| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Modulus function |
| Type | Transformations of modulus graphs from given f(x) sketch |
| Difficulty | Moderate -0.3 This is a standard C3 transformations question requiring routine application of modulus and function transformations. Part (a) involves reflecting negative portions of a graph and vertical stretching—both textbook exercises. Part (b) requires identifying standard transformations (translation, stretch) and finding intercepts using basic logarithm laws. All techniques are straightforward with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02m Graphs of functions: difference between plotting and sketching1.02w Graph transformations: simple transformations of f(x)1.06d Natural logarithm: ln(x) function and properties1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form |
| Answer | Marks | Guidance |
|---|---|---|
| Graph showing shape with coordinates \((0, b)\) and \((a, 0)\) | B1, B1 | Shape; coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Graph showing shape with coordinates \((a, 0)\) and \((0, -2b)\) | B1, B1 | Shape; coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Translation \(\begin{pmatrix} -1 \\ 0 \end{pmatrix}\); Stretch I; SF 4; II; ÷ y-axis III; All correct and no mistakes on order etc | M1, A1, M1, A1, A1, A1 | OR: I stretch M1 1+ (II or III); II SF 4; III ÷ y-axis A1; (I + II + III) Translation M1; \(\begin{pmatrix} -1 \\ -2 \end{pmatrix}\) A1 B1; both; All correct A1 |
| Alternative: \(y = 4\ln(x+1) - 2 = 4\left[\ln(x+1) - \frac{1}{2}\right]\) | B1, M1, A1, M1, A1, A1 | Translation \(\begin{pmatrix} -1 \\ -\frac{1}{2} \end{pmatrix}\); Stretch I; SF 4; II; ÷ y-axis III; All correct and no mistakes on order etc |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 4\ln(x+1) - 2\); \(x = 0\), \(y = -2\); \(y = 0\); \(4\ln(x+1) = 2\); \(\ln(x+1) = \frac{1}{2}\); \(x+1 = e^{\frac{1}{2}}\); \(x = e^{\frac{1}{2}} - 1\) o.e. | B1, M1, A1, A1 | Isolate \(\ln(x+1) = \) or \((x+1)^\dagger\); \(x + 1 = e^\dagger\); CSO isw |
**5(a)(i)**
| Graph showing shape with coordinates $(0, b)$ and $(a, 0)$ | B1, B1 | Shape; coordinates |
**5(a)(ii)**
| Graph showing shape with coordinates $(a, 0)$ and $(0, -2b)$ | B1, B1 | Shape; coordinates |
**5(b)(i)**
| Translation $\begin{pmatrix} -1 \\ 0 \end{pmatrix}$; Stretch I; SF 4; II; ÷ y-axis III; All correct and no mistakes on order etc | M1, A1, M1, A1, A1, A1 | OR: I stretch M1 1+ (II or III); II SF 4; III ÷ y-axis A1; (I + II + III) Translation M1; $\begin{pmatrix} -1 \\ -2 \end{pmatrix}$ A1 B1; both; All correct A1 |
| Alternative: $y = 4\ln(x+1) - 2 = 4\left[\ln(x+1) - \frac{1}{2}\right]$ | B1, M1, A1, M1, A1, A1 | Translation $\begin{pmatrix} -1 \\ -\frac{1}{2} \end{pmatrix}$; Stretch I; SF 4; II; ÷ y-axis III; All correct and no mistakes on order etc |
**Total: 6 marks**
**5(b)(ii)**
| $y = 4\ln(x+1) - 2$; $x = 0$, $y = -2$; $y = 0$; $4\ln(x+1) = 2$; $\ln(x+1) = \frac{1}{2}$; $x+1 = e^{\frac{1}{2}}$; $x = e^{\frac{1}{2}} - 1$ o.e. | B1, M1, A1, A1 | Isolate $\ln(x+1) = $ or $(x+1)^\dagger$; $x + 1 = e^\dagger$; CSO isw |
**Total: 14 marks**
---5
\begin{enumerate}[label=(\alph*)]
\item The diagram shows part of the curve with equation $y = \mathrm { f } ( x )$. The curve crosses the $x$-axis at the point $( a , 0 )$ and the $y$-axis at the point $( 0 , - b )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{6ce5aa0d-0a73-4bc4-aabc-314c0434e4f5-4_569_853_1206_589}
On separate diagrams, sketch the curves with the following equations. On each diagram, indicate, in terms of $a$ or $b$, the coordinates of the points where the curve crosses the coordinate axes.
\begin{enumerate}[label=(\roman*)]
\item $y = | \mathrm { f } ( x ) |$.
\item $\quad y = 2 \mathrm { f } ( x )$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Describe a sequence of geometrical transformations that maps the graph of $y = \ln x$ onto the graph of $y = 4 \ln ( x + 1 ) - 2$.
\item Find the exact values of the coordinates of the points where the graph of $y = 4 \ln ( x + 1 ) - 2$ crosses the coordinate axes.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2008 Q5 [14]}}