| 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: reading range from a graph, rearranging y = 1 - 2sin(x) to find the inverse (requiring arcsin), and using the derivative relationship f'(a) = 1/[f'^(-1)(f(a))]. All parts are routine C3 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.05a Sine, cosine, tangent: definitions for all arguments1.07b Gradient as rate of change: dy/dx notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Range is \(-1 \leq y \leq 3\) | M1 | \(-1, 3\) |
| 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 |
|---|---|---|
| Answer/Working | Mark | 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 |
| \(\sin y = \frac{1-x}{2}\) | A1 | e.g. \(\sin y = \frac{x-1}{-2}\) (or \(\sin x = \frac{y-1}{-2}\)) |
| \(y = \arcsin\left[\frac{1-x}{2}\right]\) | A1 | for \(f^{-1}(x) = \arcsin\left[\frac{1-x}{2}\right]\), not \(x\) or \(f^{-1}(y) = \arcsin\left[\frac{1-y}{2}\right]\); \(\arcsin\left[\frac{x-1}{-2}\right]\) is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f'(x) = -2\cos x\) | M1 | condone \(2\cos x\) |
| \(f'(0) = -2\) | A1 | cao |
| gradient of \(y = f^{-1}(x)\) at \((1,0) = -\frac{1}{2}\) | A1 | not \(\frac{1}{-2}\) |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Range is $-1 \leq y \leq 3$ | M1 | $-1, 3$ |
| | 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):
| Answer/Working | Mark | 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 |
| $\sin y = \frac{1-x}{2}$ | A1 | e.g. $\sin y = \frac{x-1}{-2}$ (or $\sin x = \frac{y-1}{-2}$) |
| $y = \arcsin\left[\frac{1-x}{2}\right]$ | A1 | for $f^{-1}(x) = \arcsin\left[\frac{1-x}{2}\right]$, not $x$ or $f^{-1}(y) = \arcsin\left[\frac{1-y}{2}\right]$; $\arcsin\left[\frac{x-1}{-2}\right]$ is A0 |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f'(x) = -2\cos x$ | M1 | condone $2\cos x$ |
| $f'(0) = -2$ | A1 | cao |
| gradient of $y = f^{-1}(x)$ at $(1,0) = -\frac{1}{2}$ | A1 | not $\frac{1}{-2}$ |
---3 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]{7825ba53-67eb-4050-a671-85e37a30150a-3_743_818_414_644}
\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 Q3 [8]}}