| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Proving inverse trig identities |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring chain rule with quotient rule and knowledge of arctan derivative, followed by integration to establish an identity. Part (a) is mechanical differentiation (though multi-step), and part (b) requires recognizing that integrating a constant leads to the result. More demanding than standard A-level but routine for FP2 students who know the techniques. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
(a) Given that $y = \tan^{-1}\left(\frac{1+x}{1-x}\right)$ and $x \neq 1$, show that $\frac{dy}{dx} = \frac{1}{1+x^2}$.
[4 marks]
(b) Hence, given that $x < 1$, show that $\tan^{-1}\left(\frac{1+x}{1-x}\right) - \tan^{-1} x = \frac{\pi}{4}$.
[3 marks]7
\begin{enumerate}[label=(\alph*)]
\item Given that $y = \tan ^ { - 1 } \left( \frac { 1 + x } { 1 - x } \right)$ and $x \neq 1$, show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 1 + x ^ { 2 } }$.\\[0pt]
[4 marks]
\item Hence, given that $x < 1$, show that $\tan ^ { - 1 } \left( \frac { 1 + x } { 1 - x } \right) - \tan ^ { - 1 } x = \frac { \pi } { 4 }$.\\[0pt]
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2014 Q7 [7]}}