| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find stationary points |
| Difficulty | Standard +0.3 This is a standard implicit differentiation question with routine multi-part structure. Part (a) requires substituting x=1 and solving a quadratic. Part (b)(i) is straightforward implicit differentiation. Parts (b)(ii) and (b)(iii) involve substitution and algebraic manipulation. While it requires multiple techniques, each step follows standard procedures with no novel insight needed, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation |
5 A curve is defined by the equation
$$y ^ { 2 } - x y + 3 x ^ { 2 } - 5 = 0$$
\begin{enumerate}[label=(\alph*)]
\item Find the $y$-coordinates of the two points on the curve where $x = 1$.
\item \begin{enumerate}[label=(\roman*)]
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - 6 x } { 2 y - x }$.
\item Find the gradient of the curve at each of the points where $x = 1$.
\item Show that, at the two stationary points on the curve, $33 x ^ { 2 } - 5 = 0$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2006 Q5 [14]}}