| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find equation of tangent |
| Difficulty | Moderate -0.8 This is a straightforward differentiation and tangent question requiring only basic power rule (rewriting 17/x as 17x^{-1}), substituting x=2 to find the gradient, and using y-y₁=m(x-x₁). Despite the title mentioning product/quotient rules, none are needed. This is routine P1/C1 material with no problem-solving element. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = ...x^2 + ...x + ...x^{-2}\) | M1 | Reduces the power by 1 on any term; \(...x^3 \to ...x^2\), \(...x^2 \to ...x^1\), \(...x^{-1} \to ...x^{-2}\). Do not allow just sight of \(x^2\). Index does not need to be processed. |
| \(\frac{dy}{dx} = \frac{3}{4}x^2 - 2x - 17x^{-2}\) | A1 | Two of \(+\frac{3}{4}x^2, -2x, -17x^{-2}\) or exact unsimplified equivalents. Accept \(\frac{-17}{x^2}\) but indices must be processed. Double signs e.g. \(+\frac{-17}{x^2}\) is fine. |
| \(\frac{dy}{dx} = \frac{3}{4}x^2 - 2x - 17x^{-2}\) all on one line | A1 | Allow \(x^1\). Withhold if they multiply all terms by 4 or \(a+c\) appears. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{3}{4}(2)^2 - 2(2) - 17(2)^{-2} = -\frac{21}{4}\) | M1 | Substitute \(x=2\) into their \(\frac{dy}{dx}\) which must be a changed function. |
| \(y - \frac{13}{2} = {``}-\frac{21}{4}{"}(x-2)\) | dM1 | Method for line through \(\left(2, \frac{13}{2}\right)\) using their \(\frac{dy}{dx}\) at \(x=2\). Cannot be gradient of normal. Score for sight of equation with both coordinates substituted correctly. If using \(y=mx+c\) must proceed to \(c=...\). Dependent on previous M mark. |
| \(21x + 4y - 68 = 0\) | A1 | Or any equivalent with integer coefficients, all terms on one side. |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = ...x^2 + ...x + ...x^{-2}$ | M1 | Reduces the power by 1 on any term; $...x^3 \to ...x^2$, $...x^2 \to ...x^1$, $...x^{-1} \to ...x^{-2}$. Do not allow just sight of $x^2$. Index does not need to be processed. |
| $\frac{dy}{dx} = \frac{3}{4}x^2 - 2x - 17x^{-2}$ | A1 | Two of $+\frac{3}{4}x^2, -2x, -17x^{-2}$ or exact unsimplified equivalents. Accept $\frac{-17}{x^2}$ but indices must be processed. Double signs e.g. $+\frac{-17}{x^2}$ is fine. |
| $\frac{dy}{dx} = \frac{3}{4}x^2 - 2x - 17x^{-2}$ all on one line | A1 | Allow $x^1$. Withhold if they multiply all terms by 4 or $a+c$ appears. |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{3}{4}(2)^2 - 2(2) - 17(2)^{-2} = -\frac{21}{4}$ | M1 | Substitute $x=2$ into their $\frac{dy}{dx}$ which must be a changed function. |
| $y - \frac{13}{2} = {``}-\frac{21}{4}{"}(x-2)$ | dM1 | Method for line through $\left(2, \frac{13}{2}\right)$ using their $\frac{dy}{dx}$ at $x=2$. Cannot be gradient of normal. Score for sight of equation with both coordinates substituted correctly. If using $y=mx+c$ must proceed to $c=...$. Dependent on previous M mark. |
| $21x + 4y - 68 = 0$ | A1 | Or any equivalent with integer coefficients, all terms on one side. |
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = \frac { x ^ { 3 } } { 4 } - x ^ { 2 } + \frac { 17 } { x } \quad x > 0$$
(a) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$, giving your answer in simplest form.
The point $R \left( 2 , \frac { 13 } { 2 } \right)$ lies on $C$.\\
(b) Find the equation of the tangent to $C$ at the point $R$. 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 2022 Q1 [6]}}