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Date November 2017 Marks available 2 Reference code 17N.2.sl.TZ0.10
Level SL only Paper 2 Time zone TZ0
Command term Show that Question number 10 Adapted from N/A

Question

Note: In this question, distance is in millimetres.

Let f(x)=x+asin(xπ2)+a, for x.

The graph of f passes through the origin. Let {{\text{P}}_k} be any point on the graph of f with x-coordinate 2k\pi , where k \in \mathbb{N}. A straight line L passes through all the points {{\text{P}}_k}.

Diagram 1 shows a saw. The length of the toothed edge is the distance AB.

N17/5/MATME/SP2/ENG/TZ0/10.d_01

The toothed edge of the saw can be modelled using the graph of f and the line L. Diagram 2 represents this model.

N17/5/MATME/SP2/ENG/TZ0/10.d_02

The shaded part on the graph is called a tooth. A tooth is represented by the region enclosed by the graph of f and the line L, between {{\text{P}}_k} and {{\text{P}}_{k + 1}}.

Show that f(2\pi ) = 2\pi .

[3]
a.

Find the coordinates of {{\text{P}}_0} and of {{\text{P}}_1}.

[3]
b.i.

Find the equation of L.

[3]
b.ii.

Show that the distance between the x-coordinates of {{\text{P}}_k} and {{\text{P}}_{k + 1}} is 2\pi .

[2]
c.

A saw has a toothed edge which is 300 mm long. Find the number of complete teeth on this saw.

[6]
d.

Markscheme

substituting x = 2\pi     M1

eg\,\,\,\,\,2\pi  + a\sin \left( {2\pi  - \frac{\pi }{2}} \right) + a

2\pi  + a\sin \left( {\frac{{3\pi }}{2}} \right) + a     (A1)

2\pi  - a + a     A1

f(2\pi ) = 2\pi     AG     N0

[3 marks]

a.

substituting the value of k     (M1)

{{\text{P}}_0}(0,{\text{ }}0),{\text{ }}{{\text{P}}_1}(2\pi ,{\text{ }}2\pi )     A1A1     N3

[3 marks]

b.i.

attempt to find the gradient     (M1)

eg\,\,\,\,\,\frac{{2\pi  - 0}}{{2\pi  - 0}},{\text{ }}m = 1

correct working     (A1)

eg\,\,\,\,\,\frac{{y - 2\pi }}{{x - 2\pi }} = 1,{\text{ }}b = 0,{\text{ }}y - 0 = 1(x - 0)

y = x     A1     N3

[3 marks]

b.ii.

subtracting x-coordinates of {{\text{P}}_{k + 1}} and {{\text{P}}_k} (in any order)     (M1)

eg\,\,\,\,\,2(k + 1)\pi  - 2k\pi ,{\text{ }}2k\pi  - 2k\pi  - 2\pi

correct working (must be in correct order)     A1

eg\,\,\,\,\,2k\pi  + 2\pi  - 2k\pi ,{\text{ }}\left| {2k\pi  - 2(k + 1)\pi } \right|

distance is 2\pi     AG     N0

[2 marks]

c.

METHOD 1

recognizing the toothed-edge as the hypotenuse     (M1)

eg\,\,\,\,\,{300^2} = {x^2} + {y^2}, sketch

correct working (using their equation of L     (A1)

eg\,\,\,\,\,{300^2} = {x^2} + {x^2}

x = \frac{{300}}{{\sqrt 2 }} (exact), 212.132     (A1)

dividing their value of x by 2\pi {\text{ }}\left( {{\text{do not accept }}\frac{{300}}{{2\pi }}} \right)     (M1)

eg\,\,\,\,\,\frac{{212.132}}{{2\pi }}

33.7618     (A1)

33 (teeth)     A1     N2

METHOD 2

vertical distance of a tooth is 2\pi (may be seen anywhere)     (A1)

attempt to find the hypotenuse for one tooth     (M1)

eg\,\,\,\,\,{x^2} = {(2\pi )^2} + {(2\pi )^2}

x = \sqrt {8{\pi ^2}} (exact), 8.88576     (A1)

dividing 300 by their value of x     (M1)

eg

33.7618     (A1)

33 (teeth)     A1     N2

[6 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.i.
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b.ii.
[N/A]
c.
[N/A]
d.

Syllabus sections

Topic 3 - Circular functions and trigonometry » 3.4 » The circular functions \sin x , \cos x and \tan x : their domains and ranges; amplitude, their periodic nature; and their graphs.
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