| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2021 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find normal equation at point |
| Difficulty | Standard +0.8 This question requires implicit differentiation with exponential functions (using product rule), finding a specific point by substituting x=0, then computing the normal line. While the techniques are standard P4 content, the combination of implicit differentiation with e^(-2x) and the multi-step process elevates it slightly above average difficulty for A-level. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4y^2 + 3x \rightarrow 8y\frac{\mathrm{d}y}{\mathrm{d}x}+3\) | B1 | Allow unsimplified forms such as \(4\times 2y\frac{\mathrm{d}y}{\mathrm{d}x}+3\) |
| \(6ye^{-2x} \rightarrow -12ye^{-2x}+6e^{-2x}\frac{\mathrm{d}y}{\mathrm{d}x}\) | M1 A1 | M1: uses product rule to obtain form \(Aye^{-2x}+Be^{-2x}\frac{\mathrm{d}y}{\mathrm{d}x}\) |
| \(8y\frac{\mathrm{d}y}{\mathrm{d}x}+3 = -12ye^{-2x}+6e^{-2x}\frac{\mathrm{d}y}{\mathrm{d}x} \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{12ye^{-2x}+3}{6e^{-2x}-8y}\) | M1 A1 | M1: collects two \(\frac{\mathrm{d}y}{\mathrm{d}x}\) terms and makes \(\frac{\mathrm{d}y}{\mathrm{d}x}\) subject |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets \(x=0\) in \(4y^2+3x=6ye^{-2x} \Rightarrow y = \frac{3}{2}\) | B1 | Ignore any reference to \(y=0\) |
| Substitutes \(\left(0,\frac{3}{2}\right)\) in their \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{12ye^{-2x}+3}{6e^{-2x}-8y} = \left(\frac{7}{-2}\right)\) | M1 | Must use \(x=0\) and \(y\) from substituting \(x=0\) into original equation; \(y=0\) is M0 |
| \(m_N = -1 \div \frac{7}{-2} \Rightarrow y = \frac{2}{7}x + \frac{3}{2}\) | dM1 | |
| \(y = \frac{2}{7}x + \frac{3}{2}\) e.g. \(y = \frac{6}{21}x + \frac{3}{2}\) | A1 |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4y^2 + 3x \rightarrow 8y\frac{\mathrm{d}y}{\mathrm{d}x}+3$ | B1 | Allow unsimplified forms such as $4\times 2y\frac{\mathrm{d}y}{\mathrm{d}x}+3$ |
| $6ye^{-2x} \rightarrow -12ye^{-2x}+6e^{-2x}\frac{\mathrm{d}y}{\mathrm{d}x}$ | M1 A1 | M1: uses product rule to obtain form $Aye^{-2x}+Be^{-2x}\frac{\mathrm{d}y}{\mathrm{d}x}$ |
| $8y\frac{\mathrm{d}y}{\mathrm{d}x}+3 = -12ye^{-2x}+6e^{-2x}\frac{\mathrm{d}y}{\mathrm{d}x} \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{12ye^{-2x}+3}{6e^{-2x}-8y}$ | M1 A1 | M1: collects two $\frac{\mathrm{d}y}{\mathrm{d}x}$ terms and makes $\frac{\mathrm{d}y}{\mathrm{d}x}$ subject |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $x=0$ in $4y^2+3x=6ye^{-2x} \Rightarrow y = \frac{3}{2}$ | B1 | Ignore any reference to $y=0$ |
| Substitutes $\left(0,\frac{3}{2}\right)$ in their $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{12ye^{-2x}+3}{6e^{-2x}-8y} = \left(\frac{7}{-2}\right)$ | M1 | Must use $x=0$ and $y$ from substituting $x=0$ into original equation; $y=0$ is M0 |
| $m_N = -1 \div \frac{7}{-2} \Rightarrow y = \frac{2}{7}x + \frac{3}{2}$ | dM1 | |
| $y = \frac{2}{7}x + \frac{3}{2}$ e.g. $y = \frac{6}{21}x + \frac{3}{2}$ | A1 | |6. A curve has equation
$$4 y ^ { 2 } + 3 x = 6 y \mathrm { e } ^ { - 2 x }$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
The curve crosses the $y$-axis at the origin and at the point $P$.
\item Find the equation of the normal to the curve at $P$, writing your answer in the form $y = m x + c$ where $m$ and $c$ are constants to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2021 Q6 [9]}}