| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Interpret or complete given sketch of two |linear| functions |
| Difficulty | Moderate -0.8 This is a straightforward modulus question requiring basic understanding of graph transformations and solving |x| = |x-2| + 1. Part (i) is immediate recognition that the minimum of |x-2| + 1 occurs at x=2. Part (ii) involves simple algebraic verification with the answer already given, requiring only substitution or basic case-by-case analysis. Below average difficulty as it's mostly recall and routine manipulation with minimal problem-solving. |
| 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| |
| Answer | Marks |
|---|---|
| P is \((2, 1)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \( | x | = 1\frac{1}{2}\) |
| \(\Rightarrow x = (-1\frac{1}{2})\) or \(1\frac{1}{2}\) | A1 | |
| \( | x-2 | +1 = 1\frac{1}{2} \Rightarrow |
| \(\Rightarrow x = (2\frac{1}{2})\) or \(1\frac{1}{2}\) | E1 | |
| *or* by solving equation directly: | ||
| \( | x-2 | +1 = |
| \(2-x+1 = x\) | M1 | |
| \(\Rightarrow x = 1\frac{1}{2}\) | A1 | |
| \(\Rightarrow y = | x | = 1\frac{1}{2}\) |
| [4] |
## Question 10:
### Part (i):
| P is $(2, 1)$ | B1 | |
|---|---|---|
### Part (ii):
| $|x| = 1\frac{1}{2}$ | M1 | allow $x = 1\frac{1}{2}$ unsupported |
|---|---|---|
| $\Rightarrow x = (-1\frac{1}{2})$ or $1\frac{1}{2}$ | A1 | |
| $|x-2|+1 = 1\frac{1}{2} \Rightarrow |x-2| = \frac{1}{2}$ | M1 | or $\left|1\frac{1}{2}-2\right|+1 = \frac{1}{2}+1 = 1\frac{1}{2}$ |
| $\Rightarrow x = (2\frac{1}{2})$ or $1\frac{1}{2}$ | E1 | |
| *or* by solving equation directly: | | |
| $|x-2|+1 = |x|$ | M1 | equating from graph or listing possible cases |
| $2-x+1 = x$ | M1 | |
| $\Rightarrow x = 1\frac{1}{2}$ | A1 | |
| $\Rightarrow y = |x| = 1\frac{1}{2}$ | E1 | |
| **[4]** | | |
---10 Fig. 1 shows the graphs of $y = | x |$ and $y = | x - 2 | + 1$. The point P is the minimum point of $y = | x - 2 | + 1$, and Q is the point of intersection of the two graphs.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{125b76c1-5ab3-4645-a3c4-cf167a04f453-3_491_833_503_657}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
(i) Write down the coordinates of P .\\
(ii) Verify that the $y$-coordinate of Q is $1 \frac { 1 } { 2 }$.
\hfill \mbox{\textit{OCR MEI C3 Q10 [5]}}