| Exam Board | AQA |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | Integration using inverse trig and hyperbolic functions |
| Type | Derivative of inverse trig function |
| Difficulty | Standard +0.8 This is a Further Maths FP2 question requiring understanding of inverse trig functions and their relationships. Part (a) needs insight into the relationship between sec and cos inverses (not immediately obvious to most students), while part (b) requires careful chain rule application and algebraic manipulation of surds. The multi-step nature and need to work with less familiar inverse functions places it moderately above average difficulty. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07l Derivative of ln(x): and related functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
5 The function f , where $\mathrm { f } ( x ) = \sec x$, has domain $0 \leqslant x < \frac { \pi } { 2 }$ and has inverse function $\mathrm { f } ^ { - 1 }$, where $\mathrm { f } ^ { - 1 } ( x ) = \sec ^ { - 1 } x$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\sec ^ { - 1 } x = \cos ^ { - 1 } \frac { 1 } { x }$$
\item Hence show that
$$\frac { \mathrm { d } } { \mathrm {~d} x } \left( \sec ^ { - 1 } x \right) = \frac { 1 } { \sqrt { x ^ { 4 } - x ^ { 2 } } }$$
\end{enumerate}
\hfill \mbox{\textit{AQA FP2 2012 Q5 [6]}}