Show that \({f^{ - 1}}(x) = {3^{2x}}\) .

Write down the range of \({f^{ - 1}}\) .

Let \(g(x) = {\log _3}x\) , for \(x > 0\) .

Find the value of \(({f^{ - 1}} \circ g)(2)\) , giving your answer as an integer.

interchanging x and y (seen anywhere)     (M1)

e.g. \(x = \log \sqrt y \) (accept any base)

evidence of correct manipulation     A1

e.g. \(3^x = \sqrt y \) , \({3^y} = {x^{\frac{1}{2}}}\) , \(x = \frac{1}{2}{\log _3}y\) , \(2y = {\log _3}x\)

\({f^{ - 1}}(x) = {3^{2x}}\)     AG     N0 

[2 marks]

\(y > 0\) , \({f^{ - 1}}(x) > 0\)     A1     N1

[1 mark]

METHOD 1

finding \(g(2) = lo{g_3}2\) (seen anywhere)     A1

attempt to substitute     (M1)

e.g. \(({f^{ - 1}} \circ g)(2) = {3^{2\log {_3}2}}\)

evidence of using log or index rule     (A1)

e.g. \(({f^{ - 1}} \circ g)(2) = {3^{\log {_3}4}}\) , \({3^{{{\log }_3}2^2}}\)

\(({f^{ - 1}} \circ g)(2) = 4\)     A1     N1

METHOD 2

attempt to form composite (in any order)     (M1)

e.g. \(({f^{ - 1}} \circ g)(x) = {3^{2{{\log }_3}x}}\)

evidence of using log or index rule     (A1)

e.g. \(({f^{ - 1}} \circ g)(x) = {3^{{{\log }_3}{x^2}}}\) , \({3^{{{\log }_3}{x^{}}}}^2\)

\(({f^{ - 1}} \circ g)(x) = {x^2}\)     A1

\(({f^{ - 1}} \circ g)(2) = 4\)     A1     N1

[4 marks]

Candidates were generally skilled at finding the inverse of a logarithmic function.

Few correctly gave the range of this function, often stating “all real numbers” or “ \(y \ge 0\) ”, missing the idea that the range of an inverse is the domain of the original function.

Some candidates answered part (c) correctly, although many did not get beyond \({3^{2{{\log }_3}2}}\) . Some attempted to form the composite in the incorrect order. Others interpreted \(({f^{ - 1}} \circ g)(2)\) as multiplication by 2.