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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 Hence or otherwise 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(x)=0, 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

egsketch, recognizing change in concavity

evidence of valid approach involving translation     (M1)

egx=wk, sketch, 6+2.5

w=8.5 (exact)     A1     N2

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

egg(w), πcos(π6x), max on derivative, sketch of derivative

attempt to find max value on derivative     M1

egπcos(π6(8.52.5)), 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 3— Geometry and trigonometry » SL 3.7—Circular functions: graphs, composites, transformations
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