Question 8 - A-Level Maths

OCR C1 Specimen — Question 8 14 marks

Exam Board	OCR
Module	C1 (Core Mathematics 1)
Session	Specimen
Marks	14
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 C1 stationary points question with routine differentiation, second derivative test, and curve sketching. While it has multiple parts (4 marks worth), each step follows textbook procedures with no problem-solving required. The algebraic verification in part (iii) is straightforward expansion. Slightly easier than average due to the mechanical nature of all parts.
Spec	1.02j Manipulate polynomials: expanding, factorising, division, factor theorem 1.02n Sketch curves: simple equations including polynomials 1.07n Stationary points: find maxima, minima using derivatives 1.07o Increasing/decreasing: functions using sign of dy/dx

Find the coordinates of the stationary points on the curve $y = 2 x ^ { 3 } - 3 x ^ { 2 } - 12 x - 7$.
Determine whether each stationary point is a maximum point or a minimum point.
By expanding the right-hand side, show that $$2 x ^ { 3 } - 3 x ^ { 2 } - 12 x - 7 = ( x + 1 ) ^ { 2 } ( 2 x - 7 )$$
Sketch the curve $y = 2 x ^ { 3 } - 3 x ^ { 2 } - 12 x - 7$, marking the coordinates of the stationary points and the points where the curve meets the axes.

Show mark scheme Show mark scheme source

Answer	Marks	Guidance
(i) $\frac{dy}{dx} = 6x^2 - 6x - 12$. Hence $x^2 - x - 2 = 0$, $(x-2)(x+1) = 0 \Rightarrow x = 2$ or $-1$. Stationary points are $(2, -27)$ and $(-1, 0)$	M1 A1 M1 M1 A1 A1	For differentiation with at least 1 term OK; For completely correct derivative; For equating their derivative to zero; For factorising or other solution method; For both correct $x$-coordinates; For both correct $y$-coordinates
6
(ii) $\frac{d^2y}{dx^2} = 12x - 6 = \begin{cases}+18 \text{ when } x = 2 \\ -18 \text{ when } x = -1\end{cases}$. Hence $(2, -27)$ is a min and $(-1, 0)$ is a max	M1 A1 A1	For attempt at second derivative and at least one relevant evaluation; For either one correctly identified; For both correctly identified (Alternative methods, e.g. based on gradients either side, are equally acceptable)
3
(iii) RHS $= (x^2 + 2x + 1)(2x - 7) = 2x^3 - 7x^2 + 4x^2 - 14x + 2x - 7 = 2x^3 - 3x^2 - 12x - 7$, as required	M1 A1	For squaring correctly and attempting complete expansion process; For obtaining given answer correctly
2
(iv) [Cubic sketch with correct features: cubic shape; maximum point lying on $x$-axis; $x = \frac{1}{2}$ and $y = -7$ at intersections]	B1 B1 B1	For correct cubic shape; For maximum point lying on $x$-axis; For $x = \frac{1}{2}$ and $y = -7$ at intersections
3

(i) $\frac{dy}{dx} = 6x^2 - 6x - 12$. Hence $x^2 - x - 2 = 0$, $(x-2)(x+1) = 0 \Rightarrow x = 2$ or $-1$. Stationary points are $(2, -27)$ and $(-1, 0)$ | M1 A1 M1 M1 A1 A1 | For differentiation with at least 1 term OK; For completely correct derivative; For equating their derivative to zero; For factorising or other solution method; For both correct $x$-coordinates; For both correct $y$-coordinates

| | 6 |

(ii) $\frac{d^2y}{dx^2} = 12x - 6 = \begin{cases}+18 \text{ when } x = 2 \\ -18 \text{ when } x = -1\end{cases}$. Hence $(2, -27)$ is a min and $(-1, 0)$ is a max | M1 A1 A1 | For attempt at second derivative and at least one relevant evaluation; For either one correctly identified; For both correctly identified (Alternative methods, e.g. based on gradients either side, are equally acceptable)

| | 3 |

(iii) RHS $= (x^2 + 2x + 1)(2x - 7) = 2x^3 - 7x^2 + 4x^2 - 14x + 2x - 7 = 2x^3 - 3x^2 - 12x - 7$, as required | M1 A1 | For squaring correctly and attempting complete expansion process; For obtaining given answer correctly

| | 2 |

(iv) [Cubic sketch with correct features: cubic shape; maximum point lying on $x$-axis; $x = \frac{1}{2}$ and $y = -7$ at intersections] | B1 B1 B1 | For correct cubic shape; For maximum point lying on $x$-axis; For $x = \frac{1}{2}$ and $y = -7$ at intersections

| | 3 |

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8 (i) Find the coordinates of the stationary points on the curve $y = 2 x ^ { 3 } - 3 x ^ { 2 } - 12 x - 7$.\\
(ii) Determine whether each stationary point is a maximum point or a minimum point.\\
(iii) By expanding the right-hand side, show that

$$2 x ^ { 3 } - 3 x ^ { 2 } - 12 x - 7 = ( x + 1 ) ^ { 2 } ( 2 x - 7 )$$

(iv) Sketch the curve $y = 2 x ^ { 3 } - 3 x ^ { 2 } - 12 x - 7$, marking the coordinates of the stationary points and the points where the curve meets the axes.

\hfill \mbox{\textit{OCR C1  Q8 [14]}}

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Q1 4 Q2 5 Q3 6 Q4 7 Q5 11 Q6 12 Q7 13 Q8 14

Answer	Marks	Guidance
(i) \(\frac{dy}{dx} = 6x^2 - 6x - 12\). Hence \(x^2 - x - 2 = 0\), \((x-2)(x+1) = 0 \Rightarrow x = 2\) or \(-1\). Stationary points are \((2, -27)\) and \((-1, 0)\)	M1 A1 M1 M1 A1 A1	For differentiation with at least 1 term OK; For completely correct derivative; For equating their derivative to zero; For factorising or other solution method; For both correct \(x\)-coordinates; For both correct \(y\)-coordinates
6
(ii) \(\frac{d^2y}{dx^2} = 12x - 6 = \begin{cases}+18 \text{ when } x = 2 \\ -18 \text{ when } x = -1\end{cases}\). Hence \((2, -27)\) is a min and \((-1, 0)\) is a max	M1 A1 A1	For attempt at second derivative and at least one relevant evaluation; For either one correctly identified; For both correctly identified (Alternative methods, e.g. based on gradients either side, are equally acceptable)
3
(iii) RHS \(= (x^2 + 2x + 1)(2x - 7) = 2x^3 - 7x^2 + 4x^2 - 14x + 2x - 7 = 2x^3 - 3x^2 - 12x - 7\), as required	M1 A1	For squaring correctly and attempting complete expansion process; For obtaining given answer correctly
2
(iv) [Cubic sketch with correct features: cubic shape; maximum point lying on \(x\)-axis; \(x = \frac{1}{2}\) and \(y = -7\) at intersections]	B1 B1 B1	For correct cubic shape; For maximum point lying on \(x\)-axis; For \(x = \frac{1}{2}\) and \(y = -7\) at intersections
3