| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Range or domain from dy/dx |
| Difficulty | Challenging +1.2 This question requires implicit differentiation using the product rule (part a), then analyzing where a vertical line intersects the curve exactly twice by finding the turning point using dy/dx = 0 (part b). While it involves multiple steps and connecting differentiation to curve behavior, the techniques are standard P3 material with clear signposting. The implicit differentiation is straightforward, and finding the critical point is a routine application, making this moderately above average difficulty but not requiring novel insight. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dx}{dy} = e^{2y} + 2ye^{2y}\) | M1, A1 | M1: attempt product rule on \(ye^{2y}\); A1: correct differentiation |
| \(\Rightarrow \frac{dx}{dy} = \frac{x}{y} + 2x\) | dM1 | Full method to get \(\frac{dy}{dx}\) in terms of \(x\) and \(y\) only; correct inversion (not term by term); \(e^{2y}\) replaced by \(\frac{x}{y}\), \(ye^{2y}\) replaced by \(x\) |
| \(\Rightarrow \frac{dy}{dx} = \frac{y}{x+2xy} = \frac{y}{x(1+2y)}\) * | A1* | Correct complete proof, no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Deduces \(y = -\frac{1}{2}\) | B1 | Left-hand end occurs when \(y = -\frac{1}{2}\) |
| Substitutes \(y = -\frac{1}{2} \Rightarrow x = -\frac{1}{2e}\) | M1, A1 | M1: attempt to find \(x\); may be implied by \(k \approx -0.183\) or \(-0.184\); A1: \(-\frac{1}{2e} < k < 0\) (must be range for \(k\), not \(x\); must not use "or" between two inequalities) |
# Question 10:
## Part (a) — Given $x = ye^{2y}$, show that $\frac{dy}{dx} = \frac{y}{x(1+2y)}$
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dx}{dy} = e^{2y} + 2ye^{2y}$ | M1, A1 | M1: attempt product rule on $ye^{2y}$; A1: correct differentiation |
| $\Rightarrow \frac{dx}{dy} = \frac{x}{y} + 2x$ | dM1 | Full method to get $\frac{dy}{dx}$ in terms of $x$ and $y$ only; correct inversion (not term by term); $e^{2y}$ replaced by $\frac{x}{y}$, $ye^{2y}$ replaced by $x$ |
| $\Rightarrow \frac{dy}{dx} = \frac{y}{x+2xy} = \frac{y}{x(1+2y)}$ * | A1* | Correct complete proof, no errors |
**(4 marks)**
## Part (b) — Find the range of $k$ where $x = k$ has no solution for $\frac{dy}{dx}$
| Working | Mark | Guidance |
|---------|------|----------|
| Deduces $y = -\frac{1}{2}$ | B1 | Left-hand end occurs when $y = -\frac{1}{2}$ |
| Substitutes $y = -\frac{1}{2} \Rightarrow x = -\frac{1}{2e}$ | M1, A1 | M1: attempt to find $x$; may be implied by $k \approx -0.183$ or $-0.184$; A1: $-\frac{1}{2e} < k < 0$ (must be range for $k$, not $x$; must not use "or" between two inequalities) |
**(3 marks)**
**(7 marks total)**10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f3272b4c-d8dc-4f32-add9-153de90f4d0a-30_661_743_210_603}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of the curve $C$ with equation
$$x = y \mathrm { e } ^ { 2 y } \quad y \in \mathbb { R }$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y } { x ( 1 + 2 y ) }$$
Given that the straight line with equation $x = k$, where $k$ is a constant, cuts $C$ at exactly two points,
\item find the range of possible values for $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2022 Q10 [7]}}