June 2019 Paper 1 Q9
9. Given that \(a \gt b \gt 0\) and that \(a\) and \(b\) satisfy the equation
\[\log a - \log b = \log(a - b)\]| Scheme | Marks | AO |
|---|---|---|
| States \(\log a - \log b = \log\dfrac{a}{b}\) | B1 | 1.2 |
| Proceeds from \(\dfrac{a}{b} = a - b \to \ldots\ldots \to ab - a = b^2\) | M1 | 1.1b |
| \(ab - a = b^2 \to a(b-1) = b^2 \Rightarrow a = \dfrac{b^2}{b-1}\ *\) | A1* | 2.1 |
| (3) |
Notes
B1: States or uses \(\log a - \log b = \log\dfrac{a}{b}\). This may be awarded anywhere in the question and may be implied by a starting line of \(\dfrac{a}{b} = a - b\) oe. Alternatively takes \(\log b\) to the rhs and uses the addition law \(\log(a-b) + \log b = \log(a-b)b\). Watch out for \(\log a - \log b = \dfrac{\log a}{\log b} = \log\left(\dfrac{a}{b}\right)\) which could score 010
M1: Attempts to make ‘\(a\)’ the subject. Awarded for proceeding from \(\dfrac{a}{b} = a - b\) to a point where the two terms in \(a\) are on the same side of the equation and the term in \(b\) is on the other.
A1*: CSO. Shows clear reasoning and correct mathematics leading to \(a = \dfrac{b^2}{b-1}\). Bracketing must be correct.
Allow a candidate to proceed from \(ab - a = b^2\) to \(a = \dfrac{b^2}{b-1}\) without the intermediate line.
Alt (a)
Note that it is possible to attempt part (a) by substituting \(a = \dfrac{b^2}{b-1}\) into both sides of the given identity.
\[\log a - \log b = \log(a-b) \Rightarrow \log\left(\frac{b^2}{b-1}\right) - \log b = \log\left(\frac{b^2}{b-1} - b\right)\]B1: Score for \(\log\left(\dfrac{b^2}{b-1}\right) - \log b = \log\left(\dfrac{b}{b-1}\right)\)
M1: Attempts to write \(\dfrac{b^2}{b-1} - b\) as a single fraction \(\dfrac{b^2}{b-1} - b = \dfrac{b^2 - b(b-1)}{b-1}\)
Allow as two separate fractions with the same common denominator
A1*: Achieves lhs and rhs as \(\log\left(\dfrac{b}{b-1}\right)\) and makes a comment such as "hence true"
| Scheme | Marks | AO |
|---|---|---|
| States either \(b \gt 1\) or \(b \neq 1\) with reason \(\dfrac{b^2}{b-1}\) is not defined at \(b = 1\) oe | B1 | 2.2a |
| States \(b \gt 1\) and explains that as \(a \gt 0 \Rightarrow \dfrac{b^2}{b-1} \gt 0 \Rightarrow b \gt 1\) | B1 | 2.4 |
| (2) | ||
| (5 marks) |
Notes
B1: For deducing \(b \neq 1\) as \(a \to \infty\) oe such as "you cannot divide by 0" or correctly deducing that \(b \gt 1\).
They may state that \(b\) cannot be less than 1.
B1: For \(b \gt 1\) and explaining that as \(a \gt 0 \Rightarrow \dfrac{b^2}{b-1} \gt 0 \Rightarrow b \gt 1\) (as \(b^2\) is positive)
As a minimum accept that \(b \gt 1\) as \(a\) cannot be negative.
Note that \(a \gt b \gt 1\) is a correct statement but not sufficient on its own without an explanation.