| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Simple exponential equation solving |
| Difficulty | Moderate -0.3 This is a multi-part C2 question covering standard exponential techniques: solving with logarithms, trapezium rule application, translation of curves, and algebraic manipulation of exponential equations. While part (d) requires careful algebraic manipulation with logarithms, all parts follow routine procedures taught at this level with no novel problem-solving required. Slightly easier than average due to the scaffolded structure and standard techniques. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(6 = 3 \times 12^k \Rightarrow 2 = 12^k \Rightarrow k = \log_{12}2 = \frac{\log 2}{\log 12} \approx 0.279\) | M1, A1, A1 | Must use logarithms; 3 sig figs |
| Answer | Marks | Guidance |
|---|---|---|
| \(h = 0.5\); ordinates at \(x = 0, 0.5, 1, 1.5\): \(y = 3, 3\times12^{0.5}, 36, 3\times12^{1.5}\); \(\approx \frac{0.5}{2}[3 + 2(10.39 + 36) + 124.7] \approx 55\) | B1, M1, A1, A1 | 2 sig figs required |
| Answer | Marks |
|---|---|
| Translation by \(\binom{1}{p}\) gives \(f(x) = 3\times12^{x-1} + p\); passes through \((0,0)\): \(0 = 3\times12^{-1} + p \Rightarrow p = -\frac{1}{4}\) | M1, M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(2^{2-x} = 3\times12^x\); taking \(\log_2\): \((2-x) = \log_2 3 + x\log_2 12\); \(2 - \log_2 3 = x(1 + \log_2 12)\); \(\log_2 12 = \log_2 4 + \log_2 3 = 2 + \log_2 3\); \(x = \frac{2-\log_2 3}{3 + \log_2 3}\); \(q = 3\) | M1, M1, A1, M1, A1 | Show all steps; state \(q=3\) |
## Question 9:
### Part (a):
| $6 = 3 \times 12^k \Rightarrow 2 = 12^k \Rightarrow k = \log_{12}2 = \frac{\log 2}{\log 12} \approx 0.279$ | M1, A1, A1 | Must use logarithms; 3 sig figs |
### Part (b):
| $h = 0.5$; ordinates at $x = 0, 0.5, 1, 1.5$: $y = 3, 3\times12^{0.5}, 36, 3\times12^{1.5}$; $\approx \frac{0.5}{2}[3 + 2(10.39 + 36) + 124.7] \approx 55$ | B1, M1, A1, A1 | 2 sig figs required |
### Part (c):
| Translation by $\binom{1}{p}$ gives $f(x) = 3\times12^{x-1} + p$; passes through $(0,0)$: $0 = 3\times12^{-1} + p \Rightarrow p = -\frac{1}{4}$ | M1, M1, A1 | |
### Part (d):
| $2^{2-x} = 3\times12^x$; taking $\log_2$: $(2-x) = \log_2 3 + x\log_2 12$; $2 - \log_2 3 = x(1 + \log_2 12)$; $\log_2 12 = \log_2 4 + \log_2 3 = 2 + \log_2 3$; $x = \frac{2-\log_2 3}{3 + \log_2 3}$; $q = 3$ | M1, M1, A1, M1, A1 | Show all steps; state $q=3$ |9 A curve has equation $y = 3 \times 12 ^ { x }$.
\begin{enumerate}[label=(\alph*)]
\item The point ( $k , 6$ ) lies on the curve $y = 3 \times 12 ^ { x }$. Use logarithms to find the value of $k$, giving your answer to three significant figures.
\item Use the trapezium rule with four ordinates (three strips) to find an approximate value for $\int _ { 0 } ^ { 1.5 } 3 \times 12 ^ { x } \mathrm {~d} x$, giving your answer to two significant figures.
\item The curve $y = 3 \times 12 ^ { x }$ is translated by the vector $\left[ \begin{array} { l } 1 \\ p \end{array} \right]$ to give the curve $y = \mathrm { f } ( x )$. Given that the curve $y = \mathrm { f } ( x )$ passes through the origin ( 0,0 ), find the value of the constant $p$.
\item The curve with equation $y = 2 ^ { 2 - x }$ intersects the curve $y = 3 \times 12 ^ { x }$ at the point $T$. Show that the $x$-coordinate of $T$ can be written in the form $\frac { 2 - \log _ { 2 } 3 } { q + \log _ { 2 } 3 }$, where $q$ is an integer. State the value of $q$.\\[0pt]
[5 marks]
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{30ccdbe9-0c91-4011-a3f9-3ce01862215d-20_2288_1707_221_153}
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\end{enumerate}
\hfill \mbox{\textit{AQA C2 2014 Q9 [15]}}