| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Applied context: real-world solid |
| Difficulty | Moderate -0.3 This is a straightforward application question requiring basic calculus techniques: finding a maximum by completing the square or differentiation, explaining and computing a simple definite integral (polynomial), and applying the trapezium rule with given ordinates. All steps are routine C2-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals1.09f Trapezium rule: numerical integration |
| Answer | Marks | Guidance |
|---|---|---|
| i A \(6.25\) | B2 | M1 for \(x = 5\) used to find \(y\) |
| 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| [val at \(x = 10\)] – [val at \(x = 0\)] | E1, M1, M1, A2 | Subs of correct limits into their integrand. A1 for \(166.6...\) or \(16666.6...\) or \(41.6...\)rot to 3 sf or more |
| \(4166\) to \(4167\) or \(4170\) | 5 marks | |
| ii \(52.62\) | B4 | M3 for- \(2/2×[2.15x2+2(5.64x2+6.44x2)]\)oe. Or M2 if one slip. Or M1 if 2 slips or one trap evaluated |
| Their\((5262)\) – their \((4167)\) | M1 | Must be \(>0\) |
| 5 marks | ||
| [12] |
**i A** $6.25$ | B2 | M1 for $x = 5$ used to find $y$
| | 2 marks |
**i B** (V =) area of cross-section × length
$\left(\frac{100}{4}\right)\left(\frac{10}{2}x^2 - \frac{1}{3}x^3\right)$ o.e.
[val at $x = 10$] – [val at $x = 0$] | E1, M1, M1, A2 | Subs of correct limits into their integrand. A1 for $166.6...$ or $16666.6...$ or $41.6...$rot to 3 sf or more
$4166$ to $4167$ or $4170$ | 5 marks |
**ii** $52.62$ | B4 | M3 for- $2/2×[2.15x2+2(5.64x2+6.44x2)]$oe. Or M2 if one slip. Or M1 if 2 slips or one trap evaluated
Their$(5262)$ – their $(4167)$ | M1 | Must be $>0$
| | 5 marks |
| | [12] |9
\begin{enumerate}[label=(\roman*)]
\item A tunnel is 100 m long. Its cross-section, shown in Fig. 9.1, is modelled by the curve
$$y = \frac { 1 } { 4 } \left( 10 x - x ^ { 2 } \right) ,$$
where $x$ and $y$ are horizontal and vertical distances in metres.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{72b4624f-e716-4a37-96f3-01b46e0bd0fd-5_506_812_676_653}
\captionsetup{labelformat=empty}
\caption{Figure 9.1}
\end{center}
\end{figure}
Using this model,\\
(A) find the greatest height of the tunnel,\\
(B) explain why $100 \int _ { 0 } ^ { 10 } y \mathrm {~d} x$ gives the volume, in cubic metres, of earth removed to make the tunnel. Calculate this volume.
\item The roof of the tunnel is re-shaped to allow for larger vehicles. Fig. 9.2 shows the new crosssection.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{72b4624f-e716-4a37-96f3-01b46e0bd0fd-5_513_1256_1894_575}
\captionsetup{labelformat=empty}
\caption{Fig. 9.2}
\end{center}
\end{figure}
Use the trapezium rule with 5 strips to estimate the new cross-sectional area.\\
Hence estimate the volume of earth removed when the tunnel is re-shaped.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 2005 Q9 [12]}}