| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Find intersection of exponential curves |
| Difficulty | Standard +0.3 This is a straightforward C2 exponential question requiring standard techniques: sketching y=7^x (basic recall), forming a quadratic by substitution (7^x = u), solving it to find one solution, and using logarithms to find x. The quadratic substitution is a common textbook exercise, and the logarithm application is routine. Slightly above average difficulty only due to the multi-step nature and requiring coordination of several techniques, but each individual step is standard. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.06a Exponential function: a^x and e^x graphs and properties1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Curve in 1st and 2nd quadrants only, crossing positive \(y\)-axis once, asymptotic to negative \(x\)-axis | B1 | Correct shape |
| Coordinates \((0, 1)\) | B1 (2 marks) | Accept \(y\)-intercept indicated as 1 on diagram |
| Answer | Marks | Guidance |
|---|---|---|
| \(y^2 - 12 = y\) OE; \(7^{2x} - 12 = 7^x\) OE | M1 | Eliminates either \(x\) or \(y\) correctly |
| \((y-4)(y+3) = 0\); \((7^x - 4)(7^x + 3) = 0\) | A1 | Correct factors or \(y = \frac{1 \pm \sqrt{49}}{2}\) |
| Since \(y = 7^x > 0\), \(y \neq -3\) (exactly one intersection) | E1 | Clear indication negative solution considered and rejected |
| \(y\)-coordinate is 4 | B1 (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(7^x = 4\), so \(x\log 7 = \log 4\) | M1 | OE ft on \(7^x = k\) where \(k\) is positive |
| \(x = 0.712\) to 3SF | A1 (2 marks) | Condone \(> \) three significant figures |
## Question 8:
### Part (a):
| Curve in 1st and 2nd quadrants only, crossing positive $y$-axis once, asymptotic to negative $x$-axis | B1 | Correct shape |
| Coordinates $(0, 1)$ | B1 (2 marks) | Accept $y$-intercept indicated as 1 on diagram |
### Part (b)(i):
| $y^2 - 12 = y$ OE; $7^{2x} - 12 = 7^x$ OE | M1 | Eliminates either $x$ or $y$ correctly |
| $(y-4)(y+3) = 0$; $(7^x - 4)(7^x + 3) = 0$ | A1 | Correct factors or $y = \frac{1 \pm \sqrt{49}}{2}$ |
| Since $y = 7^x > 0$, $y \neq -3$ (exactly one intersection) | E1 | Clear indication negative solution considered and rejected |
| $y$-coordinate is 4 | B1 (4 marks) | |
### Part (b)(ii):
| $7^x = 4$, so $x\log 7 = \log 4$ | M1 | OE ft on $7^x = k$ where $k$ is positive |
| $x = 0.712$ to 3SF | A1 (2 marks) | Condone $> $ three significant figures |
---8
\begin{enumerate}[label=(\alph*)]
\item Sketch the curve with equation $y = 7 ^ { x }$, indicating the coordinates of any point where the curve intersects the coordinate axes.
\item The curve $C _ { 1 }$ has equation $y = 7 ^ { x }$.
The curve $C _ { 2 }$ has equation $y = 7 ^ { 2 x } - 12$.
\begin{enumerate}[label=(\roman*)]
\item By forming and solving a quadratic equation, prove that the curves $C _ { 1 }$ and $C _ { 2 }$ intersect at exactly one point. State the $y$-coordinate of this point.
\item Use logarithms to find the $x$-coordinate of the point of intersection of $C _ { 1 }$ and $C _ { 2 }$, giving your answer to three significant figures.\\
(2 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2012 Q8 [8]}}