| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find tangent at given point (polynomial/algebraic) |
| Difficulty | Moderate -0.8 This is a straightforward AS-level differentiation question requiring only basic power rule application and point-slope form. The steps are routine: differentiate to get dy/dx = 6x² - 4, substitute x=2 to find gradient m=20, then use y-13=20(x-2) to get the tangent equation. No problem-solving insight needed, just mechanical application of standard techniques. |
| 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 Notes |
| Attempts to differentiate \(x^n \rightarrow x^{n-1}\) seen once | M1 | Score for \(x^3 \rightarrow x^2\) or \(\pm4x \rightarrow 4\) or \(+5 \rightarrow 0\) |
| \(y = 2x^3 - 4x + 5 \Rightarrow \frac{dy}{dx} = 6x^2 - 4\) | A1 | \(6x^2 - 4\) which may be unsimplified; \(6x^2 - 4 + C\) is A0 |
| For substituting \(x = 2\) into their \(\frac{dy}{dx} = 6x^2 - 4\) | dM1 | Substitutes \(x = 2\) into their \(\frac{dy}{dx}\); first M must have been awarded. Score for sight of embedded values or sight of "\(\frac{dy}{dx}\) at \(x=2\) is" or correct follow through. Note: 20 alone is insufficient. |
| For correct method of finding tangent at \(P(2,13)\); score for \(y - 13 = \text{"20"}(x-2)\) | ddM1 | Dependent on both previous M marks. If \(y = mx + c\) form used, must proceed as far as \(c = \ldots\) |
| \(y = 20x - 27\) | A1 | Completely correct \(y = 20x - 27\) (and in this form) |
| Total | (5) |
## Question 1:
| Answer/Working | Mark | Guidance Notes |
|---|---|---|
| Attempts to differentiate $x^n \rightarrow x^{n-1}$ seen once | M1 | Score for $x^3 \rightarrow x^2$ or $\pm4x \rightarrow 4$ or $+5 \rightarrow 0$ |
| $y = 2x^3 - 4x + 5 \Rightarrow \frac{dy}{dx} = 6x^2 - 4$ | A1 | $6x^2 - 4$ which may be unsimplified; $6x^2 - 4 + C$ is A0 |
| For substituting $x = 2$ into their $\frac{dy}{dx} = 6x^2 - 4$ | dM1 | Substitutes $x = 2$ into their $\frac{dy}{dx}$; first M must have been awarded. Score for sight of embedded values or sight of "$\frac{dy}{dx}$ at $x=2$ is" or correct follow through. Note: 20 alone is insufficient. |
| For correct method of finding tangent at $P(2,13)$; score for $y - 13 = \text{"20"}(x-2)$ | ddM1 | Dependent on both previous M marks. If $y = mx + c$ form used, must proceed as far as $c = \ldots$ |
| $y = 20x - 27$ | A1 | Completely correct $y = 20x - 27$ (and in this form) |
| **Total** | **(5)** | |\begin{enumerate}
\item A curve has equation
\end{enumerate}
$$y = 2 x ^ { 3 } - 4 x + 5$$
Find the equation of the tangent to the curve at the point $P ( 2,13 )$.\\
Write your answer in the form $y = m x + c$, where $m$ and $c$ are integers to be found.\\
Solutions relying on calculator technology are not acceptable.\\
(5)
\hfill \mbox{\textit{Edexcel AS Paper 1 2020 Q1 [5]}}