Use the quotient rule to show that $g'(x) = \frac{{1 - 2\ln x}}{{{x^3}}}$ .

The graph of g has a maximum point at A. Find the x-coordinate of A.

$\frac{{\rm{d}}}{{{\rm{d}}x}}\ln x = \frac{1}{x}$ , $\frac{{\rm{d}}}{{{\rm{d}}x}}{x^2} = 2x$ (seen anywhere)     A1A1

attempt to substitute into the quotient rule (do not accept product rule)     M1

e.g. $\frac{{{x^2}\left( {\frac{1}{x}} \right) - 2x\ln x}}{{{x^4}}}$

correct manipulation that clearly leads to result     A1

e.g. $\frac{{x - 2x\ln x}}{{{x^4}}}$ , $\frac{{x(1 - 2\ln x)}}{{{x^4}}}$ , $\frac{x}{{{x^4}}}$ , $\frac{{2x\ln x}}{{{x^4}}}$

$g'(x) = \frac{{1 - 2\ln x}}{{{x^3}}}$     AG     N0

[4 marks]

evidence of setting the derivative equal to zero     (M1)

e.g. $g'(x) = 0$ , $1 - 2\ln x = 0$

$\ln x = \frac{1}{2}$     A1

$x = {{\rm{e}}^{\frac{1}{2}}}$     A1     N2

[3 marks]

Many candidates clearly knew their quotient rule, although a common error was to simplify $2x\ln x$ as $2\ln {x^2}$ and then "cancel" the exponents.

For (b), those who knew to set the derivative to zero typically went on find the correct x-coordinate, which must be in terms of e, as this is the calculator-free paper. Occasionally, students would take $\frac{{1 - 2\ln x}}{{{x^3}}} = 0$ and attempt to solve from $1 - 2\ln x = {x^3}$ .