| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Parametric form dy/dx |
| Difficulty | Standard +0.3 This question requires differentiation of a product involving logarithms (using the product rule), then substituting specific points found by solving simple equations. The techniques are standard C3 content with no conceptual challenges—finding where the curve crosses axes is routine, and the product rule application is straightforward. It's slightly easier than average due to the mechanical nature of the tasks. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Attempt use of product rule | M1 | to produce expression of form (something non-zero)\(\ln(2y + 3) + \frac{\text{linear in } y}{\text{linear in } y}\); ignore what they call their derivative |
| Obtain \(\ln(2y + 3) \ldots\) | A1 | with brackets included |
| Obtain \(\ldots + \frac{2(y + 4)}{2y + 3}\) | A1 | with brackets included as necessary |
| Answer | Marks | Guidance |
|---|---|---|
| Substitute \(y = 0\) into attempt from part (i) or into their attempt (however poor) at its reciprocal | M1 | |
| Obtain 0.27 for gradient at \(A\) | A1 | or greater accuracy 0.26558…; beware of 'correct' answer coming from incorrect version \(\ln(2y + 3) + \frac{3}{8}\) of answer in part (i) |
| Attempt to find value of \(y\) for which \(x = 0\) | M1 | allowing process leading only to \(y = -4\) |
| Substitute \(y = -1\) into attempt from part (i) or into their attempt (however poor) at its reciprocal | M1 | |
| Obtain 0.17 or \(\frac{1}{6}\) for gradient at \(B\) | A1 | or greater accuracy 0.16666…; value following from correct working |
## (i)
Attempt use of product rule | M1 | to produce expression of form (something non-zero)$\ln(2y + 3) + \frac{\text{linear in } y}{\text{linear in } y}$; ignore what they call their derivative
Obtain $\ln(2y + 3) \ldots$ | A1 | with brackets included
Obtain $\ldots + \frac{2(y + 4)}{2y + 3}$ | A1 | with brackets included as necessary
**Total: [3]**
## (ii)
Substitute $y = 0$ into attempt from part (i) or into their attempt (however poor) at its reciprocal | M1 |
Obtain 0.27 for gradient at $A$ | A1 | or greater accuracy 0.26558…; beware of 'correct' answer coming from incorrect version $\ln(2y + 3) + \frac{3}{8}$ of answer in part (i)
Attempt to find value of $y$ for which $x = 0$ | M1 | allowing process leading only to $y = -4$
Substitute $y = -1$ into attempt from part (i) or into their attempt (however poor) at its reciprocal | M1 |
Obtain 0.17 or $\frac{1}{6}$ for gradient at $B$ | A1 | or greater accuracy 0.16666…; value following from correct working
**Total: [5]**
---\includegraphics{figure_7}
The diagram shows the curve with equation
$$x = (y + 4)\ln (2y + 3).$$
The curve crosses the $x$-axis at $A$ and the $y$-axis at $B$.
\begin{enumerate}[label=(\roman*)]
\item Find an expression for $\frac{dx}{dy}$ in terms of $y$. [3]
\item Find the gradient of the curve at each of the points $A$ and $B$, giving each answer correct to 2 decimal places. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR C3 2013 Q7 [8]}}