| Exam Board | AQA |
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| Module | Further Paper 1 (Further Paper 1) |
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| Year | 2021 |
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| Session | June |
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| Topic | Hyperbolic functions |
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4 Show that the solutions to the equation
$$3 \tanh ^ { 2 } x - 2 \operatorname { sech } x = 2$$
can be expressed in the form
$$x = \pm \ln ( a + \sqrt { b } )$$
where \(a\) and \(b\) are integers to be found.
You may use without proof the result \(\cosh ^ { - 1 } y = \ln \left( y + \sqrt { y ^ { 2 } - 1 } \right)\)