| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Graph y=a|bx+c|+d with unknown constants: find constants then solve |
| Difficulty | Moderate -0.3 This is a straightforward modulus function question requiring substitution to find constants from given points, then solving |x| = a|x+b| by considering cases. While it involves multiple steps and case analysis, the techniques are standard C3 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02s Modulus graphs: sketch graph of |ax+b| |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = ...\) | M1 | for attempting differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a = \frac{1}{2}\) | B1 | or 0.5 |
| \(b = 1\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{2}\ | x+1\ | = \ |
| \(\Rightarrow \frac{1}{2}(x+1) = x\) | M1 | o.e. ft their \(a(\neq 0)\), \(b\) (but allow recovery to correct values); or verified by subst \(x=1, y=1\) into \(y = \frac{1}{2}\ |
| \(\Rightarrow x=1, y=1\) | A1 | unsupported answers M0A0 |
| or \(\frac{1}{2}(x+1) = -x\) | M1 | o.e., ft their \(a\), \(b\); or verified by subst \((-\frac{1}{3}, \frac{1}{3})\) into \(y = \frac{1}{2}\ |
| \(\Rightarrow x = -\frac{1}{3}, y = \frac{1}{3}\) | A1 | or 0.33, \(-0.33\) or better; unsupported answers M0A0 |
| *or* \(\frac{1}{4}(x+1)^2 = x^2\) | M1 | ft their \(a\) and \(b\) |
| \(\Rightarrow 3x^2 - 2x - 1 = 0\) | M1ft | obtaining a quadratic \(= 0\), ft their previous line, but must have an \(x^2\) term |
| \(\Rightarrow x = -\frac{1}{3}\) or \(1\) | A1 | SC3 for \((1,1)\) \((-\frac{1}{3}, \frac{1}{3})\) and one or more additional points |
| \(y = \frac{1}{3}\) or \(1\) | A1 |
**Question 1:**
$\frac{dy}{dx} = ...$ | M1 | for attempting differentiation
etc.
# Mark Scheme Extraction - 4753/01 June 2013
---
## Question 1:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $a = \frac{1}{2}$ | B1 | or 0.5 |
| $b = 1$ | B1 | |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2}\|x+1\| = \|x\|$ | | |
| $\Rightarrow \frac{1}{2}(x+1) = x$ | M1 | o.e. ft their $a(\neq 0)$, $b$ (but allow recovery to correct values); or verified by subst $x=1, y=1$ into $y = \frac{1}{2}\|x+1\|$ and $y=\|x\|$ |
| $\Rightarrow x=1, y=1$ | A1 | unsupported answers M0A0 |
| or $\frac{1}{2}(x+1) = -x$ | M1 | o.e., ft their $a$, $b$; or verified by subst $(-\frac{1}{3}, \frac{1}{3})$ into $y = \frac{1}{2}\|x+1\|$ and $y=\|x\|$ |
| $\Rightarrow x = -\frac{1}{3}, y = \frac{1}{3}$ | A1 | or 0.33, $-0.33$ or better; unsupported answers M0A0 |
| *or* $\frac{1}{4}(x+1)^2 = x^2$ | M1 | ft their $a$ and $b$ |
| $\Rightarrow 3x^2 - 2x - 1 = 0$ | M1ft | obtaining a quadratic $= 0$, ft their previous line, but must have an $x^2$ term |
| $\Rightarrow x = -\frac{1}{3}$ or $1$ | A1 | SC3 for $(1,1)$ $(-\frac{1}{3}, \frac{1}{3})$ and one or more additional points |
| $y = \frac{1}{3}$ or $1$ | A1 | |
---1 Fig. 1 shows the graphs of $y = | x |$ and $y = a | x + b |$, where $a$ and $b$ are constants. The intercepts of $y = a | x + b |$ with the $x$-and $y$-axes are $( - 1,0 )$ and $\left( 0 , \frac { 1 } { 2 } \right)$ respectively.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{28ce1bcc-e9d5-4ae6-98c0-67b5b8c50bc6-2_624_958_468_539}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
(i) Find $a$ and $b$.\\
(ii) Find the coordinates of the two points of intersection of the graphs.
\hfill \mbox{\textit{OCR MEI C3 2013 Q1 [6]}}