| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Reciprocal Trig & Identities |
| Type | Inverse function graphs and properties |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on inverse functions requiring standard techniques: finding range from a graph, finding an inverse function algebraically (tan of both sides), and using the inverse function gradient relationship. All parts are routine applications of C3 content with no novel problem-solving required, 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 graphs1.07a Derivative as gradient: of tangent to curve |
| Answer | Marks | Guidance |
|---|---|---|
| \(-\pi/2 < \arctan x < \pi/2 \Rightarrow -\pi/4 < f(x) < \pi/4\) | M1, A1cao [2] | \(\pi/4\) or \(-\pi/4\) or \(45\) seen; not \(\leq\) |
| Range is \(-\pi/4\) to \(\pi/4\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \frac{1}{2}\arctan x \;\; x\leftrightarrow y\), \(x = \frac{1}{2}\arctan y \Rightarrow 2x = \arctan y\) | M1 | \(\tan(\arctan y \text{ or } x) = y\) or \(x\) |
| \(\Rightarrow \tan 2x = y \Rightarrow y = \tan 2x\) | A1cao | |
| Either \(\dfrac{dy}{dx} = 2\sec^2 2x\) | M1, A1cao | Derivative of \(\tan\) is \(\sec^2\) used |
| Or \(y = \dfrac{\sin 2x}{\cos 2x} \Rightarrow \dfrac{dy}{dx} = \dfrac{2\cos^2 2x + 2\sin^2 2x}{\cos^2 2x} = \dfrac{2}{\cos^2 2x}\) | M1, A1cao | Quotient rule; need not be simplified |
| When \(x=0\), \(dy/dx = 2\) | B1 [5] | www |
| Answer | Marks | Guidance |
|---|---|---|
| So 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 8:
**(i)**
| $-\pi/2 < \arctan x < \pi/2 \Rightarrow -\pi/4 < f(x) < \pi/4$ | M1, A1cao [2] | $\pi/4$ or $-\pi/4$ or $45$ seen; not $\leq$ |
|---|---|---|
| Range is $-\pi/4$ to $\pi/4$ | | |
**(ii)**
| $y = \frac{1}{2}\arctan x \;\; x\leftrightarrow y$, $x = \frac{1}{2}\arctan y \Rightarrow 2x = \arctan y$ | M1 | $\tan(\arctan y \text{ or } x) = y$ or $x$ |
|---|---|---|
| $\Rightarrow \tan 2x = y \Rightarrow y = \tan 2x$ | A1cao | |
| Either $\dfrac{dy}{dx} = 2\sec^2 2x$ | M1, A1cao | Derivative of $\tan$ is $\sec^2$ used |
| Or $y = \dfrac{\sin 2x}{\cos 2x} \Rightarrow \dfrac{dy}{dx} = \dfrac{2\cos^2 2x + 2\sin^2 2x}{\cos^2 2x} = \dfrac{2}{\cos^2 2x}$ | M1, A1cao | Quotient rule; need not be simplified |
| When $x=0$, $dy/dx = 2$ | B1 [5] | www |
**(iii)**
| So gradient of $y = \frac{1}{2}\arctan x$ is $\frac{1}{2}$ | B1ft [1] | ft their '2', but not 1 or 0 or $\infty$ |
|---|---|---|
---8 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]{11877196-83d9-4283-9eef-e617bea50c63-4_379_722_467_715}
\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 Q8 [8]}}