| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Sketch exponential graphs |
| Difficulty | Moderate -0.3 This is a multi-part question covering standard C2 exponential topics: finding y-intercepts (trivial substitution), trapezium rule (routine numerical method), transformations (standard recall), and logarithm laws (bookwork). Part (e)(ii) requires combining transformations with logarithms but follows a predictable pattern. All parts are textbook exercises requiring technique application rather than problem-solving insight, making it slightly easier than average. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.06a Exponential function: a^x and e^x graphs and properties1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b1.09f Trapezium rule: numerical integration |
| Answer | Marks |
|---|---|
| \(y = 2^0 = 1\) | B1 |
| Answer | Marks |
|---|---|
| \(h = 0.2\); ordinates: \(1, 2^{0.8}, 2^{1.6}, 2^{2.4}, 2^{3.2}, 2^4 = 1, 1.7411, 3.0314, 5.2780, 9.1895, 16\) | M1 |
| \(\approx \frac{0.2}{2}[1 + 16 + 2(1.7411 + 3.0314 + 5.2780 + 9.1895)]\) | M1 A1 |
| \(\approx 6.59\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Translation by \(\begin{pmatrix} \frac{3}{4} \\ 0 \end{pmatrix}\) (translation of \(\frac{3}{4}\) in positive \(x\)-direction) | B1 B1 | B1 for translation, B1 for correct vector |
| Answer | Marks |
|---|---|
| \(g(x) = 2^{4(x-1)} - \frac{1}{2}\) (translation by \(\begin{pmatrix}1 \\ -\frac{1}{2}\end{pmatrix}\)) | M1 |
| At \(Q\): \(2^{4(x-1)} = \frac{1}{2} = 2^{-1}\) | M1 |
| \(4(x-1) = -1\), so \(x = \frac{3}{4}\) | A1 A1 |
| Answer | Marks |
|---|---|
| \(\log_a k = 3\log_a 2 + \log_a 5 - \log_a 4\) | M1 |
| \(= \log_a 8 + \log_a 5 - \log_a 4\) | M1 |
| \(= \log_a \frac{40}{4} = \log_a 10\), so \(k = 10\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{5}{4} = 2^{4x-3}\) | M1 | |
| \((4x-3)\log_{10} 2 = \log_{10}\frac{5}{4}\) | M1 | |
| \(\log_{10}\frac{5}{4} = \log_{10}10 - \log_{10}8 = 1 - 3\log_{10}2\) (using \(k=10\)) | M1 | |
| \((4x-3) = \frac{1-3\log_{10}2}{\log_{10}2}\), giving \(x = \frac{1}{4\log_{10}2}\) | A1 | Completion shown convincingly |
# Question 8:
## Part (a):
$y = 2^0 = 1$ | B1 |
## Part (b):
$h = 0.2$; ordinates: $1, 2^{0.8}, 2^{1.6}, 2^{2.4}, 2^{3.2}, 2^4 = 1, 1.7411, 3.0314, 5.2780, 9.1895, 16$ | M1 |
$\approx \frac{0.2}{2}[1 + 16 + 2(1.7411 + 3.0314 + 5.2780 + 9.1895)]$ | M1 A1 |
$\approx 6.59$ | A1 |
## Part (c):
Translation by $\begin{pmatrix} \frac{3}{4} \\ 0 \end{pmatrix}$ (translation of $\frac{3}{4}$ in positive $x$-direction) | B1 B1 | B1 for translation, B1 for correct vector
## Part (d):
$g(x) = 2^{4(x-1)} - \frac{1}{2}$ (translation by $\begin{pmatrix}1 \\ -\frac{1}{2}\end{pmatrix}$) | M1 |
At $Q$: $2^{4(x-1)} = \frac{1}{2} = 2^{-1}$ | M1 |
$4(x-1) = -1$, so $x = \frac{3}{4}$ | A1 A1 |
## Part (e)(i):
$\log_a k = 3\log_a 2 + \log_a 5 - \log_a 4$ | M1 |
$= \log_a 8 + \log_a 5 - \log_a 4$ | M1 |
$= \log_a \frac{40}{4} = \log_a 10$, so $k = 10$ | A1 |
## Part (e)(ii):
$\frac{5}{4} = 2^{4x-3}$ | M1 |
$(4x-3)\log_{10} 2 = \log_{10}\frac{5}{4}$ | M1 |
$\log_{10}\frac{5}{4} = \log_{10}10 - \log_{10}8 = 1 - 3\log_{10}2$ (using $k=10$) | M1 |
$(4x-3) = \frac{1-3\log_{10}2}{\log_{10}2}$, giving $x = \frac{1}{4\log_{10}2}$ | A1 | Completion shown convincingly8 The diagram shows a sketch of the curve $y = 2 ^ { 4 x }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{f9a7a4dd-f7fd-4135-8872-2c1270d46a14-9_435_814_374_623}
The curve intersects the $y$-axis at the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of the $y$-coordinate of $A$.
\item Use the trapezium rule with six ordinates (five strips) to find an approximate value for $\int _ { 0 } ^ { 1 } 2 ^ { 4 x } \mathrm {~d} x$, giving your answer to two decimal places.
\item Describe the geometrical transformation that maps the graph of $y = 2 ^ { 4 x }$ onto the graph of $y = 2 ^ { 4 x - 3 }$.
\item The curve $y = 2 ^ { 4 x }$ is translated by the vector $\left[ \begin{array} { c } 1 \\ - \frac { 1 } { 2 } \end{array} \right]$ to give the curve $y = \mathrm { g } ( x )$.
The curve $y = \mathrm { g } ( x )$ crosses the $x$-axis at the point $Q$. Find the $x$-coordinate of $Q$.
\item \begin{enumerate}[label=(\roman*)]
\item Given that
$$\log _ { a } k = 3 \log _ { a } 2 + \log _ { a } 5 - \log _ { a } 4$$
show that $k = 10$.
\item The line $y = \frac { 5 } { 4 }$ crosses the curve $y = 2 ^ { 4 x - 3 }$ at the point $P$. Show that the $x$-coordinate of $P$ is $\frac { 1 } { 4 \log _ { 10 } 2 }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2010 Q8 [17]}}