Recognizing \( f'(x) = 0 \):

\( \frac{6 – 2x}{6x – x^2} = 0 \)

Correct working:

\( 6 – 2x = 0 \implies x = 3 \)

Answer: \( x = 3 \) [3 marks]

Evidence of integration:

\( f(x) = \int f'(x) \, \mathrm{d}x = \int \frac{6 – 2x}{6x – x^2} \, \mathrm{d}x \)

Using substitution, let \( u = 6x – x^2 \):

\( \frac{\mathrm{d}u}{\mathrm{d}x} = 6 – 2x \implies \mathrm{d}u = (6 – 2x) \, \mathrm{d}x \)

\( f(x) = \int \frac{6 – 2x}{6x – x^2} \, \mathrm{d}x = \int \frac{1}{u} \, \mathrm{d}u = \ln |u| + c = \ln |6x – x^2| + c \)

Since \( 0 < x < 6 \), \( 6x – x^2 > 0 \), so drop absolute value:

\( f(x) = \ln (6x – x^2) + c \)

Substitute point \( (3, \ln 27) \):

\( \ln (6 \cdot 3 – 3^2) + c = \ln 27 \implies \ln (18 – 9) + c = \ln 27 \implies \ln 9 + c = \ln 27 \)

\( c = \ln 27 – \ln 9 = \ln \left( \frac{27}{9} \right) = \ln 3 \)

EITHER

\( f(x) = \ln (6x – x^2) + \ln 3 \)

Use logarithm law: \( \ln a + \ln b = \ln (a \cdot b) \):

\( f(x) = \ln \left( 3(6x – x^2) \right) \)

OR

\( f(x) = \ln (6x – x^2) + \ln 27 – \ln 9 \)

\( \ln 27 – \ln 9 = \ln \left( \frac{27}{9} \right) = \ln 3 \)

\( f(x) = \ln (6x – x^2) + \ln 3 = \ln \left( 3(6x – x^2) \right) \)

Answer: \( f(x) = \ln \left( 3(6x – x^2) \right) \) [8 marks]

Point P has coordinates \( (3, \ln 27) \). After vertical stretch by scale factor \( \frac{1}{\ln 3} \), the new coordinates are \( (a, b) \):

\( a = 3 \) (x-coordinate unchanged)

y-coordinate transformation: \( b = \frac{\ln 27}{\ln 3} \)

Use logarithm law: \( \ln 27 = \ln (3^3) = 3 \ln 3 \):

\( b = \frac{3 \ln 3}{\ln 3} = 3 \)

Answer: \( a = 3 \), \( b = 3 \) [4 marks]