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Date November 2017 Marks available 7 Reference code 17N.1.SL.TZ0.S_7
Level Standard Level Paper Paper 1 Time zone Time zone 0
Command term Find Question number S_7 Adapted from N/A

Question

Consider f ( x ) = log k ( 6 x 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

* 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 6 x 3 x 2 = k 2

valid approach     (M1)

eg 3 x 2 + 6 x k 2 = 0 ,   3 x 2 6 x + k 2 = 0

recognizing discriminant must be zero (seen anywhere)     M1

eg Δ = 0

correct discriminant     (A1)

eg 6 2 4 ( 3 ) ( k 2 ) ,   36 12 k 2 = 0

correct working     (A1)

eg 12 k 2 = 36 ,   k 2 = 3

k = 3     A2     N2

METHOD 2 – completing the square

correct equation without logs     (A1)

eg 6 x 3 x 2 = k 2

valid approach to complete the square     (M1)

eg 3 ( x 2 2 x + 1 ) = k 2 + 3 ,   x 2 2 x + 1 1 + k 2 3 = 0

correct working     (A1)

eg 3 ( x 1 ) 2 = k 2 + 3 ,   ( x 1 ) 2 1 + k 2 3 = 0

recognizing conditions for one solution     M1

eg ( x 1 ) 2 = 0 ,   1 + k 2 3 = 0

correct working     (A1)

eg k 2 3 = 1 ,   k 2 = 3

k = 3     A2     N2

[7 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5—Calculus » SL 5.6—Stationary points, local max and min
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