Given that ${f^{ - 1}}(1) = 8$ , find the value of $k$ .

Find ${f^{ - 1}}\left( {\frac{2}{3}} \right)$ .

METHOD 1

recognizing that $f(8) = 1$     (M1)

e.g. $1 = k{\log _2}8$

recognizing that ${\log _2}8 = 3$     (A1)

e.g. $1 = 3k$

$k = \frac{1}{3}$     A1     N2

METHOD 2

attempt to find the inverse of $f(x) = k{\log _2}x$     (M1)

e.g. $x = k{\log _2}y$ , $y = {2^{\frac{x}{k}}}$

substituting 1 and 8     (M1)

e.g. $1 = k{\log _2}8$ , ${2^{\frac{1}{k}}} = 8$

$k = \frac{1}{{{{\log }_2}8}}$ $\left( {k = \frac{1}{3}} \right)$     A1     N2

[3 marks]

METHOD 1

recognizing that $f(x) = \frac{2}{3}$     (M1)

e.g. $\frac{2}{3} = \frac{1}{3}{\log _2}x$

${\log _2}x = 2$     (A1)

${f^{ - 1}}\left( {\frac{2}{3}} \right) = 4$ (accept $x = 4$)     A2     N3

METHOD 2

attempt to find inverse of $f(x) = \frac{1}{3}{\log _2}x$     (M1)

e.g. interchanging x and y , substituting $k = \frac{1}{3}$ into $y = {2^{\frac{x}{k}}}$

correct inverse     (A1)

e.g. ${f^{ - 1}}(x) = {2^{3x}}$ , ${2^{3x}}$

${f^{ - 1}}\left( {\frac{2}{3}} \right) = 4$     A2    N3

[4 marks]

A very poorly done question. Most candidates attempted to find the inverse function for $f$ and used that to answer parts (a) and (b). Few recognized that the explicit inverse function was not necessary to answer the question.

Although many candidates seem to know that they can find an inverse function by interchanging x and y, very few were able to actually get the correct inverse. Almost none recognized that if ${f^{ - 1}}(1) = 8$ , then $f(8) = 1$ . Many thought that the letters "log" could be simply "cancelled out", leaving the $2$ and the $8$.

A very poorly done question. Most candidates attempted to find the inverse function for $f$ and used that to answer parts (a) and (b). Few recognized that the explicit inverse function was not necessary to answer the question.

Although many candidates seem to know that they can find an inverse function by interchanging x and y, very few were able to actually get the correct inverse. Almost none recognized that if ${f^{ - 1}}(1) = 8$ , then $f(8) = 1$ . Many thought that the letters "log" could be simply "cancelled out", leaving the $2$ and the $8$.