Find the area of region A in terms of k .

Find the area of region B in terms of k .

Find the ratio of the area of region A to the area of region B .

\(\int_{\frac{1}{6}}^1 {\frac{k}{x} - \frac{1}{x}{\text{d}}x = (k - 1} )[\ln x]_{\frac{1}{6}}^1\)     M1     A1

Note: Award M1 for \(\int {\frac{k}{x} - \frac{1}{x}{\text{d}}x{\text{ or }}\int {\frac{1}{x} - \frac{k}{x}{\text{d}}x} } \) and A1 for \((k - 1)\ln x\) seen in part (a) or later in part (b).

\( = (1 - k)\ln \frac{1}{6}\)     A1

[3 marks]

\(\int_1^{\sqrt 6 } {\frac{k}{x} - \frac{1}{x}{\text{d}}x = (k - 1} )[\ln x]_1^{\sqrt 6 }\)     (A1)

Note: Award A1 for correct change of limits.

\( = (k - 1)\ln \sqrt 6 \)     A1

[2 marks]

\((1 - k)\ln \frac{1}{6} = (k - 1)\ln 6\)     A1

\((k - 1)\ln \sqrt 6  = \frac{1}{2}(k - 1)\ln 6\)     A1

 Note: This simplification could have occurred earlier, and marks should still be awarded.

ratio is 2 (or 2:1)     A1

[3 marks] 

Generally well answered by most candidates. Basic algebra sometimes let students down in the simplification of the ratio in part (c). It was not uncommon to see \(\frac{{\log A}}{{\log B}}\) simplified to \(\frac{A}{B}\).

Generally well answered by most candidates. Basic algebra sometimes let students down in the simplification of the ratio in part (c). It was not uncommon to see \(\frac{{\log A}}{{\log B}}\) simplified to \(\frac{A}{B}\).

Generally well answered by most candidates. Basic algebra sometimes let students down in the simplification of the ratio in part (c). It was not uncommon to see \(\frac{{\log A}}{{\log B}}\) simplified to \(\frac{A}{B}\).