| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find constant from gradient condition |
| Difficulty | Standard +0.8 This question requires implicit differentiation (standard P4 technique), then using the perpendicular gradient condition (normal gradient = -1/tangent gradient) to find that dy/dx = -1 at point P. This creates a system of equations to solve simultaneously with the original curve equation. The algebraic manipulation is non-trivial, requiring substitution and solving a cubic or careful algebraic reasoning. More challenging than routine implicit differentiation but within expected P4 scope. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(y^3 \rightarrow 3y^2\frac{dy}{dx}\) | M1 | Differentiates \(y^3 \rightarrow Py^2\frac{dy}{dx}\) |
| \(4x^2y \rightarrow 8xy + 4x^2\frac{dy}{dx}\) | M1 | Uses product rule on \(4x^2y\), obtains \(Qxy + Rx^2\frac{dy}{dx}\) |
| \(y^3 - x^2 + 4x^2y = k \Rightarrow 3y^2\frac{dy}{dx} - 2x + 8xy + 4x^2\frac{dy}{dx} = 0\) | A1 | Correct full differentiation including "\(= 0\)" |
| \(\Rightarrow \left(3y^2 + 4x^2\right)\frac{dy}{dx} = 2x - 8xy\) | M1 | Dependent on two different \(\frac{dy}{dx}\) terms (from \(y^3\) and \(4x^2y\)); collects and factorises |
| \(\Rightarrow \frac{dy}{dx} = \frac{2x - 8xy}{3y^2 + 4x^2}\) | A1 | Correct expression, e.g. \(\frac{8xy-2x}{-3y^2-4x^2}\) also acceptable |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dy}{dx} = -1\) at \(P\) | M1 | States or uses \(\frac{dy}{dx} = -1\) |
| Uses \(\frac{dy}{dx} = \pm 1\) and \(y = x\) to set up and solve equation; \(\Rightarrow -1 = \frac{2p - 8p^2}{3p^2 + 4p^2} \Rightarrow p = 2\) | M1 A1 | Uses \(y=x\) (or \(x=y\) or \(y=x="p"\)) in \(\frac{dy}{dx}=\pm1\), solves to get value; \(x\), \(y\) or "\(p\)" \(= 2\) at \(P\) |
| e.g. \(k = "2"^3 - "2"^2 + 4\times"2"^2\times"2"^3\) | ddM1 | Uses non-zero value for \(x\), \(y\) or "\(p\)" in original curve equation to find \(k\). Depends on both previous M marks |
| \(k = 36\) | A1 |
## Question 5(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $y^3 \rightarrow 3y^2\frac{dy}{dx}$ | M1 | Differentiates $y^3 \rightarrow Py^2\frac{dy}{dx}$ |
| $4x^2y \rightarrow 8xy + 4x^2\frac{dy}{dx}$ | M1 | Uses product rule on $4x^2y$, obtains $Qxy + Rx^2\frac{dy}{dx}$ |
| $y^3 - x^2 + 4x^2y = k \Rightarrow 3y^2\frac{dy}{dx} - 2x + 8xy + 4x^2\frac{dy}{dx} = 0$ | A1 | Correct full differentiation including "$= 0$" |
| $\Rightarrow \left(3y^2 + 4x^2\right)\frac{dy}{dx} = 2x - 8xy$ | M1 | Dependent on two different $\frac{dy}{dx}$ terms (from $y^3$ and $4x^2y$); collects and factorises |
| $\Rightarrow \frac{dy}{dx} = \frac{2x - 8xy}{3y^2 + 4x^2}$ | A1 | Correct expression, e.g. $\frac{8xy-2x}{-3y^2-4x^2}$ also acceptable |
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## Question 5(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = -1$ at $P$ | M1 | States or uses $\frac{dy}{dx} = -1$ |
| Uses $\frac{dy}{dx} = \pm 1$ and $y = x$ to set up and solve equation; $\Rightarrow -1 = \frac{2p - 8p^2}{3p^2 + 4p^2} \Rightarrow p = 2$ | M1 A1 | Uses $y=x$ (or $x=y$ or $y=x="p"$) in $\frac{dy}{dx}=\pm1$, solves to get value; $x$, $y$ or "$p$" $= 2$ at $P$ |
| e.g. $k = "2"^3 - "2"^2 + 4\times"2"^2\times"2"^3$ | ddM1 | Uses non-zero value for $x$, $y$ or "$p$" in original curve equation to find $k$. Depends on both previous M marks |
| $k = 36$ | A1 | |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7f5fc83d-ab7c-4edb-a2c6-7a58f1357d5a-12_678_987_248_539}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the curve $C$ with equation
$$y ^ { 3 } - x ^ { 2 } + 4 x ^ { 2 } y = k$$
where $k$ is a positive constant greater than 1
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
The point $P$ lies on $C$.\\
Given that the normal to $C$ at $P$ has equation $y = x$, as shown in Figure 2,
\item find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2023 Q5 [10]}}