| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Quadratic modelling problems |
| Difficulty | Moderate -0.3 This is a straightforward applied quadratic problem requiring basic skills: finding x-intercepts, using symmetry to find the vertex, substituting values, and solving a simple quadratic equation. All techniques are standard C1 material with clear scaffolding across multiple parts, making it slightly easier than average despite the multi-step nature. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.02z Models in context: use functions in modelling1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(10\) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([x =] 5\) or ft their (i) \(\div 2\) | 1 | Not necessarily ft from (i); e.g. they may start again with calculus to get \(x = 5\) |
| \(\text{ht} = 5[\text{m}]\) cao | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(d = \frac{7}{2}\) o.e. | M1 | Or ft their (ii) \(- 1.5\) or their (i) \(\div 2 - 1.5\) o.e. |
| \([y =] \frac{1}{5} \times 3.5 \times (10 - 3.5)\) o.e. or ft | M1 | Or \(7 - \frac{1}{5} \times 3.5^2\) or ft |
| \(= \frac{91}{20}\) o.e. cao isw | A1 | Or showing \(y - 4 = \frac{11}{20}\) o.e. cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4.5 = \frac{1}{5} \times x(10 - x)\) o.e. | M1 | |
| \(22.5 = x(10 - x)\) o.e. | M1 | e.g. \(4.5 = x(2 - 0.2x)\) etc |
| \(2x^2 - 20x + 45 [= 0]\) o.e. e.g. \(x^2 - 10x + 22.5 [= 0]\) or \((x-5)^2 = 2.5\) | A1 | cao; accept versions with fractional coefficients of \(x^2\), isw |
| \([x =] \dfrac{20 \pm \sqrt{40}}{4}\) or \(5 \pm \dfrac{1}{2}\sqrt{10}\) o.e. | M1 | Or \(x - 5 = [\pm]\sqrt{2.5}\) o.e.; ft their quadratic equation provided at least M1 gained already; condone one error in formula or substitution; need not be simplified or be real |
| Width \(= \sqrt{10}\) o.e. e.g. \(2\sqrt{2.5}\) cao | A1 | Accept simple equivalents only |
## Question 12(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $10$ | 1 | |
## Question 12(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[x =] 5$ or ft their (i) $\div 2$ | 1 | Not necessarily ft from (i); e.g. they may start again with calculus to get $x = 5$ |
| $\text{ht} = 5[\text{m}]$ cao | 1 | |
## Question 12(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $d = \frac{7}{2}$ o.e. | M1 | Or ft their (ii) $- 1.5$ or their (i) $\div 2 - 1.5$ o.e. |
| $[y =] \frac{1}{5} \times 3.5 \times (10 - 3.5)$ o.e. or ft | M1 | Or $7 - \frac{1}{5} \times 3.5^2$ or ft |
| $= \frac{91}{20}$ o.e. cao isw | A1 | Or showing $y - 4 = \frac{11}{20}$ o.e. cao |
## Question 12(iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $4.5 = \frac{1}{5} \times x(10 - x)$ o.e. | M1 | |
| $22.5 = x(10 - x)$ o.e. | M1 | e.g. $4.5 = x(2 - 0.2x)$ etc |
| $2x^2 - 20x + 45 [= 0]$ o.e. e.g. $x^2 - 10x + 22.5 [= 0]$ or $(x-5)^2 = 2.5$ | A1 | cao; accept versions with fractional coefficients of $x^2$, isw |
| $[x =] \dfrac{20 \pm \sqrt{40}}{4}$ or $5 \pm \dfrac{1}{2}\sqrt{10}$ o.e. | M1 | Or $x - 5 = [\pm]\sqrt{2.5}$ o.e.; ft their quadratic equation provided at least M1 gained already; condone one error in formula or substitution; need not be simplified or be real |
| Width $= \sqrt{10}$ o.e. e.g. $2\sqrt{2.5}$ cao | A1 | Accept simple equivalents only |12 The curve with equation $y = \frac { 1 } { 5 } x ( 10 - x )$ is used to model the arch of a bridge over a road, where $x$ and $y$ are distances in metres, with the origin as shown in Fig. 12.1. The $x$-axis represents the road surface.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ede57eaa-2645-49df-aa09-68b6d5f35a9a-4_524_885_406_628}
\captionsetup{labelformat=empty}
\caption{Fig. 12.1}
\end{center}
\end{figure}
(i) State the value of $x$ at A , where the arch meets the road.\\
(ii) Using symmetry, or otherwise, state the value of $x$ at the maximum point B of the graph.
Hence find the height of the arch.\\
(iii) Fig. 12.2 shows a lorry which is 4 m high and 3 m wide, with its cross-section modelled as a rectangle. Find the value of $d$ when the lorry is in the centre of the road. Hence show that the lorry can pass through this arch.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ede57eaa-2645-49df-aa09-68b6d5f35a9a-4_529_871_1489_678}
\captionsetup{labelformat=empty}
\caption{Fig. 12.2}
\end{center}
\end{figure}
(iv) Another lorry, also modelled as having a rectangular cross-section, has height 4.5 m and just touches the arch when it is in the centre of the road. Find the width of this lorry, giving your answer in surd form.\\[0pt]
[5]
\hfill \mbox{\textit{OCR MEI C1 2010 Q12 [11]}}