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).