| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Make x the subject |
| Difficulty | Moderate -0.3 This is a straightforward inverse function question requiring rearrangement of arcsin, differentiation using the chain rule reciprocal, and substitution. The techniques are standard C3 material with clear steps, making it slightly easier than average but not trivial due to the inverse trigonometric context. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(x - 1 = \sin y\) | M1 A1 | |
| \(\Rightarrow x = 1 + \sin y\) | E1 | www |
| \(\Rightarrow \frac{dv}{dy} = \cos y\) |
| Answer | Marks | Guidance |
|---|---|---|
| When \(x = 1.5, y = \arcsin(0.5) = \frac{\pi}{6}\) | M1 A1 | condone \(30°\) or \(0.52\) or better |
| \(\frac{dy}{dx} = \frac{1}{\cos y}\) | M1 | or \(\frac{dy}{dx} = \frac{1}{\sqrt{1-(x-1)^2}}\) |
| \(= \frac{1}{\cos \pi/6}\) | A1 | or equivalent, but must be exact |
| \(= \frac{2}{\sqrt{3}}\) | [7] |
**(i)**
$x - 1 = \sin y$ | M1 A1 |
$\Rightarrow x = 1 + \sin y$ | E1 | www
$\Rightarrow \frac{dv}{dy} = \cos y$ |
**(ii)**
When $x = 1.5, y = \arcsin(0.5) = \frac{\pi}{6}$ | M1 A1 | condone $30°$ or $0.52$ or better
$\frac{dy}{dx} = \frac{1}{\cos y}$ | M1 | or $\frac{dy}{dx} = \frac{1}{\sqrt{1-(x-1)^2}}$
$= \frac{1}{\cos \pi/6}$ | A1 | or equivalent, but must be exact
$= \frac{2}{\sqrt{3}}$ | [7] |3 Fig. 3 shows the curve defined by the equation $y = \arcsin ( x - 1 )$, for $0 \leqslant x \leqslant 2$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{72de4365-c3f0-4106-8daf-3047afeab723-2_684_551_829_758}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
(i) Find $x$ in terms of $y$, and show that $\frac { \mathrm { d } x } { \mathrm {~d} y } = \cos y$.\\
(ii) Hence find the exact gradient of the curve at the point where $x = 1.5$.
\hfill \mbox{\textit{OCR MEI C3 2006 Q3 [7]}}