| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent parallel to given line |
| Difficulty | Standard +0.8 This question requires students to work with an unspecified cubic function, deduce its properties from a graph, recognize that parallel tangents means equal gradients, differentiate to find f'(x), and prove algebraically that f'(t) = f'(-t) for all t. This goes beyond routine differentiation to require proof and insight about even/odd function properties of derivatives. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.07i Differentiate x^n: for rational n and sums |
| Answer | Marks | Guidance |
|---|---|---|
| \([y =] k(x+1)^2(x-2)\) | M1* | AO3.1a |
| Substitute \((0, -3)\) or \((1, -6)\) | M1* | AO3.1a |
| \([y =] \frac{3}{2}(x+1)^2(x-2)\) | A1dep | AO1.1 |
| \([y =] \frac{3}{2}x^3 - \frac{9}{2}x - 3\) | M1* | AO1.1 |
| gradient of tangent is \(\frac{dy}{dx} = \frac{3}{2}(3x^2 - 3)\) | A1dep | AO2.1; FT their \(f(x)\) even if the gradient property does not hold for it |
| \(t^2 = (-t)^2\) therefore the gradients are equal and the tangents are parallel | E1 [6] | AO2.2a; Not just "the gradient is the same for \(-t\)". Allow FT from their \(f(x)\) if the gradient property holds |
## Question 11:
**DR**
$[y =] k(x+1)^2(x-2)$ | M1* | AO3.1a
Substitute $(0, -3)$ or $(1, -6)$ | M1* | AO3.1a
$[y =] \frac{3}{2}(x+1)^2(x-2)$ | A1dep | AO1.1
$[y =] \frac{3}{2}x^3 - \frac{9}{2}x - 3$ | M1* | AO1.1
gradient of tangent is $\frac{dy}{dx} = \frac{3}{2}(3x^2 - 3)$ | A1dep | AO2.1; FT their $f(x)$ even if the gradient property does not hold for it
$t^2 = (-t)^2$ therefore the gradients are equal and the tangents are parallel | E1 [6] | AO2.2a; Not just "the gradient is the same for $-t$". Allow FT from their $f(x)$ if the gradient property holds
---11 In this question you must show detailed reasoning.
Fig. 11 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x )$ is a cubic function. Fig. 11 also shows the coordinates of the turning points and the points of intersection with the axes.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{05376a51-e768-4b45-9c18-c98255a4bd70-11_805_620_543_317}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{figure}
Show that the tangent to $y = \mathrm { f } ( x )$ at $x = t$ is parallel to the tangent to $y = \mathrm { f } ( x )$ at $x = - t$ for all values of $t$.
\hfill \mbox{\textit{OCR MEI AS Paper 2 Q11 [6]}}