OCR MEI AS Paper 2 2023 June — Question 11 5 marks

Exam BoardOCR MEI
ModuleAS Paper 2 (AS Paper 2)
Year2023
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStationary points and optimisation
TypeProve curve has no turning points
DifficultyModerate -0.8 This is a straightforward application of differentiation requiring students to find dy/dx = 6x² + 18x + 24, then show the discriminant is negative (b² - 4ac = 324 - 576 < 0) so no real solutions exist. While it requires showing detailed reasoning, it's a routine two-step process with no conceptual difficulty beyond standard AS-level techniques.
Spec1.07n Stationary points: find maxima, minima using derivatives

11 In this question you must show detailed reasoning.
The equation of a curve is \(y = 2 x ^ { 3 } + 9 x ^ { 2 } + 24 x - 8\).
Show that there are no stationary points on this curve.

Question 11:
AnswerMarks Guidance
\(\frac{dy}{dx} = 6x^2 + 18x + 24\)B1
*their* \(\frac{dy}{dx} = 0\)M1 The \(\frac{dy}{dx}=0\) may be implied by concluding statement e.g. 'no real roots'. Use of discriminant implies this mark.
*their* \(18^2 - 4 \times 6 \times 24\) calculatedM1 Most common quadratics: \(6x^2+18x+24=0\) or \(3x^2+9x+12=0\) or \(2x^2+6x+8=0\) or \(x^2+3x+4=0\). If no formula quoted then solutions must be correct for method mark. May need to check their quadratic. NOTE: sketch method also valid — complete the square to show TP is above \(x\)-axis.
\(-252 < 0\) o.e. for their quadraticA1 \(-252<0\) or \(-63<0\) or \(-28<0\) or \(-7<0\) etc. May be implied by correct solutions e.g. \(\frac{-3\pm\sqrt{7}\,i}{2}\)
Hence \(\frac{dy}{dx}=0\) has no solutions and therefore there are no stationary points on the curveA1\* Must give concluding statement. Depends on all previous marks. Condone SPs or 'turning points' or TPs for stationary points.
## Question 11:

$\frac{dy}{dx} = 6x^2 + 18x + 24$ | **B1** |

*their* $\frac{dy}{dx} = 0$ | **M1** | The $\frac{dy}{dx}=0$ may be implied by concluding statement e.g. 'no real roots'. Use of discriminant implies this mark.

*their* $18^2 - 4 \times 6 \times 24$ calculated | **M1** | Most common quadratics: $6x^2+18x+24=0$ or $3x^2+9x+12=0$ or $2x^2+6x+8=0$ or $x^2+3x+4=0$. If no formula quoted then solutions must be correct for method mark. May need to check their quadratic. NOTE: sketch method also valid — complete the square to show TP is above $x$-axis.

$-252 < 0$ o.e. for their quadratic | **A1** | $-252<0$ or $-63<0$ or $-28<0$ or $-7<0$ etc. May be implied by correct solutions e.g. $\frac{-3\pm\sqrt{7}\,i}{2}$

Hence $\frac{dy}{dx}=0$ has no solutions and therefore there are no stationary points on the curve | **A1\*** | Must give concluding statement. **Depends on all previous marks.** Condone SPs or 'turning points' or TPs for stationary points.

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11 In this question you must show detailed reasoning.\\
The equation of a curve is $y = 2 x ^ { 3 } + 9 x ^ { 2 } + 24 x - 8$.\\
Show that there are no stationary points on this curve.

\hfill \mbox{\textit{OCR MEI AS Paper 2 2023 Q11 [5]}}