| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Tangent/normal with axis intercepts |
| Difficulty | Standard +0.8 This P4 implicit differentiation question requires careful algebraic manipulation to derive the given derivative, then finding where the curve crosses the y-axis (requiring solving a transcendental equation), computing tangent equations at two points, and finding their intersection. The multi-step nature, combination of exponential and implicit functions, and need to handle the algebra precisely makes this moderately challenging, though the techniques are standard for P4. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2y\frac{dy}{dx} = e^{-2x}\frac{dy}{dx} - 2ye^{-2x} - 3\) | B1 M1 A1 | B1: Correct differentiation of \(y^2 \to 2y\frac{dy}{dx}\) by chain rule. M1: Product rule on \(ye^{-2x}\) giving \(e^{-2x}\frac{dy}{dx}\pm...ye^{-2x}\). A1: Correct differentiation |
| \(\left(e^{-2x}-2y\right)\frac{dy}{dx} = 2ye^{-2x}+3 \Rightarrow \frac{dy}{dx} = \frac{2ye^{-2x}+3}{e^{-2x}-2y}\) | A1* | Proceeds to given answer via intermediate line equivalent to \(\left(e^{-2x}-2y\right)\frac{dy}{dx}=2ye^{-2x}+3\) with correct bracketing |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Puts \(x=0\) into equation of curve \(\Rightarrow y=y^2 \Rightarrow y=1\) | B1 | Deduces \(x=0\), \(y=1\) at \(P\). May state or use coordinates of \(P(0,1)\) |
| Attempts tangent at \((0,0)\) or \((0,1)\): \(y=3x\) or \(y=-5x+1\) | M1 A1 | M1: Attempt to find equation of tangent at \(O\) or at \(P\). A1: Correct equation for either tangent |
| Solves \(y=3x\) with \(y=-5x+1 \Rightarrow R=\left(\frac{1}{8},\frac{3}{8}\right)\) | dM1 A1 | dM1: Correct attempt at both tangents with attempt to solve simultaneously. A1: Correct coordinates for \(R=\left(\frac{1}{8},\frac{3}{8}\right)\) oe |
# Question 5:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2y\frac{dy}{dx} = e^{-2x}\frac{dy}{dx} - 2ye^{-2x} - 3$ | B1 M1 A1 | B1: Correct differentiation of $y^2 \to 2y\frac{dy}{dx}$ by chain rule. M1: Product rule on $ye^{-2x}$ giving $e^{-2x}\frac{dy}{dx}\pm...ye^{-2x}$. A1: Correct differentiation |
| $\left(e^{-2x}-2y\right)\frac{dy}{dx} = 2ye^{-2x}+3 \Rightarrow \frac{dy}{dx} = \frac{2ye^{-2x}+3}{e^{-2x}-2y}$ | A1* | Proceeds to given answer via intermediate line equivalent to $\left(e^{-2x}-2y\right)\frac{dy}{dx}=2ye^{-2x}+3$ with correct bracketing |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Puts $x=0$ into equation of curve $\Rightarrow y=y^2 \Rightarrow y=1$ | B1 | Deduces $x=0$, $y=1$ at $P$. May state or use coordinates of $P(0,1)$ |
| Attempts tangent at $(0,0)$ **or** $(0,1)$: $y=3x$ **or** $y=-5x+1$ | M1 A1 | M1: Attempt to find equation of tangent at $O$ **or** at $P$. A1: Correct equation for **either** tangent |
| Solves $y=3x$ with $y=-5x+1 \Rightarrow R=\left(\frac{1}{8},\frac{3}{8}\right)$ | dM1 A1 | dM1: Correct attempt at **both** tangents with attempt to solve simultaneously. A1: Correct coordinates for $R=\left(\frac{1}{8},\frac{3}{8}\right)$ oe |5. A curve has equation
$$y ^ { 2 } = y \mathrm { e } ^ { - 2 x } - 3 x$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 y \mathrm { e } ^ { - 2 x } + 3 } { \mathrm { e } ^ { - 2 x } - 2 y }$$
The curve crosses the $y$-axis at the origin and at the point $P$.\\
The tangent to the curve at the origin and the tangent to the curve at $P$ meet at the point $R$.
\item Find the coordinates of $R$.\\
\includegraphics[max width=\textwidth, alt={}, center]{960fe82f-c180-422c-b409-a5cdc5fae924-17_2644_1838_121_116}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2021 Q5 [9]}}