| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve |f(x)| compared to |g(x)| with parameters: sketch then solve |
| Difficulty | Standard +0.3 This is a straightforward modulus function question requiring standard techniques: identifying intercepts from modulus graphs, sketching a simple V-shaped modulus graph, and solving by finding intersection points graphically or algebraically. The multi-part structure guides students through each step, and the methods are routine for C3/C4 level with no novel problem-solving required. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02s Modulus graphs: sketch graph of |ax+b|1.02t Solve modulus equations: graphically with modulus function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((0, 10a)\) or \(\left(-\frac{5}{2}a, 0\right)\) or \((x=0, y=10a)\) or \(\left(y=0, x=-\frac{5}{2}a\right)\) | B1 | One correct coordinate pair. Allow as separate coordinates or clear sight of the "0"s and allow \( |
| \((0,10a)\) and \(\left(-\frac{5}{2}a, 0\right)\) or \((x=0, y=10a)\) and \(\left(y=0, x=-\frac{5}{2}a\right)\) | B1 | Two correct coordinate pairs. Allow as separate coordinates or clear sight of the "0"s and allow \( |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| V (or tick) shape with vertex on the negative \(y\)-axis and branches pointing upwards with one branch to the left and one to the right of the \(y\)-axis, with part of the V in all 4 quadrants | B1, M1 on ePEN | Ignore gradient as long as it is a V shape. Do not be overly concerned by lack of symmetry. Allow the diagram above the question to be adapted or a separate sketch. |
| Intersections at \((-a,0)\), \((a,0)\) and \((0,-a)\) only | B1, A1 on ePEN | Can be seen as coordinates or as shown in the diagram. If the coordinates are shown away from the sketch they must appear as \((-a,0),(a,0)\) and \((0,-a)\) and must correspond with the sketch. If there is any ambiguity the sketch has precedence. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-x-a = 4x+10a \Rightarrow x = \ldots\) or \(-x-a = -4x-10a \Rightarrow x = \ldots\) | M1 | Attempts to solve \(-x-a=4x+10a\) or \(-x-a=-4x-10a\) or equivalent equations to obtain \(x\) in terms of \(a\) |
| \(x = -\frac{11}{5}a\) or \(-3a\) | A1 | One correct. Allow \(-\frac{9}{3}a\) for \(-3a\) |
| \(x = -\frac{11}{5}a\) and \(-3a\) | A1 | Both correct and no other values. Allow \(-\frac{9}{3}a\) for \(-3a\). Note that attempts to square both sides and solve the resulting quadratic generally scores no marks. However if you think such attempts deserve credit then use Review. |
## Question 6(a)(i)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(0, 10a)$ or $\left(-\frac{5}{2}a, 0\right)$ or $(x=0, y=10a)$ or $\left(y=0, x=-\frac{5}{2}a\right)$ | B1 | One correct coordinate pair. Allow as separate coordinates or clear sight of the "0"s and allow $|10a|$ for $10a$ and allow equivalents for $-\frac{5}{2}a$ e.g. $-\frac{10}{4}a$. **Ignore labelling of parts and points** |
| $(0,10a)$ **and** $\left(-\frac{5}{2}a, 0\right)$ or $(x=0, y=10a)$ **and** $\left(y=0, x=-\frac{5}{2}a\right)$ | B1 | Two correct coordinate pairs. Allow as separate coordinates or clear sight of the "0"s and allow $|10a|$ for $10a$ and allow equivalents for $-\frac{5}{2}a$ e.g. $-\frac{10}{4}a$. **Ignore labelling of parts and points** |
---
## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| V (or tick) shape with vertex on the negative $y$-axis and branches pointing upwards with one branch to the left and one to the right of the $y$-axis, with part of the V in all 4 quadrants | B1, M1 on ePEN | Ignore gradient as long as it is a V shape. Do not be overly concerned by lack of symmetry. Allow the diagram above the question to be adapted or a separate sketch. |
| Intersections at $(-a,0)$, $(a,0)$ and $(0,-a)$ only | B1, A1 on ePEN | Can be seen as coordinates or as shown in the diagram. If the coordinates are shown away from the sketch they must appear as $(-a,0),(a,0)$ and $(0,-a)$ and must correspond with the sketch. If there is any ambiguity the sketch has precedence. |
---
## Question 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-x-a = 4x+10a \Rightarrow x = \ldots$ or $-x-a = -4x-10a \Rightarrow x = \ldots$ | M1 | Attempts to solve $-x-a=4x+10a$ or $-x-a=-4x-10a$ or equivalent equations **to obtain $x$ in terms of $a$** |
| $x = -\frac{11}{5}a$ **or** $-3a$ | A1 | One correct. Allow $-\frac{9}{3}a$ for $-3a$ |
| $x = -\frac{11}{5}a$ **and** $-3a$ | A1 | Both correct and no other values. Allow $-\frac{9}{3}a$ for $-3a$. **Note that attempts to square both sides and solve the resulting quadratic generally scores no marks. However if you think such attempts deserve credit then use Review.** |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a9870c94-0910-46ec-a54a-44a431cb324e-14_988_1120_123_395}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the graph with equation $y = | 4 x + 10 a |$, where $a$ is a positive constant.
The graph cuts the $y$-axis at the point $P$ and meets the $x$-axis at the point $Q$ as shown.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item State the coordinates of $P$.
\item State the coordinates of $Q$.
\end{enumerate}\item A copy of Figure 1 is shown on page 15. On this copy, sketch the graph with equation
$$y = | x | - a$$
Show on the sketch the coordinates of each point where your graph cuts or meets the coordinate axes.
\item Hence, or otherwise, solve the equation
$$| 4 x + 10 a | = | x | - a$$
giving your answers in terms of $a$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a9870c94-0910-46ec-a54a-44a431cb324e-15_860_1128_447_392}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
$\_\_\_\_$ 7
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2019 Q6 [7]}}