Consider $f(x) = \log k(6x - 3{x^2})$, for $0 < x < 2$, where $k > 0$.

The equation $f(x) = 2$ has exactly one solution. Find the value of $k$.

METHOD 1 – using discriminant

correct equation without logs     (A1)

eg$\,\,\,\,\,$$6x - 3{x^2} = {k^2}$

valid approach     (M1)

eg$\,\,\,\,\,$$- 3{x^2} + 6x - {k^2} = 0,{\text{ }}3{x^2} - 6x + {k^2} = 0$

recognizing discriminant must be zero (seen anywhere)     M1

eg$\,\,\,\,\,$$\Delta  = 0$

correct discriminant     (A1)

eg$\,\,\,\,\,$${6^2} - 4( - 3)( - {k^2}),{\text{ }}36 - 12{k^2} = 0$

correct working     (A1)

eg$\,\,\,\,\,$$12{k^2} = 36,{\text{ }}{k^2} = 3$

$k = \sqrt 3$     A2     N2

METHOD 2 – completing the square

correct equation without logs     (A1)

eg$\,\,\,\,\,$$6x - 3{x^2} = {k^2}$

valid approach to complete the square     (M1)

eg$\,\,\,\,\,$$3({x^2} - 2x + 1) =  - {k^2} + 3,{\text{ }}{x^2} - 2x + 1 - 1 + \frac{{{k^2}}}{3} = 0$

correct working     (A1)

eg$\,\,\,\,\,$$3{(x - 1)^2} =  - {k^2} + 3,{\text{ }}{(x - 1)^2} - 1 + \frac{{{k^2}}}{3} = 0$

recognizing conditions for one solution     M1

eg$\,\,\,\,\,$${(x - 1)^2} = 0,{\text{ }} - 1 + \frac{{{k^2}}}{3} = 0$

correct working     (A1)

eg$\,\,\,\,\,$$\frac{{{k^2}}}{3} = 1,{\text{ }}{k^2} = 3$

$k = \sqrt 3$     A2     N2

[7 marks]