Show that $f'(x) = \frac{{\ln x}}{x}$.

There is a minimum on the graph of $f$. Find the $x$-coordinate of this minimum.

Write down the value of $p$.

The graph of $g$ intersects the graph of $f'$ when $x = q$.

Find the value of $q$.

The graph of $g$ intersects the graph of $f'$ when $x = q$.

Let $R$ be the region enclosed by the graph of $f'$, the graph of $g$ and the line $x = p$.

Show that the area of $R$ is $\frac{1}{2}$.

METHOD 1

correct use of chain rule     A1A1

eg     $\frac{{2\ln x}}{2} \times \frac{1}{x},{\text{ }}\frac{{2\ln x}}{{2x}}$

Note: Award A1 for $\frac{{2\ln x}}{{2x}}$, A1 for $\times \frac{1}{x}$.

$f'(x) = \frac{{\ln x}}{x}$     AG     N0

[2 marks]

METHOD 2

correct substitution into quotient rule, with derivatives seen     A1

eg     $\frac{{2 \times 2\ln x \times \frac{1}{x} - 0 \times {{(\ln x)}^2}}}{4}$

correct working     A1

eg     $\frac{{4\ln x \times \frac{1}{x}}}{4}$

$f'(x) = \frac{{\ln x}}{x}$     AG     N0

[2 marks]

setting derivative $= 0$     (M1)

eg     $f'(x) = 0,{\text{ }}\frac{{\ln x}}{x} = 0$

correct working     (A1)

eg     $\ln x = 0,{\text{ }}x = {{\text{e}}^0}$

$x = 1$     A1     N2

[3 marks] 

intercept when $f'(x) = 0$     (M1)

$p = 1$     A1     N2

[2 marks]

equating functions     (M1)

eg     $f' = g,{\text{ }}\frac{{\ln x}}{x} = \frac{1}{x}$

correct working     (A1)

eg     $\ln x = 1$

$q = {\text{e   (accept }}x = {\text{e)}}$     A1     N2

[3 marks]

evidence of integrating and subtracting functions (in any order, seen anywhere)     (M1)

eg     $\int_q^e {\left( {\frac{1}{x} - \frac{{\ln x}}{x}} \right){\text{d}}x{\text{, }}\int {f' - g} }$

correct integration $\ln x - \frac{{{{(\ln x)}^2}}}{2}$     A2

substituting limits into their integrated function and subtracting (in any order)     (M1)

eg     $(\ln {\text{e}} - \ln 1) - \left( {\frac{{{{(\ln {\text{e}})}^2}}}{2} - \frac{{{{(\ln 1)}^2}}}{2}} \right)$

Note: Do not award M1 if the integrated function has only one term.

correct working     A1

eg     $(1 - 0) - \left( {\frac{1}{2} - 0} \right),{\text{ }}1 - \frac{1}{2}$

${\text{area}} = \frac{1}{2}$     AG     N0

Notes: Candidates may work with two separate integrals, and only combine them at the end. Award marks in line with the markscheme.

[5 marks]