Given that \(f(x) = \ln x\), use the mean value theorem to show that, for \(0 < a < b\), \(\frac{{b - a}}{b} < \ln \frac{b}{a} < \frac{{b - a}}{a}\).

Hence show that \(\ln (1.2)\) lies between \(\frac{1}{m}\) and \(\frac{1}{n}\), where \(m\), \(n\) are consecutive positive integers to be determined.

\(f'(x) = \frac{1}{x}\)    (A1)

using the MVT \(f'(c) = \frac{{f(b) - f(a)}}{{b - a}}\) (where \(c\) lies between \(a\) and \(b\))     (M1)

\(f'(c) = \frac{{\ln b - \ln a}}{{b - a}}\)    A1

\(\ln \frac{b}{a} = \ln b - \ln a\)    (M1)

\(f'(c) = \frac{{\ln \frac{b}{a}}}{{b - a}}\)

since \(f'(x)\) is a decreasing function or \(a < c < b \Rightarrow \frac{1}{b} < \frac{1}{c} < \frac{1}{a}\)     R1

\(f'(b) < f'(c) < f'(a)\)    (M1)

\(\frac{1}{b} < \frac{{\ln \frac{b}{a}}}{{b - a}} < \frac{1}{a}\)    A1

\(\frac{{b - a}}{b} < \ln \frac{b}{a} < \frac{{b - a}}{a}\)    AG

[7 marks]

putting \(b = 1.2,{\text{ }}a = 1\), or equivalent     M1

\(\frac{1}{6} < \ln 1.2 < \frac{1}{5}\)    A1

\((m = 6,{\text{ }}n = 5)\)

[2 marks]

Although many candidates achieved at least a few marks in this question, the answers revealed difficulties in setting up a proof. The Mean value theorem was poorly quoted and steps were often skipped. The conditions under which the Mean value theorem is valid were largely ignored, as were the reasoned steps towards the answer.

There were inequalities everywhere, without a great deal of meaning or showing progress. A number of candidates attempted to work backwards and presented the work in a way that made it difficult to follow their reasoning; in part (b) many candidates ignored the instruction ‘hence’ and just used GDC to find the required values; candidates that did notice the link to part a) answered this question well in general. A number of candidates guessed the answer and did not present an analytical derivation as required.