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 $\frac{A}{B}$.