Challenging +1.2 This is a Further Maths hyperbolic functions question requiring algebraic manipulation using the identity sech²x + tanh²x = 1, solving a quadratic in cosh x, and applying the given inverse hyperbolic formula. While it involves multiple steps and careful algebra, the path is relatively standard for Further Maths students: convert to cosh x, solve quadratic, apply formula. The 5 marks and structured approach make it moderately challenging but not exceptional.
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 \(\cosh^{-1} y = \ln(y + \sqrt{y^2 - 1})\)
[5 marks]
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 $\cosh^{-1} y = \ln(y + \sqrt{y^2 - 1})$
[5 marks]
\hfill \mbox{\textit{AQA Further Paper 1 2021 Q4 [5]}}