The functions f and g are defined as:

$$f(x) = {{\text{e}}^{{x^2}}},{\text{ }}x \geqslant 0$$

$$g(x) = \frac{1}{{x + 3}},{\text{ }}x \ne - 3.$$

(a)     Find $h(x){\text{ where }}h(x) = g \circ f(x)$ .

(b)     State the domain of ${h^{ - 1}}(x)$ .

(c)     Find ${h^{ - 1}}(x)$ .

(a)     $h(x) = g \circ f(x) = \frac{1}{{{{\text{e}}^{{x^2}}} + 3}},{\text{ }}(x \geqslant 0)$     (M1)A1

(b)     $0 < x \leqslant \frac{1}{4}$     A1A1

Note: Award A1 for limits and A1 for correct inequality signs.

(c)     $y = \frac{1}{{{{\text{e}}^{{x^2}}} + 3}}$

$y{{\text{e}}^{{x^2}}} + 3y = 1$     M1

${{\text{e}}^{{x^2}}} = \frac{{1 - 3y}}{y}$     A1

${x^2} = \ln \frac{{1 - 3y}}{y}$     M1

$x = \pm \sqrt {\ln \frac{{1 - 3y}}{y}}$

$\Rightarrow {h^{ - 1}}(x) = \sqrt {\ln \frac{{1 - 3x}}{x}} {\text{ }}\left( { = \sqrt {\ln \left( {\frac{1}{x} - 3} \right)} } \right)$     A1

[8 marks]

Part (a) was correctly done by the vast majority of candidates. In contrast, only the very best students gave the correct answer to part (b). Part (c) was correctly started by a majority of candidates, but many did not realise that they needed to use logarithms and were careless about the use of notation