| 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 inverse function question requiring standard techniques: evaluating arcsin at a simple value (π/6), finding the inverse function g(x) = sin(x/2), differentiating it, and applying the reflection property that gradients at corresponding points are reciprocals. All steps are routine C3 material with no novel problem-solving required. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07b Gradient as rate of change: dy/dx notation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 2\arcsin\frac{1}{2} = 2 \times \frac{\pi}{6}\) | M1 | \(y = 2\arcsin\frac{1}{2}\); must be in terms of \(\pi\) – can isw approximate answers |
| \(= \frac{\pi}{3}\) | A1 | \(1.047\ldots\) implies M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 2\arcsin x\), \(x \leftrightarrow y\): \(x = 2\arcsin y\) | ||
| \(\frac{x}{2} = \arcsin y\) | M1 | or \(\frac{y}{2} = \arcsin x\); but must interchange \(x\) and \(y\) at some stage |
| \(y = \sin\left(\frac{x}{2}\right)\) [so \(g(x) = \sin\left(\frac{x}{2}\right)\)] | A1cao | |
| \(\frac{dy}{dx} = \frac{1}{2}\cos\left(\frac{x}{2}\right)\) | A1cao | |
| At Q, \(x = \frac{\pi}{3}\) | M1 | substituting their \(\frac{\pi}{3}\) into their derivative; must be exact, with \(\cos\left(\frac{\pi}{6}\right)\) evaluated |
| \(\frac{dy}{dx} = \frac{1}{2}\cos\frac{\pi}{6} = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}\) | A1 | e.g. \(\frac{4\sqrt{3}}{3}\) but must be exact; ft their \(\frac{\sqrt{3}}{4}\) unless 1 |
| gradient at P \(= \frac{4}{\sqrt{3}}\) | B1ft | or \(f'(x) = \frac{2}{\sqrt{1-x^2}}\), \(f'\left(\frac{1}{2}\right) = \frac{2}{\sqrt{3}/4} = \frac{4}{\sqrt{3}}\) cao |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 2\arcsin\frac{1}{2} = 2 \times \frac{\pi}{6}$ | M1 | $y = 2\arcsin\frac{1}{2}$; must be in terms of $\pi$ – can isw approximate answers |
| $= \frac{\pi}{3}$ | A1 | $1.047\ldots$ implies M1 |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 2\arcsin x$, $x \leftrightarrow y$: $x = 2\arcsin y$ | | |
| $\frac{x}{2} = \arcsin y$ | M1 | or $\frac{y}{2} = \arcsin x$; but must interchange $x$ and $y$ at some stage |
| $y = \sin\left(\frac{x}{2}\right)$ [so $g(x) = \sin\left(\frac{x}{2}\right)$] | A1cao | |
| $\frac{dy}{dx} = \frac{1}{2}\cos\left(\frac{x}{2}\right)$ | A1cao | |
| At Q, $x = \frac{\pi}{3}$ | M1 | substituting their $\frac{\pi}{3}$ into their derivative; must be exact, with $\cos\left(\frac{\pi}{6}\right)$ evaluated |
| $\frac{dy}{dx} = \frac{1}{2}\cos\frac{\pi}{6} = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}$ | A1 | e.g. $\frac{4\sqrt{3}}{3}$ but must be exact; ft their $\frac{\sqrt{3}}{4}$ unless 1 |
| gradient at P $= \frac{4}{\sqrt{3}}$ | B1ft | or $f'(x) = \frac{2}{\sqrt{1-x^2}}$, $f'\left(\frac{1}{2}\right) = \frac{2}{\sqrt{3}/4} = \frac{4}{\sqrt{3}}$ cao |
---4 Fig. 6 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = 2 \arcsin x , - 1 \nless \leqslant 1$.\\
Fig. 6 also shows the curve $y = \mathrm { g } ( x )$, where $\mathrm { g } ( x )$ is the inverse function of $\mathrm { f } ( x )$.\\
P is the point on the curve $y = \mathrm { f } ( x )$ with $x$-coordinate $\frac { 1 } { 2 }$.
\begin{tikzpicture}[>=stealth, scale=0.85]
% --- Axes ---
\draw[->] (-3.5,0) -- (3.5,0) node[right] {$x$};
\draw[->] (0,-4) -- (0,4.5) node[above] {$y$};
% --- Tick marks ---
\draw (1,0.1) -- (1,-0.1) node[below] {$1$};
\draw (-1,0.1) -- (-1,-0.1) node[below] {$-1$};
\draw (0.1,3.1416) -- (-0.1,3.1416) node[left] {$\pi$};
\draw (0.1,-3.1416) -- (-0.1,-3.1416) node[left] {$-\pi$};
% --- y = f(x): S-curve with asymptotes at +-pi ---
\draw[thick, domain=-.999:.999, samples=200]
plot ({\x}, {2*asin(\x)*pi/180});
\node[above left] at (1.8, 3.1416) {$y = \mathrm{f}(x)$};
% --- y = g(x): dotted curve through origin, gentle slope ---
\draw[thick, dotted, domain=-3:3.0, samples=150]
plot ({\x}, {sin(deg(\x/2))});
\node[right] at (3.0, {sin(deg(1.5))}) {$y = \mathrm{g}(x)$};
% --- Point P on f(x) ---
\filldraw (0.7, {2*asin(0.7)*pi/180}) circle (2pt);
\node[above left] at (0.7, {2*asin(0.7)*pi/180}) {P};
% --- Point Q on x-axis below P ---
\filldraw ({2*asin(0.7)*pi/180}, 0.7) circle (2pt);
\node[below right] at ({2*asin(0.7)*pi/180}, 0.7) {Q};
\end{tikzpicture}
(i) Find the $y$-coordinate of P , giving your answer in terms of $\pi$.
The point Q is the reflection of P in $y = x$.\\
(ii) Find $\mathrm { g } ( x )$ and its derivative $\mathrm { g } ^ { \prime } ( x )$. Hence determine the exact gradient of the curve $y = \mathrm { g } ( x )$ at the point Q .
Write down the exact gradient of $y = \mathrm { f } ( x )$ at the point P .
\hfill \mbox{\textit{OCR MEI C3 Q4 [8]}}