| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Tangent with given gradient |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question with standard algebraic manipulation. Part (a) requires routine application of implicit differentiation rules and algebraic rearrangement to reach a given result. Part (b) involves solving simultaneous equations (the curve equation and y=2x), which is algebraically simple. The question is slightly above average difficulty due to the implicit differentiation technique and multi-step nature, but follows a predictable pattern with no novel insight required. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(6x - 2y\frac{dy}{dx} + \left(y + x\frac{dy}{dx}\right) = 0\) | M1 | Differentiates implicitly to include \(\pm ky\frac{dy}{dx}\) or \(x\frac{dy}{dx}\) |
| Correct application of product rule | B1 | |
| \((3x^2 - y^2) \to \left(6x - 2y\frac{dy}{dx}\right)\) and \((4 \to 0)\) | A1 | |
| \(\frac{dy}{dx} = \frac{-6x-y}{x-2y}\) or \(\frac{6x+y}{2y-x}\) | Not necessarily required | |
| Substituting \(\frac{dy}{dx} = \frac{8}{3}\): \(\frac{-6x-y}{x-2y} = \frac{8}{3}\) | M1* | |
| \(13y = 26x\) | dM1* | Attempt to combine terms in \(x\) or \(y\) to give \(ax\) or \(by\) |
| \(y - 2x = 0\) | A1 cso | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substituting \(y = 2x\) into equation: \(3x^2 - (2x)^2 + x(2x) = 4\) | M1 | Attempt replacing \(y\) by \(2x\) in at least one \(y\) term |
| \(x^2 = 4 \Rightarrow x = \pm 2\) | A1 | Either \(x=2\) or \(x=-2\) |
| Coordinates \((2,4)\) and \((-2,-4)\) | A1 | Both \((2,4)\) and \((-2,-4)\) |
# Question 4:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6x - 2y\frac{dy}{dx} + \left(y + x\frac{dy}{dx}\right) = 0$ | M1 | Differentiates implicitly to include $\pm ky\frac{dy}{dx}$ or $x\frac{dy}{dx}$ |
| Correct application of product rule | B1 | |
| $(3x^2 - y^2) \to \left(6x - 2y\frac{dy}{dx}\right)$ and $(4 \to 0)$ | A1 | |
| $\frac{dy}{dx} = \frac{-6x-y}{x-2y}$ or $\frac{6x+y}{2y-x}$ | | Not necessarily required |
| Substituting $\frac{dy}{dx} = \frac{8}{3}$: $\frac{-6x-y}{x-2y} = \frac{8}{3}$ | M1* | |
| $13y = 26x$ | dM1* | Attempt to combine terms in $x$ or $y$ to give $ax$ or $by$ |
| $y - 2x = 0$ | A1 cso | AG |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substituting $y = 2x$ into equation: $3x^2 - (2x)^2 + x(2x) = 4$ | M1 | Attempt replacing $y$ by $2x$ in at least one $y$ term |
| $x^2 = 4 \Rightarrow x = \pm 2$ | A1 | Either $x=2$ or $x=-2$ |
| Coordinates $(2,4)$ and $(-2,-4)$ | A1 | Both $(2,4)$ and $(-2,-4)$ |
---4. A curve has equation $3 x ^ { 2 } - y ^ { 2 } + x y = 4$. The points $P$ and $Q$ lie on the curve. The gradient of the tangent to the curve is $\frac { 8 } { 3 }$ at $P$ and at $Q$.
\begin{enumerate}[label=(\alph*)]
\item Use implicit differentiation to show that $y - 2 x = 0$ at $P$ and at $Q$.
\item Find the coordinates of $P$ and $Q$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2008 Q4 [9]}}