| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find curve equation from derivative |
| Difficulty | Standard +0.3 This is a straightforward integration problem requiring two successive integrations with constants determined from given conditions. Part (a) is routine (finding normal from tangent), and part (b) involves standard integration of polynomial and power terms. While multi-step, it requires only direct application of basic calculus techniques with no novel insight or complex problem-solving. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.07m Tangents and normals: gradient and equations1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| States or implies that the gradient of the normal is \(-\dfrac{1}{3}\) | M1 | |
| Correct equation of normal e.g. \(y - 16 = -\dfrac{1}{3}(x-4)\) | A1 | Finds the equation of the normal \(y - 16 = -\dfrac{1}{3}(x-4)\), o.e. e.g. \(y = -\dfrac{1}{3}x + \dfrac{52}{3}\), \(3y + x - 52 = 0\). Note: \(\dfrac{y-16}{x-4} = -\dfrac{1}{3}\) is not sufficient. Requires a full correct equation. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(f''(x) = 4x + x^{-\frac{1}{2}} \Rightarrow f'(x) = 2x^2 + 2x^{\frac{1}{2}} + c\) | M1 | Attempts to integrate \(f''(x)\) once with one index correct. E.g. \(4x \to \ldots x^2\) or \(\dfrac{1}{\sqrt{x}} \to \ldots x^{\frac{1}{2}}\) |
| \(x = 4, f'(x) = 3 \Rightarrow 3 = 32 + 4 + c \Rightarrow c = \ldots(-33)\) | dM1 | Applies \(f'(4) = 3\) and solves to find constant of integration. Depends on first M1 |
| \(f'(x) = 2x^2 + 2x^{\frac{1}{2}} - 33\) | A1 | \((f'(x) =) 2x^2 + 2x^{\frac{1}{2}} - 33\) or obtains \((f'(x) =) 2x^2 + 2x^{\frac{1}{2}} + c\) with \(c\) correctly calculated as \(-33\). Ignore labelling. |
| \(f'(x) = 2x^2 + 2x^{\frac{1}{2}} - 33 \Rightarrow f(x) = \dfrac{2}{3}x^3 + \dfrac{4}{3}x^{\frac{3}{2}} - 33x + d\) | dM1 | Dependent on first M1 only. For an attempt to integrate \(f''(x)\) twice and achieve a form \((f(x) =) ax^3 + bx^{\frac{3}{2}} + \ldots\) where \(\ldots\) could be 0. |
| \(x = 4, f(x) = 16 \Rightarrow \left(f(x) =\right)\dfrac{2}{3}x^3 + \dfrac{4}{3}x^{\frac{3}{2}} - 33x + \dfrac{284}{3}\) | ddM1A1 | Dependent on all previous M's. Uses \(f(4) = 16\) (may be implied) to find constant of integration. A1: \(\left(f(x) =\right)\dfrac{2}{3}x^3 + \dfrac{4}{3}x^{\frac{3}{2}} - 33x + \dfrac{284}{3}\) or exact equivalent. Also allow with \(c\) correctly calculated as \(\dfrac{284}{3}\). |
## Question 11:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| States or implies that the gradient of the normal is $-\dfrac{1}{3}$ | M1 | |
| Correct equation of normal e.g. $y - 16 = -\dfrac{1}{3}(x-4)$ | A1 | Finds the equation of the normal $y - 16 = -\dfrac{1}{3}(x-4)$, o.e. e.g. $y = -\dfrac{1}{3}x + \dfrac{52}{3}$, $3y + x - 52 = 0$. Note: $\dfrac{y-16}{x-4} = -\dfrac{1}{3}$ is **not** sufficient. Requires a full correct equation. |
**Total: 2 marks**
---
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $f''(x) = 4x + x^{-\frac{1}{2}} \Rightarrow f'(x) = 2x^2 + 2x^{\frac{1}{2}} + c$ | M1 | Attempts to integrate $f''(x)$ once with one index correct. E.g. $4x \to \ldots x^2$ or $\dfrac{1}{\sqrt{x}} \to \ldots x^{\frac{1}{2}}$ |
| $x = 4, f'(x) = 3 \Rightarrow 3 = 32 + 4 + c \Rightarrow c = \ldots(-33)$ | dM1 | Applies $f'(4) = 3$ and solves to find constant of integration. **Depends on first M1** |
| $f'(x) = 2x^2 + 2x^{\frac{1}{2}} - 33$ | A1 | $(f'(x) =) 2x^2 + 2x^{\frac{1}{2}} - 33$ or obtains $(f'(x) =) 2x^2 + 2x^{\frac{1}{2}} + c$ with $c$ correctly calculated as $-33$. Ignore labelling. |
| $f'(x) = 2x^2 + 2x^{\frac{1}{2}} - 33 \Rightarrow f(x) = \dfrac{2}{3}x^3 + \dfrac{4}{3}x^{\frac{3}{2}} - 33x + d$ | dM1 | **Dependent on first M1 only**. For an attempt to integrate $f''(x)$ twice and achieve a form $(f(x) =) ax^3 + bx^{\frac{3}{2}} + \ldots$ where $\ldots$ could be 0. |
| $x = 4, f(x) = 16 \Rightarrow \left(f(x) =\right)\dfrac{2}{3}x^3 + \dfrac{4}{3}x^{\frac{3}{2}} - 33x + \dfrac{284}{3}$ | ddM1A1 | **Dependent on all previous M's**. Uses $f(4) = 16$ (may be implied) to find constant of integration. A1: $\left(f(x) =\right)\dfrac{2}{3}x^3 + \dfrac{4}{3}x^{\frac{3}{2}} - 33x + \dfrac{284}{3}$ or exact equivalent. Also allow with $c$ correctly calculated as $\dfrac{284}{3}$. |
**Total: 6 marks**
**Overall Total: 8 marks**\begin{enumerate}
\item A curve $C$ has equation $y = \mathrm { f } ( x ) , \quad x > 0$
\end{enumerate}
Given that
\begin{itemize}
\item $\mathrm { f } ^ { \prime \prime } ( x ) = 4 x + \frac { 1 } { \sqrt { x } }$
\item the point $P$ has $x$ coordinate 4 and lies on $C$
\item the tangent to $C$ at $P$ has equation $y = 3 x + 4$\\
(a) find an equation of the normal to $C$ at $P$\\
(b) find $\mathrm { f } ( x )$, writing your answer in simplest form.
\end{itemize}
\hfill \mbox{\textit{Edexcel P1 2023 Q11 [8]}}