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.