| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find inverse function |
| Difficulty | Moderate -0.3 This is a straightforward inverse function question requiring standard technique: rearranging y = 1 + 2sin(x) to find x in terms of y, then swapping variables. Finding the domain requires understanding range of the original function. Part (ii) involves identifying key points using symmetry about y=x, which is routine for inverse function questions. Slightly easier than average due to the 'show that' structure and minimal algebraic manipulation required. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.05a Sine, cosine, tangent: definitions for all arguments1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 1 + 2\sin x\), swap \(x \leftrightarrow y\) | M1 | Attempt to invert |
| \(x = 1 + 2\sin y \Rightarrow x - 1 = 2\sin y\) | ||
| \(\Rightarrow (x-1)/2 = \sin y\) | A1 | |
| \(\Rightarrow y = \arcsin\left(\frac{x-1}{2}\right)\) | E1 | |
| Domain is \(-1 \leq x \leq 3\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(A\) is \((\pi/2, 3)\) | B1cao | Allow \(\pi/2 = 1.57\) or better, ft on their A |
| \(B\) is \((1, 0)\) | B1cao | |
| \(C\) is \((3, \pi/2)\) | B1ft |
## Question 4:
### Part (i)
| $y = 1 + 2\sin x$, swap $x \leftrightarrow y$ | M1 | Attempt to invert |
|---|---|---|
| $x = 1 + 2\sin y \Rightarrow x - 1 = 2\sin y$ | | |
| $\Rightarrow (x-1)/2 = \sin y$ | A1 | |
| $\Rightarrow y = \arcsin\left(\frac{x-1}{2}\right)$ | E1 | |
| Domain is $-1 \leq x \leq 3$ | B1 | |
### Part (ii)
| $A$ is $(\pi/2, 3)$ | B1cao | Allow $\pi/2 = 1.57$ or better, ft on their A |
|---|---|---|
| $B$ is $(1, 0)$ | B1cao | |
| $C$ is $(3, \pi/2)$ | B1ft | |
---4 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 $\mathrm { f } ^ { - 1 } ( x ) = \arcsin \left( \frac { x \mathrm { r } } { 2 } \right)$ 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]{d4aa92fb-d21b-4387-b711-b0a6b0d57baa-2_499_562_779_785}
\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 Q4 [7]}}