| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Solve |linear| = linear (non-modulus) |
| Difficulty | Moderate -0.8 This is a straightforward modulus equation requiring students to split into two cases (3-2x ≥ 0 and 3-2x < 0), solve two linear equations, and check solutions against the case conditions. It's a standard textbook exercise with clear methodology and minimal problem-solving required, making it easier than average but not trivial since students must remember to check validity of solutions. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02t Solve modulus equations: graphically with modulus function |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Attempt to solve either equation: \(3-2x=7+x \Rightarrow x=\ldots\) or \(2x-3=7+x \Rightarrow x=\ldots\) | M1 | Attempts to solve either correct equation. Allow equivalent forms e.g. \(3-2x=-7-x \Rightarrow x=\ldots\) |
| Either \(x=-\dfrac{4}{3}\) or \(x=10\) | A1 | One correct solution. Allow exact equivalents e.g. \(-1\dfrac{1}{3}\) or \(-1.\dot{3}\) but not e.g. \(-1.33\) |
| Attempt to solve both equations: \(3-2x=7+x \Rightarrow x=\ldots\) and \(2x-3=7+x \Rightarrow x=\ldots\) | dM1 | Depends on first M mark. Allow equivalent equations |
| Both \(x=-\dfrac{4}{3}\) and \(x=10\) with no extra solutions | A1 | Ignore subsequent attempts to find \(y\) coordinates; isw if values placed in inequality |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \((3-2x)^2=(7+x)^2 \Rightarrow 9-12x+4x^2=49+14x+x^2\) | M1 | |
| \(3x^2-26x-40=0\) | A1 | |
| \(3x^2-26x-40=0 \Rightarrow x=\ldots\) | dM1 | |
| Both \(x=-\dfrac{4}{3}\) and \(x=10\) with no extra solutions | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| > Note: Working must be shown — not just answers. But correct equations followed by correct answers can score full marks. If \(x=-\dfrac{4}{3}\) is obtained and candidate states \(x=\left | -\dfrac{4}{3}\right | \), score A0. |
## Question 1:
**Main Method:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempt to solve either equation: $3-2x=7+x \Rightarrow x=\ldots$ **or** $2x-3=7+x \Rightarrow x=\ldots$ | M1 | Attempts to solve either correct equation. Allow equivalent forms e.g. $3-2x=-7-x \Rightarrow x=\ldots$ |
| **Either** $x=-\dfrac{4}{3}$ **or** $x=10$ | A1 | One correct solution. Allow exact equivalents e.g. $-1\dfrac{1}{3}$ or $-1.\dot{3}$ but not e.g. $-1.33$ |
| Attempt to solve **both** equations: $3-2x=7+x \Rightarrow x=\ldots$ **and** $2x-3=7+x \Rightarrow x=\ldots$ | dM1 | Depends on first M mark. Allow equivalent equations |
| **Both** $x=-\dfrac{4}{3}$ **and** $x=10$ with no extra solutions | A1 | Ignore subsequent attempts to find $y$ coordinates; isw if values placed in inequality |
**Alternative (squaring method):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(3-2x)^2=(7+x)^2 \Rightarrow 9-12x+4x^2=49+14x+x^2$ | M1 | |
| $3x^2-26x-40=0$ | A1 | |
| $3x^2-26x-40=0 \Rightarrow x=\ldots$ | dM1 | |
| **Both** $x=-\dfrac{4}{3}$ **and** $x=10$ with no extra solutions | A1 | |
**Total: 4 marks**
> Note: Working must be shown — not just answers. But correct equations followed by correct answers can score full marks. If $x=-\dfrac{4}{3}$ is obtained and candidate states $x=\left|-\dfrac{4}{3}\right|$, score A0.\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying entirely on calculator technology are not acceptable.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{824d73c5-525c-4876-ad66-33c8f1664277-02_671_759_383_653}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the graph with equation $y = | 3 - 2 x |$\\
Solve
$$| 3 - 2 x | = 7 + x$$
\hfill \mbox{\textit{Edexcel Paper 2 2022 Q1 [4]}}