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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 \(f(x) = a\sin bx + c\), for \(0 \leqslant x \leqslant 12\).

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

Syllabus sections

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