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.