| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find stationary points |
| Difficulty | Standard +0.8 This question requires implicit differentiation with product rule, then solving a system involving the original cubic equation and the derivative condition. Part (a) is standard technique, but part (b) requires algebraic manipulation of simultaneous equations where one is cubic, demanding more problem-solving than typical stationary point questions. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Differentiating implicitly to obtain \(\pm ay^2\frac{dy}{dx}\) and/or \(\pm bx^2\frac{dy}{dx}\) | M1 | |
| \(48y^2\frac{dy}{dx}+\ldots-54\ldots\) | A1 | |
| \(9x^2y \rightarrow 9x^2\frac{dy}{dx}+18xy\) | B1 | or equivalent |
| \(\left(48y^2+9x^2\right)\frac{dy}{dx}+18xy-54=0\) | M1 | |
| \(\frac{dy}{dx}=\frac{54-18xy}{48y^2+9x^2}\left(=\frac{18-6xy}{16y^2+3x^2}\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(18-6xy=0\) | M1 | |
| Using \(x=\frac{3}{y}\) or \(y=\frac{3}{x}\) | ||
| \(16y^3+9\left(\frac{3}{y}\right)^2 y-54\left(\frac{3}{y}\right)=0\) or \(16\left(\frac{3}{x}\right)^3+9x^2\left(\frac{3}{x}\right)-54x=0\) | M1 | |
| \(16y^4+81-162=0\) or \(16+x^4-2x^4=0\) | M1 | |
| \(y^4=\frac{81}{16}\) or \(x^4=16\) | ||
| \(y=\frac{3}{2},\ -\frac{3}{2}\) or \(x=2,\ -2\) | A1, A1 | |
| Substitute into \(xy=3\) to obtain other variable | M1 | |
| \(\left(2,\frac{3}{2}\right),\ \left(-2,-\frac{3}{2}\right)\) | A1 | both |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Differentiating implicitly to obtain $\pm ay^2\frac{dy}{dx}$ and/or $\pm bx^2\frac{dy}{dx}$ | M1 | |
| $48y^2\frac{dy}{dx}+\ldots-54\ldots$ | A1 | |
| $9x^2y \rightarrow 9x^2\frac{dy}{dx}+18xy$ | B1 | or equivalent |
| $\left(48y^2+9x^2\right)\frac{dy}{dx}+18xy-54=0$ | M1 | |
| $\frac{dy}{dx}=\frac{54-18xy}{48y^2+9x^2}\left(=\frac{18-6xy}{16y^2+3x^2}\right)$ | A1 | |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $18-6xy=0$ | M1 | |
| Using $x=\frac{3}{y}$ or $y=\frac{3}{x}$ | | |
| $16y^3+9\left(\frac{3}{y}\right)^2 y-54\left(\frac{3}{y}\right)=0$ or $16\left(\frac{3}{x}\right)^3+9x^2\left(\frac{3}{x}\right)-54x=0$ | M1 | |
| $16y^4+81-162=0$ or $16+x^4-2x^4=0$ | M1 | |
| $y^4=\frac{81}{16}$ or $x^4=16$ | | |
| $y=\frac{3}{2},\ -\frac{3}{2}$ or $x=2,\ -2$ | A1, A1 | |
| Substitute into $xy=3$ to obtain other variable | M1 | |
| $\left(2,\frac{3}{2}\right),\ \left(-2,-\frac{3}{2}\right)$ | A1 | both |
---6. The curve $C$ has equation
$$16 y ^ { 3 } + 9 x ^ { 2 } y - 54 x = 0$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
\item Find the coordinates of the points on $C$ where $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 Q6 [12]}}