g. $x+5a=-3x-a+a$ or $x+5a=3x+a+a$ | M1 | |
| $x=-\frac{5}{4}a$ **or** $x=\frac{3}{2}a$ | A1 | |
| $x=-\frac{5}{4}a$ **and** $x=\frac{3}{2}a$ | A1 | |
| $x=\ldots \Rightarrow y=\ldots$ using either equation | dM1 | |
| $\left(-\frac{5}{4}a,\frac{15}{4}a\right)$ and $\left(\frac{3}{2}a,\frac{13}{2}a\right)$ | A1 | |
# Question (Modulus/Absolute Value Question):
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| One of $x = -\frac{a}{3}$ or $y = a$ | B1 | |
| Both coordinates correct: $x = -\frac{a}{3}$ **and** $y = a$ | B1 | May be written separately |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct shape in quadrants 1 and 2, vertex on negative $x$-axis | B1 | Condone asymmetric graph; ignore "dotted" lines |
| Correct intercepts as coordinates or marked on axes | B1 | Condone $(5a, 0)$ for $(0, 5a)$ if on correct axis; if graph only in Q2, can score for meeting axes at $(-5a, 0)$ and $(0, 5a)$ |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempts to solve either correct equation (no modulus involved) | M1 | Allow $x+5a=-3x$; $x+5a=-3x-a+a$; $x+5a=3x+2a$; $x+5a=3x+a+a$. Do not condone incorrect combining of modulus terms |
| One correct value: $x = -\frac{5}{4}a$ or $x = \frac{3}{2}a$ | A1 | Following correct non-modulus equation; allow even if solution subsequently rejected |
| Both values correct $x = -\frac{5}{4}a$ and $x = \frac{3}{2}a$, no additional values | A1 | Cannot score if either solution rejected |
| Correct method to obtain at least one $y$ value | dM1 | Evidence of embedded values leading to $y=$ |
| Both coordinates $\left(-\frac{5}{4}a, \frac{15}{4}a\right)$ and $\left(\frac{3}{2}a, \frac{13}{2}a\right)$ | A1 | May be given as $x=\ldots, y=\ldots$; no extras |
**Special case:** B1 for either correct coordinate pair with no working or following incorrect modulus combining. Scored 1 0 0 0 0
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{624e9e2f-b6b8-47ce-accc-31dcd5f0554e-10_646_762_264_593}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the graph with equation $y = \mathrm { f } ( x )$, where
$$\mathrm { f } ( x ) = | 3 x + a | + a$$
and where $a$ is a positive constant.
The graph has a vertex at the point $P$, as shown in Figure 2 .
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $a$, the coordinates of $P$.
\item Sketch the graph with equation $y = g ( x )$, where
$$g ( x ) = | x + 5 a |$$
On your sketch, show the coordinates, in terms of $a$, of each point where the graph cuts or meets the coordinate axes.
The graph with equation $y = \mathrm { g } ( x )$ intersects the graph with equation $y = \mathrm { f } ( x )$ at two points.
\item Find, in terms of $a$, the coordinates of the two points.
\includegraphics[max width=\textwidth, alt={}, center]{624e9e2f-b6b8-47ce-accc-31dcd5f0554e-11_2255_50_314_34}\\
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
VIXV SIHIANI III IM IONOO & VIAV SIHI NI JYHAM ION OO & VI4V SIHI NI JLIYM ION OO \\
\hline
\end{tabular}
\end{center}
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2021 Q4 [9]}}