| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Second derivative test justification |
| Difficulty | Moderate -0.8 This is a straightforward C2 question testing standard procedures: finding stationary points using first and second derivatives, and finding a normal line equation. Both parts require only routine application of well-practiced techniques with no problem-solving insight needed. The calculations are simple (cubic differentiation, basic arithmetic), making this easier than average for A-level. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y' = 3x^2 - 12x\) | B1B1 | |
| use of \(y' = 0\) | M1 | |
| \(x = 0\) and \(4\) | A1 | |
| \((0,\ 12)\) and \((4,\ -20)\) | A1 | Allow \(y=12\) and \(y=-20\) |
| \(y'' = 6x - 12\) used | M1 | \(y'\) used each side of TP or good sketch |
| max when \(x=0\), min when \(x=4\) | A1 | Both stated, only one needs testing |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| when \(x=2\), \(y' = -12\) | B1 | |
| grad of normal \(= 1/12\) | B1ft | from their \(y'\) |
| \(y + 4 = \frac{1}{12}(x-2)\) | M1ft | accept any numerical m; or \(-4 = \text{their}(m) \times 2 + c\) |
| \(y = \frac{1}{12}x - 4\frac{1}{6}\) | A1 | Any recognisable \(25/6\), at worst \(4.1\) |
| [4] [11] |
## Question 2:
**Part i:**
| Answer | Mark | Guidance |
|--------|------|----------|
| $y' = 3x^2 - 12x$ | B1B1 | |
| use of $y' = 0$ | M1 | |
| $x = 0$ and $4$ | A1 | |
| $(0,\ 12)$ and $(4,\ -20)$ | A1 | Allow $y=12$ and $y=-20$ |
| $y'' = 6x - 12$ used | M1 | $y'$ used each side of TP or good sketch |
| max when $x=0$, min when $x=4$ | A1 | Both stated, only one needs testing |
| | | **[7]** |
**Part ii:**
| Answer | Mark | Guidance |
|--------|------|----------|
| when $x=2$, $y' = -12$ | B1 | |
| grad of normal $= 1/12$ | B1ft | from their $y'$ |
| $y + 4 = \frac{1}{12}(x-2)$ | M1ft | accept any numerical m; or $-4 = \text{their}(m) \times 2 + c$ |
| $y = \frac{1}{12}x - 4\frac{1}{6}$ | A1 | Any recognisable $25/6$, at worst $4.1$ |
| | | **[4] [11]** |2 A curve has equation $y = x ^ { 3 } - 6 x ^ { 2 } + 12$.\\
(i) Use calculus to find the coordinates of the turning points of this curve. Determine also the nature of these turning points.\\
(ii) Find, in the form $y = m x + c$, the equation of the normal to the curve at the point $( 2 , - 4 )$.
\hfill \mbox{\textit{OCR MEI C2 Q2 [11]}}