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.