| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| 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 (standard A-level technique) followed by recognizing that the curve is undefined where dy/dx would be undefined, leading to a discriminant condition. The implicit differentiation is routine, but part (b) requires insight to connect the undefined interval to where the derivative's denominator vanishes and then solve a quadratic inequality. This is moderately above average difficulty due to the conceptual leap needed in part (b), though the algebraic manipulation remains straightforward. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y^2=2x^2+15x+10y \Rightarrow 2y\frac{dy}{dx}=4x+15+10\frac{dy}{dx}\) | M1 A1 | M1: Correct differentiation of one \(y\) term i.e. \(y^2\to 2y\frac{dy}{dx}\) or \(10y\to10\frac{dy}{dx}\); A1: Fully correct differentiation |
| \((2y-10)\frac{dy}{dx}=4x+15 \Rightarrow \frac{dy}{dx}=\frac{4x+15}{2y-10}\) o.e. | M1, A1 | M1: Rearranges to make \(\frac{dy}{dx}\) subject; expression must contain exactly two \(\frac{dy}{dx}\) terms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Deduces \(2y-10=0 \Rightarrow y=5\) | B1ft | Follow through on denominator of form \(ay+b\), \(a,b\neq0\) |
| Substitutes \(y=5\) into \(y^2=2x^2+15x+10y \Rightarrow 2x^2+15x+25=0\) and solves for \(x\) | M1 | Usual rules apply; if no working shown must give correct values for their quadratic |
| \((p=)-5,\ (q=)-\frac{5}{2}\) | A1 | Correct values; accept if given as endpoints \(\left(-5,-\frac{5}{2}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(y^2 = 2x^2 + 15x + 10y \Rightarrow y^2 - 10y - (2x^2 + 15x) = 0\) | ||
| Deduces roots for \(x\) don't exist when \(100 + 4 \times (2x^2 + 15x) < 0\) | B1 | |
| Solves \(2x^2 + 15x + 25 < 0\) | M1 | |
| \((p =) -5,\ (q =) -\frac{5}{2}\) | A1 | Correct values, see note on main scheme |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(y = 0 \Rightarrow 2x^2 + 15x = 0 \Rightarrow x = 0, -\frac{15}{2}\) | B1 | Correct values found for \(x\) when \(y = 0\) |
| Full method to use symmetry to deduce required values of \(x\); values are \(x = \frac{1}{3} \times -\frac{15}{2},\ \frac{2}{3} \times -\frac{15}{2}\) | M1 | E.g. by symmetry, values required are one third and two thirds way between these |
| \((p =) -5,\ (q =) -\frac{5}{2}\) | A1 | Correct values, see note on main scheme |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y^2=2x^2+15x+10y \Rightarrow 2y\frac{dy}{dx}=4x+15+10\frac{dy}{dx}$ | **M1 A1** | M1: Correct differentiation of one $y$ term i.e. $y^2\to 2y\frac{dy}{dx}$ or $10y\to10\frac{dy}{dx}$; A1: Fully correct differentiation |
| $(2y-10)\frac{dy}{dx}=4x+15 \Rightarrow \frac{dy}{dx}=\frac{4x+15}{2y-10}$ o.e. | **M1, A1** | M1: Rearranges to make $\frac{dy}{dx}$ subject; expression must contain exactly two $\frac{dy}{dx}$ terms |
---
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces $2y-10=0 \Rightarrow y=5$ | **B1ft** | Follow through on denominator of form $ay+b$, $a,b\neq0$ |
| Substitutes $y=5$ into $y^2=2x^2+15x+10y \Rightarrow 2x^2+15x+25=0$ and solves for $x$ | **M1** | Usual rules apply; if no working shown must give correct values for their quadratic |
| $(p=)-5,\ (q=)-\frac{5}{2}$ | **A1** | Correct values; accept if given as endpoints $\left(-5,-\frac{5}{2}\right)$ |
## Question (b) Alt Method 1:
| Working | Mark | Guidance |
|---------|------|----------|
| $y^2 = 2x^2 + 15x + 10y \Rightarrow y^2 - 10y - (2x^2 + 15x) = 0$ | | |
| Deduces roots for $x$ don't exist when $100 + 4 \times (2x^2 + 15x) < 0$ | B1 | |
| Solves $2x^2 + 15x + 25 < 0$ | M1 | |
| $(p =) -5,\ (q =) -\frac{5}{2}$ | A1 | Correct values, see note on main scheme |
## Question (b) Alt Method 2:
| Working | Mark | Guidance |
|---------|------|----------|
| $y = 0 \Rightarrow 2x^2 + 15x = 0 \Rightarrow x = 0, -\frac{15}{2}$ | B1 | Correct values found for $x$ when $y = 0$ |
| Full method to use symmetry to deduce required values of $x$; values are $x = \frac{1}{3} \times -\frac{15}{2},\ \frac{2}{3} \times -\frac{15}{2}$ | M1 | E.g. by symmetry, values required are one third and two thirds way between these |
| $(p =) -5,\ (q =) -\frac{5}{2}$ | A1 | Correct values, see note on main scheme |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c46ca445-cf59-4664-931e-add9f2f81851-14_940_881_251_593}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the curve with equation
$$y ^ { 2 } = 2 x ^ { 2 } + 15 x + 10 y$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
The curve is not defined for values of $x$ in the interval ( $p , q$ ), as shown in Figure 2.
\item Using your answer to part (a) or otherwise, find the value of $p$ and the value of $q$.\\
(Solutions relying entirely on calculator technology are not acceptable.)
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2023 Q5 [7]}}