A function is defined by \(h(x) = 2{{\text{e}}^x} - \frac{1}{{{{\text{e}}^x}}},{\text{ }}x \in \mathbb{R}\) . Find an expression for \({h^{ - 1}}(x)\) .

\(x = 2{{\text{e}}^y} - \frac{1}{{{{\text{e}}^y}}}\)     M1

Note: The M1 is for switching the variables and may be awarded at any stage in the process and is awarded independently. Further marks do not rely on this mark being gained.

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

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

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

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

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

since ln is undefined for the second solution     R1

Note: Accept \(y = \ln \left( {\frac{{x + \sqrt {{x^2} + 8} }}{4}} \right)\).

Note: The R1 may be gained by an appropriate comment earlier.

[6 marks]

A significant number of candidates did not recognise the need for the quadratic formula in order to find the inverse. Even when they did most candidates who got this far did not recognise the need to limit the solution to the positive only. This question was done well by a very limited number of candidates.