Show that $f$ is an even function.

By considering limits, show that the graph of $y=f\left(x\right)$ has a horizontal asymptote and state its equation.

Show that $f\text{'}\left(x\right)=\frac{2x}{\sqrt{x^2}\left(x^2+1\right)}$ for $x\in \mathrm{ℝ}, x\ne 0$.

By using the expression for $f\text{'}\left(x\right)$ and the result $\sqrt{x^2}=\left|x\right|$, show that $f$ is decreasing for $x<0$.

Find an expression for $g^{-1}\left(x\right)$, justifying your answer.

State the domain of $g^{-1}$.

Sketch the graph of $y=g^{-1}\left(x\right)$, clearly indicating any asymptotes with their equations and stating the values of any axes intercepts.

EITHER

$f\left(-x\right)=\text{arcsin}\left(\frac{{\left(-x\right)}^2-1}{{\left(-x\right)}^2+1}\right)=\text{arcsin}\left(\frac{x^2-1}{x^2+1}\right)=f\left(x\right)$            R1

OR

a sketch graph of $y=f\left(x\right)$ with line symmetry in the $y$-axis indicated            R1

THEN

so $f\left(x\right)$ is an even function.            AG

[1 mark]

as $x\to \pm \infty , f\left(x\right)\to \text{arcsin} 1\left(\to \frac{\mathrm{\pi }}{2}\right)$            A1

so the horizontal asymptote is $y=\frac{\mathrm{\pi }}{2}$            A1 

[2 marks]

attempting to use the quotient rule to find $\frac{\text{d}}{dx}\left(\frac{x^2-1}{x^2+1}\right)$            M1

$\frac{\text{d}}{dx}\left(\frac{x^2-1}{x^2+1}\right)=\frac{2x\left(x^2+1\right)-2x\left(x^2-1\right)}{{\left(x^2+1\right)}^2} \left(=\frac{4x}{{\left(x^2+1\right)}^2}\right)$            A1

attempting to use the chain rule to find $\frac{\text{d}}{dx}\left(\text{arcsin}\left(\frac{x^2-1}{x^2+1}\right)\right)$            M1

let $u=\frac{x^2-1}{x^2+1}$ and so $y=\text{arcsin} u$ and $\frac{dy}{du}=\frac{1}{\sqrt{1-u^2}}$

$f\text{'}\left(x\right)=\frac{1}{\sqrt{1-{\left({\displaystyle \frac{x^2-1}{x^2+1}}\right)}^2}}\times \frac{4x}{{\left(x^2+1\right)}^2}$            M1

$=\frac{4x}{\sqrt{{\left(x^2+1\right)}^2-{\left(x^2-1\right)}^2}}\times \frac{1}{\left(x^2+1\right)}$            A1

$=\frac{4x}{\sqrt{4x^2}}\times \frac{1}{\left(x^2+1\right)}$            A1

$=\frac{2x}{\sqrt{x^2}\left(x^2+1\right)}$            AG

[6 marks]

$f\text{'}\left(x\right)=\frac{2x}{\left|x\right|\left(x^2+1\right)}$

EITHER

for $x<0, \left|x\right|=-x$            (A1)

so $f\text{'}\left(x\right)=-\frac{2x}{x^2+1}$            A1

OR

$\left|x\right|>0$ and $x^2+1>0$            A1

$2x<0, x<0$            A1

THEN

$f\text{'}\left(x\right)<0$              R1

Note: Award R1 for stating that in $f\text{'}\left(x\right)$, the numerator is negative, and the denominator is positive.

so $f$ is decreasing for $x<0$            AG

Note: Do not accept a graphical solution

[3 marks]

$x=\text{arcsin}\left(\frac{y^2-1}{y^2+1}\right)$            M1

$\text{sin} x=\frac{y^2-1}{y^2+1}\Rightarrow y^2 \text{sin} x+\text{sin} x=y^2-1$            A1

$y^2=\frac{1+\text{sin} x}{1-\text{sin} x}$            A1

domain of $g$ is $x\in \mathrm{ℝ}, x\ge 0$ and so the range of $g^{-1}$ must be $y\in \mathrm{ℝ}, y\ge 0$

hence the positive root is taken (or the negative root is rejected)              R1

Note: The R1 is dependent on the above A1.

so $\left(g^{-1}\left(x\right)=\right)\sqrt{\frac{1+\text{sin} x}{1-\text{sin} x}}$            A1

Note: The final A1 is not dependent on R1 mark.

[5 marks]

domain is $-\frac{\mathrm{\pi }}{2}\le x<\frac{\mathrm{\pi }}{2}$            A1

Note: Accept correct alternative notations, for example, $\lfloor -\frac{\mathrm{\pi }}{2}, \frac{\mathrm{\pi }}{2}\lfloor$  or $\lfloor -\frac{\mathrm{\pi }}{2}, \frac{\mathrm{\pi }}{2})$.
Accept $[-1.57, 1.57[$  if correct to $3$ s.f.

[1 mark]

          A1A1A1

Note: A1 for correct domain and correct range and $y$-intercept at $y=1$
         A1 for asymptotic behaviour $x\to \frac{\mathrm{\pi }}{2}$
         A1 for $x=\frac{\mathrm{\pi }}{2}$
         Coordinates are not required. 
         Do not accept $x=1.57$ or other inexact values.

[3 marks]