| 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 | Standard +0.3 This is a straightforward inverse function question involving arctan. Part (i) requires knowledge of arctan range, part (ii) is standard algebraic manipulation to find the inverse (tan of both sides), and part (iii) uses the derivative relationship between inverse functions. All techniques are routine for C3 level, 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 existence1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(-\frac{\pi}{2} < \arctan x < \frac{\pi}{2}\) | M1 | \(\frac{\pi}{4}\) or \(-\frac{\pi}{4}\) or 45 seen |
| \(\Rightarrow -\frac{\pi}{4} < f(x) < \frac{\pi}{4}\) | A1cao [2] | not \(\leq\) |
| Range is \(-\frac{\pi}{4}\) to \(\frac{\pi}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = \frac{1}{2}\arctan x\), swap \(x \leftrightarrow y\) | M1 | \(\tan(\arctan y \text{ or } x) = y\) or \(x\) |
| \(x = \frac{1}{2}\arctan y \Rightarrow 2x = \arctan y\) | ||
| \(\Rightarrow \tan 2x = y \Rightarrow y = \tan 2x\) | A1cao | |
| Either \(\frac{dy}{dx} = 2\sec^2 2x\) | M1 A1cao | derivative of tan is \(\sec^2\) used |
| Or \(y = \frac{\sin 2x}{\cos 2x} \Rightarrow \frac{dy}{dx} = \frac{2\cos^2 2x + 2\sin^2 2x}{\cos^2 2x}\) | M1 | quotient rule |
| \(= \frac{2}{\cos^2 2x}\) | A1cao | need not be simplified but mark final answer |
| When \(x = 0\), \(\frac{dy}{dx} = 2\) | B1 [5] | www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Gradient of \(y = \frac{1}{2}\arctan x\) is \(\frac{1}{2}\) | B1ft [1] | ft their '2', but not 1 or 0 or \(\infty\) |
## Question 2:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $-\frac{\pi}{2} < \arctan x < \frac{\pi}{2}$ | M1 | $\frac{\pi}{4}$ or $-\frac{\pi}{4}$ or 45 seen |
| $\Rightarrow -\frac{\pi}{4} < f(x) < \frac{\pi}{4}$ | A1cao [2] | not $\leq$ |
| Range is $-\frac{\pi}{4}$ to $\frac{\pi}{4}$ | | |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = \frac{1}{2}\arctan x$, swap $x \leftrightarrow y$ | M1 | $\tan(\arctan y \text{ or } x) = y$ or $x$ |
| $x = \frac{1}{2}\arctan y \Rightarrow 2x = \arctan y$ | | |
| $\Rightarrow \tan 2x = y \Rightarrow y = \tan 2x$ | A1cao | |
| Either $\frac{dy}{dx} = 2\sec^2 2x$ | M1 A1cao | derivative of tan is $\sec^2$ used |
| Or $y = \frac{\sin 2x}{\cos 2x} \Rightarrow \frac{dy}{dx} = \frac{2\cos^2 2x + 2\sin^2 2x}{\cos^2 2x}$ | M1 | quotient rule |
| $= \frac{2}{\cos^2 2x}$ | A1cao | need not be simplified but mark final answer |
| When $x = 0$, $\frac{dy}{dx} = 2$ | B1 [5] | www |
### Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Gradient of $y = \frac{1}{2}\arctan x$ is $\frac{1}{2}$ | B1ft [1] | ft their '2', but not 1 or 0 or $\infty$ |
---2 Fig. 6 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = \frac { 1 } { 2 } \arctan x$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6555136d-0444-41f6-9063-21960352089d-2_388_727_434_701}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
(i) Find the range of the function $\mathrm { f } ( x )$, giving your answer in terms of $\pi$.\\
(ii) Find the inverse function $\mathrm { f } ^ { - 1 } ( x )$. Find the gradient of the curve $y = \mathrm { f } ^ { - 1 } ( x )$ at the origin.\\
(iii) Hence write down the gradient of $y = \frac { 1 } { 2 } \arctan x$ at the origin.
\hfill \mbox{\textit{OCR MEI C3 Q2 [8]}}