Find the equation of the tangent to C at the point (2, e).

Find $f(x)$.

Determine the largest possible domain of f.

Show that the equation $f(x) = f'(x)$ has no solution.

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{\text{e}}}{{\ln {\text{e}}}}(2 + 2) = 4{\text{e}}$     A1

at (2, e) the tangent line is $y - {\text{e}} = 4{\text{e}}(x - 2)$     M1

hence $y = 4{\text{e}}x - 7{\text{e}}$     A1

[3 marks]

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{{\ln y}}(x + 2) \Rightarrow \frac{{\ln y}}{y}{\text{d}}y = (x + 2){\text{d}}x$     M1

$\int {\frac{{\ln y}}{y}{\text{d}}y = \int {(x + 2){\text{d}}x} }$

using substitution $u = \ln y;{\text{ d}}u = \frac{1}{y}{\text{d}}y$     (M1)(A1)

$\Rightarrow \int {\frac{{\ln y}}{y}{\text{d}}y = \int {u{\text{d}}u = \frac{1}{2}{u^2}} }$     (A1)

$\Rightarrow \frac{{{{(\ln y)}^2}}}{2} = \frac{{{x^2}}}{2} + 2x + c$     A1A1

at (2, e), $\frac{{{{(\ln {\text{e}})}^2}}}{2} = 6 + c$     M1

$\Rightarrow c = - \frac{{11}}{2}$     A1

$\Rightarrow \frac{{{{(\ln y)}^2}}}{2} = \frac{{{x^2}}}{2} + 2x - \frac{{11}}{2} \Rightarrow {(\ln y)^2} = {x^2} + 4x - 11$

$\ln y = \pm \sqrt {{x^2} + 4x - 11}  \Rightarrow y = {{\text{e}}^{ \pm \sqrt {{x^2} + 4x - 11} }}$     M1A1

since y > 1, $f(x) = {{\text{e}}^{\sqrt {{x^2} + 4x - 11} }}$     R1

Note:M1 for attempt to make y the subject.

[11 marks]

EITHER

${x^2} + 4x - 11 > 0$     A1

using the quadratic formula     M1

critical values are $\frac{{ - 4 \pm \sqrt {60} }}{2}{\text{ }}\left( { = - 2 \pm \sqrt {15} } \right)$     A1

using a sign diagram or algebraic solution     M1

$x < - 2 - \sqrt {15} ;{\text{ }}x > - 2 + \sqrt {15}$     A1A1

OR

${x^2} + 4x - 11 > 0$     A1

by methods of completing the square     M1

${(x + 2)^2} > 15$     A1

$\Rightarrow x + 2 < - \sqrt {15} {\text{ or }}x + 2 > \sqrt {15}$     (M1)

$x < - 2 - \sqrt {15} ;{\text{ }}x > - 2 + \sqrt {15}$     A1A1

[6 marks]

$f(x) = f'(x) \Rightarrow f(x) = \frac{{f(x)}}{{\ln f(x)}}(x + 2)$     M1

$\Rightarrow \ln \left( {f(x)} \right) = x + 2\,\,\,\,\,\left( { \Rightarrow x + 2 = \sqrt {{x^2} + 4x - 11} } \right)$     A1

$\Rightarrow {(x + 2)^2} = {x^2} + 4x - 11 \Rightarrow {x^2} + 4x + 4 = {x^2} + 4x - 11$     A1

$\Rightarrow 4 = - 11,{\text{ hence }}f(x) \ne f'(x)$     R1AG

[4 marks]

Nearly always correctly answered.

Most candidates separated the variables and attempted the integrals. Very few candidates made use of the condition y > 1, so losing 2 marks.

Part (c) was often well answered, sometimes with follow through.

Only the best candidates were successful on part (d).