| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Inverse function graphs and properties |
| Difficulty | Moderate -0.3 This question tests standard inverse function properties with arctan, requiring knowledge of its range, finding the inverse (which is just tan(2x)), and using the reciprocal gradient relationship. All parts are routine applications of C3 inverse function theory with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs |
| Answer | Marks | Guidance |
|---|---|---|
| Range: \(-\frac{\pi}{4} < f(x) < \frac{\pi}{4}\) | B1 | Correct bounds |
| — | B1 | [2] Correct strict inequalities in terms of \(\pi\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \frac{1}{2}\arctan x \Rightarrow x = \tan 2y\) | M1 | Attempt to find inverse |
| \(f^{-1}(x) = \tan 2x\) | A1 | Correct inverse function |
| Gradient of \(y=f^{-1}(x)\) at origin: \(\frac{d}{dx}(\tan 2x) = 2\sec^2 2x\) | M1 | Differentiating |
| At \(x=0\): gradient \(= 2\) | A1 | Correct gradient |
| — | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient of \(y = \frac{1}{2}\arctan x\) at origin \(= \frac{1}{2}\) | B1 | [1] Reciprocal of 2 |
# Question 6:
**Part (i):**
| Range: $-\frac{\pi}{4} < f(x) < \frac{\pi}{4}$ | B1 | Correct bounds |
| — | B1 | [2] Correct strict inequalities in terms of $\pi$ |
**Part (ii):**
| $y = \frac{1}{2}\arctan x \Rightarrow x = \tan 2y$ | M1 | Attempt to find inverse |
| $f^{-1}(x) = \tan 2x$ | A1 | Correct inverse function |
| Gradient of $y=f^{-1}(x)$ at origin: $\frac{d}{dx}(\tan 2x) = 2\sec^2 2x$ | M1 | Differentiating |
| At $x=0$: gradient $= 2$ | A1 | Correct gradient |
| — | A1 | [5] |
**Part (iii):**
| Gradient of $y = \frac{1}{2}\arctan x$ at origin $= \frac{1}{2}$ | B1 | [1] Reciprocal of 2 |
---6 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]{0ee3d87a-0d9e-4fa5-b8f5-8b28489e65b5-3_378_725_367_669}
\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 2007 Q6 [8]}}