| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Equation of normal line |
| Difficulty | Easy -1.2 This is a straightforward application of the chain rule to differentiate a composite function, followed by routine substitution to find a gradient and then the normal equation. Both parts require only standard procedures with no problem-solving insight, making it easier than average for A-level. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = 6 \times 3(3x-2)^5 = 18(3x-2)^5\) | M1 | Applies chain rule to achieve form \(A(3x-2)^5\); index does not need to be processed |
| \(\frac{dy}{dx} = 18(3x-2)^5\) or unsimplified equivalent e.g. \(6 \times 3(3x-2)^5\) | A1 | Index must be processed; if expanded accept \(-576+4320x-12960x^2+19440x^3-14580x^4+4374x^5\); accept \(x^1\) for \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{"18"}(3 \times \frac{1}{3}-2)^{\text{"5"}} = -18\) | M1 | Substitutes \(x=\frac{1}{3}\) into \(\frac{dy}{dx}\) of form \(A(3x-2)^5\); must find a value |
| \(-18 \rightarrow \frac{1}{18}\) | M1 | Finds negative reciprocal of gradient; cannot be scored for setting \(\frac{dy}{dx}=0\) |
| \(y-1 = \text{"}\frac{1}{18}\text{"}\left(x-\frac{1}{3}\right)\) | dM1 | Attempts equation of normal; must be changed gradient from tangent; if using \(y=mx+c\) must reach \(c=\ldots\left(=\text{"}\frac{53}{54}\text{"}\right)\) |
| \(3x - 54y + 53 = 0\) | A1 | Or any integer multiple with all terms on one side; must see \(=0\) |
## Question 1:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 6 \times 3(3x-2)^5 = 18(3x-2)^5$ | M1 | Applies chain rule to achieve form $A(3x-2)^5$; index does not need to be processed |
| $\frac{dy}{dx} = 18(3x-2)^5$ or unsimplified equivalent e.g. $6 \times 3(3x-2)^5$ | A1 | Index must be processed; if expanded accept $-576+4320x-12960x^2+19440x^3-14580x^4+4374x^5$; accept $x^1$ for $x$ |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{"18"}(3 \times \frac{1}{3}-2)^{\text{"5"}} = -18$ | M1 | Substitutes $x=\frac{1}{3}$ into $\frac{dy}{dx}$ of form $A(3x-2)^5$; must find a value |
| $-18 \rightarrow \frac{1}{18}$ | M1 | Finds negative reciprocal of gradient; cannot be scored for setting $\frac{dy}{dx}=0$ |
| $y-1 = \text{"}\frac{1}{18}\text{"}\left(x-\frac{1}{3}\right)$ | dM1 | Attempts equation of normal; must be changed gradient from tangent; if using $y=mx+c$ must reach $c=\ldots\left(=\text{"}\frac{53}{54}\text{"}\right)$ |
| $3x - 54y + 53 = 0$ | A1 | Or any integer multiple with all terms on one side; must see $=0$ |
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = ( 3 x - 2 ) ^ { 6 }$$
(a) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$
Given that the point $P \left( \frac { 1 } { 3 } , 1 \right)$ lies on $C$,\\
(b) find the equation of the normal to $C$ at $P$. 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 P3 2022 Q1 [6]}}