17N.1.SL.TZ0.S_7

pestleMathematicsAISLPaper 117N· sl-5-6-stationary-points-local-max-and-minsource ↗

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 $k$ .

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

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]

Examiners’ report

[N/A]