The function $f:[0,{\text{ }}\infty [ \to [0,{\text{ }}\infty [$ is defined by $f(x) = 2{{\text{e}}^x} + {{\text{e}}^{ - x}} - 3$ .

(a)     Find $f'(x)$ .

(b)     Show that f is a bijection.

(c)     Find an expression for ${f^{ - 1}}(x)$ .

(a)     $f'(x) = 2{{\text{e}}^x} - {{\text{e}}^{ - x}}$     A1

[1 mark]

(b)     f is an injection because $f'(x) > 0$ for $x \in [0,{\text{ }}\infty [$     R2

(accept GDC solution backed up by a correct graph)

since $f(0) = 0$ and $f(x) \to \infty$ as $x \to \infty$ , (and f is continuous) it is a surjection     R1

hence it is a bijection     AG

[3 marks]

(c)     let $y = 2{{\text{e}}^x} + {{\text{e}}^{ - x}} - 3$     M1

so $2{{\text{e}}^{2x}} - (y + 3){{\text{e}}^x} + 1 = 0$     A1

${{\text{e}}^x} = \frac{{y + 3 \pm \sqrt {{{(y + 3)}^2} - 8} }}{4}$     A1

$x = \ln \left( {\frac{{y + 3 \pm \sqrt {{{(y + 3)}^2} - 8} }}{4}} \right)$     A1

since $x \geqslant 0$ we must take the positive square root     (R1)

${f^{ - 1}}(x) = \ln \left( {\frac{{x + 3 + \sqrt {{{(x + 3)}^2} - 8} }}{4}} \right)$     A1

[6 marks]

Total [10 marks]

In many cases the attempts at showing that f is a bijection were unconvincing. The candidates were guided towards showing that f is an injection by noting that $f'(x) > 0$ for all x, but some candidates attempted to show that $f(x) = f(y) \Rightarrow x = y$ which is much more difficult. Solutions to (c) were often disappointing, with the algebra defeating many candidates.