| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Inverse function graphs and properties |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on inverse functions and their properties. Part (i) requires simple substitution into arcsin. Part (ii) involves finding the inverse function (which reverses the operations), differentiating using the chain rule, and applying the reflection property that inverse functions are reflections in y=x. The gradient relationship between a function and its inverse at corresponding points is a standard C3 result. While it requires understanding of inverse trig functions and their derivatives, all steps follow routine procedures without requiring novel insight. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07l Derivative of ln(x): and related functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y=2\arcsin\tfrac{1}{2}=2\times\frac{\pi}{6}\) | M1 | \(y=2\arcsin\tfrac{1}{2}\) |
| \(=\pi/3\) | A1 [2] | must be in terms of \(\pi\) – can isw approximate answers; \(1.047\ldots\) implies M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y=2\arcsin x \;\; x\leftrightarrow y\) | ||
| \(\Rightarrow x=2\arcsin y\) | M1 | or \(y/2=\arcsin x\); but must interchange \(x\) and \(y\) at some stage |
| \(\Rightarrow x/2=\arcsin y\) | A1 | |
| \(\Rightarrow y=\sin(x/2)\) [so \(g(x)=\sin(x/2)\)] | A1cao | |
| \(\Rightarrow dy/dx=\tfrac{1}{2}\cos(\tfrac{1}{2}x)\) | M1 | substituting their \(\pi/3\) into their derivative; must be exact, with \(\cos(\pi/6)\) evaluated |
| At Q, \(x=\pi/3\): \(dy/dx=\tfrac{1}{2}\cos\frac{\pi}{6}=\tfrac{1}{2}\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}\) | A1 | o.e. e.g. \(4\sqrt{3}/3\) but must be exact |
| \(\Rightarrow\) gradient at P \(=4/\sqrt{3}\) | B1ft [6] | ft their \(\sqrt{3}/4\) unless 1; or \(f'(x)=2/\sqrt{(1-x^2)}\), \(f'(\tfrac{1}{2})=2/\sqrt{\tfrac{3}{4}}=4/\sqrt{3}\) cao |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y=2\arcsin\tfrac{1}{2}=2\times\frac{\pi}{6}$ | M1 | $y=2\arcsin\tfrac{1}{2}$ |
| $=\pi/3$ | A1 **[2]** | must be in terms of $\pi$ – can isw approximate answers; $1.047\ldots$ implies M1 |
---
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y=2\arcsin x \;\; x\leftrightarrow y$ | | |
| $\Rightarrow x=2\arcsin y$ | M1 | or $y/2=\arcsin x$; but must interchange $x$ and $y$ at some stage |
| $\Rightarrow x/2=\arcsin y$ | A1 | |
| $\Rightarrow y=\sin(x/2)$ [so $g(x)=\sin(x/2)$] | A1cao | |
| $\Rightarrow dy/dx=\tfrac{1}{2}\cos(\tfrac{1}{2}x)$ | M1 | substituting their $\pi/3$ into their derivative; must be exact, with $\cos(\pi/6)$ evaluated |
| At Q, $x=\pi/3$: $dy/dx=\tfrac{1}{2}\cos\frac{\pi}{6}=\tfrac{1}{2}\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}$ | A1 | o.e. e.g. $4\sqrt{3}/3$ but must be exact |
| $\Rightarrow$ gradient at P $=4/\sqrt{3}$ | B1ft **[6]** | ft their $\sqrt{3}/4$ unless 1; or $f'(x)=2/\sqrt{(1-x^2)}$, $f'(\tfrac{1}{2})=2/\sqrt{\tfrac{3}{4}}=4/\sqrt{3}$ cao |
---6 Fig. 6 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = 2 \arcsin x , - 1 \leqslant x \leqslant 1$.\\
Fig. 6 also shows the curve $y = \mathrm { g } ( x )$, where $\mathrm { g } ( x )$ is the inverse function of $\mathrm { f } ( x )$.\\
P is the point on the curve $y = \mathrm { f } ( x )$ with $x$-coordinate $\frac { 1 } { 2 }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7b77c646-2bc5-4166-b22e-3c1229abd722-3_711_693_466_685}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
(i) Find the $y$-coordinate of P , giving your answer in terms of $\pi$.
The point Q is the reflection of P in $y = x$.\\
(ii) Find $\mathrm { g } ( x )$ and its derivative $\mathrm { g } ^ { \prime } ( x )$. Hence determine the exact gradient of the curve $y = \mathrm { g } ( x )$ at the point Q .
Write down the exact gradient of $y = \mathrm { f } ( x )$ at the point P .
\hfill \mbox{\textit{OCR MEI C3 2012 Q6 [8]}}