Find \((g \circ f)(x)\) .

Write down \({g^{ - 1}}(x)\) .

Find \((f \circ {g^{ - 1}})(5)\) .

attempt to form composite     (M1)

e.g. \(g(7 - 2x)\) , \(7 - 2x + 3\)

\((g \circ f)(x) = 10 - 2x\)     A1     N2

[2 marks]

\({g^{ - 1}}(x) = x - 3\)     A1     N1

[1 mark]

METHOD 1

valid approach     (M1)

e.g. \({g^{ - 1}}(5)\) , \(2\) , \(f(5)\)

\(f(2) = 3\)     A1     N2

METHOD 2

attempt to form composite of f and \({g^{ - 1}}\)     (M1)

e.g. \((f \circ {g^{ - 1}})(x) = 7 - 2(x - 3)\) , \(13 - 2x\)

\((f \circ {g^{ - 1}})(5) = 3\)     A1     N2

[2 marks]

A majority of candidates found success in the opening question. Common errors in (a) were to give \(f \circ g\) or to multiply f by g. 

For (b) some gave the inverse as the reciprocal function \(\frac{1}{{x + 3}}\) , or wrote \(x = y + 3\) .

Most candidates chose to find a composite in (c), sometimes making simple errors when working with brackets and a negative sign. Only a handful used the more efficient \(f(2) = 3\) . Additionally, it was not uncommon for candidates to give a correct substitution but not complete the result. Simple expressions such as \((7 - 2x) + 3\) should be finished as \(10 - 2x\) .