| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| 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 C3 inverse function question with standard techniques: finding range from a graph, algebraically finding the inverse of a trigonometric function (requiring rearrangement and arcsin), and applying the derivative relationship between a function and its inverse. All parts are routine applications of core syllabus material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence1.05a Sine, cosine, tangent: definitions for all arguments1.07k Differentiate trig: sin(kx), cos(kx), tan(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | 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 | 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 |
| \(\Rightarrow \sin y = \frac{(1-x)}{2}\) | A1 | o.e. e.g. \(\sin y = \frac{(x-1)}{(-2)}\) (or \(\sin x = \frac{(y-1)}{(-2)}\)) |
| \(\Rightarrow y = \arcsin\left[\frac{(1-x)}{2}\right]\) | A1 | or \(f^{-1}(x) = \arcsin\left[\frac{(1-x)}{2}\right]\), not \(x\) or \(f^{-1}(y) = \arcsin\left[\frac{1-y}{2}\right]\) (viz must have swapped \(x\) and \(y\) for final 'A' mark); \(\arcsin\left[\frac{(x-1)}{-2}\right]\) is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | 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 3:
### Part (i):
| Answer | 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 | 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 |
| $\Rightarrow \sin y = \frac{(1-x)}{2}$ | A1 | o.e. e.g. $\sin y = \frac{(x-1)}{(-2)}$ (or $\sin x = \frac{(y-1)}{(-2)}$) |
| $\Rightarrow y = \arcsin\left[\frac{(1-x)}{2}\right]$ | A1 | or $f^{-1}(x) = \arcsin\left[\frac{(1-x)}{2}\right]$, not $x$ or $f^{-1}(y) = \arcsin\left[\frac{1-y}{2}\right]$ (viz must have swapped $x$ and $y$ for final 'A' mark); $\arcsin\left[\frac{(x-1)}{-2}\right]$ is A0 |
### Part (iii):
| Answer | Mark | 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 $1/-2$ |
---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]{28ce1bcc-e9d5-4ae6-98c0-67b5b8c50bc6-3_732_807_349_612}
\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 2013 Q3 [8]}}