| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Complex Numbers Argand & Loci |
| Type | Argument calculations and identities |
| Difficulty | Standard +0.8 This question requires sketching inverse trig functions (moderate), then solving a system involving inverse trig functions by using the identity sin(arcsin x) = x and cos(arccos x) = x, followed by manipulating tan b = 1/2 to find the exact value of a using right-angled triangle relationships and surds. The multi-step reasoning connecting inverse functions, trigonometric identities, and exact value manipulation elevates this above routine C3 exercises. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 |
| Answer | Marks |
|---|---|
| Sketch showing \(y = \cos^{-1}(2x)\) and \(y = \sin^{-1}x\) intersecting at origin | B3 |
| Answer | Marks |
|---|---|
| \(b = \sin^{-1} a \Rightarrow a = \sin b\) | M1 |
| \(b = \cos^{-1} 2a \Rightarrow 2a = \cos b\) | M1 |
| \(\therefore 2 \sin b = \cos b\) | |
| \(\frac{\sin b}{\cos b} = \frac{1}{2}\) | |
| \(\tan b = \frac{1}{2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan^2 b = \frac{1}{4}\) | ||
| \(\sec^2 b = 1 + \frac{1}{4} = \frac{5}{4}\) | M1 | |
| \(\cos^2 b = \frac{4}{5}\) | ||
| \(\cos b = \pm \frac{2}{\sqrt{5}}\) | A1 | |
| \(a = \frac{1}{2} \cos b = \pm \frac{1}{\sqrt{5}}\) | M1 | |
| from diagram, \(a > 0 \therefore a = \frac{1}{\sqrt{5}} = \frac{1}{5}\sqrt{5}\) | A1 | (10) |
## (i)
Sketch showing $y = \cos^{-1}(2x)$ and $y = \sin^{-1}x$ intersecting at origin | B3 |
## (ii)
$b = \sin^{-1} a \Rightarrow a = \sin b$ | M1 |
$b = \cos^{-1} 2a \Rightarrow 2a = \cos b$ | M1 |
$\therefore 2 \sin b = \cos b$ | |
$\frac{\sin b}{\cos b} = \frac{1}{2}$ | |
$\tan b = \frac{1}{2}$ | A1 |
## (iii)
$\tan^2 b = \frac{1}{4}$ | |
$\sec^2 b = 1 + \frac{1}{4} = \frac{5}{4}$ | M1 |
$\cos^2 b = \frac{4}{5}$ | |
$\cos b = \pm \frac{2}{\sqrt{5}}$ | A1 |
$a = \frac{1}{2} \cos b = \pm \frac{1}{\sqrt{5}}$ | M1 |
from diagram, $a > 0 \therefore a = \frac{1}{\sqrt{5}} = \frac{1}{5}\sqrt{5}$ | A1 | (10)
---\begin{enumerate}[label=(\roman*)]
\item Sketch on the same diagram the graphs of
$$y = \sin^{-1} x, \quad -1 \leq x \leq 1$$
and
$$y = \cos^{-1} (2x), \quad -\frac{1}{2} \leq x \leq \frac{1}{2}.$$ [3]
\end{enumerate}
Given that the graphs intersect at the point with coordinates $(a, b)$,
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item show that $\tan b = \frac{1}{2}$, [3]
\item find the value of $a$ in the form $k\sqrt{5}$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR C3 Q8 [10]}}