OCR MEI C2 2006 January — Question 10 12 marks

Exam BoardOCR MEI
ModuleC2 (Core Mathematics 2)
Year2006
SessionJanuary
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStationary points and optimisation
TypeFind stationary point then sketch curve
DifficultyModerate -0.8 This is a straightforward multi-part question testing routine differentiation and integration techniques. Part (i) requires finding dy/dx, setting it to zero, and finding axis intercepts—all standard procedures. Part (ii) is direct polynomial integration. Part (iii) uses a simple area subtraction method. All steps are textbook exercises with no problem-solving insight required, making it easier than average for A-level.
Spec1.02n Sketch curves: simple equations including polynomials1.07n Stationary points: find maxima, minima using derivatives1.08d Evaluate definite integrals: between limits1.08f Area between two curves: using integration

10 The equation of a curve is \(y = 7 + 6 x - x ^ { 2 }\).
  1. Use calculus to find the coordinates of the turning point on this curve. Find also the coordinates of the points of intersection of this curve with the axes, and sketch the curve.
  2. Find \(\int _ { 1 } ^ { 5 } \left( 7 + 6 x - x ^ { 2 } \right) \mathrm { d } x\), showing your working.
  3. The curve and the line \(y = 12\) intersect at ( 1,12 ) and ( 5,12 ). Using your answer to part (ii), find the area of the finite region between the curve and the line \(y = 12\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{15b8f97b-c058-409f-907f-cb0a6102abc4-5_643_1034_331_513} \captionsetup{labelformat=empty} \caption{Fig. 11}
    \end{figure} The equation of the curve shown in Fig. 11 is \(y = x ^ { 3 } - 6 x + 2\).

AnswerMarks Guidance
(i) \(y' = 6 - 2x\)M1 condone one error
\(y' = 0\) usedM1
\(x = 3\)A1
\(y = 16\)A1
\((0, 7)\) \((-1, 0)\) and \((7,0)\) found or marked on graph3 marks 1 each
sketch of correct shape1 mark must reach pos. y - axis
(ii) 58.6 to 58.73 marks B1 for \(7x + 3x^2 - \frac{x^3}{3}\) [their value at 5] – [their value at 1] dependent on integration attempted; M1
(iii) using his (ii) and 481 mark 1 mark total
(i) $y' = 6 - 2x$ | M1 | condone one error
$y' = 0$ used | M1 |
$x = 3$ | A1 |
$y = 16$ | A1 |

$(0, 7)$ $(-1, 0)$ and $(7,0)$ found or marked on graph | 3 marks | 1 each

sketch of correct shape | 1 mark | must reach pos. y - axis | 8 marks total

(ii) 58.6 to 58.7 | 3 marks | B1 for $7x + 3x^2 - \frac{x^3}{3}$ [their value at 5] – [their value at 1] dependent on integration attempted; M1 |

(iii) using his (ii) and 48 | 1 mark | 1 mark total
10 The equation of a curve is $y = 7 + 6 x - x ^ { 2 }$.\\
(i) Use calculus to find the coordinates of the turning point on this curve.

Find also the coordinates of the points of intersection of this curve with the axes, and sketch the curve.\\
(ii) Find $\int _ { 1 } ^ { 5 } \left( 7 + 6 x - x ^ { 2 } \right) \mathrm { d } x$, showing your working.\\
(iii) The curve and the line $y = 12$ intersect at ( 1,12 ) and ( 5,12 ). Using your answer to part (ii), find the area of the finite region between the curve and the line $y = 12$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{15b8f97b-c058-409f-907f-cb0a6102abc4-5_643_1034_331_513}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{figure}

The equation of the curve shown in Fig. 11 is $y = x ^ { 3 } - 6 x + 2$.\\

\hfill \mbox{\textit{OCR MEI C2 2006 Q10 [12]}}