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.