Inverse hyperbolic functionsEdexcel A-Level Further Maths: Subtopic test

1. 1
   The inverse hyperbolic functions have the logarithmic forms $\operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right)$, $\operatorname{arcosh}x=\ln\left(x+\sqrt{x^2-1}\right)$ and $\operatorname{artanh}x=\frac12\ln\frac{1+x}{1-x}$.

   (a)

   Find the exact value of $\operatorname{arsinh}\frac34$.

   [1 mark]

   • A$\ln\frac54$
   • B$\ln2$
   • C$\ln\frac12$
   • D$\ln\frac74$

   (b)

   State the largest possible domain of $\operatorname{arcosh}x$.

   [1 mark]

   • Aall real $x$
   • B$-1<x<1$
   • C$x\ge1$
   • D$x\ge0$

   (c)

   Find the exact value of $\operatorname{artanh}\frac13$, giving your answer as a multiple of $\ln2$.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   Let $y=\operatorname{arcosh}x$ for $x\ge1$, so that $x=\cosh y$ with $y\ge0$.

   (a)

   Find the exact value of $\operatorname{arcosh}\frac53$.

   [1 mark]

   • A$\ln3$
   • B$\ln\frac13$
   • C$\ln\frac53$
   • D$\ln\frac{5+\sqrt{34}}{3}$

   (b)

   Solving $x=\frac{e^{y}+e^{-y}}{2}$ for $e^{y}$ gives $e^{y}=x\pm\sqrt{x^2-1}$. Which statement explains why the positive sign is chosen?

   [1 mark]

   • AThe root with the negative sign is complex.
   • BThe root with the negative sign is negative, but $e^y>0$.
   • C$\operatorname{arcosh}$ is an odd function, so either sign may be used.
   • D$y\ge0$ needs $e^y\ge1$, and only the positive sign guarantees this.

   (c)

   Find the exact value of $\operatorname{arcosh}\frac{13}{5}$.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   Let $y=\operatorname{arsinh}x$, so that $x=\sinh y$, where $\sinh y=\frac{e^{y}-e^{-y}}{2}$.

   (a)

   Derive the logarithmic form $\operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right)$.

   [3 marks]

   (b)

   Solve $2\cosh^2x-3\sinh x=4$, giving your answers as exact logarithms.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   In this question use $\sinh u=\frac{e^{u}-e^{-u}}{2}$, $\tanh u=\frac{\sinh u}{\cosh u}$ and $\cosh^2u-\sinh^2u=1$.

   (a)

   (i) Show that $\operatorname{artanh}x=\frac12\ln\frac{1+x}{1-x}$ for $-1<x<1$.
   (ii) Hence solve $\operatorname{artanh}x=\ln2$.

   [6 marks]

   (b)

   Use the substitution $x=3\sinh u$ to show that $\int_0^4\frac{1}{\sqrt{x^2+9}}\,dx=\ln3$.

   [6 marks]

   Total for question 4: 12 marks

End of questions