| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Tangent/normal with axis intercepts |
| Difficulty | Standard +0.3 This is a standard implicit differentiation question with straightforward steps: substitute to find a coordinate, differentiate implicitly (involving 2^x and product rule), then find a tangent line equation and its x-intercept. While it requires multiple techniques, each step follows routine procedures taught in P4/Further Pure, with no novel insight required. The final answer form is given, making it slightly easier than average. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.06f Laws of logarithms: addition, subtraction, power rules1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=2 \Rightarrow 4-8y+y^2=13 \Rightarrow (y^2-8y-9=0) \Rightarrow y=...\) | M1 | Substitutes \(x=2\), forms quadratic in \(y\), solves to obtain at least one value |
| \(y=9\) | A1 | \(y=9\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2^x \rightarrow 2^x\ln 2\) | B1 | Correct differentiation of \(2^x\); also allow \(2^x = e^{x\ln 2} \rightarrow e^{x\ln 2}\ln 2\) |
| \(-4xy \rightarrow -4x\frac{dy}{dx}-4y \quad \text{OR} \quad y^2 \rightarrow 2y\frac{dy}{dx}\) | M1 | Differentiates \(-4xy \rightarrow \pm Ax\frac{dy}{dx}\pm By\) OR differentiates \(y^2 \rightarrow ky\frac{dy}{dx}\) |
| \(2^x\ln 2 - 4x\frac{dy}{dx}-4y+2y\frac{dy}{dx}=0\) | A1 | Fully correct differentiation; allow versions such as \(2^x\ln 2\,dx - 4x\,dy - 4y\,dx + 2y\,dy = 0\) |
| \(\frac{dy}{dx}(2y-4x) = 4y - 2^x\ln 2 \Rightarrow \frac{dy}{dx}=...\) | M1 | Attempts to make \(\frac{dy}{dx}\) the subject with 2 terms in \(\frac{dy}{dx}\) coming from differentiating \(y^2\) and \(-4xy\) |
| \(\frac{dy}{dx} = \frac{4y-2^x\ln 2}{2y-4x}\) or \(\frac{2^x\ln 2-4y}{4x-2y}\) or \(\frac{\ln 2e^{x\ln 2}-4y}{4x-2y}\) | A1 | Any correct expression; note \(\frac{2y-2^{x-1}\ln 2}{y-2x}\) is correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((2,9) \rightarrow \frac{dy}{dx} = \frac{4(9)-2^2\ln 2}{2(9)-4(2)} \Rightarrow y-"9" = "\frac{36-4\ln 2}{10}"(x-2)\) | M1 | Uses \(x=2\) and \(y\) value from (a) to find gradient at \(P\) and attempts to form equation of tangent; \(\frac{dy}{dx}\) must have both \(x\) and \(y\) terms |
| \(y=0 \Rightarrow 0-"9" = "\frac{36-4\ln 2}{10}"(x-2) \Rightarrow x=...\) | dM1 | Substitutes \(y=0\) into tangent equation and rearranges to find \(x\); dependent on previous M |
| \(x = \frac{4\ln 2+9}{2\ln 2-18}\) o.e. e.g. \(x=\frac{-8\ln 2-18}{-4\ln 2+36}\) | A1 | Correct expression in required form; must come from correct \(\frac{dy}{dx}\) |
# Question 2:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=2 \Rightarrow 4-8y+y^2=13 \Rightarrow (y^2-8y-9=0) \Rightarrow y=...$ | M1 | Substitutes $x=2$, forms quadratic in $y$, solves to obtain at least one value |
| $y=9$ | A1 | $y=9$ only |
**Total: 2 marks**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2^x \rightarrow 2^x\ln 2$ | B1 | Correct differentiation of $2^x$; also allow $2^x = e^{x\ln 2} \rightarrow e^{x\ln 2}\ln 2$ |
| $-4xy \rightarrow -4x\frac{dy}{dx}-4y \quad \text{OR} \quad y^2 \rightarrow 2y\frac{dy}{dx}$ | M1 | Differentiates $-4xy \rightarrow \pm Ax\frac{dy}{dx}\pm By$ OR differentiates $y^2 \rightarrow ky\frac{dy}{dx}$ |
| $2^x\ln 2 - 4x\frac{dy}{dx}-4y+2y\frac{dy}{dx}=0$ | A1 | Fully correct differentiation; allow versions such as $2^x\ln 2\,dx - 4x\,dy - 4y\,dx + 2y\,dy = 0$ |
| $\frac{dy}{dx}(2y-4x) = 4y - 2^x\ln 2 \Rightarrow \frac{dy}{dx}=...$ | M1 | Attempts to make $\frac{dy}{dx}$ the subject with 2 terms in $\frac{dy}{dx}$ coming from differentiating $y^2$ and $-4xy$ |
| $\frac{dy}{dx} = \frac{4y-2^x\ln 2}{2y-4x}$ or $\frac{2^x\ln 2-4y}{4x-2y}$ or $\frac{\ln 2e^{x\ln 2}-4y}{4x-2y}$ | A1 | Any correct expression; note $\frac{2y-2^{x-1}\ln 2}{y-2x}$ is correct |
**Total: 5 marks**
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(2,9) \rightarrow \frac{dy}{dx} = \frac{4(9)-2^2\ln 2}{2(9)-4(2)} \Rightarrow y-"9" = "\frac{36-4\ln 2}{10}"(x-2)$ | M1 | Uses $x=2$ and $y$ value from (a) to find gradient at $P$ and attempts to form equation of tangent; $\frac{dy}{dx}$ must have both $x$ and $y$ terms |
| $y=0 \Rightarrow 0-"9" = "\frac{36-4\ln 2}{10}"(x-2) \Rightarrow x=...$ | dM1 | Substitutes $y=0$ into tangent equation and rearranges to find $x$; dependent on previous M |
| $x = \frac{4\ln 2+9}{2\ln 2-18}$ o.e. e.g. $x=\frac{-8\ln 2-18}{-4\ln 2+36}$ | A1 | Correct expression in required form; must come from correct $\frac{dy}{dx}$ |
**Total: 3 marks**
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2bacec90-3b67-4307-9608-246ecdb6b5e2-06_695_700_251_683}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve $C$ with equation
$$2 ^ { x } - 4 x y + y ^ { 2 } = 13 \quad y \geqslant 0$$
The point $P$ lies on $C$ and has $x$ coordinate 2
\begin{enumerate}[label=(\alph*)]
\item Find the $y$ coordinate of $P$.
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
The tangent to $C$ at $P$ crosses the $x$-axis at the point $Q$.
\item Find the $x$ coordinate of $Q$, giving your answer in the form $\frac { a \ln 2 + b } { c \ln 2 + d }$ where $a , b , c$ and $d$ are integers to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2023 Q2 [10]}}