The function $f$ is defined as $f(x) = \sqrt {\frac{{1 - x}}{{1 + x}}} ,{\text{ }} - 1 < x \leqslant 1$.

Find the inverse function, ${f^{ - 1}}$ stating its domain and range.

$x = \sqrt {\frac{{1 - y}}{{1 + y}}}$    M1

Note:     Award M1 for interchanging $x$ and $y$ (can be done at a later stage).

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

${x^2} + {x^2}y = 1 - y$    M1

Note:     Award M1 for attempting to make $y$ the subject.

$y(1 + {x^2}) = 1 - {x^2}$    (A1)

${f^{ - 1}}(x) = \frac{{1 - {x^2}}}{{1 + {x^2}}},{\text{ }}x \geqslant 0$    A1A1

Note:     Award A1 only if ${f^{ - 1}}(x)$ is seen. Award A1 for the domain.

the range of ${f^{ - 1}}$ is $- 1 < {f^{ - 1}}(x) \leqslant 1$     A1

Note:     Accept correct alternative notation eg. $- 1 < y \leqslant 1$.

[6 marks]

Most candidates were able to find an expression for the inverse function. A large number of candidates however were unable to determine the domain and range of the inverse.