| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Simple exponential equation solving |
| Difficulty | Standard +0.3 This is a straightforward C2 question involving sketching exponential curves (routine recall of graph shapes and intercepts) and solving a^x = 2b^x algebraically using logarithms. The proof requires standard log manipulation but follows a predictable path with no novel insight needed. Slightly easier than average due to its structured guidance and standard techniques. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.06a Exponential function: a^x and e^x graphs and properties1.06c Logarithm definition: log_a(x) as inverse of a^x1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Sketch showing exponential growth; intersection with \(y\)-axis is \((0,1)\) | M1, A1 | 2 For correct shape in at least 1st quadrant; for 1st and 2nd quadrants, and \(y\)-coordinate 1 stated |
| Answer | Marks | Guidance |
|---|---|---|
| Sketch showing exponential decay; intersection with \(y\)-axis is \((0,2)\) | M1, A1 | 2 For correct shape in at least 1st quadrant; for 1st and 2nd quadrants, and \(y\)-coordinate 2 stated |
| Answer | Marks | Guidance |
|---|---|---|
| \(a^x = 2b^x\) | B1 | For stating the equation in \(x\) |
| Hence \(x\log_2 a = \log_2 2 + x\log_2 b\) | M1 | For taking logs (any base) |
| M1 | For use of one log law | |
| M1 | For use of a second log law | |
| i.e. \(x = \dfrac{1}{\log_2 a - \log_2 b}\) | A1 | 5 For showing the given answer correctly |
# Question 8:
## Part (i)(a):
| Sketch showing exponential growth; intersection with $y$-axis is $(0,1)$ | M1, A1 | **2** For correct shape in at least 1st quadrant; for 1st and 2nd quadrants, and $y$-coordinate 1 stated |
|---|---|---|
## Part (i)(b):
| Sketch showing exponential decay; intersection with $y$-axis is $(0,2)$ | M1, A1 | **2** For correct shape in at least 1st quadrant; for 1st and 2nd quadrants, and $y$-coordinate 2 stated |
|---|---|---|
## Part (ii):
| $a^x = 2b^x$ | B1 | For stating the equation in $x$ |
|---|---|---|
| Hence $x\log_2 a = \log_2 2 + x\log_2 b$ | M1 | For taking logs (any base) |
| | M1 | For use of one log law |
| | M1 | For use of a second log law |
| i.e. $x = \dfrac{1}{\log_2 a - \log_2 b}$ | A1 | **5** For showing the given answer correctly |
---8 (i) On a single diagram, sketch the curves with the following equations. In each case state the coordinates of any points of intersection with the axes.
\begin{enumerate}[label=(\alph*)]
\item $y = a ^ { x }$, where $a$ is a constant such that $a > 1$.
\item $y = 2 b ^ { x }$, where $b$ is a constant such that $0 < b < 1$.\\
(ii) The curves in part (i) intersect at the point $P$. Prove that the $x$-coordinate of $P$ is
$$\frac { 1 } { \log _ { 2 } a - \log _ { 2 } b } .$$
\end{enumerate}
\hfill \mbox{\textit{OCR C2 2005 Q8 [9]}}