AQA FP2 2014 June — Question 8 11 marks

Exam BoardAQA
ModuleFP2 (Further Pure Mathematics 2)
Year2014
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHyperbolic functions
TypeIntegrate using hyperbolic substitution
DifficultyChallenging +1.8 This is a multi-part Further Maths question requiring arc length formula derivation, hyperbolic identity manipulation, and integration via hyperbolic substitution. While the steps are guided, executing the substitution x = cosh²θ correctly, simplifying the resulting integral, and evaluating limits using the given identity requires solid technical facility with hyperbolic functions beyond standard A-level. The guidance structure prevents it from being extremely difficult, but it's substantially harder than typical FP2 integration questions.
Spec1.08h Integration by substitution4.07f Inverse hyperbolic: logarithmic forms4.08f Integrate using partial fractions
8 A curve has equation \(y = 2 \sqrt { x - 1 }\), where \(x > 1\). The length of the arc of the curve between the points on the curve where \(x = 2\) and \(x = 9\) is denoted by \(s\).
  1. Show that \(s = \int _ { 2 } ^ { 9 } \sqrt { \frac { x } { x - 1 } } \mathrm {~d} x\).
    1. Show that \(\cosh ^ { - 1 } 3 = 2 \ln ( 1 + \sqrt { 2 } )\).
    2. Use the substitution \(x = \cosh ^ { 2 } \theta\) to show that $$s = m \sqrt { 2 } + \ln ( 1 + \sqrt { 2 } )$$ where \(m\) is an integer.
      [0pt] [6 marks]
      \includegraphics[max width=\textwidth, alt={}]{5287255f-5ac4-401a-b850-758257412ff7-20_1638_1709_1069_153}
      \includegraphics[max width=\textwidth, alt={}, center]{5287255f-5ac4-401a-b850-758257412ff7-24_2489_1728_221_141}
A curve has equation \(y = 2\sqrt{x-1}\), where \(x > 1\). The length of the arc of the curve between the points on the curve where \(x = 2\) and \(x = 9\) is denoted by \(s\).
(a) Show that \(s = \int_2^9 \sqrt{\frac{x}{x-1}} \, dx\).
[3 marks]
(b)(i) Show that \(\cosh^{-1} 3 = 2\ln(1 + \sqrt{2})\).
[2 marks]
(ii) Use the substitution \(x = \cosh^2 y\) to show that \(s = m(2 + \ln(1 + \sqrt{2}))\) where \(m\) is an integer.
[6 marks]
A curve has equation $y = 2\sqrt{x-1}$, where $x > 1$. The length of the arc of the curve between the points on the curve where $x = 2$ and $x = 9$ is denoted by $s$.

(a) Show that $s = \int_2^9 \sqrt{\frac{x}{x-1}} \, dx$.
[3 marks]

(b)(i) Show that $\cosh^{-1} 3 = 2\ln(1 + \sqrt{2})$.
[2 marks]

(ii) Use the substitution $x = \cosh^2 y$ to show that $s = m(2 + \ln(1 + \sqrt{2}))$ where $m$ is an integer.
[6 marks]
8 A curve has equation $y = 2 \sqrt { x - 1 }$, where $x > 1$. The length of the arc of the curve between the points on the curve where $x = 2$ and $x = 9$ is denoted by $s$.
\begin{enumerate}[label=(\alph*)]
\item Show that $s = \int _ { 2 } ^ { 9 } \sqrt { \frac { x } { x - 1 } } \mathrm {~d} x$.
\item \begin{enumerate}[label=(\roman*)]
\item Show that $\cosh ^ { - 1 } 3 = 2 \ln ( 1 + \sqrt { 2 } )$.
\item Use the substitution $x = \cosh ^ { 2 } \theta$ to show that

$$s = m \sqrt { 2 } + \ln ( 1 + \sqrt { 2 } )$$

where $m$ is an integer.\\[0pt]
[6 marks]

\begin{center}
\includegraphics[max width=\textwidth, alt={}]{5287255f-5ac4-401a-b850-758257412ff7-20_1638_1709_1069_153}
\end{center}

\includegraphics[max width=\textwidth, alt={}, center]{5287255f-5ac4-401a-b850-758257412ff7-24_2489_1728_221_141}
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA FP2 2014 Q8 [11]}}
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Q1 7 Q2 8 Q3 7 Q4 14 Q5 9 Q6 12 Q7 7 Q8 11