| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Solve exponential equation by substitution |
| Difficulty | Moderate -0.8 This is a routine C2 exponential equation question with clear scaffolding. Part (c)(i) explicitly gives the substitution and target form, requiring only algebraic manipulation of exponential laws (9^x = (3^2)^x = (3^x)^2 = Y^2 and 3^(x+1) = 3ยท3^x = 3Y). Part (c)(ii) is straightforward: substitute Y values back and apply logarithms. The scaffolding and standard techniques make this easier than average. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.06a Exponential function: a^x and e^x graphs and properties1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Shape (graph goes below intersection point) | B1 | Shape must clearly go below the intersection pt. Condone if \(x\)-axis is a tangent |
| Only intersection with \(y\)-axis at \((0,1)\) stated/indicated | B1 | Accept 1 on \(y\)-axis as equivalent; total 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Stretch (I) in \(x\)-direction (II), scale factor \(0.5\) (III) | M1, A1 | Need (I) & one of (II),(III); M0 if \(>1\) transformation; total 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Translation | B1 | Must be 'Translation' or 'translate(d)' for 1st B mark |
| \(\begin{bmatrix}-1\\0\end{bmatrix}\) | B1 | Accept full equivalent in words linked to 'translation/move/shift' and negative \(x\)-direction. Note: B0 B1 is possible; total 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(9^x = (3^2)^x = 3^{2x} = (3^x)^2 = Y^2\); \(3^{x+1} = 3^x \times 3^1 = 3Y\) | M1 | Justifying either \(9^x = Y^2\) or \(3^{x+1} = 3Y\) |
| \(9^x - 3^{x+1} + 2 = 0 \Rightarrow Y^2 - 3Y + 2 = 0 \Rightarrow (Y-1)(Y-2)=0\) | A1 | AG; total 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(Y=1 \Rightarrow 3^x = 1 \Rightarrow x=0\) | B1 | AG (Accept direct substitution if convinced) |
| \(Y=2 \Rightarrow 3^x = 2\); \(\log_{10} 3^x = \log_{10} 2\) | M1 | Takes logs of both, PI by 'correct' value(s) later; or \(x = \log_3 2\) seen |
| \(x\log_{10} 3 = \log_{10} 2\) | m1 | Use of \(\log 3^x = x\log 3\); or \(\log_3 2 = \frac{\lg 2}{\lg 3}\) OE (PI by \(\log_3 2 = 0.630\) or better) |
| \(x = \frac{\lg 2}{\lg 3} = 0.630929\ldots = 0.6309\) to 4dp | A1 | Must show that logarithms have been used otherwise 0/3; total 4 marks |
## Question 8:
**Part (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Shape (graph goes below intersection point) | B1 | Shape must clearly go below the intersection pt. Condone if $x$-axis is a tangent |
| Only intersection with $y$-axis at $(0,1)$ stated/indicated | B1 | Accept 1 on $y$-axis as equivalent; total 2 marks |
**Part (b)(i):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Stretch (I) in $x$-direction (II), scale factor $0.5$ (III) | M1, A1 | Need (I) & one of (II),(III); M0 if $>1$ transformation; total 2 marks |
**Part (b)(ii):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Translation | B1 | Must be 'Translation' or 'translate(d)' for 1st B mark |
| $\begin{bmatrix}-1\\0\end{bmatrix}$ | B1 | Accept full equivalent in words linked to 'translation/move/shift' and **negative** $x$-direction. Note: B0 B1 is possible; total 2 marks |
**Part (c)(i):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $9^x = (3^2)^x = 3^{2x} = (3^x)^2 = Y^2$; $3^{x+1} = 3^x \times 3^1 = 3Y$ | M1 | Justifying either $9^x = Y^2$ or $3^{x+1} = 3Y$ |
| $9^x - 3^{x+1} + 2 = 0 \Rightarrow Y^2 - 3Y + 2 = 0 \Rightarrow (Y-1)(Y-2)=0$ | A1 | AG; total 2 marks |
**Part (c)(ii):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $Y=1 \Rightarrow 3^x = 1 \Rightarrow x=0$ | B1 | AG (Accept direct substitution if convinced) |
| $Y=2 \Rightarrow 3^x = 2$; $\log_{10} 3^x = \log_{10} 2$ | M1 | Takes logs of both, PI by 'correct' value(s) later; or $x = \log_3 2$ seen |
| $x\log_{10} 3 = \log_{10} 2$ | m1 | Use of $\log 3^x = x\log 3$; or $\log_3 2 = \frac{\lg 2}{\lg 3}$ OE (PI by $\log_3 2 = 0.630$ or better) |
| $x = \frac{\lg 2}{\lg 3} = 0.630929\ldots = 0.6309$ to 4dp | A1 | Must show that logarithms have been used otherwise 0/3; total 4 marks |
---8
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of $y = 3 ^ { x }$, stating the coordinates of the point where the graph crosses the $y$-axis.
\item Describe a single geometrical transformation that maps the graph of $y = 3 ^ { x }$ :
\begin{enumerate}[label=(\roman*)]
\item onto the graph of $y = 3 ^ { 2 x }$;
\item onto the graph of $y = 3 ^ { x + 1 }$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Using the substitution $Y = 3 ^ { x }$, show that the equation
$$9 ^ { x } - 3 ^ { x + 1 } + 2 = 0$$
can be written as
$$( Y - 1 ) ( Y - 2 ) = 0$$
\item Hence show that the equation $9 ^ { x } - 3 ^ { x + 1 } + 2 = 0$ has a solution $x = 0$ and, by using logarithms, find the other solution, giving your answer to four decimal places.\\
(4 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2008 Q8 [12]}}