Question

The general value of ${\text{cosh}}^{-1}\text{x}$ is

• A. $2\mathrm{\pi ri}+\log \left[\mathrm{x}+\sqrt{{\mathrm{x}}^2+1}\right]$
• B. $2\mathrm{\pi ri}+\log \left[\mathrm{x}+\sqrt{{\mathrm{x}}^2-1}\right]$
• C. $\mathrm{\pi ri}+{(-1)}^r\log \left[\mathrm{x}+\sqrt{\mathrm{x}}\right]$
• D. $2\mathrm{\pi ri}+{(-1)}^r\log \left[\mathrm{x}+\sqrt{\mathrm{x}-1}\right]$

Answer 1

The correct option is B

$2\mathrm{\pi ri}+\log \left[\mathrm{x}+\sqrt{{\mathrm{x}}^2-1}\right]$

Explanation for the correct answer:

Compute the inverse of $\text{cosh}\left(\text{x}\right)$.

We know that, $\cos hx=\cos h(2r\mathrm{\pi i}+\mathrm{x})$.

Also, $\cos h\left(x\right)=\frac{e^x+e^{-x}}{2}$.

Therefore, $\cos h(2r\mathrm{\pi i}+\mathrm{x})=\frac{e^{2r\mathrm{\pi i}+\mathrm{x}}+e^{-(2r\mathrm{\pi i}+\mathrm{x})}}{2}$

Assume that, $\mathrm{y}=\frac{e^{2r\mathrm{\pi i}+\mathrm{x}}+e^{-(2r\mathrm{\pi i}+\mathrm{x})}}{2}$.

Now, Find the inverse of the function as follows:

$$\begin{gathered} 2\mathrm{y}=e^{2r\mathrm{\pi i}+\mathrm{x}}+\frac{1}{e^{(2r\mathrm{\pi i}+\mathrm{x})}} \\ \Rightarrow 2{\mathrm{ye}}^{(2\mathrm{r\pi i}+\mathrm{x})}={\left(e^{2r\mathrm{\pi i}+\mathrm{x}}\right)}^2+1 \\ \Rightarrow {\left(e^{2r\mathrm{\pi i}+\mathrm{x}}\right)}^2-2{\mathrm{ye}}^{(2\mathrm{r\pi i}+\mathrm{x})}+1=0 \end{gathered}$$

Assume that, $e^{2r\mathrm{\pi i}+\mathrm{x}}=z$.

Therefore, $z^2-2yz+1=0$.

$$\begin{gathered} z=\frac{2y\pm \sqrt{4y^2-4}}{2} \\ \Rightarrow z=y\pm \sqrt{y^2-1} \end{gathered}$$

So, $e^{2r\mathrm{\pi i}+\mathrm{x}}=y\pm \sqrt{y^2-1}$

Take $\log$ on both sides.

$2r\mathrm{\pi i}+\mathrm{x}=\log \left(y+\sqrt{y^2-1}\right)$

Now, replace $x$ and $y$.

$y=2r\mathrm{\pi i}+\log (\mathrm{x}+\sqrt{{\mathrm{x}}^2+1})$.

Therefore, the value of $\cos h^{-1}\left(x\right)$ is $2r\mathrm{\pi i}+\log (\mathrm{x}+\sqrt{{\mathrm{x}}^2+1})$.

Hence, option $B$ is the correct answer.