Pre-U Pre-U 9795/1 Specimen — Question 9 16 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
SessionSpecimen
Marks16
TopicHyperbolic functions
TypeDifferentiate inverse hyperbolic functions
DifficultyStandard +0.8 This is a multi-part question requiring differentiation of inverse hyperbolic functions, proving the logarithmic form, and evaluating a non-trivial integral using partial fractions and inverse hyperbolic/trigonometric substitutions. While the techniques are standard for Further Maths, the integration in part (ii) requires careful manipulation and multiple steps, making it moderately challenging but within expected scope.
Spec4.07e Inverse hyperbolic: definitions, domains, ranges4.07f Inverse hyperbolic: logarithmic forms4.08h Integration: inverse trig/hyperbolic substitutions
9
  1. (a) Given that \(y = \tanh ^ { - 1 } x , - 1 < x < 1\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\).
    (b) Show that \(y = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)\).
  2. Show that \(\int _ { 0 } ^ { 1 / \sqrt { 3 } } \frac { 2 } { 1 - x ^ { 4 } } \mathrm {~d} x = \ln ( 1 + \sqrt { 3 } ) - \frac { 1 } { 2 } \ln 2 + \frac { 1 } { 6 } \pi\).
(i)(a) \(\tanh y = x\), including differentiation attempt: M1
\(\Rightarrow \text{sech}^2 y\,\frac{dy}{dx} = 1\): A1
\(\Rightarrow \frac{dy}{dx} = \frac{1}{1-\tanh^2 y} = \frac{1}{1-x^2}\), from correct working: A1 [3]
(i)(b) \(\tanh y = x\): M1
\(\Rightarrow x = \frac{e^{2y}-1}{e^{2y}+1}\): B1
\(\Rightarrow e^{2y} = \frac{1+x}{1-x}\): M1
\(\Rightarrow y = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\), legitimately: A1 [4]
*Alternative (b):* \(y = \int \frac{1}{1-x^2}\,dx\), attempt to integrate their (a): M1(dep)
\(= \frac{1}{2}\int\left(\frac{1}{1+x} + \frac{1}{1-x}\right)dx\), use of partial fractions: M1(dep)
\(= \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)\), ignore missing '\(+c\)' here: A1
\(c = 0\) using \(\tanh^{-1}0 = 0\): B1 [4]
(ii) \(\int_0^{1/\sqrt{3}} \frac{2}{1-x^4}\,dx = \int_0^{1/\sqrt{3}} \frac{2}{(1-x^2)(1+x^2)}\,dx = \int_0^{1/\sqrt{3}}\left(\frac{1}{1-x^2}+\frac{1}{1+x^2}\right)dx\): M1, A1
\(= \left[\tanh^{-1}x + \tan^{-1}x\right]_0^{1/\sqrt{3}}\) or equivalent: A1
\(= \frac{1}{2}\ln\left(\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}\right) + \frac{1}{6}\pi\), for use of log form for \(\tanh^{-1}x\): M1
Showing \(\frac{1}{2}\ln\left(\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}\right) = \frac{1}{2}\ln\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}\right) = \frac{1}{2}\ln(\sqrt{3}+1)^2 - \frac{1}{2}\ln 2\)
and getting given answer \(\ln(\sqrt{3}+1) - \frac{1}{2}\ln 2 + \frac{1}{6}\pi\): A1 [5]
**(i)(a)** $\tanh y = x$, including differentiation attempt: M1

$\Rightarrow \text{sech}^2 y\,\frac{dy}{dx} = 1$: A1

$\Rightarrow \frac{dy}{dx} = \frac{1}{1-\tanh^2 y} = \frac{1}{1-x^2}$, from correct working: A1 **[3]**

**(i)(b)** $\tanh y = x$: M1

$\Rightarrow x = \frac{e^{2y}-1}{e^{2y}+1}$: B1

$\Rightarrow e^{2y} = \frac{1+x}{1-x}$: M1

$\Rightarrow y = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$, legitimately: A1 **[4]**

*Alternative (b):* $y = \int \frac{1}{1-x^2}\,dx$, attempt to integrate their (a): M1(dep)

$= \frac{1}{2}\int\left(\frac{1}{1+x} + \frac{1}{1-x}\right)dx$, use of partial fractions: M1(dep)

$= \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$, ignore missing '$+c$' here: A1

$c = 0$ using $\tanh^{-1}0 = 0$: B1 **[4]**

**(ii)** $\int_0^{1/\sqrt{3}} \frac{2}{1-x^4}\,dx = \int_0^{1/\sqrt{3}} \frac{2}{(1-x^2)(1+x^2)}\,dx = \int_0^{1/\sqrt{3}}\left(\frac{1}{1-x^2}+\frac{1}{1+x^2}\right)dx$: M1, A1

$= \left[\tanh^{-1}x + \tan^{-1}x\right]_0^{1/\sqrt{3}}$ or equivalent: A1

$= \frac{1}{2}\ln\left(\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}\right) + \frac{1}{6}\pi$, for use of log form for $\tanh^{-1}x$: M1

Showing $\frac{1}{2}\ln\left(\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}\right) = \frac{1}{2}\ln\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}\right) = \frac{1}{2}\ln(\sqrt{3}+1)^2 - \frac{1}{2}\ln 2$

and getting given answer $\ln(\sqrt{3}+1) - \frac{1}{2}\ln 2 + \frac{1}{6}\pi$: A1 **[5]**
9
\begin{enumerate}[label=(\roman*)]
\item (a) Given that $y = \tanh ^ { - 1 } x , - 1 < x < 1$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$.\\
(b) Show that $y = \frac { 1 } { 2 } \ln \left( \frac { 1 + x } { 1 - x } \right)$.
\item Show that $\int _ { 0 } ^ { 1 / \sqrt { 3 } } \frac { 2 } { 1 - x ^ { 4 } } \mathrm {~d} x = \ln ( 1 + \sqrt { 3 } ) - \frac { 1 } { 2 } \ln 2 + \frac { 1 } { 6 } \pi$.
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9795/1  Q9 [16]}}
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