| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find stationary points |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring product rule and algebraic manipulation, followed by solving simultaneous equations. While it involves multiple steps, the techniques are standard C3/C4 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y^3 + x^2y - 6x = 0 \Rightarrow 3y^2\frac{dy}{dx} + x^2\frac{dy}{dx} + 2xy - 6 = 0\) | B1 M1 A1 | B1: Product rule on \(x^2y\) giving \(x^2\frac{dy}{dx} + 2xy\); M1: Chain rule on \(y^3\) giving \(Ay^2\frac{dy}{dx}\); A1: \(y^3 - 6x = 0 \Rightarrow 3y^2\frac{dy}{dx} - 6 = 0\) correct with "= 0" seen |
| \(\Rightarrow \frac{dy}{dx} = \frac{6-2xy}{x^2+3y^2}\) | M1 A1 | M1: Attempts to make \(\frac{dy}{dx}\) subject with two \(\frac{dy}{dx}\) terms; A1: Correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(6 - 2xy = 0 \Rightarrow y = \frac{3}{x}\) | M1 | Sets numerator = 0, writes \(x\) in terms of \(y\) or vice versa |
| Substitute \(y = \frac{3}{x}\) into \(y^3 + x^2y - 6x = 0 \Rightarrow \frac{27}{x^3} + \frac{3x^2}{x} - 6x = 0\) | dM1 | Substitutes into original equation; dependent on first M1 |
| \(\Rightarrow x^4 = 9\) | ddM1 A1 | Reaches form \(Ax^m = Bx^n\); correct equation |
| Points \((\sqrt{3}, \sqrt{3})(-\sqrt{3}, -\sqrt{3})\) | A1 A1 | First A1: two correct values; Second A1: all four values correct and simplified |
# Question 2:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y^3 + x^2y - 6x = 0 \Rightarrow 3y^2\frac{dy}{dx} + x^2\frac{dy}{dx} + 2xy - 6 = 0$ | B1 M1 A1 | B1: Product rule on $x^2y$ giving $x^2\frac{dy}{dx} + 2xy$; M1: Chain rule on $y^3$ giving $Ay^2\frac{dy}{dx}$; A1: $y^3 - 6x = 0 \Rightarrow 3y^2\frac{dy}{dx} - 6 = 0$ correct with "= 0" seen |
| $\Rightarrow \frac{dy}{dx} = \frac{6-2xy}{x^2+3y^2}$ | M1 A1 | M1: Attempts to make $\frac{dy}{dx}$ subject with two $\frac{dy}{dx}$ terms; A1: Correct expression |
**(5 marks)**
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $6 - 2xy = 0 \Rightarrow y = \frac{3}{x}$ | M1 | Sets numerator = 0, writes $x$ in terms of $y$ or vice versa |
| Substitute $y = \frac{3}{x}$ into $y^3 + x^2y - 6x = 0 \Rightarrow \frac{27}{x^3} + \frac{3x^2}{x} - 6x = 0$ | dM1 | Substitutes into original equation; dependent on first M1 |
| $\Rightarrow x^4 = 9$ | ddM1 A1 | Reaches form $Ax^m = Bx^n$; correct equation |
| Points $(\sqrt{3}, \sqrt{3})(-\sqrt{3}, -\sqrt{3})$ | A1 A1 | First A1: two correct values; Second A1: all four values correct and simplified |
**(6 marks total, 11 marks total)**
---2. The curve $C$ has equation
$$y ^ { 3 } + x ^ { 2 } y - 6 x = 0$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
\item Hence find the exact coordinates of the points on $C$ for which $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2017 Q2 [11]}}