| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 8 |
| 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 with standard techniques: identifying range from a given graph, finding inverse of a trigonometric function (arcsin), and using the derivative relationship between a function and its inverse. All parts are routine C3 material requiring recall and direct application rather than problem-solving insight. |
| 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 graphs1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Range is \(-1 \leq y \leq 3\) | M1, A1 | \(-1 \leq y \leq 3\) or \(-1 \leq f(x) \leq 3\) or \([-1, 3]\); (not \(-1\) to \(3\), \(-1 \leq x \leq 3\), \(-1 < y < 3\) etc) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 1 - 2\sin x\), \(x \leftrightarrow y\) | Can interchange \(x\) and \(y\) at any stage | |
| \(x = 1 - 2\sin y \Rightarrow x - 1 = -2\sin y\) | M1 | Attempt to re-arrange |
| \(\Rightarrow \sin y = \frac{1-x}{2}\) | A1 | |
| \(\Rightarrow y = \arcsin\left[\frac{1-x}{2}\right]\) | A1 | or \(f^{-1}(x) = \arcsin\left[\frac{1-x}{2}\right]\); \(\arcsin\left[\frac{x-1}{-2}\right]\) is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(f'(x) = -2\cos x\) | M1 | Condone \(2\cos x\) |
| \(\Rightarrow f'(0) = -2\) | A1 | cao |
| \(\Rightarrow\) gradient of \(y = f^{-1}(x)\) at \((1, 0) = -\frac{1}{2}\) | A1 | Not \(1/-2\) |
## Question 8:
### Part (i)
| Range is $-1 \leq y \leq 3$ | M1, A1 | $-1 \leq y \leq 3$ or $-1 \leq f(x) \leq 3$ or $[-1, 3]$; (not $-1$ to $3$, $-1 \leq x \leq 3$, $-1 < y < 3$ etc) |
|---|---|---|
### Part (ii)
| $y = 1 - 2\sin x$, $x \leftrightarrow y$ | | Can interchange $x$ and $y$ at any stage |
|---|---|---|
| $x = 1 - 2\sin y \Rightarrow x - 1 = -2\sin y$ | M1 | Attempt to re-arrange |
| $\Rightarrow \sin y = \frac{1-x}{2}$ | A1 | |
| $\Rightarrow y = \arcsin\left[\frac{1-x}{2}\right]$ | A1 | or $f^{-1}(x) = \arcsin\left[\frac{1-x}{2}\right]$; $\arcsin\left[\frac{x-1}{-2}\right]$ is A0 |
### Part (iii)
| $f'(x) = -2\cos x$ | M1 | Condone $2\cos x$ |
|---|---|---|
| $\Rightarrow f'(0) = -2$ | A1 | cao |
| $\Rightarrow$ gradient of $y = f^{-1}(x)$ at $(1, 0) = -\frac{1}{2}$ | A1 | Not $1/-2$ |8 The function $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = 1 - 2 \sin x$ for $- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi$. Fig. 3 shows the curve $y = \mathrm { f } ( x )$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d4aa92fb-d21b-4387-b711-b0a6b0d57baa-4_736_809_419_653}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
(i) Write down the range of the function $\mathrm { f } ( x )$.\\
(ii) Find the inverse function $\mathrm { f } ^ { - 1 } ( x )$.\\
(iii) Find $\mathrm { f } ^ { \prime } ( 0 )$. Hence write down the gradient of $y = \mathrm { f } ^ { - 1 } ( x )$ at the point $( 1,0 )$.
\hfill \mbox{\textit{OCR MEI C3 Q8 [8]}}