The function \(f:\mathbb{R} \to \mathbb{R}\) is defined by

\[f(x) = 2{{\text{e}}^x} - {{\text{e}}^{ - x}}.\]

(a)     Show that f is a bijection.

(b)     Find an expression for \({f^{ - 1}}(x)\).

(a)     EITHER

consider

\(f'(x) = 2{{\text{e}}^x} - {{\text{e}}^{ - x}} > 0\) for all x     M1A1

so f is an injection     A1

OR

let \(2{{\text{e}}^x} - {{\text{e}}^{ - x}} = 2{{\text{e}}^y} - {{\text{e}}^{ - y}}\)     M1

\(2({{\text{e}}^x} - {{\text{e}}^y}) + {{\text{e}}^{ - y}} - {{\text{e}}^{ - x}} = 0\)

\(2({{\text{e}}^x} - {{\text{e}}^y}) + {{\text{e}}^{ - (x + y)}}({{\text{e}}^x} - {{\text{e}}^y}) = 0\)

\(\left( {2 + {{\text{e}}^{ - (x + y)}}} \right)({{\text{e}}^x} - {{\text{e}}^y}) = 0\)

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

\(x = y\)     A1

Note: Sufficient working must be shown to gain the above A1.

so f is an injection     A1

Note: Accept a graphical justification i.e. horizontal line test.

THEN

it is also a surjection (accept any justification including graphical)     R1

therefore it is a bijection     AG

[4 marks]

(b)     let \(y = 2{{\text{e}}^x} - {{\text{e}}^{ - x}}\)     M1

\(2{{\text{e}}^{2x}} - y{{\text{e}}^x} - 1 = 0\)     A1

\({{\text{e}}^x} = \frac{{y \pm \sqrt {{y^2} + 8} }}{4}\)     M1A1

since \({{\text{e}}^x}\) is never negative, we take the + sign     R1

\({f^{ - 1}}(x) = \ln \left( {\frac{{x + \sqrt {{x^2} + 8} }}{4}} \right)\)     A1

[6 marks]

Total [10 marks]

Solutions to (a) were often disappointing. Many candidates tried to use the result that, for an injection, \(f(a) = f(b) \Rightarrow a = b\) – although this is the definition, it is often much easier to proceed by showing that the derivative is everywhere positive or everywhere negative or even to use a horizontal line test. Although (b) is based on core material, solutions were often disappointing with some very poor use of algebra seen.