AQA Further AS Paper 1 2018 June — Question 6

Exam BoardAQA
ModuleFurther AS Paper 1 (Further AS Paper 1)
Year2018
SessionJune
TopicHyperbolic functions

6
  1. Matthew is finding a formula for the inverse function \(\operatorname { arsinh } x\). He writes his steps as follows: $$\begin{gathered} \text { Let } y = \sinh x
    y = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } - \mathrm { e } ^ { - x } \right)
    2 y = \mathrm { e } ^ { x } - \mathrm { e } ^ { - x }
    0 = \mathrm { e } ^ { x } - 2 y - \mathrm { e } ^ { - x }
    0 = \left( \mathrm { e } ^ { x } \right) ^ { 2 } - 2 y \mathrm { e } ^ { x } - 1
    0 = \left( \mathrm { e } ^ { x } - y \right) ^ { 2 } - y ^ { 2 } - 1
    y ^ { 2 } + 1 = \left( \mathrm { e } ^ { x } - y \right) ^ { 2 }
    \pm \sqrt { y ^ { 2 } + 1 } = \mathrm { e } ^ { x } - y
    y \pm \sqrt { y ^ { 2 } + 1 } = \mathrm { e } ^ { x } \end{gathered}$$ To find the inverse function, swap \(x\) and \(y : x \pm \sqrt { x ^ { 2 } + 1 } = \mathrm { e } ^ { y }\) $$\begin{gathered} \ln \left( x \pm \sqrt { x ^ { 2 } + 1 } \right) = y
    \operatorname { arsinh } x = \ln \left( x \pm \sqrt { x ^ { 2 } + 1 } \right) \end{gathered}$$ Identify, and explain, the error in Matthew's proof. 6
  2. Solve \(\ln \left( x + \sqrt { x ^ { 2 } + 1 } \right) = 3\)