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Date May 2016 Marks available [N/A] Reference code 16M.1.sl.TZ1.9
Level SL only Paper 1 Time zone TZ1
Command term Question number 9 Adapted from N/A

Question

Let f(x)=62x6xx2, for 0<x<6.

The graph of f has a maximum point at P.

The y-coordinate of P is ln27.

Find the x-coordinate of P.

[3]
a.

Find f(x), expressing your answer as a single logarithm.

[8]
b.

The graph of f is transformed by a vertical stretch with scale factor 1ln3. The image of P under this transformation has coordinates (a, b).

Find the value of a and of b, where a, bN.

[[N/A]]
c.

Markscheme

recognizing f(x)=0     (M1)

correct working     (A1)

eg62x=0

x=3    A1     N2

[3 marks]

a.

evidence of integration     (M1)

egf, 62x6xx2dx

using substitution     (A1)

eg1udu where u=6xx2

correct integral     A1

egln(u)+c, ln(6xx2)

substituting (3, ln27) into their integrated expression (must have c)     (M1)

egln(6×332)+c=ln27, ln(189)+lnk=ln27

correct working     (A1)

egc=ln27ln9

EITHER

c=ln3    (A1)

attempt to substitute their value of c into f(x)     (M1)

egf(x)=ln(6xx2)+ln3     A1     N4

OR

attempt to substitute their value of c into f(x)     (M1)

egf(x)=ln(6xx2)+ln27ln9

correct use of a log law     (A1)

egf(x)=ln(6xx2)+ln(279), f(x)=ln(27(6xx2))ln9

f(x)=ln(3(6xx2))    A1     N4

[8 marks]

b.

a=3    A1     N1

correct working     A1

egln27ln3

correct use of log law     (A1)

eg3ln3ln3, log327

b=3    A1     N2

[4 marks]

c.

Examiners report

Part a) was well answered.

a.

In part b) most candidates realised that integration was required but fewer recognised the need to use integration by substitution. Quite a number of candidates who integrated correctly omitted finding the constant of integration.

b.

In part c) many candidates showed good understanding of transformations and could apply them correctly, however, correct use of the laws of logarithms was challenging for many. In particular, a common error was ln27ln3=ln9.

c.

Syllabus sections

Topic 6 - Calculus » 6.3 » Local maximum and minimum points.
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