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Date November 2017 Marks available 3 Reference code 17N.2.SL.TZ0.S_10
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number S_10 Adapted from N/A

Question

Note: In this question, distance is in millimetres.

Let f ( x ) = x + a sin ( x π 2 ) + a , for x 0 .

The graph of f passes through the origin. Let P k be any point on the graph of f with x -coordinate 2 k π , where k N . A straight line L passes through all the points 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 P k and P k + 1 .

Show that f ( 2 π ) = 2 π .

[3]
a.

Find the coordinates of P 0 and of P 1 .

[3]
b.i.

Find the equation of L .

[3]
b.ii.

Show that the distance between the x -coordinates of P k and P k + 1 is 2 π .

[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

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

substituting x = 2 π     M1

eg 2 π + a sin ( 2 π π 2 ) + a

2 π + a sin ( 3 π 2 ) + a     (A1)

2 π a + a     A1

f ( 2 π ) = 2 π     AG     N0

[3 marks]

a.

substituting the value of k     (M1)

P 0 ( 0 ,   0 ) ,   P 1 ( 2 π ,   2 π )     A1A1     N3

[3 marks]

b.i.

attempt to find the gradient     (M1)

eg 2 π 0 2 π 0 ,   m = 1

correct working     (A1)

eg y 2 π x 2 π = 1 ,   b = 0 ,   y 0 = 1 ( x 0 )

y = x     A1     N3

[3 marks]

b.ii.

subtracting x -coordinates of P k + 1 and P k (in any order)     (M1)

eg 2 ( k + 1 ) π 2 k π ,   2 k π 2 k π 2 π

correct working (must be in correct order)     A1

eg 2 k π + 2 π 2 k π ,   | 2 k π 2 ( k + 1 ) π |

distance is 2 π     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 = 300 2 (exact), 212.132     (A1)

dividing their value of x by 2 π   ( do not accept  300 2 π )     (M1)

eg 212.132 2 π

33.7618     (A1)

33 (teeth)     A1     N2

METHOD 2

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

attempt to find the hypotenuse for one tooth     (M1)

eg x 2 = ( 2 π ) 2 + ( 2 π ) 2

x = 8 π 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.
[N/A]
b.ii.
[N/A]
c.
[N/A]
d.

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.7—Circular functions: graphs, composites, transformations
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Topic 3— Geometry and trigonometry

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