| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find dy/dx at a point |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question with standard techniques: (a) requires substitution to find k (basic algebra with logarithms), (b) requires applying the product rule and chain rule to differentiate implicitly, then substituting the point. While it involves exponentials and logarithms, the methods are routine C4 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute \(x = \ln 2\), \(y = \frac{1}{2}\): \(\left(\frac{1}{2}\right)^3 + 2e^{-3\ln 2}\cdot\frac{1}{2} - \ln 2 = k\) | ||
| \(\frac{1}{8} + 2\cdot\frac{1}{8}\cdot\frac{1}{2} - \ln 2 = \frac{1}{8} + \frac{1}{8} - \ln 2 = \frac{1}{4} - \ln 2\) | B1 | So \(q = \frac{1}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiate implicitly: \(3y^2\frac{dy}{dx} + 2(-3)e^{-3x}y + 2e^{-3x}\frac{dy}{dx} - 1 = 0\) | M1 A1 | M1 for attempt at implicit differentiation, A1 for \(3y^2\frac{dy}{dx}\) |
| \(-6e^{-3x}y\) term | A1 | |
| \(2e^{-3x}\frac{dy}{dx}\) term | A1 | |
| Substitute \(x=\ln 2\), \(y=\frac{1}{2}\): \(e^{-3\ln 2} = \frac{1}{8}\) | M1 | Substituting point |
| \(3\cdot\frac{1}{4}\cdot\frac{dy}{dx} + 2\cdot(-3)\cdot\frac{1}{8}\cdot\frac{1}{2} + 2\cdot\frac{1}{8}\cdot\frac{dy}{dx} - 1 = 0\) | ||
| \(\frac{3}{4}\frac{dy}{dx} - \frac{3}{8} + \frac{1}{4}\frac{dy}{dx} = 1\) | ||
| \(\frac{dy}{dx} = \frac{11}{8}\) | A1 |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $x = \ln 2$, $y = \frac{1}{2}$: $\left(\frac{1}{2}\right)^3 + 2e^{-3\ln 2}\cdot\frac{1}{2} - \ln 2 = k$ | | |
| $\frac{1}{8} + 2\cdot\frac{1}{8}\cdot\frac{1}{2} - \ln 2 = \frac{1}{8} + \frac{1}{8} - \ln 2 = \frac{1}{4} - \ln 2$ | B1 | So $q = \frac{1}{4}$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiate implicitly: $3y^2\frac{dy}{dx} + 2(-3)e^{-3x}y + 2e^{-3x}\frac{dy}{dx} - 1 = 0$ | M1 A1 | M1 for attempt at implicit differentiation, A1 for $3y^2\frac{dy}{dx}$ |
| $-6e^{-3x}y$ term | A1 | |
| $2e^{-3x}\frac{dy}{dx}$ term | A1 | |
| Substitute $x=\ln 2$, $y=\frac{1}{2}$: $e^{-3\ln 2} = \frac{1}{8}$ | M1 | Substituting point |
| $3\cdot\frac{1}{4}\cdot\frac{dy}{dx} + 2\cdot(-3)\cdot\frac{1}{8}\cdot\frac{1}{2} + 2\cdot\frac{1}{8}\cdot\frac{dy}{dx} - 1 = 0$ | | |
| $\frac{3}{4}\frac{dy}{dx} - \frac{3}{8} + \frac{1}{4}\frac{dy}{dx} = 1$ | | |
| $\frac{dy}{dx} = \frac{11}{8}$ | A1 | |
I can see these are answer space pages (pages 17-20) from a P/Jun15/MPC4 exam paper. These pages contain only blank lined answer spaces for Questions 7 and 8 - there is no mark scheme content visible on these pages.
To get the mark scheme content you're looking for, you would need to provide images of the actual mark scheme document, which is a separate publication from the question paper.
Would you like me to instead:
1. **Work through the solutions** to Question 8 (the differential equations question visible on page 18) showing full working and likely mark allocations based on standard A-level marking conventions?
2. Help with any other aspect of these questions?7 A curve has equation $y ^ { 3 } + 2 \mathrm { e } ^ { - 3 x } y - x = k$, where $k$ is a constant.\\
The point $P \left( \ln 2 , \frac { 1 } { 2 } \right)$ lies on this curve.
\begin{enumerate}[label=(\alph*)]
\item Show that the exact value of $k$ is $q - \ln 2$, where $q$ is a rational number.
\item Find the gradient of the curve at $P$.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2015 Q7 [7]}}