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]