| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find stationary points and nature |
| Difficulty | Moderate -0.3 This is a straightforward differentiation and algebra question requiring polynomial differentiation (not actually chain rule despite the topic label), solving a linear equation, and finding an intersection point. The steps are routine and mechanical with no conceptual challenges, making it slightly easier than average for A-level. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.07i Differentiate x^n: for rational n and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f(x) = 2x^3 - kx^2 + 14x + 24 \Rightarrow f'(x) = 6x^2 - 2kx + 14\) | M1A1 | M1: For \(x^3 \to ...x^2\) or \(x^2 \to ...x\) or \(14x \to 14\). A1: Correct simplified first derivative. Do not allow \(x^1\) for \(x\). Penalise inclusion of "\(+c\)" on either derivative only once on first occurrence. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f''(x) = 12x - 2k\) | A1ft | Correct simplified second derivative. Follow through from their first derivative. Must follow M1. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(6x^2 - 2kx + 14 = 12x - 2k \Rightarrow 6(5)^2 - 2k(5) + 14 = 12(5) - 2k \Rightarrow k = ...\) | M1 | Sets \(f'(x) = f''(x)\), substitutes \(x=5\) and solves linear equation in \(k\) |
| \(k = 13\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(k = 13 \Rightarrow 6x^2 - 38x + 40 = 0 \Rightarrow x = ...\) | M1 | Substitutes their \(k\) into equated \(f'(x) = f''(x)\) and solves resulting 3TQ to obtain at least one value for \(x\) |
| \(x = "\frac{4}{3}" \Rightarrow y = 12\left("\frac{4}{3}"\right) - 2\times13\) or \(y = 6\left("\frac{4}{3}"\right)^2 - 2\times13\times\left("\frac{4}{3}"\right) + 14\) | M1 | Substitutes their \(x\) (not 5) and their \(k\) into \(f'(x)\) or \(f''(x)\) to find value for \(y\) |
| \(x = \frac{4}{3},\ y = -10\) | A1 | Correct coordinates. May be seen as \(x = ...,\ y = ...\) or as coordinate pair. Allow missing brackets e.g. \(\frac{4}{3}, -10\). Allow "made up" \(k\) in (c). |
## Question 7:
### Part (a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(x) = 2x^3 - kx^2 + 14x + 24 \Rightarrow f'(x) = 6x^2 - 2kx + 14$ | M1A1 | M1: For $x^3 \to ...x^2$ or $x^2 \to ...x$ or $14x \to 14$. A1: Correct simplified first derivative. Do not allow $x^1$ for $x$. Penalise inclusion of "$+c$" on either derivative only once on first occurrence. |
### Part (a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f''(x) = 12x - 2k$ | A1ft | Correct simplified second derivative. Follow through from their first derivative. Must follow M1. |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $6x^2 - 2kx + 14 = 12x - 2k \Rightarrow 6(5)^2 - 2k(5) + 14 = 12(5) - 2k \Rightarrow k = ...$ | M1 | Sets $f'(x) = f''(x)$, substitutes $x=5$ and solves linear equation in $k$ |
| $k = 13$ | A1 | |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $k = 13 \Rightarrow 6x^2 - 38x + 40 = 0 \Rightarrow x = ...$ | M1 | Substitutes their $k$ into equated $f'(x) = f''(x)$ and solves resulting 3TQ to obtain at least one value for $x$ |
| $x = "\frac{4}{3}" \Rightarrow y = 12\left("\frac{4}{3}"\right) - 2\times13$ or $y = 6\left("\frac{4}{3}"\right)^2 - 2\times13\times\left("\frac{4}{3}"\right) + 14$ | M1 | Substitutes their $x$ (not 5) and their $k$ into $f'(x)$ or $f''(x)$ to find value for $y$ |
| $x = \frac{4}{3},\ y = -10$ | A1 | Correct coordinates. May be seen as $x = ...,\ y = ...$ or as coordinate pair. Allow missing brackets e.g. $\frac{4}{3}, -10$. Allow "made up" $k$ in (c). |
---\begin{enumerate}
\item The curve $C$ has equation $y = \mathrm { f } ( x )$ where
\end{enumerate}
$$f ( x ) = 2 x ^ { 3 } - k x ^ { 2 } + 14 x + 24$$
and $k$ is a constant.\\
(a) Find, in simplest form,\\
(i) $\mathrm { f } ^ { \prime } ( x )$\\
(ii) $\mathrm { f } ^ { \prime \prime } ( x )$
The curve with equation $y = \mathrm { f } ^ { \prime } ( x )$ intersects the curve with equation $y = \mathrm { f } ^ { \prime \prime } ( x )$ at the points $A$ and $B$.
Given that the $x$ coordinate of $A$ is 5\\
(b) find the value of $k$.\\
(c) Hence find the coordinates of $B$.
\hfill \mbox{\textit{Edexcel P1 2024 Q7 [8]}}