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]