| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2024 |
| Session | January |
| Marks | 9 |
| 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 application, followed by substituting known coordinates and the stationary point condition (dy/dx = 0) to find two constants. The steps are mechanical and follow a standard template for this topic, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y^2x \to ...\times y\frac{dy}{dx}(+...)\) or \(3y\to3\frac{dy}{dx}\) | M1 | Some evidence of implicit differentiation; one term differentiated with \(3\frac{dy}{dx}\) or a term with \(kxy\frac{dy}{dx}\) or \(y^2\to ky\frac{dy}{dx}\) as part of product rule |
| \(y^2x\to 2xy\frac{dy}{dx}+y^2\) | A1 | \(y^2x\) differentiated correctly |
| \(...+3y\to...+3\frac{dy}{dx}\) and \(4x^2+k\to8x\) | A1 | Both remaining terms differentiated correctly |
| \(2xy\frac{dy}{dx}+y^2+3\frac{dy}{dx}=8x\Rightarrow\frac{dy}{dx}=...\) | dM1 | Make \(\frac{dy}{dx}\) the subject with correct terms factored; at least two \(\frac{dy}{dx}\) terms in expression |
| \(\frac{dy}{dx}=\frac{8x-y^2}{2xy+3}\) (oe) | A1 | Correct expression, accept any equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx}=0\Rightarrow 8x-4=0\Rightarrow x=...\) | M1 | Sets numerator of their \(\frac{dy}{dx}\) to zero, substitutes \(y=2\), solves for \(x\); must have a \(y\) term in numerator |
| \(p=\frac{1}{2}\) | A1 | Correct value for \(p\) or \(x\) |
| \(k=2^2\times\frac{1}{2}+3\times2-4\times\frac{1}{4}=...\) | M1 | Substitutes back into curve to find \(k\) |
| \(k=7\) | A1 | Correct value for \(k\) |
# Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y^2x \to ...\times y\frac{dy}{dx}(+...)$ or $3y\to3\frac{dy}{dx}$ | M1 | Some evidence of implicit differentiation; one term differentiated with $3\frac{dy}{dx}$ or a term with $kxy\frac{dy}{dx}$ or $y^2\to ky\frac{dy}{dx}$ as part of product rule |
| $y^2x\to 2xy\frac{dy}{dx}+y^2$ | A1 | $y^2x$ differentiated correctly |
| $...+3y\to...+3\frac{dy}{dx}$ and $4x^2+k\to8x$ | A1 | Both remaining terms differentiated correctly |
| $2xy\frac{dy}{dx}+y^2+3\frac{dy}{dx}=8x\Rightarrow\frac{dy}{dx}=...$ | dM1 | Make $\frac{dy}{dx}$ the subject with correct terms factored; at least two $\frac{dy}{dx}$ terms in expression |
| $\frac{dy}{dx}=\frac{8x-y^2}{2xy+3}$ (oe) | A1 | Correct expression, accept any equivalent |
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# Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx}=0\Rightarrow 8x-4=0\Rightarrow x=...$ | M1 | Sets numerator of their $\frac{dy}{dx}$ to zero, substitutes $y=2$, solves for $x$; must have a $y$ term in numerator |
| $p=\frac{1}{2}$ | A1 | Correct value for $p$ or $x$ |
| $k=2^2\times\frac{1}{2}+3\times2-4\times\frac{1}{4}=...$ | M1 | Substitutes back into curve to find $k$ |
| $k=7$ | A1 | Correct value for $k$ |3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6f577461-24b7-4615-b58b-e67597fd9675-08_815_849_248_607}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The curve $C$, shown in Figure 1, has equation
$$y ^ { 2 } x + 3 y = 4 x ^ { 2 } + k \quad y > 0$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$
The point $P ( p , 2 )$, where $p$ is a constant, lies on $C$.\\
Given that $P$ is the minimum turning point on $C$,
\item find
\begin{enumerate}[label=(\roman*)]
\item the value of $p$
\item the value of $k$
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2024 Q3 [9]}}