Show that $f'(x) = \frac{{1 - \ln x}}{{{x^2}}}$.

Find the coordinates of B, at which the curve reaches its maximum value.

Find the coordinates of C, the point of inflexion on the curve.

The graph of $y = {\text{ }}f(x)$ crosses the $x$-axis at the point A.

Find the equation of the tangent to the graph of $f$ at the point A.

The graph of $y = {\text{ }}f(x)$ crosses the $x$-axis at the point A.

Find the area enclosed by the curve $y = f(x)$, the tangent at A, and the line $x = {\text{e}}$.

$f'(x) = \frac{{x \times \frac{1}{x} - \ln x}}{{{x^2}}}$     M1A1

$= \frac{{1 - \ln x}}{{{x^2}}}$     AG

[2 marks]

$\frac{{1 - \ln x}}{{{x^2}}} = 0$ has solution $x = {\text{e}}$     M1A1

$y = \frac{1}{{\text{e}}}$     A1

hence maximum at the point $\left( {{\text{e, }}\frac{1}{{\text{e}}}} \right)$

[3 marks]

$f''(x) = \frac{{{x^2}\left( { - \frac{1}{x}} \right) - 2x(1 - \ln x)}}{{{x^4}}}$     M1A1

$= \frac{{2\ln x - 3}}{{{x^3}}}$

Note:     The M1A1 should be awarded if the correct working appears in part (b).

point of inflexion where $f''(x) = 0$     M1

so $x = {{\text{e}}^{\frac{3}{2}}},{\text{ }}y = \frac{3}{2}{{\text{e}}^{\frac{{ - 3}}{2}}}$     A1A1

C has coordinates $\left( {{{\text{e}}^{\frac{3}{2}}},{\text{ }}\frac{3}{2}{{\text{e}}^{\frac{{ - 3}}{2}}}} \right)$

[5 marks]

$f(1) = 0$     A1

$f'(1) = 1$     (A1)

$y = x + c$     (M1)

through (1, 0)

equation is $y = x - 1$     A1

[4 marks]

METHOD 1

area $= \int_1^{\text{e}} {x - 1 - \frac{{\ln x}}{x}{\text{d}}x}$     M1A1A1

Note:     Award M1 for integration of difference between line and curve, A1 for correct limits, A1 for correct expressions in either order.

$\int {\frac{{\ln x}}{x}{\text{d}}x = \frac{{{{(\ln x)}^2}}}{2}} ( + c)$     (M1)A1

$\int {(x - 1){\text{d}}x = \frac{{{x^2}}}{2} - x( + c)}$     A1

$= \left[ {\frac{1}{2}{x^2} - x - \frac{1}{2}{{(\ln x)}^2}} \right]_1^{\text{e}}$

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

$= \frac{1}{2}{{\text{e}}^2} - {\text{e}}$     A1

METHOD 2

area = area of triangle $- \int_1^e {\frac{{\ln x}}{x}{\text{d}}x}$     M1A1

Note:     A1 is for correct integral with limits and is dependent on the M1.

$\int {\frac{{\ln x}}{x}{\text{d}}x = \frac{{{{(\ln x)}^2}}}{2}( + c)}$     (M1)A1

area of triangle $= \frac{1}{2}(e - 1)(e - 1)$     M1A1

$\frac{1}{2}(e - 1)(e - 1) - \left( {\frac{1}{2}} \right) = \frac{1}{2}{{\text{e}}^2} - {\text{e}}$     A1

[7 marks]