Consider the functions \(f(x) = {(\ln x)^2},{\text{ }}x > 1\) and \(g(x) = \ln \left( {f(x)} \right),{\text{ }}x > 1\).

(i)     Find \(f'(x)\).

(ii)     Find \(g'(x)\).

(iii)     Hence, show that \(g(x)\) is increasing on \(\left] {1,{\text{ }}\infty } \right[\).

Consider the differential equation

\[(\ln x)\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{2}{x}y = \frac{{2x - 1}}{{(\ln x)}},{\text{ }}x > 1.\]

(i)     Find the general solution of the differential equation in the form \(y = h(x)\).

(ii)     Show that the particular solution passing through the point with coordinates \(\left( {{\text{e, }}{{\text{e}}^2}} \right)\) is given by \(y = \frac{{{x^2} - x + {\text{e}}}}{{{{(\ln x)}^2}}}\).

(iii)     Sketch the graph of your solution for \(x > 1\), clearly indicating any asymptotes and any maximum or minimum points.

(i)     attempt at chain rule     (M1)

\(f'(x) = \frac{{2\ln x}}{x}\)     A1

(ii)     attempt at chain rule     (M1)

\(g'(x) = \frac{2}{{x\ln x}}\)     A1

(iii)     \(g'(x)\) is positive on \(\left] {1,{\text{ }}\infty } \right[\)     A1

so \(g(x)\) is increasing on \(\left] {1,{\text{ }}\infty } \right[\)     AG

[5 marks]

(i)     rearrange in standard form:

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{2}{{x\ln x}}y = \frac{{2x - 1}}{{{{(\ln x)}^2}}},{\text{ }}x > 1\)     (A1)

integrating factor:

\({{\text{e}}^{\int {\frac{2}{{x\ln x}}{\text{d}}x} }}\)     (M1)

\( = {{\text{e}}^{\ln \left( {{{(\ln x)}^2}} \right)}}\)

\( = {(\ln x)^2}\)     (A1)

multiply by integrating factor     (M1)

\({(\ln x)^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + \frac{{2\ln x}}{x}y = 2x - 1\)

\(\frac{{\text{d}}}{{{\text{d}}x}}\left( {y{{(\ln x)}^2}} \right) = 2x - 1{\text{ }}\left( {{\text{or }}y{{(\ln x)}^2} = \int {2x - 1{\text{d}}x} } \right)\)     M1

attempt to integrate:     M1

\({(\ln x)^2}y = {x^2} - x + c\)

\(y = \frac{{{x^2} - x + c}}{{{{(\ln x)}^2}}}\)     A1

(ii)     attempt to use the point \(\left( {{\text{e, }}{{\text{e}}^2}} \right)\) to determine c:     M1

eg, \({(\ln {\text{e}})^2}{{\text{e}}^2} = {{\text{e}}^2} - {\text{e}} + {\text{c}}\) or \({{\text{e}}^2} = \frac{{{{\text{e}}^2} - {\text{e}} + {\text{c}}}}{{{{(\ln {\text{e}})}^2}}}\) or \({{\text{e}}^2} = {{\text{e}}^2} - {\text{e}} + {\text{c}}\)

\({\text{c}} = {\text{e}}\)     A1

\(y = \frac{{{x^2} - x + {\text{e}}}}{{{{(\ln x)}^2}}}\)     AG

(iii)     

graph with correct shape     A1

minimum at \(x = 3.1\) (accept answers to a minimum of 2 s.f)     A1

asymptote shown at \(x = 1\)     A1

Note: y-coordinate of minimum not required for A1;

     Equation of asymptote not required for A1 if VA appears on the sketch.

     Award A0 for asymptotes if more than one asymptote are shown

[12 marks]