| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2006 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Simple exponential equation solving |
| Difficulty | Moderate -0.8 Part (a) is a straightforward logarithmic equation requiring only taking logs of both sides and rearranging. Part (b)(i) involves simple manipulation of the sum to infinity formula (S = a/(1-r)), and part (b)(ii) requires solving an inequality with geometric sequences. All parts are routine C2-level exercises with standard techniques and no problem-solving insight required, making this easier than average but not trivial. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04j Sum to infinity: convergent geometric series |r|<11.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\log 0.8^x = \log 0.05\) or \(x = \log_{0.8} 0.05\) | M1 | NMS: SC B2 for 13.425 or better (B1 for 13.4 or 13.43; 13.42) |
| \(x \log_{10} 0.8 = \log_{10} 0.05\) oe | A1 | |
| \(x = 13.425\) to 3dp; \(13.425\) (A2) | A1 | Condone greater accuracy |
| (else A1 for 1 or 2dp) | ||
| Total (a) | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{a}{1-r}\) | M1 | \(S_\infty = \frac{a}{1-r}\) used |
| \(\frac{a}{1-r} = 5a \Rightarrow a = 5a(1-r)\) | A1 | Or better |
| \(\Rightarrow 1 = 5(1-r) \Rightarrow r = \frac{4}{5} = 0.8\) | A1 | AG (be convinced) |
| Total (b)(i) | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(n^{\text{th}}\) term \(= 20 \times (0.8)^{n-1}\) | M1 | Condone \(20\times(0.8)^n\). \(0.8^{n-1} < 0.05\) or \(0.8^{n-1} = k\), where \(k=0.05\) or \(k\) rounds up to \(0.050\) |
| \(n^{\text{th}}\) term \(< 1 \Rightarrow 0.8^{n-1} < \frac{1}{20}\) oe | A1 | |
| Least \(n\) is 15 | A1F | If not 15, ft on integer part of [answer (a)+2] provided \(n>2\). SC 3/3 for 15 if no error. SC \(n^{\text{th}}\) term\(=16^{n-1}\) M1A0A0 |
| Total (b)(ii) | 3 |
## Question 3(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\log 0.8^x = \log 0.05$ or $x = \log_{0.8} 0.05$ | M1 | **NMS**: SC B2 for 13.425 or better (B1 for 13.4 or 13.43; 13.42) |
| $x \log_{10} 0.8 = \log_{10} 0.05$ oe | A1 | |
| $x = 13.425$ to 3dp; $13.425$ **(A2)** | A1 | Condone greater accuracy |
| (else A1 for 1 or 2dp) | | |
| **Total (a)** | **3** | |
## Question 3(b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{a}{1-r}$ | M1 | $S_\infty = \frac{a}{1-r}$ **used** |
| $\frac{a}{1-r} = 5a \Rightarrow a = 5a(1-r)$ | A1 | Or better |
| $\Rightarrow 1 = 5(1-r) \Rightarrow r = \frac{4}{5} = 0.8$ | A1 | AG (be convinced) |
| **Total (b)(i)** | **3** | |
## Question 3(b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $n^{\text{th}}$ term $= 20 \times (0.8)^{n-1}$ | M1 | Condone $20\times(0.8)^n$. $0.8^{n-1} < 0.05$ or $0.8^{n-1} = k$, where $k=0.05$ or $k$ rounds up to $0.050$ |
| $n^{\text{th}}$ term $< 1 \Rightarrow 0.8^{n-1} < \frac{1}{20}$ oe | A1 | |
| Least $n$ is 15 | A1F | If not 15, ft on integer part of [answer (a)+2] provided $n>2$. **SC** 3/3 for 15 if no error. **SC** $n^{\text{th}}$ term$=16^{n-1}$ M1A0A0 |
| **Total (b)(ii)** | **3** | |
---3
\begin{enumerate}[label=(\alph*)]
\item Use logarithms to solve the equation $0.8 ^ { x } = 0.05$, giving your answer to three decimal places.
\item An infinite geometric series has common ratio $r$. The sum to infinity of the series is five times the first term of the series.
\begin{enumerate}[label=(\roman*)]
\item Show that $r = 0.8$.
\item Given that the first term of the series is 20 , find the least value of $n$ such that the $n$th term of the series is less than 1 .
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2006 Q3 [9]}}