| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | January |
| 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 differentiation techniques: finding turning points using first and second derivatives, and finding a normal line equation. Both parts are routine textbook exercises requiring only direct application of learned procedures with no problem-solving insight needed. The calculations are simple (cubic differentiation, basic arithmetic) making this easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| \((0, 12)\) and \((4, -20)\) | B1B1, M1, A1, A1 | Allow \(y = 12\) and \(y = -20\) |
| Answer | Marks | Guidance |
|---|---|---|
| when \(x = 2\) \(y' = -12\) | M1, A1, B1 | \(y'\) used each side of TP or good sketch. Both stated, only one needs testing |
| \(y + 4 = \frac{1}{12}(x - 2)\) | M1ft | from their \(y'\). accept any numerical m. Or \(-4 = \) their\((m) x 2 + c\) |
| \(y = \frac{1}{12}x - 4\frac{1}{6}\) | A1 | Any recognisable \(\frac{25}{6}\), at worst \(4.1\) |
| 7 marks | ||
| 4 marks | ||
| [11] |
**i** $y' = 3x^2 - 12x$
use of $y' = 0$
$x = 0$ and $4$
$(0, 12)$ and $(4, -20)$ | B1B1, M1, A1, A1 | Allow $y = 12$ and $y = -20$
$y'' = 6x - 12$ used
max when $x = 0$, min when $x = 4$
when $x = 2$ $y' = -12$ | M1, A1, B1 | $y'$ used each side of TP or good sketch. Both stated, only one needs testing
$y + 4 = \frac{1}{12}(x - 2)$ | M1ft | from their $y'$. accept any numerical m. Or $-4 = $ their$(m) x 2 + c$
$y = \frac{1}{12}x - 4\frac{1}{6}$ | A1 | Any recognisable $\frac{25}{6}$, at worst $4.1$
| | 7 marks |
| | 4 marks |
| | [11] |
**ii**10 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 2005 Q10 [11]}}