| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Determine nature of stationary points |
| Difficulty | Standard +0.3 Finding points of inflection requires computing the second derivative, setting it equal to zero, solving a quadratic equation, and verifying the sign change. This is a standard A-level procedure with straightforward algebra, slightly above average difficulty due to the multi-step process and verification requirement. |
| Spec | 1.07f Convexity/concavity: points of inflection1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = 4x^3 - 12x + 4\) | M1 | 1.1 |
| (first derivative correct) | A1 | 1.1 |
| \(\frac{d^2y}{dx^2} = 12x^2 - 12 = 0\) | M1 | 1.2 |
| \(x = \pm 1\) | A1 | 1.1 |
| \((-1, -4)\) and \((1, 4)\) | A1 | 2.1 |
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = 4x^3 - 12x + 4$ | M1 | 1.1 | Differentiating once |
| (first derivative correct) | A1 | 1.1 | First derivative |
| $\frac{d^2y}{dx^2} = 12x^2 - 12 = 0$ | M1 | 1.2 | Differentiating a second time and equating to zero |
| $x = \pm 1$ | A1 | 1.1 |
| $(-1, -4)$ and $(1, 4)$ | A1 | 2.1 | **[5]** |
---6 Fig. 6 shows the curve with equation $y = x ^ { 4 } - 6 x ^ { 2 } + 4 x + 5$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b4e10fd2-4144-4019-bf00-070f93a2b05d-06_869_750_370_242}
\captionsetup{labelformat=empty}
\caption{Fig. 6\\
Find the coordinates of the points of inflection.}
\end{center}
\end{figure}
\hfill \mbox{\textit{OCR MEI Paper 3 Q6 [5]}}