(a)     Find the range of f .

(b)     Prove that f is an injection.

(c)     Taking the codomain of f to be equal to the range of f , find an expression for \({f^{ - 1}}(x)\) .

 

(a)     \(\left] { - 1,1} \right[\)     A1A1

Note: Award A1 for the values –1, 1 and A1 for the open interval.

 

[2 marks]

 

(b)     EITHER

Let \(\frac{{1 - {{\text{e}}^{ - x}}}}{{1 + {{\text{e}}^{ - x}}}} = \frac{{1 - {{\text{e}}^{ - y}}}}{{1 + {{\text{e}}^{ - y}}}}\)     M1

\(1 - {{\text{e}}^{ - x}} + {{\text{e}}^{ - y}} - {{\text{e}}^{ - (x + y)}} = 1 + {{\text{e}}^{ - x}} - {{\text{e}}^{ - y}} - {{\text{e}}^{ - (x + y)}}\)     A1

\({{\text{e}}^{ - x}} = {{\text{e}}^{ - y}}\)

\(x = y\)     A1

Therefore f is an injection     AG

OR

Consider

\(f'(x) = \frac{{{{\text{e}}^{ - x}}(1 + {{\text{e}}^{ - x}}) + {{\text{e}}^{ - x}}(1 - {{\text{e}}^{ - x}})}}{{{{(1 + {{\text{e}}^{ - x}})}^2}}}\)     M1

\( = \frac{{2{{\text{e}}^{ - x}}}}{{{{(1 + {{\text{e}}^{ - x}})}^2}}}\)     A1

\( > 0\) for all x.     A1

Therefore f is an injection.     AG

Note: Award M1A1A0 for a graphical solution.

 

[3 marks]

 

(c)     Let \(y = \frac{{1 - {{\text{e}}^{ - x}}}}{{1 + {{\text{e}}^{ - x}}}}\)     M1

\(y(1 + {{\text{e}}^{ - x}}) = 1 - {{\text{e}}^{ - x}}\)     A1

\({{\text{e}}^{ - x}}(1 + y) = 1 - y\)     A1

\({{\text{e}}^{ - x}} = \frac{{1 - y}}{{1 + y}}\)

\(x = \ln \left( {\frac{{1 + y}}{{1 - y}}} \right)\)     A1

\({f^{ - 1}}(x) = \ln \left( {\frac{{1 + x}}{{1 - x}}} \right)\)     A1

[5 marks]

Total [10 marks]

 

Most candidates found the range of f correctly. Two algebraic methods were seen for solving (b), either showing that the derivative of f is everywhere positive or showing that \(f(a) = f(b) \Rightarrow a = b\) . Candidates who based their ‘proof’ on a graph produced on their graphical calculators were given only partial credit on the grounds that the whole domain could not be shown and, in any case, it was not clear from the graph that f was an injection.