| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Differentiate inverse trigonometric functions |
| Difficulty | Standard +0.3 This is a structured, multi-part question that guides students through deriving the derivative of an inverse trig function using implicit differentiation and the chain rule. While it requires knowledge of inverse functions, trigonometric identities, and the reciprocal relationship between dy/dx and dx/dy, the question provides significant scaffolding through its parts (a), (b), and (c). The techniques are standard C3 material with no novel insight required—students follow the prompted steps to reach the given result. Slightly easier than average due to the guided structure. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = \frac{1}{2}\); \(y = \frac{\pi}{2}\) (or \(1.57, \sin^{-1}\)) | B1 | ignore 90°; 1 mark total |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \sin^{-1} 2x\); \(\sin y = 2x\) and \(\frac{1}{2}\sin y = x\) | B1 | AG (be convinced); 1 mark total |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dx}{dy} = \frac{1}{2}\cos y\) | B1 | 1 mark total |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{2}{\cos y}\); \(\sin y = 2x\) and \(\sin^2 + \cos^2 = 1\); \(\cos y = \sqrt{1-4x^2}\); \(\frac{dy}{dx} = \frac{2}{\sqrt{1-4x^2}}\) | M1A1 M1 A1 | M1 for \(\frac{k}{\cos y}\); use of to get \(\cos y\) or \(\cos^2 y\); AG; condone omission of proof of sign; 4 marks total |
### 9(a)
$x = \frac{1}{2}$; $y = \frac{\pi}{2}$ (or $1.57, \sin^{-1}$) | B1 | ignore 90°; 1 mark total
### 9(b)(i)
$y = \sin^{-1} 2x$; $\sin y = 2x$ and $\frac{1}{2}\sin y = x$ | B1 | AG (be convinced); 1 mark total
### 9(b)(ii)
$\frac{dx}{dy} = \frac{1}{2}\cos y$ | B1 | 1 mark total
### 9(c)
$\frac{dy}{dx} = \frac{2}{\cos y}$; $\sin y = 2x$ and $\sin^2 + \cos^2 = 1$; $\cos y = \sqrt{1-4x^2}$; $\frac{dy}{dx} = \frac{2}{\sqrt{1-4x^2}}$ | M1A1 M1 A1 | M1 for $\frac{k}{\cos y}$; use of to get $\cos y$ or $\cos^2 y$; AG; condone omission of proof of sign; 4 marks total
**Total for Question 9:** 7 marks
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**GRAND TOTAL:** 75 marks9 The diagram shows the curve with equation $y = \sin ^ { - 1 } 2 x$, where $- \frac { 1 } { 2 } \leqslant x \leqslant \frac { 1 } { 2 }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{0ab0e757-270b-4c15-9202-9df2f02dddf3-4_790_752_906_644}
\begin{enumerate}[label=(\alph*)]
\item Find the $y$-coordinate of the point $A$, where $x = \frac { 1 } { 2 }$.
\item \begin{enumerate}[label=(\roman*)]
\item Given that $y = \sin ^ { - 1 } 2 x$, show that $x = \frac { 1 } { 2 } \sin y$.
\item Given that $x = \frac { 1 } { 2 } \sin y$, find $\frac { \mathrm { d } x } { \mathrm {~d} y }$ in terms of $y$.
\end{enumerate}\item Using the answers to part (b) and a suitable trigonometrical identity, show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { \sqrt { 1 - 4 x ^ { 2 } } }$$
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2006 Q9 [7]}}