AQA FP2 2013 June — Question 7 12 marks

Exam BoardAQA
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionJune
Marks12
PaperDownload PDF ↗
TopicHyperbolic functions
TypeSurface area of revolution with hyperbolics
DifficultyChallenging +1.8 This is a challenging Further Maths question combining hyperbolic functions, integration techniques, and surface area of revolution. Part (a)(i) requires careful differentiation of a composite expression involving inverse hyperbolics; part (a)(ii) applies this as a reverse chain rule integration. Part (b) requires deriving the surface area formula, then executing a substitution that connects back to part (a). While the structure is guided and uses standard FP2 techniques, it demands multiple sophisticated steps, careful algebraic manipulation, and the ability to recognize how the parts connect—placing it well above average difficulty but not at the extreme end for Further Maths.
Spec1.08h Integration by substitution4.07e Inverse hyperbolic: definitions, domains, ranges4.07f Inverse hyperbolic: logarithmic forms4.08e Mean value of function: using integral

7
    1. Show that $$\frac { \mathrm { d } } { \mathrm {~d} u } \left( 2 u \sqrt { 1 + 4 u ^ { 2 } } + \sinh ^ { - 1 } 2 u \right) = k \sqrt { 1 + 4 u ^ { 2 } }$$ where \(k\) is an integer.
    2. Hence show that $$\int _ { 0 } ^ { 1 } \sqrt { 1 + 4 u ^ { 2 } } \mathrm {~d} u = p \sqrt { 5 } + q \sinh ^ { - 1 } 2$$ where \(p\) and \(q\) are rational numbers.
  1. The arc of the curve with equation \(y = \frac { 1 } { 2 } \cos 4 x\) between the points where \(x = 0\) and \(x = \frac { \pi } { 8 }\) is rotated through \(2 \pi\) radians about the \(x\)-axis.
    1. Show that the area \(S\) of the curved surface formed is given by $$S = \pi \int _ { 0 } ^ { \frac { \pi } { 8 } } \cos 4 x \sqrt { 1 + 4 \sin ^ { 2 } 4 x } \mathrm {~d} x$$
    2. Use the substitution \(u = \sin 4 x\) to find the exact value of \(S\).

7
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that

$$\frac { \mathrm { d } } { \mathrm {~d} u } \left( 2 u \sqrt { 1 + 4 u ^ { 2 } } + \sinh ^ { - 1 } 2 u \right) = k \sqrt { 1 + 4 u ^ { 2 } }$$

where $k$ is an integer.
\item Hence show that

$$\int _ { 0 } ^ { 1 } \sqrt { 1 + 4 u ^ { 2 } } \mathrm {~d} u = p \sqrt { 5 } + q \sinh ^ { - 1 } 2$$

where $p$ and $q$ are rational numbers.
\end{enumerate}\item The arc of the curve with equation $y = \frac { 1 } { 2 } \cos 4 x$ between the points where $x = 0$ and $x = \frac { \pi } { 8 }$ is rotated through $2 \pi$ radians about the $x$-axis.
\begin{enumerate}[label=(\roman*)]
\item Show that the area $S$ of the curved surface formed is given by

$$S = \pi \int _ { 0 } ^ { \frac { \pi } { 8 } } \cos 4 x \sqrt { 1 + 4 \sin ^ { 2 } 4 x } \mathrm {~d} x$$
\item Use the substitution $u = \sin 4 x$ to find the exact value of $S$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA FP2 2013 Q7 [12]}}