| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Find intersection of exponential curves |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic exponential function skills: substituting x=0, applying the trapezium rule (a standard numerical method), and solving a simple exponential equation using logarithms. All parts are routine C2-level exercises requiring direct application of standard techniques with no problem-solving insight needed. |
| Spec | 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 | Guidance |
|---|---|---|
| 6(a) | \(y_d = 3(2^0 + 1) = 6\) | M1, A1 |
| 6(b) | \(h = 2\) | B1 |
| Integral \(= h/2 \{\ldots, \ldots\}\) | M1 | |
| \(\{...\} = f(0) + 2[f(2) + f(4)] + f(6)\) | ||
| \(\{\} = 6 + 2[3 \times 5 + 3 \times 17] + 3 \times 65\) | ||
| \(= 6 + 2[15 + 51] + 195\) | ||
| Integral \(= 333\) | A1 | 4 marks |
| 6(c)(i) | \(21 = 3(2^x + 1) \Rightarrow 2^x = 6\) | B1 |
| 6(c)(ii) | \(\log_{10} 2^x = \log_{10} 6\) | M1 |
| \(x \log_{10} 2 = \log_{10} 6\) | m1 | |
| \(x = \frac{\lg 6}{\lg 2} = 2.5849\ldots = 2.58 \text{ to 3sf}\) | A1 | 3 marks |
**6(a)** | $y_d = 3(2^0 + 1) = 6$ | M1, A1 | 2 marks | Substituting $x = 0$; PI |
**6(b)** | $h = 2$ | B1 | | PI |
| | Integral $= h/2 \{\ldots, \ldots\}$ | M1 | | OE summing of areas of the three traps |
| | $\{...\} = f(0) + 2[f(2) + f(4)] + f(6)$ | | |
| | $\{\} = 6 + 2[3 \times 5 + 3 \times 17] + 3 \times 65$ | | | Condone 1 numerical slip {if not recovered} |
| | $= 6 + 2[15 + 51] + 195$ | | | [Sum of 3 traps. = 21 + 66 + 246] |
| | Integral $= 333$ | A1 | 4 marks | CAO |
**6(c)(i)** | $21 = 3(2^x + 1) \Rightarrow 2^x = 6$ | B1 | 1 mark | AG (be convinced) |
**6(c)(ii)** | $\log_{10} 2^x = \log_{10} 6$ | M1 | | Take ln or $\log_{10}$ of both sides of $a^x = b$ or other relevant base if clear. The equation $a^x = b$ used must be correct. |
| | $x \log_{10} 2 = \log_{10} 6$ | m1 | | Use of $\log 2^x = x \log 2$ OE |
| | $x = \frac{\lg 6}{\lg 2} = 2.5849\ldots = 2.58 \text{ to 3sf}$ | A1 | 3 marks | Both method marks must have been awarded |
---6 The diagram shows a sketch of the curve with equation $y = 3 \left( 2 ^ { x } + 1 \right)$.\\
\includegraphics[max width=\textwidth, alt={}, center]{ad574bde-3bf1-45be-a454-9c723088b357-5_465_851_390_607}
The curve $y = 3 \left( 2 ^ { x } + 1 \right)$ intersects the $y$-axis at the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Find the $y$-coordinate of the point $A$.
\item Use the trapezium rule with four ordinates (three strips) to find an approximate value for $\int _ { 0 } ^ { 6 } 3 \left( 2 ^ { x } + 1 \right) d x$.
\item The line $y = 21$ intersects the curve $y = 3 \left( 2 ^ { x } + 1 \right)$ at the point $P$.
\begin{enumerate}[label=(\roman*)]
\item Show that the $x$-coordinate of $P$ satisfies the equation
$$2 ^ { x } = 6$$
\item Use logarithms to find the $x$-coordinate of $P$, giving your answer to three significant figures.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2007 Q6 [10]}}