| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Single transformation between given equations |
| Difficulty | Standard +0.3 This is a multi-part question covering standard C2 topics: trapezium rule (routine calculation), simple translation vector (basic recall), logarithm manipulation (standard identities), and finding intersection points. While part (c)(iii) requires combining several steps, all techniques are straightforward applications of A-level methods with no novel insight required. Slightly easier than average due to heavily scaffolded structure. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.06f Laws of logarithms: addition, subtraction, power rules1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(h = 0.25\) | B1 | PI |
| \(I \approx \frac{h}{2}\{f(0) + f(1) + 2[f(0.25) + f(0.5) + f(0.75)]\}\) | M1 | OE summing areas of trapezoids |
| \(= \log 1 + \log 2 + 2\left[\log\frac{17}{16} + \log\frac{5}{4} + \log\frac{25}{16}\right]\) | A1 | OE accept 1sf evidence |
| \(I \approx 0.125[0.935147...] = 0.117\) (to 3SF) | A1 (4 marks) | CAO Must be 0.117 |
| Answer | Marks |
|---|---|
| \(\begin{bmatrix} 0 \\ 1 \end{bmatrix}\) | B1 (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\log_{10}(10x^2) = \log_{10}10 + \log_{10}x^2 = 1 + 2\log_{10}x\) | M1, A1 (2 marks) | Bases must be included or statement \(\log_{10}10 = 1\) given |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 1 + 2\log_{10}x = \log_{10}(10x^2)\) | M1 | PI |
| \(y = 2\log_{10}(\sqrt{10}\, x)\) | A1 | Writing in correct form for stretch |
| Stretch parallel to \(x\)-axis, scale factor \(\frac{1}{\sqrt{10}}\) | B2,1,0 (4 marks) | B2 for correct direction and scale factor ACF |
| Answer | Marks | Guidance |
|---|---|---|
| \(\log_{10}(10x^2) = \log_{10}(x^2 + 1)\), so \(10x^2 = x^2 + 1\), \(9x^2 = 1\) | M1 | PI by \(10x^2 = x^2 + 1\) or correct \(x\) |
| Since \(x > 0\), \(x = \frac{1}{3}\) | A1 | OE stated or used |
| \(y\)-coordinate of \(P\): \(y = \log_{10}\frac{10}{9}\) | A1 | PI by \(3\log_{10}\frac{10}{9}\) OE for gradient of \(OP\) |
| Gradient of \(OP = 3\log_{10}\frac{10}{9} = \log_{10}\frac{1000}{729}\) | A1 (4 marks) | Accept '\(a=1000\), \(b=729\)' |
## Question 9:
### Part (a):
| $h = 0.25$ | B1 | PI |
| $I \approx \frac{h}{2}\{f(0) + f(1) + 2[f(0.25) + f(0.5) + f(0.75)]\}$ | M1 | OE summing areas of trapezoids |
| $= \log 1 + \log 2 + 2\left[\log\frac{17}{16} + \log\frac{5}{4} + \log\frac{25}{16}\right]$ | A1 | OE accept 1sf evidence |
| $I \approx 0.125[0.935147...] = 0.117$ (to 3SF) | A1 (4 marks) | CAO Must be 0.117 |
### Part (b):
| $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ | B1 (1 mark) | |
### Part (c)(i):
| $\log_{10}(10x^2) = \log_{10}10 + \log_{10}x^2 = 1 + 2\log_{10}x$ | M1, A1 (2 marks) | Bases must be included or statement $\log_{10}10 = 1$ given |
### Part (c)(ii):
| $y = 1 + 2\log_{10}x = \log_{10}(10x^2)$ | M1 | PI |
| $y = 2\log_{10}(\sqrt{10}\, x)$ | A1 | Writing in correct form for stretch |
| Stretch parallel to $x$-axis, scale factor $\frac{1}{\sqrt{10}}$ | B2,1,0 (4 marks) | B2 for correct direction and scale factor ACF |
### Part (c)(iii):
| $\log_{10}(10x^2) = \log_{10}(x^2 + 1)$, so $10x^2 = x^2 + 1$, $9x^2 = 1$ | M1 | PI by $10x^2 = x^2 + 1$ or correct $x$ |
| Since $x > 0$, $x = \frac{1}{3}$ | A1 | OE stated or used |
| $y$-coordinate of $P$: $y = \log_{10}\frac{10}{9}$ | A1 | PI by $3\log_{10}\frac{10}{9}$ OE for gradient of $OP$ |
| Gradient of $OP = 3\log_{10}\frac{10}{9} = \log_{10}\frac{1000}{729}$ | A1 (4 marks) | Accept '$a=1000$, $b=729$' |9 The diagram shows part of a curve whose equation is $y = \log _ { 10 } \left( x ^ { 2 } + 1 \right)$.\\
\includegraphics[max width=\textwidth, alt={}, center]{a5fa3066-e330-46d0-98e3-92d438ed6f61-5_355_451_367_799}
\begin{enumerate}[label=(\alph*)]
\item Use the trapezium rule with five ordinates (four strips) to find an approximate value for
$$\int _ { 0 } ^ { 1 } \log _ { 10 } \left( x ^ { 2 } + 1 \right) d x$$
giving your answer to three significant figures.
\item The graph of $y = 2 \log _ { 10 } x$ can be transformed into the graph of $y = 1 + 2 \log _ { 10 } x$ by means of a translation. Write down the vector of the translation.
\item \begin{enumerate}[label=(\roman*)]
\item Show that $\log _ { 10 } \left( 10 x ^ { 2 } \right) = 1 + 2 \log _ { 10 } x$.
\item Show that the graph of $y = 2 \log _ { 10 } x$ can also be transformed into the graph of $y = 1 + 2 \log _ { 10 } x$ by means of a stretch, and describe the stretch.
\item The curve with equation $y = 1 + 2 \log _ { 10 } x$ intersects the curve $y = \log _ { 10 } \left( x ^ { 2 } + 1 \right)$ at the point $P$. Given that the $x$-coordinate of $P$ is positive, find the gradient of the line $O P$, where $O$ is the origin. Give your answer in the form $\log _ { 10 } \left( \frac { a } { b } \right)$, where $a$ and $b$ are integers.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2012 Q9 [15]}}