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Date November 2016 Marks available 6 Reference code 16N.2.SL.TZ0.S_10
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Show that and Find Question number S_10 Adapted from N/A

Question

The following diagram shows the graph of f(x)=asinbx+c, for 0x12.

N16/5/MATME/SP2/ENG/TZ0/10

The graph of f has a minimum point at (3, 5) and a maximum point at (9, 17).

The graph of g is obtained from the graph of f by a translation of (k0). The maximum point on the graph of g has coordinates (11.5, 17).

The graph of g changes from concave-up to concave-down when x=w.

(i)     Find the value of c.

(ii)     Show that b=π6.

(iii)     Find the value of a.

[6]
a.

(i)     Write down the value of k.

(ii)     Find g(x).

[3]
b.

(i)     Find w.

(ii)     Hence or otherwise, find the maximum positive rate of change of g.

[6]
c.

Markscheme

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

(i)     valid approach     (M1)

eg5+172

c=11    A1     N2

(ii)     valid approach     (M1)

egperiod is 12, per =2πb, 93

b=2π12    A1

b=π6     AG     N0

(iii)     METHOD 1

valid approach     (M1)

eg5=asin(π6×3)+11, substitution of points

a=6     A1     N2

METHOD 2

valid approach     (M1)

eg1752, amplitude is 6

a=6     A1     N2

[6 marks]

a.

(i)     k=2.5     A1     N1

(ii)     g(x)=6sin(π6(x2.5))+11     A2     N2

[3 marks]

b.

(i)     METHOD 1 Using g

recognizing that a point of inflexion is required     M1

egsketch, recognizing change in concavity

evidence of valid approach     (M1)

egg, sketch, coordinates of max/min on  g

w = 8.5 (exact)     A1     N2

METHOD 2 Using f

recognizing that a point of inflexion is required     M1

eg sketch, recognizing change in concavity

evidence of valid approach involving translation     (M1)

eg x = w k , sketch,  6 + 2.5

w = 8.5  (exact)     A1     N2

(ii)     valid approach involving the derivative of g or f (seen anywhere)     (M1)

eg g ( w ) ,   π cos ( π 6 x ) , max on derivative, sketch of derivative

attempt to find max value on derivative     M1

eg π cos ( π 6 ( 8.5 2.5 ) ) ,   f ( 6 ) , dot on max of sketch

3.14159

max rate of change = π  (exact), 3.14     A1     N2

[6 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

Topic 4—Statistics and probability » SL 4.1—Concepts, reliability and sampling techniques
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Topic 5—Calculus » SL 5.4—Tangents and normal
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Topic 5—Calculus » SL 5.7—Optimisation
Topic 2—Functions
Topic 4—Statistics and probability
Topic 5—Calculus

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