| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Find range of k for number of roots |
| Difficulty | Standard +0.3 This is a straightforward modulus function question requiring graph interpretation and basic transformations. Part (a) involves reading the range from a given graph for one root, part (b) is a routine modulus equation solved by cases, and part (c) applies standard transformation rules to find a vertex. All techniques are standard C3 material 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 function1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks |
|---|---|
| Either \(k > 13\) or \(k = 3\) | B1 |
| Both \(k > 13\) \(k = 3\) | B1 |
| Answer | Marks |
|---|---|
| Smaller solution: \(2(5-x)+3=\frac{1}{2}x+10 \Rightarrow x=\frac{6}{5}\) | M1 A1 |
| Larger solution: \(-2(5-x)+3=\frac{1}{2}x+10 \Rightarrow x=\frac{34}{3}\) | M1 A1 |
| Answer | Marks |
|---|---|
| \((6,12)\) | B1B1 |
| Answer | Marks | Guidance |
|---|---|---|
| - M1: \(2 | 5-x | +3=\frac{1}{2}x+10 \Rightarrow 4(5-x)^2=\left(\frac{1}{2}x+7\right)^2\) oe. In the main scheme the equation must be correct of the correct form but in this case you may condone '2' not being squared |
**(a)**
| Either $k > 13$ or $k = 3$ | B1 |
| Both $k > 13$ $k = 3$ | B1 |
(2 marks)
**(b)**
| Smaller solution: $2(5-x)+3=\frac{1}{2}x+10 \Rightarrow x=\frac{6}{5}$ | M1 A1 |
| Larger solution: $-2(5-x)+3=\frac{1}{2}x+10 \Rightarrow x=\frac{34}{3}$ | M1 A1 |
(4 marks)
**(c)**
| $(6,12)$ | B1B1 |
(2 marks)
(8 marks total)
**Guidance notes:**
**(a)**
- B1: Either $k > 13$ or $k = 3$. Condone $k \geq 13$ instead of $k > 13$ for this mark only. Also condone $y \leftrightarrow k$. Do not accept $k \geq 3$ for B1
- B1: Both $k > 13, k = 3$ with no other restrictions. Accept and/or $/$, between the two solutions
**(b)**
- M1: An acceptable method of finding the **smaller intersection**. The initial equation must be of the correct form and it must lead to a value of $x$. For example $2(5-x)+3=\frac{1}{2}x+10 \Rightarrow x = ...$ or $5 - x = \left(\frac{1}{4}x + \frac{7}{2}\right)$
- A1: For $x = \frac{6}{5}$ or equivalent such as $1.2$. Ignore any reference to the $y$ coordinate
- M1: An acceptable method of finding the **larger intersection**. The initial equation must be of the correct form and it must lead to a value of $x$. For example $-2(5-x)+3=\frac{1}{2}x+10 \Rightarrow x = ...$ or $5 - x = -\left(\frac{1}{4}x + \frac{7}{2}\right)$
- A1: For $x = \frac{34}{3}$ or equivalent such as 11.3. Ignore any reference to the $y$ coordinate
If there are any extra solutions in addition to the correct two, then withhold the final A1 mark. ISW if the candidate then refers back to the range in (a) and deletes a solution.
**Alt method by squaring:**
- M1: $2|5-x|+3=\frac{1}{2}x+10 \Rightarrow 4(5-x)^2=\left(\frac{1}{2}x+7\right)^2$ oe. In the main scheme the equation must be correct of the correct form but in this case you may condone '2' not being squared
- A1: Correct 3TQ. The $= 0$ may be implied by subsequent work. $15x^2-47x+51=0$ oe
- M1: Solves using an appropriate method $15x^2-188x+204=0 \Rightarrow (5x-6)(3x-34)=0 \Rightarrow x = ...$
- A1: Both $x = \frac{6}{5} x = \frac{34}{3}$ and no others.
**(c)**
- B1: Accept $p = 6$ or $q = 12$. Allow in coordinates as $x = 6$ or $y = 12$
- B1: For both $p = 6$ **and** $q = 12$. Allow in coordinates as $x = 6$ **and** $y = 12$
Allow embedded within a single coordinate $(6,12)$. So for example $(2,12)$ is scored B1 B0.
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{42aff260-e734-48ff-a92a-674032cb0377-16_561_848_214_699}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows part of the graph with equation $y = \mathrm { f } ( x )$, where
$$\mathrm { f } ( x ) = 2 | 5 - x | + 3 , \quad x \geqslant 0$$
Given that the equation $\mathrm { f } ( x ) = k$, where $k$ is a constant, has exactly one root,
\begin{enumerate}[label=(\alph*)]
\item state the set of possible values of $k$.
\item Solve the equation $\mathrm { f } ( x ) = \frac { 1 } { 2 } x + 10$
The graph with equation $y = \mathrm { f } ( x )$ is transformed onto the graph with equation $y = 4 \mathrm { f } ( x - 1 )$. The vertex on the graph with equation $y = 4 \mathrm { f } ( x - 1 )$ has coordinates $( p , q )$.
\item State the value of $p$ and the value of $q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2018 Q5 [8]}}