| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Equation of normal line |
| Difficulty | Moderate -0.3 This is a straightforward differentiation and tangent/normal question requiring standard techniques: rewriting terms as powers, applying power rule, evaluating at a point, and finding perpendicular gradient. While it involves multiple terms and fractional/negative powers, these are routine P1 skills with no conceptual challenges or novel problem-solving required. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = ...x^1 + ...x^{-\frac{3}{2}} + ...x^{-2}\) | M1 | Reduces the power by 1 on any term; \(...x^2 \to ...x^1\), \(...x^{-\frac{1}{2}} \to ...x^{-\frac{3}{2}}\), \(...x^{-1} \to ...x^{-2}\), \(5 \to 0\) |
| \(\frac{dy}{dx} = \frac{2}{3}x - 2x^{-\frac{3}{2}} - \frac{8}{3}x^{-2}\) | A1A1A1 | One correct term for first A1; two correct terms for second A1; all three correct on one line for third A1. Condone \(\frac{2}{3}x^1\). Do not allow double signs. Exact simplified equivalents accepted e.g. \(\frac{-2}{(\sqrt{x})^3}\) or \(-\frac{8}{3x^2}\) or \(-2.6x^{-2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{2}{3}\times4 - 2\times4^{-\frac{3}{2}} - \frac{8}{3}\times4^{-2} = \frac{9}{4}\) | M1 | Substitutes \(x=4\) into their \(\frac{dy}{dx}\) to find numerical gradient. Condone slips; may be implied by further work |
| \(\frac{9}{4} \to -\frac{4}{9}\) | M1 | Correct attempt using \(m_N = -\frac{1}{m_T}\) to find gradient of perpendicular |
| \(y - 3 = -\frac{4}{9}(x-4)\) | dM1 | Finding line through \((4,3)\) with a changed gradient. Dependent on first M1. Both brackets correct. Alternatively \(y=mx+c\) proceeding to find \(c\) |
| \(4x + 9y - 43 = 0\) | A1 | All terms on one side \(= 0\). Accept \(\pm A(4x+9y-43=0)\) where \(A \in \mathbb{N}\) |
# Question 1:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = ...x^1 + ...x^{-\frac{3}{2}} + ...x^{-2}$ | M1 | Reduces the power by 1 on any term; $...x^2 \to ...x^1$, $...x^{-\frac{1}{2}} \to ...x^{-\frac{3}{2}}$, $...x^{-1} \to ...x^{-2}$, $5 \to 0$ |
| $\frac{dy}{dx} = \frac{2}{3}x - 2x^{-\frac{3}{2}} - \frac{8}{3}x^{-2}$ | A1A1A1 | One correct term for first A1; two correct terms for second A1; all three correct on one line for third A1. Condone $\frac{2}{3}x^1$. Do not allow double signs. Exact simplified equivalents accepted e.g. $\frac{-2}{(\sqrt{x})^3}$ or $-\frac{8}{3x^2}$ or $-2.6x^{-2}$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{2}{3}\times4 - 2\times4^{-\frac{3}{2}} - \frac{8}{3}\times4^{-2} = \frac{9}{4}$ | M1 | Substitutes $x=4$ into their $\frac{dy}{dx}$ to find numerical gradient. Condone slips; may be implied by further work |
| $\frac{9}{4} \to -\frac{4}{9}$ | M1 | Correct attempt using $m_N = -\frac{1}{m_T}$ to find gradient of perpendicular |
| $y - 3 = -\frac{4}{9}(x-4)$ | dM1 | Finding line through $(4,3)$ with a changed gradient. Dependent on first M1. Both brackets correct. Alternatively $y=mx+c$ proceeding to find $c$ |
| $4x + 9y - 43 = 0$ | A1 | All terms on one side $= 0$. Accept $\pm A(4x+9y-43=0)$ where $A \in \mathbb{N}$ |
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = \frac { x ^ { 2 } } { 3 } + \frac { 4 } { \sqrt { x } } + \frac { 8 } { 3 x } - 5 \quad x > 0$$
(a) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$, giving your answer in simplest form.
The point $P ( 4,3 )$ lies on $C$.\\
(b) Find the equation of the normal to $C$ at the point $P$. Write your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers to be found.
\hfill \mbox{\textit{Edexcel P1 2021 Q1 [8]}}