| 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 involving arctan. Part (i) requires knowing the range of arctan (standard knowledge), and part (ii) involves algebraic rearrangement to find the inverse. The sketching component is routine reflection in y=x. Slightly easier than average due to the direct nature of the inverse calculation. |
| 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 |
|---|---|---|
| Bounds \(-\pi+1\), \(\pi+1 \Rightarrow -\pi+1 < f(x) < \pi+1\) | B1 B1 B1cao [3] | Or \(\ldots < y < \ldots\) or \((-\pi+1,\, \pi+1)\); not \(\ldots < x < \ldots\), not 'between...' |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 2\arctan x + 1 \;\;x \leftrightarrow y\), \(x = 2\arctan y + 1\) | M1 | Attempt to invert formula; one step is enough |
| \(\Rightarrow \dfrac{x-1}{2} = \arctan y\) | A1 | Or \(\dfrac{y-1}{2} = \arctan x\); need not have interchanged \(x\) and \(y\) at this stage |
| \(\Rightarrow y = \tan\!\left(\dfrac{x-1}{2}\right) \Rightarrow f^{-1}(x) = \tan\!\left(\dfrac{x-1}{2}\right)\) | A1 | Allow \(y = \ldots\) |
| Reasonable reflection in \(y = x\) | B1 | Curves must cross on \(y=x\) line; curves shouldn't touch or cross in third quadrant |
| \((1, 0)\) intercept indicated | B1 [5] |
## Question 5:
**(i)**
| Bounds $-\pi+1$, $\pi+1 \Rightarrow -\pi+1 < f(x) < \pi+1$ | B1 B1 B1cao [3] | Or $\ldots < y < \ldots$ or $(-\pi+1,\, \pi+1)$; not $\ldots < x < \ldots$, not 'between...' |
|---|---|---|
**(ii)**
| $y = 2\arctan x + 1 \;\;x \leftrightarrow y$, $x = 2\arctan y + 1$ | M1 | Attempt to invert formula; one step is enough |
|---|---|---|
| $\Rightarrow \dfrac{x-1}{2} = \arctan y$ | A1 | Or $\dfrac{y-1}{2} = \arctan x$; need not have interchanged $x$ and $y$ at this stage |
| $\Rightarrow y = \tan\!\left(\dfrac{x-1}{2}\right) \Rightarrow f^{-1}(x) = \tan\!\left(\dfrac{x-1}{2}\right)$ | A1 | Allow $y = \ldots$ |
| Reasonable reflection in $y = x$ | B1 | Curves must cross on $y=x$ line; curves shouldn't touch or cross in third quadrant |
| $(1, 0)$ intercept indicated | B1 [5] | |
---5 Fig. 7 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = 1 + 2 \arctan x , x \in \mathbb { R }$. The scales on the $x$ - and $y$-axes are the same.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{11877196-83d9-4283-9eef-e617bea50c63-2_855_838_1028_688}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
(i) Find the range of f , giving your answer in terms of $\pi$.\\
(ii) Find $\mathrm { f } ^ { - 1 } ( x )$, and add a sketch of the curve $y = \mathrm { f } ^ { - 1 } ( x )$ to the copy of Fig. 7.
\hfill \mbox{\textit{OCR MEI C3 Q5 [8]}}