| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Determine domain for composite |
| Difficulty | Standard +0.3 This is a multi-part question covering standard C3 topics (domain, differentiation using chain rule, even functions, and function transformations). While it has several parts, each step is routine: finding domain from a square root constraint, applying quotient/chain rule differentiation, verifying even function property, and recognizing horizontal translation. The algebraic manipulation is straightforward and the question guides students through the reasoning with clear prompts. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02w Graph transformations: simple transformations of f(x)1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| so \(x = 2\) | M1, A1, B1ft | [3] or by verification; \(x > 0\) and \(x < 2\), not \(\le\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = (2x - x^2)^{-1/2}\) | M1, B1 | chain rule (or within correct quotient rule); \(-\frac{1}{2}u^{-3/2}\) or \(-\frac{1}{2}(2x - x^2)^{-3/2}\) or \(\frac{1}{2}(2x - x^2)^{-1/2}\) in quotient rule |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\frac{1}{2}(2x - x^2)^{-3/2} \cdot (2 - 2x)\) | A1 | \(\times (2 - 2x)\) |
| \(= \frac{x-1}{(2x - x^2)^{3/2}}\) * | E1 | [8] www – penalise missing brackets here |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow x = 1\), \(y = 1/\sqrt{(2 - 1)} = 1\) | M1, A1, B1 | extraneous solutions M0 |
| Range is \(y \ge 1\) | B1ft | [8] |
| Answer | Marks | Guidance |
|---|---|---|
| \(g(-x) = \frac{1}{\sqrt{1 - (-x)^2}} = \frac{1}{\sqrt{1 - x^2}} = g(x)\) | M1, E1 | Expression for \(g(-x)\) – must have \(g(-x) = g(x)\) seen |
| Answer | Marks | Guidance |
|---|---|---|
| \(g(x - 1) = \frac{1}{\sqrt{1 - (x-1)^2}}\) | M1 | |
| \(= \frac{1}{\sqrt{1 - x^2 + 2x - 1}} = \frac{1}{\sqrt{2x - x^2}} = f(x)\) | E1 | must expand bracket |
| Answer | Marks | Guidance |
|---|---|---|
| So \(f(x)\) is symmetrical about \(x = 1\). | M1, M1, A1 | dep both M1s |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow f(x)\) is symmetrical about \(x = 1\). | M1 | or \(f(1-x) = \frac{1}{\sqrt{2-2x-(1-x)^2}} = \frac{1}{\sqrt{2-2x-1+2x-x^2}} = \frac{1}{\sqrt{1-x^2}}\); \(f(1+x) = \frac{1}{\sqrt{2+2x-(1+x)^2}} = \frac{1}{\sqrt{2+2x-1-2x-x^2}} = \frac{1}{\sqrt{1-x^2}}\) |
## Part (i)
Asymptotes when $\sqrt{y}(2x - x^2) = 0$
$\Rightarrow x(2 - x) = 0$
$\Rightarrow x = 0$ or $2$
so $x = 2$ | M1, A1, B1ft | [3] or by verification; $x > 0$ and $x < 2$, not $\le$
Domain is $0 < x < 2$
## Part (ii)
$y = (2x - x^2)^{-1/2}$ | M1, B1 | chain rule (or within correct quotient rule); $-\frac{1}{2}u^{-3/2}$ or $-\frac{1}{2}(2x - x^2)^{-3/2}$ or $\frac{1}{2}(2x - x^2)^{-1/2}$ in quotient rule
Let $u = 2x - x^2$, $y = u^{-1/2}$
$\Rightarrow \frac{dy}{du} = -\frac{1}{2}u^{-3/2}$, $\frac{du}{dx} = 2 - 2x$
$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = -\frac{1}{2}(2x - x^2)^{-3/2} \cdot (2 - 2x)$ | A1 | $\times (2 - 2x)$
$= \frac{x-1}{(2x - x^2)^{3/2}}$ * | E1 | [8] www – penalise missing brackets here
$\frac{dy}{dx} = 0$ when $x - 1 = 0$
$\Rightarrow x = 1$, $y = 1/\sqrt{(2 - 1)} = 1$ | M1, A1, B1 | extraneous solutions M0
Range is $y \ge 1$ | B1ft | [8]
## Part (iii) (A)
$g(-x) = \frac{1}{\sqrt{1 - (-x)^2}} = \frac{1}{\sqrt{1 - x^2}} = g(x)$ | M1, E1 | Expression for $g(-x)$ – must have $g(-x) = g(x)$ seen
## Part (iii) (B)
$g(x - 1) = \frac{1}{\sqrt{1 - (x-1)^2}}$ | M1 |
$= \frac{1}{\sqrt{1 - x^2 + 2x - 1}} = \frac{1}{\sqrt{2x - x^2}} = f(x)$ | E1 | must expand bracket
## Part (iii) (C)
$f(x)$ is $g(x)$ translated 1 unit to the right. But $g(x)$ is symmetrical about Oy
So $f(x)$ is symmetrical about $x = 1$. | M1, M1, A1 | dep both M1s
or $f(1 - x) = g(-x)$, $f(1 + x) = g(x)$
$\Rightarrow f(1 + x) = f(1 - x)$
$\Rightarrow f(x)$ is symmetrical about $x = 1$. | M1 | or $f(1-x) = \frac{1}{\sqrt{2-2x-(1-x)^2}} = \frac{1}{\sqrt{2-2x-1+2x-x^2}} = \frac{1}{\sqrt{1-x^2}}$; $f(1+x) = \frac{1}{\sqrt{2+2x-(1+x)^2}} = \frac{1}{\sqrt{2+2x-1-2x-x^2}} = \frac{1}{\sqrt{1-x^2}}$ | E1, A1 | [7]9 Fig. 9 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = \frac { 1 } { \sqrt { 2 x - x ^ { 2 } } }$.\\
The curve has asymptotes $x = 0$ and $x = a$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{56672660-b7dc-4e10-8039-1c041e75b598-4_655_800_431_669}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Find $a$. Hence write down the domain of the function.
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x - 1 } { \left( 2 x - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }$.
Hence find the coordinates of the turning point of the curve, and write down the range of the function.
The function $\mathrm { g } ( x )$ is defined by $\mathrm { g } ( x ) = \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }$.
\item (A) Show algebraically that $\mathrm { g } ( x )$ is an even function.\\
(B) Show that $\mathrm { g } ( x - 1 ) = \mathrm { f } ( x )$.\\
(C) Hence prove that the curve $y = \mathrm { f } ( x )$ is symmetrical, and state its line of symmetry.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 2009 Q9 [18]}}