| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Show no stationary points exist |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question with standard follow-up analysis. Part (i) requires applying the product rule and chain rule systematically then rearranging—routine C4 technique. Part (ii) simply requires recognizing that vertical tangents occur when dx/dy = 0 (denominator = 0) and showing this leads to a contradiction. Both parts are mechanical applications of standard methods with no novel insight required, making this slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) For \(\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}\) | B1, M1, A1, M1 | Using \(d(uv) = u \, dv + v \, du\) |
| \(2xy \frac{dy}{dx} + y^2 = 2 + 3\frac{dy}{dx}\) | ||
| \(\frac{dy}{dx} = \frac{(2 - y^2)}{(2xy - 3)}\) | A1 5 | Solving an equation, with at least 2 dy/dx terms, for dy/dx; dy/dx on one side, non dy/dx on other. AG |
| (ii) Stating/using \(2xy - 3 = 0\) | B1, M1, A1 3 | No use of \(2 - y^2\) in this part; Between \(2xy - 3 = 0\) & eqn of curve; Together with suitable finish |
| Attempt to eliminate x or y | ||
| \(8x^2 = -9\) or \(y^2 = -2\) |
(i) For $\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$ | B1, M1, A1, M1 | Using $d(uv) = u \, dv + v \, du$
$2xy \frac{dy}{dx} + y^2 = 2 + 3\frac{dy}{dx}$ | | | $2xy \frac{dy}{dx} + y^2 = 2 + 3\frac{dy}{dx}$
$\frac{dy}{dx} = \frac{(2 - y^2)}{(2xy - 3)}$ | A1 5 | Solving an equation, with at least 2 dy/dx terms, for dy/dx; dy/dx on one side, non dy/dx on other. **AG**
(ii) Stating/using $2xy - 3 = 0$ | B1, M1, A1 3 | No use of $2 - y^2$ in this part; Between $2xy - 3 = 0$ & eqn of curve; Together with suitable finish | **8**
Attempt to eliminate x or y | |
$8x^2 = -9$ or $y^2 = -2$ | |The equation of a curve is $xy^2 = 2x + 3y$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\frac{dy}{dx} = \frac{2 - y^2}{2xy - 3}$. [5]
\item Show that the curve has no tangents which are parallel to the $y$-axis. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 2005 Q6 [8]}}