| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Curve with parametric or implicit features |
| Difficulty | Challenging +1.2 This question involves algebraic manipulation of an implicit equation (squaring to eliminate square roots), curve sketching with domain restrictions, symmetry recognition, and area calculation using integration. While it requires multiple techniques and careful attention to domains, the individual steps are standard C3 material with no particularly novel insights needed. The multi-part structure and need to track transformations elevates it slightly above average difficulty. |
| Spec | 1.02b Surds: manipulation and rationalising denominators1.02n Sketch curves: simple equations including polynomials1.02w Graph transformations: simple transformations of f(x)1.07s Parametric and implicit differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\sqrt{y} = 1 - \sqrt{x}\) | M1 | Or: squaring introduces also the negative arm |
| \(\Rightarrow y = (1-\sqrt{x})^2 = 1 + x - 2\sqrt{x}\) | A1, A1 | |
| Undefined for \(x < 0\) but defined for \(x > 1\) so that part is missing | B1, B1 | \(x<0\), \(x>0\) |
| Sketch of correct shape with correct range | B1, B1 | Extra range, shape |
| Total: 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| A is \(\sqrt{-x}+\sqrt{y}=1\) for \(-1 \le x \le 0\) | B1, B1 | |
| B is \(\sqrt{-x}+\sqrt{-y}=1\) for \(-1 \le x \le 0\) | B1, B1 | |
| C is \(\sqrt{x}+\sqrt{-y}=1\) for \(0 \le x \le 1\) | B1, B1 | |
| Total: 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Area in Fig. 9.1 is \(\int_0^1\left(1+x-2\sqrt{x}\right)dx\) | M1 | |
| \(= \left[x + \frac{1}{2}x^2 - \frac{4}{3}x^{\frac{3}{2}}\right]_0^1 = 1 + \frac{1}{2} - \frac{4}{3} = \frac{1}{6}\) | A1, A1 | |
| \(\Rightarrow\) Area of shape is \(\frac{1}{6}+\frac{1}{6}+2\times\frac{1}{6}+2\times\frac{1}{6} = 1\) | M1, A1 | |
| Total: 5 |
## Question 9:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\sqrt{y} = 1 - \sqrt{x}$ | M1 | Or: squaring introduces also the negative arm |
| $\Rightarrow y = (1-\sqrt{x})^2 = 1 + x - 2\sqrt{x}$ | A1, A1 | |
| Undefined for $x < 0$ but defined for $x > 1$ so that part is missing | B1, B1 | $x<0$, $x>0$ |
| Sketch of correct shape with correct range | B1, B1 | Extra range, shape |
| **Total: 7** | | |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| A is $\sqrt{-x}+\sqrt{y}=1$ for $-1 \le x \le 0$ | B1, B1 | |
| B is $\sqrt{-x}+\sqrt{-y}=1$ for $-1 \le x \le 0$ | B1, B1 | |
| C is $\sqrt{x}+\sqrt{-y}=1$ for $0 \le x \le 1$ | B1, B1 | |
| **Total: 6** | | |
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Area in Fig. 9.1 is $\int_0^1\left(1+x-2\sqrt{x}\right)dx$ | M1 | |
| $= \left[x + \frac{1}{2}x^2 - \frac{4}{3}x^{\frac{3}{2}}\right]_0^1 = 1 + \frac{1}{2} - \frac{4}{3} = \frac{1}{6}$ | A1, A1 | |
| $\Rightarrow$ Area of shape is $\frac{1}{6}+\frac{1}{6}+2\times\frac{1}{6}+2\times\frac{1}{6} = 1$ | M1, A1 | |
| **Total: 5** | | |9 The curve in Fig. 9.1 has equation $\sqrt { x } + \sqrt { y } = 1$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2f403099-2813-40d8-a9ae-1f7e64d41f80-4_426_647_299_667}
\captionsetup{labelformat=empty}
\caption{Fig. 9.1}
\end{center}
\end{figure}
(i) Show that this is part, but not all of the curve $y = 1 - 2 \sqrt { x } + x$.
Sketch the full curve $y = 1 - 2 \sqrt { x } + x$.\\
(ii) Fig.9.2 shows a star shape made up of four parts, one of which is given in part (i) above.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2f403099-2813-40d8-a9ae-1f7e64d41f80-4_380_681_1197_651}
\captionsetup{labelformat=empty}
\caption{Fig. 9.2}
\end{center}
\end{figure}
For each of the sections of the shape labelled $\mathrm { A } , \mathrm { B }$ and C , state the equation of the curve and the domain.\\
(iii) The shape shown in Fig.9.2 is made into that in Fig. 10.3 by stretching the part of the figure for which $y > 0$ by a scale factor of 2 .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2f403099-2813-40d8-a9ae-1f7e64d41f80-4_405_686_1996_605}
\captionsetup{labelformat=empty}
\caption{Fig. 9.3}
\end{center}
\end{figure}
Find the area of this shape.
\hfill \mbox{\textit{OCR MEI C3 Q9 [18]}}