(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 \(y = \sqrt[3]{{\frac{{x - 3}}{2}}}\) .