| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Find stationary points of standard polynomial |
| Difficulty | Moderate -0.3 This is a standard C2 curve sketching question covering routine techniques: expanding brackets, finding stationary points via differentiation, and sketching a cubic. Part (iv) adds mild challenge by requiring interpretation of the sketch to determine when a horizontal line intersects once, but overall this is slightly easier than average due to the straightforward algebraic manipulation and the 'show that' scaffold in part (ii). |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Multiply out to get \(x^3 - 3x + 2\) | M1, A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = x^3 - 3x + 2 \Rightarrow \frac{dy}{dx} = 3x^2 - 3\) | M1, A1 | |
| \(\frac{dy}{dx} = 0\) when \(x = \pm1\) | M1 | |
| \(\frac{d^2y}{dx^2} = 6x\) | M1 | |
| When \(x = -1\), \(\frac{d^2y}{dx^2} < 0\) so maximum | E1 | |
| When \(x = 1\), \(\frac{d^2y}{dx^2} > 0\) so minimum | M1, B1 | |
| When \(x = 1\), \(y = 0\) | [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correct sketch of cubic curve | B1 | [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| For values of \(k\) within max and min values of \(y\) there are three roots | M1, B1 | |
| When \(x = -1\), \(y = 4\) | A1 | |
| i.e. \(k < 0\) and \(k > 4\) | [3] |
## Question 9:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Multiply out to get $x^3 - 3x + 2$ | M1, A1 | **[2]** |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = x^3 - 3x + 2 \Rightarrow \frac{dy}{dx} = 3x^2 - 3$ | M1, A1 | |
| $\frac{dy}{dx} = 0$ when $x = \pm1$ | M1 | |
| $\frac{d^2y}{dx^2} = 6x$ | M1 | |
| When $x = -1$, $\frac{d^2y}{dx^2} < 0$ so maximum | E1 | |
| When $x = 1$, $\frac{d^2y}{dx^2} > 0$ so minimum | M1, B1 | |
| When $x = 1$, $y = 0$ | | **[7]** |
### Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct sketch of cubic curve | B1 | **[1]** |
### Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| For values of $k$ within max and min values of $y$ there are three roots | M1, B1 | |
| When $x = -1$, $y = 4$ | A1 | |
| i.e. $k < 0$ and $k > 4$ | | **[3]** |
---9 The equation of a curve is given by $y = ( x - 1 ) ^ { 2 } ( x + 2 )$.\\
(i) Write $( x - 1 ) ^ { 2 } ( x + 2 )$ in the form $x ^ { 3 } + p x ^ { 2 } + q x + r$ where $p , q$ and $r$ are to be determined.\\
(ii) Show that the curve $y = ( x - 1 ) ^ { 2 } ( x + 2 )$ has a maximum point when $x = - 1$ and find the coordinates of the minimum point.\\
(iii) Sketch the curve $y = ( x - 1 ) ^ { 2 } ( x + 2 )$.\\
(iv) For what values of $k$ does $( x - 1 ) ^ { 2 } ( x + 2 ) = k$ have exactly one root.
\hfill \mbox{\textit{OCR MEI C2 Q9 [13]}}