OCR MEI C3 — Question 8 3 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeGraph y = a|bx+c| + d: identify vertex and intercepts
DifficultyEasy -1.3 This is a straightforward substitution question requiring students to find intercepts of a given modulus function by setting x=0 and y=0. It involves only basic algebraic manipulation and understanding that |x-1|=0 when x=1, with no problem-solving or conceptual challenge beyond direct application of the modulus definition.
Spec1.02l Modulus function: notation, relations, equations and inequalities1.02s Modulus graphs: sketch graph of |ax+b|

8 Fig. 4 shows a sketch of the graph of \(y = 2 | x - 1 |\). It meets the \(x\) - and \(y\)-axes at ( \(a , 0\) ) and ( \(0 , b\) ) respectively. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{125b76c1-5ab3-4645-a3c4-cf167a04f453-2_478_546_1299_834} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure} Find the values of \(a\) and \(b\).

Question 8:
AnswerMarks Guidance
\(g(x) = 2x-1 \)
\(b = 20-1 = 2\) or \((0, 2)\)
\(2x-1 = 0\)
\(\Rightarrow x=1\), so \(a=1\) or \((1,0)\)A1 www
[3]
## Question 8:

| $g(x) = 2|x-1|$ | | Allow unsupported answers |
|---|---|---|
| $b = 2|0-1| = 2$ or $(0, 2)$ | B1 | www |
| $2|x-1| = 0$ | M1 | $|x|=1$ is A0 |
| $\Rightarrow x=1$, so $a=1$ or $(1,0)$ | A1 | www |
| **[3]** | | |

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8 Fig. 4 shows a sketch of the graph of $y = 2 | x - 1 |$. It meets the $x$ - and $y$-axes at ( $a , 0$ ) and ( $0 , b$ ) respectively.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{125b76c1-5ab3-4645-a3c4-cf167a04f453-2_478_546_1299_834}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}

Find the values of $a$ and $b$.

\hfill \mbox{\textit{OCR MEI C3  Q8 [3]}}