OCR MEI C3 — Question 4 18 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComposite & Inverse Functions
TypeDetermine domain for composite
DifficultyStandard +0.3 This is a multi-part question requiring standard techniques: finding domain from asymptotes (solving 2x - x² = 0), differentiation using chain rule, identifying turning points, and proving function properties (even functions, translations, symmetry). While it has many parts (7 marks typical), each step uses routine C3 methods without requiring novel insight or complex problem-solving.
Spec1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates
4 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]
\includegraphics[alt={},max width=\textwidth]{2437cecc-f084-4e49-ab36-1c132ba13267-2_652_795_876_717} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure}
  1. Find \(a\). Hence write down the domain of the function.
  2. 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 \(\mathrm { g } ( x ) = \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }\).
  3. (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.
Question 4:
(i) Asymptotes when \(\sqrt{2x - x^2} = 0\)
\(\Rightarrow x(2-x) = 0\)
\(\Rightarrow x = 0\) or \(2\), so \(a = 2\)
AnswerMarks Guidance
Domain is \(0 < x < 2\)M1, A1, B1ft [3] Or by verification; \(x > 0\) and \(x < 2\), not \(\leq\)
(ii) \(y = (2x - x^2)^{-1/2}\), let \(u = 2x - x^2\), \(y = u^{-1/2}\)
\(\Rightarrow \frac{dy}{du} = -\frac{1}{2}u^{-3/2}\), \(\frac{du}{dx} = 2 - 2x\)
\(\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} = -\frac{1}{2}(2x-x^2)^{-3/2}(2-2x)\)
AnswerMarks Guidance
\(= \frac{x-1}{(2x-x^2)^{3/2}}\)M1, B1, A1, E1 [part of 8] Chain rule; \(-\frac{1}{2}u^{-3/2}\); \(\times(2-2x)\); penalise missing brackets
\(\frac{dy}{dx} = 0\) when \(x - 1 = 0 \Rightarrow x = 1\)
\(y = 1/\sqrt{2-1} = 1\)
AnswerMarks Guidance
Range is \(y \geq 1\)M1, A1, B1, B1ft [8] Extraneous solutions M0
(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
(iii)(B) \(g(x-1) = \frac{1}{\sqrt{1-(x-1)^2}} = \frac{1}{\sqrt{1-x^2+2x-1}} = \frac{1}{\sqrt{2x-x^2}} = f(x)\)M1, E1 Must expand bracket
(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* \(f(1-x) = g(-x)\), \(f(1+x) = g(x)\)
\(\Rightarrow f(1+x) = f(1-x)\)
AnswerMarks
\(\Rightarrow f(x)\) is symmetrical about \(x = 1\)M1, E1, A1 [7] 
## Question 4:

**(i)** Asymptotes when $\sqrt{2x - x^2} = 0$

$\Rightarrow x(2-x) = 0$

$\Rightarrow x = 0$ or $2$, so $a = 2$

Domain is $0 < x < 2$ | M1, A1, B1ft [3] | Or by verification; $x > 0$ and $x < 2$, not $\leq$

**(ii)** $y = (2x - x^2)^{-1/2}$, let $u = 2x - x^2$, $y = u^{-1/2}$

$\Rightarrow \frac{dy}{du} = -\frac{1}{2}u^{-3/2}$, $\frac{du}{dx} = 2 - 2x$

$\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx} = -\frac{1}{2}(2x-x^2)^{-3/2}(2-2x)$

$= \frac{x-1}{(2x-x^2)^{3/2}}$ | M1, B1, A1, E1 [part of 8] | Chain rule; $-\frac{1}{2}u^{-3/2}$; $\times(2-2x)$; penalise missing brackets

$\frac{dy}{dx} = 0$ when $x - 1 = 0 \Rightarrow x = 1$

$y = 1/\sqrt{2-1} = 1$

Range is $y \geq 1$ | M1, A1, B1, B1ft [8] | Extraneous solutions M0

**(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

**(iii)(B)** $g(x-1) = \frac{1}{\sqrt{1-(x-1)^2}} = \frac{1}{\sqrt{1-x^2+2x-1}} = \frac{1}{\sqrt{2x-x^2}} = f(x)$ | M1, E1 | Must expand bracket

**(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, E1, A1 [7] |
4 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]{2437cecc-f084-4e49-ab36-1c132ba13267-2_652_795_876_717}
\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  Q4 [18]}}
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