| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find stationary point then sketch curve |
| Difficulty | Moderate -0.3 This is a standard C2 differentiation question covering routine techniques: finding derivatives, equation of tangent, stationary points via dy/dx=0, and curve sketching. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average but not trivial due to the multi-part nature and algebraic manipulation involved. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks |
|---|---|
| (i) \(\frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 18x + 12\) | M1 A1 |
| At \(x=3\): \(\frac{\mathrm{d}y}{\mathrm{d}x} = 54 - 54 + 12 = 12\) | A1 |
| \(y = 2(27) - 9(9) + 36 - 2 = 7\); equation: \(y - 7 = 12(x-3)\) | M1 |
| Check \((-1, -41)\): \(-41 - 7 = 12(-1-3) = -48\) ✓ | A1 |
| (ii) \(6x^2 - 18x + 12 = 0 \Rightarrow x^2 - 3x + 2 = 0\) | M1 |
| \((x-1)(x-2) = 0\), \(x = 1\) or \(x = 2\) | A1 |
| \((1, 3)\) and \((2, 2)\) | A1 A1 |
| (iii) Correct cubic shape, one real root near \(x = 0.2\), turning points at \((1,3)\) and \((2,2)\) | B1 B1 B1 |
## Question 9:
**(i)** $\frac{\mathrm{d}y}{\mathrm{d}x} = 6x^2 - 18x + 12$ | M1 A1 |
At $x=3$: $\frac{\mathrm{d}y}{\mathrm{d}x} = 54 - 54 + 12 = 12$ | A1 |
$y = 2(27) - 9(9) + 36 - 2 = 7$; equation: $y - 7 = 12(x-3)$ | M1 |
Check $(-1, -41)$: $-41 - 7 = 12(-1-3) = -48$ ✓ | A1 |
**(ii)** $6x^2 - 18x + 12 = 0 \Rightarrow x^2 - 3x + 2 = 0$ | M1 |
$(x-1)(x-2) = 0$, $x = 1$ or $x = 2$ | A1 |
$(1, 3)$ and $(2, 2)$ | A1 A1 |
**(iii)** Correct cubic shape, one real root near $x = 0.2$, turning points at $(1,3)$ and $(2,2)$ | B1 B1 B1 |
---9 The equation of a cubic curve is $y = 2 x ^ { 3 } - 9 x ^ { 2 } + 12 x - 2$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and show that the tangent to the curve when $x = 3$ passes through the point $( - 1 , - 41 )$.\\
(ii) Use calculus to find the coordinates of the turning points of the curve. You need not distinguish between the maximum and minimum.\\
(iii) Sketch the curve, given that the only real root of $2 x ^ { 3 } - 9 x ^ { 2 } + 12 x - 2 = 0$ is $x = 0.2$ correct to 1 decimal place.
\hfill \mbox{\textit{OCR MEI C2 2007 Q9 [12]}}