| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Specific function transformation description |
| Difficulty | Standard +0.3 This is a slightly easier than average C3 question. Part (i) tests standard transformations of logarithmic functions (translation and stretch). Part (ii) requires solving ln x = 2ln(x-6), which simplifies to a quadratic using log laws. Part (iii) is routine application of Simpson's rule. All parts follow standard procedures with no novel problem-solving required, making it marginally easier than the typical C3 question. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.06d Natural logarithm: ln(x) function and properties1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| Refer to translation and stretch | M1 | in either order; allow here equiv informal terms such as 'move', … |
| State translation in \(x\) direction by 6 | A1 | or equiv; now with correct terminology |
| State stretch in \(y\) direction by 2 | A1\3 | or equiv; now with correct terminology |
| Answer | Marks | Guidance |
|---|---|---|
| State \(2\ln(x - 6) = \ln x\) | B1 | or \(2\ln(a - 6) = \ln a\) or equiv |
| Show correct use of logarithm property | *M1 | |
| Attempt solution of 3-term quadratic | M1 | dep *M |
| Obtain 9 only | A1\4 | following correct solution of equation |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt evaluation of form \(k(y_0 + 4y_1 + y_2)\) | M1 | any constant \(k\); maybe with \(y_0 = 0\) implied |
| Obtain \(\frac{1}{2} \times \ln(2 \ln 1 + 8 \ln 2 + 2 \ln 3)\) | A1 | |
| Obtain 2.58 | A1\3 | or greater accuracy 2.5808… |
## Part (i)
Refer to translation and stretch | M1 | in either order; allow here equiv informal terms such as 'move', …
State translation in $x$ direction by 6 | A1 | or equiv; now with correct terminology
State stretch in $y$ direction by 2 | A1\3 | or equiv; now with correct terminology
[SC: if M0 but one transformation completely correct, give B1]
## Part (ii)
State $2\ln(x - 6) = \ln x$ | B1 | or $2\ln(a - 6) = \ln a$ or equiv
Show correct use of logarithm property | *M1 |
Attempt solution of 3-term quadratic | M1 | dep *M
Obtain 9 only | A1\4 | following correct solution of equation
## Part (iii)
Attempt evaluation of form $k(y_0 + 4y_1 + y_2)$ | M1 | any constant $k$; maybe with $y_0 = 0$ implied
Obtain $\frac{1}{2} \times \ln(2 \ln 1 + 8 \ln 2 + 2 \ln 3)$ | A1 |
Obtain 2.58 | A1\3 | or greater accuracy 2.5808…
---\includegraphics{figure_4}
The diagram shows the curves $y = \ln x$ and $y = 2 \ln(x - 6)$. The curves meet at the point $P$ which has $x$-coordinate $a$. The shaded region is bounded by the curve $y = 2 \ln(x - 6)$ and the lines $x = a$ and $y = 0$.
\begin{enumerate}[label=(\roman*)]
\item Give details of the pair of transformations which transforms the curve $y = \ln x$ to the curve $y = 2 \ln(x - 6)$. [3]
\item Solve an equation to find the value of $a$. [4]
\item Use Simpson's rule with two strips to find an approximation to the area of the shaded region. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR C3 2009 Q8 [10]}}