(i)     Find $g(0)$ .

(ii)    Find $(f \circ g)(0)$ .

Find ${f^{ - 1}}(x)$ .

(i) $g(0) = {{\rm{e}}^0} - 2$     (A1)

$= - 1$     A1     N2

(ii) METHOD 1

substituting answer from (i)     (M1)

e.g. $(f \circ g)(0) = f( - 1)$

correct substitution $f( - 1) = 2{( - 1)^3} + 3$     (A1)

$f( - 1) = 1$     A1     N3

METHOD 2

attempt to find $(f \circ g)(x)$     (M1)

e.g. $(f \circ g)(x) = f({{\rm{e}}^{3x}} - 2)$ $= 2{({{\rm{e}}^{3x}} - 2)^3} + 3$

correct expression for $(f \circ g)(x)$     (A1)

e.g. $2{({{\rm{e}}^{3x}} - 2)^3} + 3$

$(f \circ g)(0) = 1$     A1     N3

[5 marks]

interchanging x and y (seen anywhere)     (M1)

e.g. $x = 2{y^3} + 3$

attempt to solve     (M1)

e.g. ${y^3} = \frac{{x - 3}}{2}$

${f^{ - 1}}(x) = \sqrt[3]{{\frac{{x - 3}}{2}}}$     A1     N3

[3 marks]

This question was generally done well, although some students consider ${{\rm{e}}^0}$ to be 0, losing them a mark.

A few candidates composed in the wrong order. Most found the formula of the inverse correctly, even if in some cases there were errors when trying to isolate x (or y). A common incorrect solution found was to find $y = \sqrt[3]{{\frac{{x - 3}}{2}}}$ .