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Date November 2016 Marks available 6 Reference code 16N.2.sl.TZ0.10
Level SL only Paper 2 Time zone TZ0
Command term Find and Show that Question number 10 Adapted from N/A

Question

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

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

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

The graph of g is obtained from the graph of f by a translation of \left( {\begin{array}{*{20}{c}} k \\ 0 \end{array}} \right). The maximum point on the graph of g has coordinates (11.5,{\text{ }}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 = \frac{\pi }{6}.

(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

(i)     valid approach     (M1)

eg\,\,\,\,\,\frac{{5 + 17}}{2}

c = 11    A1     N2

(ii)     valid approach     (M1)

eg\,\,\,\,\,period is 12, per  = \frac{{2\pi }}{b},{\text{ }}9 - 3

b = \frac{{2\pi }}{{12}}    A1

b = \frac{\pi }{6}     AG     N0

(iii)     METHOD 1

valid approach     (M1)

eg\,\,\,\,\,5 = a\sin \left( {\frac{\pi }{6} \times 3} \right) + 11, substitution of points

a =  - 6     A1     N2

METHOD 2

valid approach     (M1)

eg\,\,\,\,\,\frac{{17 - 5}}{2}, amplitude is 6

a =  - 6     A1     N2

[6 marks]

a.

(i)     k = 2.5     A1     N1

(ii)     g(x) =  - 6\sin \left( {\frac{\pi }{6}(x - 2.5)} \right) + 11     A2     N2

[3 marks]

b.

(i)     METHOD 1 Using g

recognizing that a point of inflexion is required     M1

eg\,\,\,\,\,sketch, recognizing change in concavity

evidence of valid approach     (M1)

eg\,\,\,\,\,g''(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

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),{\text{ }} - \pi \cos \left( {\frac{\pi }{6}x} \right), max on derivative, sketch of derivative

attempt to find max value on derivative     M1

eg\,\,\,\,\, - \pi \cos \left( {\frac{\pi }{6}(8.5 - 2.5)} \right),{\text{ }}f'(6), dot on max of sketch

3.14159

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

[6 marks]

c.

Examiners report

[N/A]
a.
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b.
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c.

Syllabus sections

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