| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Find intersection of exponential curves |
| Difficulty | Standard +0.3 This question involves standard transformations of exponential functions and solving a straightforward exponential equation. Part (a) requires recognizing that 2^(3x+2) = 4·2^(3x), giving a vertical stretch factor 4. Part (b) involves setting up 2^(3x+2) - 2^(3x) = 36, factoring to 3·2^(3x) = 36, then using logarithms to express the answer in the required form. While it requires multiple steps and logarithm manipulation, these are routine A-level techniques with no novel problem-solving required. |
| 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 | Marks | Guidance |
| scale factor 4 | B1 | Allow s.f. or 'factor' but not just 4 or "4 units" or "scale factor 4 units". Comments MUST refer to a stretch otherwise B0B0 |
| parallel to the \(y\)-axis | B1 | Allow: in the \(y\)-direction, vertically, vertically upwards, vertical stretch, in the vertical direction, positive \(y\)-direction, parallel to the positive \(y\)-axis. Not: just upwards or in/on/about/across/through/along/towards the \(y\)-axis. If B0B0 then SCB1 for re-writing as \((y=)4\left(2^{3x}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2^{3x+2} - 2^{3x} = 36\) | B1 | For the correct equation in \(x\). Or \(4y_A - y_A = 36 \Rightarrow y_A = 12\) |
| \(2^{3x} = 12 \Rightarrow \log\left(2^{3x}\right) = \log 12\) | M1 | Simplify to \(2^{3x} = k\) where \(k>0\) and take logs of both sides (any base) |
| \(x = \frac{1}{3}\log_2 12\) | A1 | oe, e.g. \(x = \log_2(12^{1/3})\) |
## Question 5:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| scale factor 4 | B1 | Allow s.f. or 'factor' but not just 4 or "4 units" or "scale factor 4 units". Comments MUST refer to a stretch otherwise B0B0 |
| parallel to the $y$-axis | B1 | Allow: in the $y$-direction, vertically, vertically upwards, vertical stretch, in the vertical direction, positive $y$-direction, parallel to the positive $y$-axis. Not: just upwards or in/on/about/across/through/along/towards the $y$-axis. If B0B0 then SCB1 for re-writing as $(y=)4\left(2^{3x}\right)$ |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2^{3x+2} - 2^{3x} = 36$ | B1 | For the correct equation in $x$. Or $4y_A - y_A = 36 \Rightarrow y_A = 12$ |
| $2^{3x} = 12 \Rightarrow \log\left(2^{3x}\right) = \log 12$ | M1 | Simplify to $2^{3x} = k$ where $k>0$ and take logs of both sides (any base) |
| $x = \frac{1}{3}\log_2 12$ | A1 | oe, e.g. $x = \log_2(12^{1/3})$ |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{8c0b68bd-2257-4994-b444-def0b3f64334-4_591_547_262_242}
The diagram shows the graphs of $y = 2 ^ { 3 x }$ and $y = 2 ^ { 3 x + 2 }$. The graph of $y = 2 ^ { 3 x }$ can be transformed to the graph of $y = 2 ^ { 3 x + 2 }$ by means of a stretch.
\begin{enumerate}[label=(\alph*)]
\item Give details of the stretch.
The point $A$ lies on $y = 2 ^ { 3 x }$ and the point $B$ lies on $y = 2 ^ { 3 x + 2 }$. The line segment $A B$ is parallel to the $y$-axis and the difference between the $y$-coordinates of $A$ and $B$ is 36 .
\item Determine the $x$-coordinate of $A$. Give your answer in the form $m \log _ { 2 } n$ where $m$ and $n$ are constants to be determined.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q5 [5]}}