| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Curve with parametric or implicit features |
| Difficulty | Standard +0.3 This is a structured multi-part question covering standard C3 topics (circle equations, differentiation of surds, transformations). Part (i) is algebraic manipulation, part (ii) combines geometric and calculus approaches to tangents (routine but good practice), parts (iii-iv) are standard transformations and algebraic verification. All parts follow predictable patterns with clear scaffolding, making it slightly easier than average. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = \sqrt{4-x^2}\), squaring: \(y^2 = 4 - x^2\) | M1 | squaring |
| \(\Rightarrow x^2 + y^2 = 4\) | A1 | \(x^2 + y^2 = 4\) + comment (correct) |
| Square root does not give negative values, so this is only a semi-circle | B1 | oe, e.g. f is a function and therefore single valued |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| (A) Grad of \(OP = \frac{b}{a}\), grad of tangent \(= -\frac{a}{b}\) | M1 A1 | |
| (B) \(f'(x) = \frac{1}{2}(4-x^2)^{-1/2} \cdot (-2x)\) | M1 | chain rule or implicit differentiation |
| \(= -\frac{x}{\sqrt{4-x^2}}\) | A1 | oe |
| \(\Rightarrow f'(a) = -\frac{a}{\sqrt{4-a^2}}\) | B1 | substituting \(a\) into \(f'(x)\) |
| (C) \(b = \sqrt{4-a^2}\), so \(f'(a) = -\frac{a}{b}\) as before | E1 | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Translation through \(\begin{pmatrix}2\\0\end{pmatrix}\) | M1 | Translation in \(x\)-direction |
| A1 | through \(\begin{pmatrix}2\\0\end{pmatrix}\) or 2 to right ('shift', 'move' M1 A0) | |
| \(\begin{pmatrix}2\\0\end{pmatrix}\) alone is SC1 | ||
| followed by stretch scale factor 3 in \(y\)-direction | M1 | stretch in \(y\)-direction (condone \(y\) 'axis') |
| A1 | (scale) factor 3 | |
| M1 | elliptical (or circular) shape through \((0,0)\) and \((4,0)\) | |
| A1 | and \((2,6)\) (soi); \(-1\) if whole ellipse shown | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = 3f(x-2) = 3\sqrt{4-(x-2)^2}\) | M1 | or substituting \(3\sqrt{4-(x-2)^2}\) oe for \(y\) in \(9x^2+y^2\) |
| \(= 3\sqrt{4-x^2+4x-4} = 3\sqrt{4x-x^2}\) | A1 | \(4x - x^2\) |
| \(y^2 = 9(4x-x^2)\) | ||
| \(\Rightarrow 9x^2 + y^2 = 36x\) * | E1 | www |
| [3] |
# Question 9:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = \sqrt{4-x^2}$, squaring: $y^2 = 4 - x^2$ | M1 | squaring |
| $\Rightarrow x^2 + y^2 = 4$ | A1 | $x^2 + y^2 = 4$ + comment (correct) |
| Square root does not give negative values, so this is only a semi-circle | B1 | oe, e.g. f is a function and therefore single valued |
| | [3] | |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| **(A)** Grad of $OP = \frac{b}{a}$, grad of tangent $= -\frac{a}{b}$ | M1 A1 | |
| **(B)** $f'(x) = \frac{1}{2}(4-x^2)^{-1/2} \cdot (-2x)$ | M1 | chain rule or implicit differentiation |
| $= -\frac{x}{\sqrt{4-x^2}}$ | A1 | oe |
| $\Rightarrow f'(a) = -\frac{a}{\sqrt{4-a^2}}$ | B1 | substituting $a$ into $f'(x)$ |
| **(C)** $b = \sqrt{4-a^2}$, so $f'(a) = -\frac{a}{b}$ as before | E1 | |
| | [6] | |
## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Translation through $\begin{pmatrix}2\\0\end{pmatrix}$ | M1 | Translation in $x$-direction |
| | A1 | through $\begin{pmatrix}2\\0\end{pmatrix}$ or 2 to right ('shift', 'move' M1 A0) |
| | | $\begin{pmatrix}2\\0\end{pmatrix}$ alone is SC1 |
| followed by stretch scale factor 3 in $y$-direction | M1 | stretch in $y$-direction (condone $y$ 'axis') |
| | A1 | (scale) factor 3 |
| | M1 | elliptical (or circular) shape through $(0,0)$ and $(4,0)$ |
| | A1 | and $(2,6)$ (soi); $-1$ if whole ellipse shown |
| | [6] | |
## Part (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = 3f(x-2) = 3\sqrt{4-(x-2)^2}$ | M1 | or substituting $3\sqrt{4-(x-2)^2}$ oe for $y$ in $9x^2+y^2$ |
| $= 3\sqrt{4-x^2+4x-4} = 3\sqrt{4x-x^2}$ | A1 | $4x - x^2$ |
| $y^2 = 9(4x-x^2)$ | | |
| $\Rightarrow 9x^2 + y^2 = 36x$ * | E1 | www |
| | [3] | |9 The function $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = \sqrt { 4 - x ^ { 2 } }$ for $- 2 \leqslant x \leqslant 2$.
\begin{enumerate}[label=(\roman*)]
\item Show that the curve $y = \sqrt { 4 - x ^ { 2 } }$ is a semicircle of radius 2 , and explain why it is not the whole of this circle.
Fig. 9 shows a point $\mathrm { P } ( a , b )$ on the semicircle. The tangent at P is shown.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8feffafd-4eba-4968-b4d2-88fa364d6170-4_625_933_589_607}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
\item (A) Use the gradient of OP to find the gradient of the tangent at P in terms of $a$ and $b$.\\
(B) Differentiate $\sqrt { 4 - x ^ { 2 } }$ and deduce the value of $\mathrm { f } ^ { \prime } ( a )$.\\
(C) Show that your answers to parts ( $A$ ) and ( $B$ ) are equivalent.
The function $\mathrm { g } ( x )$ is defined by $\mathrm { g } ( x ) = 3 \mathrm { f } ( x - 2 )$, for $0 \leqslant x \leqslant 4$.
\item Describe a sequence of two transformations that would map the curve $y = \mathrm { f } ( x )$ onto the curve $y = \mathrm { g } ( x )$.
Hence sketch the curve $y = \mathrm { g } ( x )$.
\item Show that if $y = \mathrm { g } ( x )$ then $9 x ^ { 2 } + y ^ { 2 } = 36 x$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 2008 Q9 [18]}}