| Exam Board | OCR MEI |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2021 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points and nature |
| Difficulty | Moderate -0.3 This is a straightforward application of the product rule with polynomial factors, followed by standard stationary point analysis. The factorisation is guided by the structure, the second derivative is provided, and all steps follow routine procedures. Slightly easier than average due to the given second derivative and simple polynomial structure. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(2x(x-2)^3 + x^2 \times 3(x-2)^2\) | M1 | AO3.1a |
| A1 | AO1.1 | |
| \((x-2)\) identified as factor | M1 | AO2.1 |
| \(x(x-2)^2(5x-4)\) cao | A1 | AO1.1 |
| Answer | Marks |
|---|---|
| \(5x^4 - 24x^3 + 36x^2 - 16x\) | M1 |
| A1 | |
| \((x-2)\) identified as factor | M1 |
| \(x(x-2)^2(5x-4)\) cao | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Their \(\frac{dy}{dx} = 0\) soi | M1 | AO2.1 |
| \(x = 0, 2, \frac{4}{5}\) | A1 | AO1.1 |
| \((0,0)\) and \((2,0)\) | A1 | AO1.1 |
| \(\left(0.8, -1.10592\right)\) or \(\left(0.8, -\frac{3456}{3125}\right)\) | A1 | AO2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| 2nd derivative \(= -16\) at \((0,0)\) so max | B1 | AO1.1 |
| 2nd derivative \(= 5.76\) at \((0.8, -1.10592)\) so min | B1 | AO1.1 |
| e.g. gradient \(= 1\) at \(x=1\) and \(33\) at \(x=3\) so inflection at \(x=2\); e.g. \(y=-1\) at \(x=1\) and \(y=9\) at \(x=3\) so inflection at \(x=2\); NB 2nd derivative test is indecisive at \(x=2\) | B1 | AO3.1a |
| Answer | Marks | Guidance |
|---|---|---|
| Shape of curve correct with max, min and inflection | M1 | AO1.1 |
| Correct intercepts marked on sketch or identified next to graph | A1 | AO1.1 |
# Question 14:
## Part (a):
$2x(x-2)^3 + x^2 \times 3(x-2)^2$ | **M1** | AO3.1a | Product rule & chain rule; allow one error
| **A1** | AO1.1 |
$(x-2)$ identified as factor | **M1** | AO2.1 |
$x(x-2)^2(5x-4)$ cao | **A1** | AO1.1 |
**[4 marks]**
*Alternative:* From $x^5 - 6x^4 + 12x^3 - 8x^2$, expand brackets and differentiate; allow one error
$5x^4 - 24x^3 + 36x^2 - 16x$ | **M1** | |
| **A1** |
$(x-2)$ identified as factor | **M1** |
$x(x-2)^2(5x-4)$ cao | **A1** |
**[4 marks]**
## Part (b):
Their $\frac{dy}{dx} = 0$ soi | **M1** | AO2.1 |
$x = 0, 2, \frac{4}{5}$ | **A1** | AO1.1 |
$(0,0)$ and $(2,0)$ | **A1** | AO1.1 |
$\left(0.8, -1.10592\right)$ or $\left(0.8, -\frac{3456}{3125}\right)$ | **A1** | AO2.2a | Accept $-1.10592$ to 2 sf or better
**[4 marks]**
## Part (c):
2nd derivative $= -16$ at $(0,0)$ so max | **B1** | AO1.1 | Or for any of the three points: considers $y$ or $\frac{dy}{dx}$ or $\frac{d^2y}{dx^2}$ either side of correct stationary point accompanied by suitable commentary; must see numerical values
2nd derivative $= 5.76$ at $(0.8, -1.10592)$ so min | **B1** | AO1.1 |
e.g. gradient $= 1$ at $x=1$ and $33$ at $x=3$ so inflection at $x=2$; e.g. $y=-1$ at $x=1$ and $y=9$ at $x=3$ so inflection at $x=2$; NB 2nd derivative test is indecisive at $x=2$ | **B1** | AO3.1a |
**[3 marks]**
## Part (d):
Shape of curve correct with max, min and inflection | **M1** | AO1.1 |
Correct intercepts marked on sketch or identified next to graph | **A1** | AO1.1 |
**[2 marks]**
---14 The equation of a curve is\\
$y = x ^ { 2 } ( x - 2 ) ^ { 3 }$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { dy } } { \mathrm { dx } }$, giving your answer in factorised form.
\item Determine the coordinates of the stationary points on the curve.
In part (c) you may use the result $\frac { d ^ { 2 } y } { d x ^ { 2 } } = 4 ( x - 2 ) \left( 5 x ^ { 2 } - 8 x + 2 \right)$.
\item Determine the nature of the stationary points on the curve.
\item Sketch the curve.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 2 2021 Q14 [13]}}