| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Sketch y=|linear| and y=linear, solve inequality: numeric coefficients |
| Difficulty | Moderate -0.3 This is a standard modulus function question covering basic sketching and solving modulus inequalities. Part (a) is routine graph sketching, part (b) is a straightforward two-case split giving x < -1 or x > 6, and part (c) requires comparing to a linear function but follows standard techniques. Slightly easier than average due to the linear nature of all expressions involved and being a typical textbook exercise with no novel problem-solving required. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02s Modulus graphs: sketch graph of |ax+b| |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct graph in quadrant 1 and quadrant 2 with V on the \(x\)-axis | B1 | AO1.1b |
| States \((0, 5)\) and \(\left(\frac{5}{2}, 0\right)\); or \(\frac{5}{2}\) marked in correct position on \(x\)-axis and 5 marked in correct position on \(y\)-axis | B1 | AO1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \( | 2x-5 | > 7\); \(2x-5=7 \Rightarrow x = \ldots\) and \(-(2x-5)=7 \Rightarrow x = \ldots\) |
| Critical values \(x=6,\ x=-1\); giving \(x < -1\) or \(x > 6\) | A1 | AO1.1b; Correct answer e.g. \(x<-1\) or \(x>6\); \(x<-1 \cup x>6\); \(\{x: x<-1\}\cup\{x: x>6\}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \( | 2x-5 | > x - \frac{5}{2}\); Solves \(2x-5 = x-\frac{5}{2}\) to give \(x=\frac{5}{2}\), and solves \(-(2x-5)=x-\frac{5}{2}\) to also give \(x=\frac{5}{2}\); or sketches \(y= |
| \(\left\{x:\ x<\frac{5}{2}\right\} \cup \left\{x:\ x>\frac{5}{2}\right\}\); or \(\left\{x\in\mathbb{R},\ x\neq\frac{5}{2}\right\}\); or \(\mathbb{R} - \left\{\frac{5}{2}\right\}\) | A1 | AO2.5; Final answer must be expressed using set notation |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct graph in quadrant 1 and quadrant 2 with V on the $x$-axis | B1 | AO1.1b |
| States $(0, 5)$ and $\left(\frac{5}{2}, 0\right)$; or $\frac{5}{2}$ marked in correct position on $x$-axis **and** 5 marked in correct position on $y$-axis | B1 | AO1.1b |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $|2x-5| > 7$; $2x-5=7 \Rightarrow x = \ldots$ **and** $-(2x-5)=7 \Rightarrow x = \ldots$ | M1 | AO1.1b |
| Critical values $x=6,\ x=-1$; giving $x < -1$ or $x > 6$ | A1 | AO1.1b; Correct answer e.g. $x<-1$ or $x>6$; $x<-1 \cup x>6$; $\{x: x<-1\}\cup\{x: x>6\}$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $|2x-5| > x - \frac{5}{2}$; Solves $2x-5 = x-\frac{5}{2}$ to give $x=\frac{5}{2}$, and solves $-(2x-5)=x-\frac{5}{2}$ to also give $x=\frac{5}{2}$; or sketches $y=|2x-5|$ and $y=x-\frac{5}{2}$ and indicates they meet at $\left(\frac{5}{2},0\right)$ | M1 | AO3.1a; A complete process of finding that $y=|2x-5|$ and $y=x-\frac{5}{2}$ meet at **only** one point, either algebraically or graphically |
| $\left\{x:\ x<\frac{5}{2}\right\} \cup \left\{x:\ x>\frac{5}{2}\right\}$; or $\left\{x\in\mathbb{R},\ x\neq\frac{5}{2}\right\}$; or $\mathbb{R} - \left\{\frac{5}{2}\right\}$ | A1 | AO2.5; Final answer must be expressed using set notation |
---\begin{enumerate}
\item (a) Sketch the graph with equation
\end{enumerate}
$$y = | 2 x - 5 |$$
stating the coordinates of any points where the graph cuts or meets the coordinate axes.\\
(b) Find the values of $x$ which satisfy
$$| 2 x - 5 | > 7$$
(c) Find the values of $x$ which satisfy
$$| 2 x - 5 | > x - \frac { 5 } { 2 }$$
Write your answer in set notation.
\hfill \mbox{\textit{Edexcel Paper 2 Q4 [6]}}