| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Moderate -0.8 This is a straightforward inverse function question requiring standard algebraic manipulation (swap x and y, rearrange) and understanding that domain of f becomes range of f^(-1). Part (ii) involves reading key points from a graph using the reflection property y=x. Both parts are routine C3 exercises with no problem-solving insight required. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (i) \(y = 1+2\sin x \leftrightarrow x\): swap to get \(x = 1+2\sin y\) | M1 | Attempt to invert |
| \(\frac{x-1}{2} = \sin y \Rightarrow y = \arcsin\!\left(\frac{x-1}{2}\right)\) * | A1, E1 | |
| Domain is \(-1 \le x \le 3\) | B1 | |
| (ii) A is \(\left(\frac{\pi}{2}, 3\right)\) | B1cao | Allow \(\pi/2 = 1.57\) or better |
| B is \((1, 0)\) | B1cao | |
| C is \(\left(3, \frac{\pi}{2}\right)\) | B1ft | ft on their A |
## Question 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(i)** $y = 1+2\sin x \leftrightarrow x$: swap to get $x = 1+2\sin y$ | M1 | Attempt to invert |
| $\frac{x-1}{2} = \sin y \Rightarrow y = \arcsin\!\left(\frac{x-1}{2}\right)$ * | A1, E1 | |
| Domain is $-1 \le x \le 3$ | B1 | |
| **(ii)** A is $\left(\frac{\pi}{2}, 3\right)$ | B1cao | Allow $\pi/2 = 1.57$ or better |
| B is $(1, 0)$ | B1cao | |
| C is $\left(3, \frac{\pi}{2}\right)$ | B1ft | ft on their A |
---
# Section B6 The function $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = 1 + 2 \sin x$ for $- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi$.\\
(i) Show that $\mathrm { f } ^ { - 1 } ( x ) = \arcsin \left( \frac { x - 1 } { 2 } \right)$ and state the domain of this function.
Fig. 6 shows a sketch of the graphs of $y = \mathrm { f } ( x )$ and $y = \mathrm { f } ^ { - 1 } ( x )$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6a4c3f3b-a298-4b13-b97e-b52f8d9d527b-4_506_561_705_751}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
(ii) Write down the coordinates of the points $\mathrm { A } , \mathrm { B }$ and C .
\hfill \mbox{\textit{OCR MEI C3 2006 Q6 [7]}}