ib math sl: trig practice problems alei - desert...

$: IB Math SL: Trig Practice Problems Alei - Desert …users.desertacademy.org/balei/Math/SL/SLTrigPractice.pdfIB Math – SL: Trig Practice Problems Alei - Desert Academy C:\Users\Bob\Documents\Dropbox\Desert\SL\3Trig\TestsQuizzesPractice\SLTrigPractice.docx$
IB Math – SL: Trig Practice Problems Alei - Desert Academy

C:\Users\Bob\Documents\Dropbox\Desert\SL\3Trig\TestsQuizzesPractice\SLTrigPractice.docx on 3/18/14 at 8:50 AM Page 1 of 23

Circular Functions and Trig - Practice Problems (to 07) 1. In the triangle PQR, PR = 5 cm, QR = 4 cm and PQ = 6 cm.

Calculate(a) the size of ; (b) the area of triangle PQR.

(Total 6 marks) 2. The following diagram shows a triangle ABC, where is 90, AB = 3, AC = 2 and is .

(a) Show that sin = .

(b) Show that sin 2 = .

(c) Find the exact value of cos 2. (Total 6 marks)

3. The following diagram shows a sector of a circle of radius r cm, and angle at the centre. The perimeter of thesector is 20 cm.

(a) Show that = .

(b) The area of the sector is 25 cm2. Find the value of r. (Total 6 marks)

4. The following diagram shows the triangle AOP, where OP = 2 cm, AP = 4 cm and AO = 3 cm.

(a) Calculate , giving your answer in radians. (3)

The following diagram shows two circles which intersect at the points A and B. The smaller circle C1 has centre O and radius 3 cm, the larger circle C2 has centre P and radius 4 cm, and OP = 2 cm. The point D lies on the circumference of C1 and E on the circumference of C2.Triangle AOP is the same as triangle AOP in the diagram above.

RQP

BCA CAB

35

954

rr220

O

A

P

diagram not toscale

POA



(b) Find , giving your answer in radians. (2)

(c) Given that is 1.63 radians, calculate the area of (i) sector PAEB; (ii) sector OADB.

(5) (d) The area of the quadrilateral AOBP is 5.81 cm2.

(i) Find the area of AOBE. (ii) Hence find the area of the shaded region AEBD.

(4) (Total 14 marks)

5. The following diagram shows a pentagon ABCDE, with AB = 9.2 cm, BC = 3.2 cm, BD = 7.1 cm, =110, = 52 and = 60.

(a) Find AD. (4)

(b) Find DE. (4)

(c) The area of triangle BCD is 5.68 cm2. Find . (4)

(d) Find AC. (4)

(e) Find the area of quadrilateral ABCD. (5)

(Total 21 marks)

6. The diagram below shows the graph of f (x) = 1 + tan for −360 x 360.

(a) On the same diagram, draw the asymptotes. (2)

diagram not toscale

A

B

D E O P

C

C

1

2

BOA

BPA

DEAEDA DBA

CBD

2x



(b) Write down (i) the period of the function; (ii) the value of f (90).

(2) (c) Solve f (x) = 0 for −360 x 360.

(2) (Total 6 marks)

7. (a) Consider the equation 4x2 + kx + 1 = 0. For what values of k does this equation have two equal roots? (3)

Let f be the function f ( ) = 2 cos 2 + 4 cos + 3, for −360 360.

(b) Show that this function may be written as f ( ) = 4 cos2 + 4 cos + 1. (1)

(c) Consider the equation f ( ) = 0, for −360 360. (i) How many distinct values of cos satisfy this equation? (ii) Find all values of which satisfy this equation.

(5) (d) Given that f ( ) = c is satisfied by only three values of , find the value of c.


8. A Ferris wheel with centre O and a radius of 15 metres is represented in the diagram below. Initially seat A is at

ground level. The next seat is B, where = .

(a) Find the length of the arc AB.

(2) (b) Find the area of the sector AOB.

(2)

(c) The wheel turns clockwise through an angle of . Find the height of A above the ground.

(3) The height, h metres, of seat C above the ground after t minutes, can be modelled by the function

h (t) = 15 − 15 cos .

(d) (i) Find the height of seat C when t = .

(ii) Find the initial height of seat C. (iii) Find the time at which seat C first reaches its highest point.

(8) (e) Find h′ (t).

(2) (f) For 0 t ,

(i) sketch the graph of h′; (ii) find the time at which the height is changing most rapidly.


9. Let f (x) = a (x − 4)2 + 8. (a) Write down the coordinates of the vertex of the curve of f. (b) Given that f (7) = −10, find the value of a. (c) Hence find the y-intercept of the curve of f.

(Total 6 marks)

BOA6π

32π

4π2t

4π



10. The following diagram shows a circle with radius r and centre O. The points A, B and C are on the circle and =.

The area of sector OABC is and the length of arc ABC is .

Find the value of r and of . (Total 6 marks)

11. Let ƒ (x) = a sin b (x − c). Part of the graph of ƒ is given below.

Given that a, b and c are positive, find the value of a, of b and of c.

(Total 6 marks) 12. The points P(−2, 4), Q (3, 1) and R (1, 6) are shown in the diagram below.

(a) Find the vector . (b) Find a vector equation for the line through R parallel to the line (PQ).

(Total 6 marks) 13. The diagram below shows a circle of radius r and centre O. The angle = .

The length of the arc AB is 24 cm. The area of the sector OAB is 180 cm2. Find the value of r and of .

(Total 6 marks) 14. The diagram below shows a quadrilateral ABCD. AB = 4, AD = 8, CD =12, B D = 25, =.

COA

34

32

PQ

BOA

C DAB



(a) Use the cosine rule to show that BD = .

(2) Let = 40. (b) (i) Find the value of sin .

(ii) Find the two possible values for the size of . (iii) Given that is an acute angle, find the perimeter of ABCD.

(12) (c) Find the area of triangle ABD.


15. (a) Let y = –16x2 + 160x –256. Given that y has a maximum value, find (i) the value of x giving the maximum value of y; (ii) this maximum value of y.

The triangle XYZ has XZ = 6, YZ = x, XY = z as shown below. The perimeter of triangle XYZ is 16. (4)

(b) (i) Express z in terms of x.

(ii) Using the cosine rule, express z2 in terms of x and cos Z.

(iii) Hence, show that cos Z = .

(7) Let the area of triangle XYZ be A.

(c) Show that A2 = 9x2 sin2 Z. (2)

(d) Hence, show that A2 = –16x2 + 160x – 256. (4)

(e) (i) Hence, write down the maximum area for triangle XYZ. (ii) What type of triangle is the triangle with maximum area?


16. The function f is defined by f : x 30 sin 3x cos 3x, 0 x .

(a) Write down an expression for f (x) in the form a sin 6x, where a is an integer. (b) Solve f (x) = 0, giving your answers in terms of .

(Total 6 marks) 17. The following diagram shows two semi-circles. The larger one has centre O and radius 4 cm. The smaller one has

centre P, radius 3 cm, and passes through O. The line (OP) meets the larger semi-circle at S. The semi-circles intersect at Q.

cos454

DBCDBC

DBC

xx3

165

3π



(a) (i) Explain why OPQ is an isosceles triangle.

(ii) Use the cosine rule to show that cos = .

(iii) Hence show that sin = .

(iv) Find the area of the triangle OPQ. (7)

(b) Consider the smaller semi-circle, with centre P. (i) Write down the size of (ii) Calculate the area of the sector OPQ.

(3) (c) Consider the larger semi-circle, with centre O. Calculate the area of the sector QOS.

(3) (d) Hence calculate the area of the shaded region.


18. The graph of a function of the form y = p cos qx is given in the diagram below.

(a) Write down the value of p. (b) Calculate the value of q.

(Total 6 marks) 19. A farmer owns a triangular field ABC. One side of the triangle, [AC], is 104 m, a second side, [AB], is 65 m and

the angle between these two sides is 60°. (a) Use the cosine rule to calculate the length of the third side of the field.

(3)

(b) Given that sin 60° = find the area of the field in the form where p is an integer.

(3) Let D be a point on [BC] such that [AD] bisects the 60° angle. The farmer divides the field into two parts A1 and

A2 by constructing a straight fence [AD] of length x metres, as shown on the diagram below.

(c) (i) Show that the area of Al is given by .

(ii) Find a similar expression for the area of A2.

QPO91

QPO980

Q.PO

40

30

20

10

–10

–20

–30

–40

/2 x

y

,23 3p

104 m

A

A

A

B

C

65 m

30°

30°

2

1

xD

465x



(iii) Hence, find the value of x in the form , where q is an integer. (7)

(d) (i) Explain why . (ii) Use the result of part (i) and the sine rule to show that

.


20. The following diagram shows a circle of centre O, and radius r. The shaded sector OACB has an area of 27 cm2. Angle = θ = 1.5 radians.

(a) Find the radius. (b) Calculate the length of the minor arc ACB.

Working: Answers:

(a) .................................................................. (b) ..................................................................

(Total 6 marks)

21. Consider y = sin .

(a) The graph of y intersects the x-axis at point A. Find the x-coordinate of A, where 0 x π.

(b) Solve the equation sin = – , for 0 x 2.

Working: Answers:

(a) .................................................................. (b) ..................................................................

(Total 6 marks) 22. The diagram shows a triangular region formed by a hedge [AB], a part of a river bank [AC] and a fence [BC]. The

hedge is 17 m long and is 29°. The end of the fence, point C, can be positioned anywhere along the river bank. (a) Given that point C is 15 m from A, find the length of the fence [BC].

(3)

(b) The farmer has another, longer fence. It is possible for him to enclose two different triangular regions with this fence. He places the fence so that is 85°. (i) Find the distance from A to C. (ii) Find the area of the region ABC with the fence in this position.

(5) (c) To form the second region, he moves the fencing so that point C is closer to point A.

Find the new distance from A to C. (4)

(d) Find the minimum length of fence [BC] needed to enclose a triangular region ABC.

3q

BDAsinCDAsin

85

DCBD

BOA

Or

AC

B

9x

9x

21

CAB

29°A

B

C15 m

17 m

river bank

CBA




23. Let f (x) = sin 2x + cos x for 0 x 2.

(a) (i) Find f (x).

One way of writing f (x) is –2 sin2 x – sin x + 1.

(ii) Factorize 2 sin2 x + sin x – 1. (iii) Hence or otherwise, solve f (x) = 0.

(6) The graph of y = f (x) is shown below.

There is a maximum point at A and a minimum point at B.

(b) Write down the x-coordinate of point A. (1)

(c) The region bounded by the graph, the x-axis and the lines x = a and x = b is shaded in the diagram above. (i) Write down an expression that represents the area of this shaded region. (ii) Calculate the area of this shaded region.


24. In triangle PQR, PQ is 10 cm, QR is 8 cm and angle PQR is acute. The area of the triangle is 20 cm2. Find the size of angle

(Total 6 marks) 25. Let f (x) = 6 sin x , and g (x) = 6e–x – 3 , for 0 x 2. The graph of f is shown on the diagram below. There is a

maximum value at B (0.5, b).

(a) Write down the value of b. (b) On the same diagram, sketch the graph of g. (c) Solve f (x) = g (x) , 0.5 x 1.5.

(Total 6 marks) 26. Consider the equation 3 cos 2x + sin x = 1

(a) Write this equation in the form f (x) = 0 , where f (x) = p sin2 x + q sin x + r , and p , q , r . (b) Factorize f (x). (c) Write down the number of solutions of f (x) = 0, for 0 x 2.

(Total 6 marks) 27. The diagram below shows two circles which have the same centre O and radii 16 cm and 10 cm respectively. The

two arcs AB and CD have the same sector angle = 1.5 radians.

21

A

B

ab

x

y

0 2

R.QP

0 1 2

B

x

y



Find the area of the shaded region. (Total 6 marks) 28. Let f (x) = sin (2x + 1), 0 x π.

(a) Sketch the curve of y = f (x) on the grid below.

(b) Find the x-coordinates of the maximum and minimum points of f (x), giving your answers correct to one

decimal place. (Total 6 marks)

29. In a triangle ABC, AB = 4 cm, AC = 3 cm and the area of the triangle is 4.5 cm2. Find the two possible values of the angle .

Working: Answer:

………………………………………….. (Total 6 marks)

30. Solve the equation 2 cos2 x = sin 2x for 0 x π, giving your answers in terms of π. Working: Answer:

………………………………………….. (Total 6 marks)

31. The depth y metres of water in a harbour is given by the equation

y = 10 + 4 sin ,

where t is the number of hours after midnight. (a) Calculate the depth of the water

(i) when t = 2; (ii) at 2100.

(3) The sketch below shows the depth y, of water, at time t, during one day (24 hours).

A B

C D

O

2

1.5

1

0.5

0

–0.5

–1

–1.5

–2

0.5 1 1.5 2 2.5 3 3.5 x

y

CAB

2t



(b) (i) Write down the maximum depth of water in the harbour.

(ii) Calculate the value of t when the water is first at its maximum depth during the day. (3)

The harbour gates are closed when the depth of the water is less than seven metres. An alarm rings when the gates are opened or closed. (c) (i) How many times does the alarm sound during the day?

(ii) Find the value of t when the alarm sounds first. (iii) Use the graph to find the length of time during the day when the harbour gates are closed. Give

your answer in hours, to the nearest hour. (7)

(Total 13 marks) 32. The following diagram shows a triangle ABC, where BC = 5 cm, = 60°, = 40°.

(a) Calculate AB. (b) Find the area of the triangle.

Working: Answers:

(a) ………………………………………….. (b) …………………………………………..

(Total 6 marks) 33. The diagram below shows a circle of radius 5 cm with centre O. Points A and B are on the circle, and is

0.8 radians. The point N is on [OB] such that [AN] is perpendicular to [OB].

Find the area of the shaded region.

Working: Answer:

…………………………………………........ (Total 6 marks)

34. Let f (x) = 1 + 3 cos (2x) for 0 x π, and x is in radians.

151413121110987654321

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24time (hours)

depth (metres)

t

y

B C

60° 40°

A

B C5 cm

BOA

0.8

5 cm

N

A

BO



(a) (i) Find f (x). (ii) Find the values for x for which f (x) = 0, giving your answers in terms of .

(6)

The function g (x) is defined as g (x) = f (2x) – 1, 0 x .

(b) (i) The graph of f may be transformed to the graph of g by a stretch in the x-direction with scale factor followed by another transformation. Describe fully this other transformation.

(ii) Find the solution to the equation g (x) = f (x) (4)

(Total 10 marks) 35. Part of the graph of y = p + q cos x is shown below. The graph passes through the points (0, 3) and (, –1).

Find the value of

(a) p; (b) q.

Working: Answers:

(a) .................................................................. (b) ..................................................................

(Total 6 marks) 36. Find all solutions of the equation cos 3x = cos (0.5x), for 0 x .

Working: Answer:

.................................................................. (Total 6 marks)

37. The diagram below shows a triangle and two arcs of circles. The triangle ABC is a right-angled isosceles triangle, with AB = AC = 2. The point P is the midpoint of [BC]. The arc BDC is part of a circle with centre A. The arc BEC is part of a circle with centre P.

(a) Calculate the area of the segment BDCP. (b) Calculate the area of the shaded region BECD.

(Total 6 marks)

38. The diagram shows a parallelogram OPQR in which = , =

2π

21

y

x

3

2

1

0

–1

2

A

B

C

D

E

P2

2

OP

37

OQ .1

10



(a) Find the vector .

(3)

(b) Use the scalar product of two vectors to show that cos = –

(4) (c) (i) Explain why cos = –cos

(ii) Hence show that sin = .

(iii) Calculate the area of the parallelogram OPQR, giving your answer as an integer. (7)

(Total 14 marks) 39. The points P, Q, R are three markers on level ground, joined by straight paths PQ, QR, PR as shown in the

diagram. QR = 9 km, = 35°, = 25°.

Diagram not to scale

(a) Find the length PR. (3)

(b) Tom sets out to walk from Q to P at a steady speed of 8 km h–1. At the same time, Alan sets out to jog

from R to P at a steady speed of a km h–1. They reach P at the same time. Calculate the value of a. (7)

(c) The point S is on [PQ], such that RS = 2QS, as shown in the diagram.

Find the length QS.


40. Consider the function f (x) = cos x + sin x.

(a) (i) Show that f (– ) = 0.

(ii) Find in terms of , the smallest positive value of x which satisfies f (x) = 0. (3)

The diagram shows the graph of y = ex (cos x + sin x), – 2 x 3. The graph has a maximum turning point at C(a, b) and a point of inflexion at D.

y

x

P

OQ

R

OR

QPO .75415

RQP Q.PO

RQP75423

RQP QRP

35° 25°9 km

P

Q R

P

Q R

S

4π



(b) Find .

(3) (c) Find the exact value of a and of b.

(4)

(d) Show that at D, y = . (5)

(e) Find the area of the shaded region. (2)

(Total 17 marks) 41. The graph of the function f (x) = 3x – 4 intersects the x-axis at A and the y-axis at B.

(a) Find the coordinates of (i) A; (ii) B.

(b) Let O denote the origin. Find the area of triangle OAB. Working: Answers:

(a) (i) ........................................................... (ii) ........................................................... (b) ..................................................................

(Total 6 marks) 42. (a) Factorize the expression 3 sin2 x – 11 sin x + 6.

(b) Consider the equation 3 sin2 x – 11 sin x + 6 = 0. (i) Find the two values of sin x which satisfy this equation, (ii) Solve the equation, for 0° x 180°.

Working: Answers:

(a) .................................................................. (b) (i) ........................................................... (ii) ...........................................................

(Total 6 marks) 43. The diagram below shows a circle, centre O, with a radius 12 cm. The chord AB subtends at an angle of 75° at the

centre. The tangents to the circle at A and at B meet at P.

6

4

2

1 2 3–2 –1

D

C(a, b)

y

x

xy

dd

4π

e2



(a) Using the cosine rule, show that the length of AB is 12 .

(2) (b) Find the length of BP.

(3) (c) Hence find

(i) the area of triangle OBP; (ii) the area of triangle ABP.

(4) (d) Find the area of sector OAB.

(2) (e) Find the area of the shaded region.


44. Note: Radians are used throughout this question. A mass is suspended from the ceiling on a spring. It is pulled down to point P and then released. It oscillates up

and down.

Its distance, s cm, from the ceiling, is modelled by the function s = 48 + 10 cos 2πt where t is the time in seconds

from release. (a) (i) What is the distance of the point P from the ceiling?

(ii) How long is it until the mass is next at P? (5)

(b) (i) Find .

(ii) Where is the mass when the velocity is zero? (7)

A second mass is suspended on another spring. Its distance r cm from the ceiling is modelled by the function r = 60 + 15 cos 4t. The two masses are released at the same instant. (c) Find the value of t when they are first at the same distance below the ceiling.

(2) (d) In the first three seconds, how many times are the two masses at the same height?


12 cmA

P

B

75ºO diagram not toscale

75cos–12

P

diagram not toscale

ts

dd



45. The following diagram shows a circle of centre O, and radius 15 cm. The arc ACB subtends an angle of 2 radians at the centre O.

Find

(a) the length of the arc ACB; (b) the area of the shaded region.

Working: Answers:

(a) .................................................................. (b) ..................................................................

(Total 6 marks) 46. Two boats A and B start moving from the same point P. Boat A moves in a straight line at 20 km h–1 and boat B

moves in a straight line at 32 km h–1. The angle between their paths is 70°. Find the distance between the boats after 2.5 hours.

(Total 6 marks) 47. Let f (x) = sin 2x and g (x) = sin (0.5x).

(a) Write down (i) the minimum value of the function f ; (ii) the period of the function g.

(b) Consider the equation f (x) = g (x).

Find the number of solutions to this equation, for 0 x .

Working: Answers:

(a) (i) .......................................................... (ii) .......................................................... (b) .................................................................

(Total 6 marks) 48. Consider the following statements

A: log10 (10x) > 0. B: –0.5 cos (0.5x) 0.5.

C: – arctan x .

(a) Determine which statements are true for all real numbers x. Write your answers (yes or no) in the table below. Statement (a) Is the statement true for all

real numbers x? (Yes/No) (b) If not true, example

A B C

(b) If a statement is not true for all x, complete the last column by giving an example of one value of x for

which the statement is false. Working:

(Total 6 marks)

O

B

C

A

2 rad

15 cm Diagram not to scale

A B = 2 radiansOA = 15 cmÔ

2π3

2π

2π



49. The diagram shows a triangle ABC in which AC = 7 , BC = 6, = 45°.

(a) Use the fact that sin 45° = to show that sin = .

(2)

The point D is on (AB), between A and B, such that sin = .

(b) (i) Write down the value of + . (ii) Calculate the angle BCD. (iii) Find the length of [BD].

(6)

(c) Show that = .


50. In triangle ABC, AC = 5, BC = 7, = 48°, as shown in the diagram.

Find giving your answer correct to the nearest degree.

Working: Answer:

...................................................................... (Total 6 marks)

51. Given that sin x = , where x is an acute angle, find the exact value of

(a) cos x; (b) cos 2x.

Working: Answers:

(a) .................................................................. (b) ..................................................................

(Total 6 marks) 52. Consider the trigonometric equation 2 sin2 x = 1 + cos x.

(a) Write this equation in the form f (x) = 0, where f (x) = a cos2 x + b cos x + c, and a, b, c .

(b) Factorize f (x). (c) Solve f (x) = 0 for 0° x 360°.

Working: Answers:

(a) ..................................................................

22 CBA

A

B C645°

7 22

Diagramnot to scale

22 CAB

76

CDB76

CDB CAB

BAC of AreaBDC of Area

BABD

A

A B

C

5 7

48°

diagram not to scale

,B

31



(b) .................................................................. (c) ..................................................................

(Total 6 marks) 53. The following diagram shows a triangle with sides 5 cm, 7 cm, 8 cm.


Find (a) the size of the smallest angle, in degrees; (b) the area of the triangle.

Working: Answers:

(a) .................................................................. (b) ..................................................................

(Total 4 marks) 54. (a) Write the expression 3 sin2 x + 4 cos x in the form a cos2 x + b cos x + c.

(b) Hence or otherwise, solve the equation

3 sin2 x + 4 cos x – 4 = 0, 0 x 90. Working: Answers:

(a) .................................................................. (b) ..................................................................

(Total 4 marks) 55. In the following diagram, O is the centre of the circle and (AT) is the tangent to the circle at T.

Diagram not to scale If OA = 12 cm, and the circle has a radius of 6 cm, find the area of the shaded region.

Working: Answer:

....................................................................... (Total 4 marks)

56. In the diagram below, the points O(0, 0) and A(8, 6) are fixed. The angle varies as the point P(x, 10) moves along the horizontal line y = 10.

Diagram to scale

(a) (i) Show that

5 7

8

O

T

A

APO

y

x

A(8, 6)

O(0, 0)

y=10P( , 10)x

.8016–AP 2 xx



(ii) Write down a similar expression for OP in terms of x. (2)

(b) Hence, show that

(3) (c) Find, in degrees, the angle when x = 8.

(2) (d) Find the positive value of x such that .

(4) Let the function f be defined by

(e) Consider the equation f (x) = 1. (i) Explain, in terms of the position of the points O, A, and P, why this

equation has a solution. (ii) Find the exact solution to the equation.


57. The diagram below shows a sector AOB of a circle of radius 15 cm and centre O. The angle at the centre of the circle is 2 radians.


(a) Calculate the area of the sector AOB. (b) Calculate the area of the shaded region.

Working: Answers:

(a) .................................................................. (b) ..................................................................

(Total 4 marks) 58. The diagrams below show two triangles both satisfying the conditions

AB = 20 cm, AC = 17 cm, = 50°. Diagrams not

to scale

(a) Calculate the size of in Triangle 2. (b) Calculate the area of Triangle 1.

Working: Answers:

(a) ..................................................................

,)}100)(8016–{(

408–APOcos 22

2

xxxxx

APO

60APO

.150,)}100)(8016–{(

408–APOcos)( 22

2

x

xxxxxxf

A B

O

CBA

A

B C

A

B C

Triangle 1 Triangle 2

BCA



(b) .................................................................. (Total 4 marks)

59. The depth, y metres, of sea water in a bay t hours after midnight may be represented by the function

, where a, b and k are constants.

The water is at a maximum depth of 14.3 m at midnight and noon, and is at a minimum depth of 10.3 m at 06:00 and at 18:00.

Write down the value of (a) a; (b) b; (c) k.

Working: Answers:

(a) .................................................................. (b) .................................................................. (c) ..................................................................

(Total 4 marks) 60. Town A is 48 km from town B and 32 km from town C as shown in the diagram.

Given that town B is 56 km from town C, find the size of angle to the nearest degree.

(Total 4 marks) 61. (a) Express 2 cos2 x + sin x in terms of sin x only.

(b) Solve the equation 2 cos2 x + sin x = 2 for x in the interval 0 x , giving your answers exactly. (Total 4 marks)

62. Note: Radians are used throughout this question. (a) Draw the graph of y = + x cos x, 0 x 5, on millimetre square graph paper, using a scale of 2 cm per

unit. Make clear (i) the integer values of x and y on each axis; (ii) the approximate positions of the x-intercepts and the turning points.

(5) (b) Without the use of a calculator, show that is a solution of the equation

+ x cos x = 0. (3)

(c) Find another solution of the equation + x cos x = 0 for 0 x 5, giving your answer to six significant figures.

(2) (d) Let R be the region enclosed by the graph and the axes for 0 x . Shade R on your diagram, and write

down an integral which represents the area of R . (2)

(e) Evaluate the integral in part (d) to an accuracy of six significant figures. (If you consider it necessary, you

can make use of the result


63. A formula for the depth d metres of water in a harbour at a time t hours after midnight is

where P and Q are positive constants. In the following graph the point (6, 8.2) is a minimum point and the point (12, 14.6) is a maximum point.

t

kbay 2cos

AB

C

32km

48kmBAC

.)cos)cossin(dd xxxxxx

,240,6

cos

ttQPd



(a) Find the value of

(i) Q; (ii) P.

(3) (b) Find the first time in the 24-hour period when the depth of the water is 10 metres.

(3) (c) (i) Use the symmetry of the graph to find the next time when the depth of the water is 10 metres.

(ii) Hence find the time intervals in the 24-hour period during which the water is less than 10 metres deep.

(4) 64. Solve the equation 3 cos x = 5 sin x, for x in the interval 0° x 360°, giving your answers to the nearest degree.

(Total 4 marks)

65. If A is an obtuse angle in a triangle and sin A = , calculate the exact value of sin 2A.

(Total 4 marks)

66. (a) Sketch the graph of y = sin x – x, –3 x 3, on millimetre square paper, using a scale of 2 cm per unit on each axis.

Label and number both axes and indicate clearly the approximate positions of the x-intercepts and the local maximum and minimum points.

(5) (b) Find the solution of the equation

sin x – x = 0, x > 0. (1)

(c) Find the indefinite integral

and hence, or otherwise, calculate the area of the region enclosed by the graph, the x-axis and the line x = 1.


67. Given that sin θ = , cos θ = – and 0° ≤ θ ≤ 360°,

(a) find the value of θ; (b) write down the exact value of tan θ.

(Total 4 marks) 68. The diagram shows a vertical pole PQ, which is supported by two wires fixed to the horizontal ground at A and B.

= 40 m

0 6 12 18 24

15

10.

5

d

t

(6, 8.2)

(12, 14.6)

135

xxx )d sin (

21

23

Q

P

A

B3630

70

BQ



= 36°

= 70°

= 30° Find

(a) the height of the pole, PQ; (b) the distance between A and B.

(Total 4 marks) 69. The diagram shows a circle of radius 5 cm.

Find the perimeter of the shaded region.

(Total 4 marks)

70. f (x) = 4 sin .

For what values of k will the equation f (x) = k have no solutions? (Total 4 marks)

71. In this question you should note that radians are used throughout. (a) (i) Sketch the graph of y = x2 cos x, for 0 x 2 making clear the approximate positions of the

positive x-intercept, the maximum point and the end-points. (ii) Write down the approximate coordinates of the positive x-intercept, the maximum point and the

end-points. (7)

(b) Find the exact value of the positive x-intercept for 0 x 2. (2)

Let R be the region in the first quadrant enclosed by the graph and the x-axis. (c) (i) Shade R on your diagram.

(ii) Write down an integral which represents the area of R. (3)

(d) Evaluate the integral in part (c)(ii), either by using a graphic display calculator, or by using the following information.

(x2 sin x + 2x cos x – 2 sin x) = x2 cos x.


72. In this part of the question, radians are used throughout. The function f is given by

f (x) = (sin x)2 cos x. The following diagram shows part of the graph of y = f (x).

QBPQABQBA

1 radian

23 x

xdd



The point A is a maximum point, the point B lies on the x-axis, and the point C is a point of inflexion.

(a) Give the period of f. (1)

(b) From consideration of the graph of y = f (x), find to an accuracy of one significant figure the range of f. (1)

(c) (i) Find f (x).

(ii) Hence show that at the point A, cos x = .

(iii) Find the exact maximum value. (9)

(d) Find the exact value of the x-coordinate at the point B. (1)

(e) (i) Find dx.

(ii) Find the area of the shaded region in the diagram. (4)

(f) Given that f (x) = 9(cos x)3 – 7 cos x, find the x-coordinate at the point C. (4)

(Total 20 marks) 73. A triangle has sides of length 4, 5, 7 units. Find, to the nearest tenth of a degree, the size of the largest angle.

(Total 4 marks) 74. O is the centre of the circle which has a radius of 5.4 cm.

The area of the shaded sector OAB is 21.6 cm2. Find the length of the minor arc AB.

(Total 4 marks)

75. The circle shown has centre O and radius 6. is the vector , is the vector and is the

vector .

A

C

B

y

xO

31

)(xf

O

A B

OA

06

OB

06

OC

115

AB

C

O x

y



(a) Verify that A, B and C lie on the circle. (3)

(b) Find the vector . (2)

(c) Using an appropriate scalar product, or otherwise, find the cosine of angle . (3)

(d) Find the area of triangle ABC, giving your answer in the form a , where a . (4)

(Total 12 marks) 76. Solve the equation 3 sin2 x = cos2 x, for 0° x 180°.

(Total 4 marks) 77. The diagrams show a circular sector of radius 10 cm and angle θ radians which is formed into a cone of slant

height 10 cm. The vertical height h of the cone is equal to the radius r of its base. Find the angle θ radians.

(Total 4 marks)

78. The diagram shows the graph of the function f given by

f (x) = A sin + B,

for 0 x 5, where A and B are constants, and x is measured in radians.

The graph includes the points (1, 3) and (5, 3), which are maximum points of the graph.

(a) Write down the values of f (1) and f (5). (2)

(b) Show that the period of f is 4. (2)

The point (3, –1) is a minimum point of the graph. (c) Show that A = 2, and find the value of B.

(5)

(d) Show that f (x) = cos .

(4) The line y = k – x is a tangent line to the graph for 0 x 5.

(e) Find (i) the point where this tangent meets the curve; (ii) the value of k.

(6) (f) Solve the equation f (x) = 2 for 0 x 5.


AC

CAO ˆ

11

10cm

10cmh

r

x

2

0 1 2 3 4 5

2

y

x(0, 1)

(1,3)

(3, –1)

(5, 3)

x

2

IB Math – SL: Trig Practice Problems: MarkScheme Alei - Desert Academy


Circular Functions and Trig - Practice Problems (to 07) MarkScheme 1. (a) Evidence of using the cosine rule (M1)

eg cos = cos

Correct substitution A1

eg

cos = (A1)

= 55.8 (0.973 radians) A1N2

(b) Area = sin

For substituting correctly sin 55.8 A1

= 9.92 (cm2) A1N1 [6]

2. Note: Throughout this question, do not accept methods which involve finding .

(a) Evidence of correct approach A1

eg sin =

sin = AG N0

(b) Evidence of using sin 2 = 2 sin cos (M1)

= A1

= AGN0

(c) Evidence of using an appropriate formula for cos 2 M1

eg

cos 2 = A2 N2

[6] 3. (a) For using perimeter = r + r + arc length (M1)

20 = 2r + r A1

AG N0

(b) Finding A = (A1)

For setting up equation in r M1 Correct simplified equation, or sketch

eg 10r – r2 = 25, r2 – 10r + 25 = 0 (A1) r = 5 cm A1 N2

[6] 4. Notes: Candidates may have differing answers due to using approximate answers from previous parts or using

answers from the GDC. Some leeway is provided to accommodate this.

RQP prrpqpr

qrp 2,2

222222

RQP

Qcos642645,642564 222

222

RQP 5625.04827

RQP

pr21 RQP

6421

523BC,ABBC 22

35

32

352

954

81801,

9521,1

942,

95

94

91

rr220

22 1022021 rr

rrr



(a) METHOD 1 Evidence of using the cosine rule (M1)

eg cos C =

Correct substitution

eg cos = A1

cos = 0.25

= 1.82 (radians) A1 N2

METHOD 2 Area of AOBP = 5.81 (from part (d)) Area of triangle AOP = 2.905 (M1) 2.9050 = 0.5 2 3 sin A1

= 1.32 or 1.82

= 1.82 (radians) A1 N2

(b) = 2( 1.82) (= 2 3.64) (A1)

= 2.64 (radians) A1N2

(c) (i) Appropriate method of finding area (M1)

eg area =

Area of sector PAEB = A1

= 13.0 (cm2) (accept the exact value 13.04) A1N2

(ii) Area of sector OADB = A1

= 11.9 (cm2) A1N1 (d) (i) Area AOBE = Area PAEB Area AOBP (= 13.0 5.81) M1

= 7.19 (accept 7.23 from the exact answer for PAEB) A1N1 (ii) Area shaded = Area OADB Area AOBE (= 11.9 7.19) M1

= 4.71 (accept answers between 4.63 and 4.72) A1N1 [14]

5. (a) Evidence of choosing cosine rule (M1)

eg a2 = b2 + c2 2bc cos A Correct substitution A1

eg (AD)2 = 7.12 + 9.22 2(7.1) (9.2) cos 60

(AD)2 = 69.73 (A1) AD = 8.35 (cm) A1N2

(b) 180 162 = 18 (A1) Evidence of choosing sine rule (M1) Correct substitution A1

eg =

DE = 2.75 (cm) A1 N2 (c) Setting up equation (M1)

Abccbaab

cba cos2,2

222222

POA POAcos232234,232423 222

222

POA

POA

4526

POAPOA

POA

4526

BOA

4538

2

21 θr

63.1421 2

64.2321 2

18sin DE

110sin35.8



eg ab sin C = 5.68, bh = 5.68

Correct substitution A1

eg 5.68 = (3.2) (7.1) sin , 3.2 h = 5.68, (h = 3.55)

sin = 0.5 (A1) 30 and/or 150 A1 N2

(d) Finding A C (60 + D C) (A1) Using appropriate formula (M1)

eg (AC)2 = (AB)2 + (BC)2, (AC)2 = (AB)2 + (BC)2 2 (AB) (BC) cos ABC Correct substitution (allow FT on their seen )

eg (AC)2 = 9.22 + 3.22 A1 AC = 9.74 (cm) A1 N3

(e) For finding area of triangle ABD (M1)

Correct substitution Area = 9.2 7.1 sin 60 A1

= 28.28... A1 Area of ABCD = 28.28... + 5.68 (M1)

= 34.0 (cm2) A1N3 [21]

6. (a)

Correct asymptotes A1A1 N2

(b) (i) Period = 360 (accept 2) A1 N1 (ii) f (90) = 2 A1 N1

(c) 270, 90 A1A1 N1N1 Notes: Penalize 1 mark for any additional values. Penalize 1 mark for correct answers given

in radians

[6] 7. (a) METHOD 1

Using the discriminant = 0 (M1)

k2 = 4 4 1 k = 4, k = 4 A1A1 N3 METHOD 2 Factorizing (M1)

(2x 1)2 k = 4, k = 4 A1A1 N3

21

21

21

CBD21

CBDCBDB B

CBA

21

360°180°–180°–360° 0

10

5

–5

–10

y

x

.57.1,71.4or,2

,2

3



(b) Evidence of using cos 2 = 2 cos2 1 M1

eg 2(2 cos2 1) + 4 cos + 3

f () = 4 cos2 + 4 cos + 1 AG N0

(c) (i) 1 A1 N1 (ii) METHOD 1

Attempting to solve for cos M1

cos = (A1)

= 240, 120, 240, 120 (correct four values only) A2N3 METHOD 2 Sketch of y = 4 cos2

+ 4 cos + 1 M1

Indicating 4 zeros (A1) = 240, 120, 240, 120 (correct four values only) A2N3

(d) Using sketch (M1) c = 9 A1 N2

[11] 8. Note: Accept exact answers given in terms of .

(a) Evidence of using l = r (M1) arc AB = 7.85 (m) A1 N2

(b) Evidence of using (M1)

Area of sector AOB = 58.9 (m2) A1 N2 (c) METHOD 1

angle = (A1)

attempt to find 15 sin M1

height = 15 + 15 sin

= 22.5 (m) A1 N2 METHOD 2

21

y

x–360 –180 180 360

9

θrA 2

21

π 62

3π

306

6

6

π3



angle = (A1)

attempt to find 15 cos M1

height = 15 + 15 cos

= 22.5 (m) A1 N2

(d) (i) (M1)

= 25.6 (m) A1N2

(ii) h(0) = 15 15 cos (M1)

= 4.39(m) A1N2 (iii) METHOD 1

Highest point when h = 30 R1

30 = 15 15 cos M1

cos = 1 (A1)

t = 1.18 A1N2

METHOD 2

Sketch of graph of h M2 Correct maximum indicated (A1) t = 1.18 A1N2 METHOD 3 Evidence of setting h(t) = 0 M1

sin (A1)

Justification of maximum R1 eg reasoning from diagram, first derivative test, second

derivative test

t = 1.18 A1N2

(e) h(t) = 30 sin (may be seen in part (d)) A1A1 N2

(f) (i)

603

3

3

42cos1515

4h

40

42t

42t

83accept

h30

t2π

04

2

t

8

3accept

42t

h(t)30

–30

tππ2



A1A1A1 N3 Notes: Award A1 for range 30 to 30, A1 for two zeros. Award A1 for approximate

correct sinusoidal shape. (ii) METHOD 1

Maximum on graph of h (M1) t = 0.393 A1N2 METHOD 2 Minimum on graph of h (M1) t = 1.96 A1N2 METHOD 3 Solving h(t) = 0 (M1) One or both correct answers A1 t = 0.393, t = 1.96 N2

[22] 9. (a) Vertex is (4, 8) A1A1 N2

(b) Substituting 10 = a(7 4)2 + 8 M1 a = 2 A1N1

(c) For y-intercept, x = 0 (A1) y = 24 A1 N2

[6] 10. METHOD 1

Evidence of correctly substituting into A = A1

Evidence of correctly substituting into l = r A1 For attempting to eliminate one variable … (M1) leading to a correct equation in one variable A1

r = 4 = (= 0.524, 30) A1A1 N3

METHOD 2 Setting up and equating ratios (M1)

A1A1

Solving gives r = 4 A1

r = A1

= A1

r = 4 = N3

[6]

11. a = 4, b = 2, c = A2A2A2 N6

[6]

12. (a) = A1A1 N2

(b) Using r = a + tb

A2A1A1 N4

[6] 13. METHOD 1

Evidence of correctly substituting into l = r A1

θr 2

21

6

rr

232

34

2

34

21or

32 2r

30,524.06

30,524.06

etc2

3or2

PQ

35

35

61

tyx



Evidence of correctly substituting into A = A1

For attempting to solve these equations (M1) eliminating one variable correctly A1 r = 15 = 1.6 (= 91.7) A1A1 N3 METHOD 2 Setting up and equating ratios (M1)

A1A1

Solving gives r = 15 A1

r = 24 A1

= 1.6 (= 91.7) A1 r = 15 = 1.6 (= 91.7) N3

[6] 14. (a) For correct substitution into cosine rule A1

BD =

For factorizing 16, BD = A1

= AGN0 (b) (i) BD = 5.5653 ... (A1)

M1A1

sin = 0.911 (accept 0.910, subject to AP) A1N2 (ii) = 65.7 A1 N1

Or = 180 their acute angle (M1) = 114 A1N2

(iii) = 89.3 (A1)

(or cosine rule) M1A1

BC = 13.2 (accept 13.17…) A1 Perimeter = 4 + 8 + 12 + 13.2 = 37.2 A1N2

(c) Area = 4 8 sin 40 A1

= 10.3 A1 N1 [16]

15. (a) METHOD 1 Note: There are many valid algebraic approaches to this problem (eg completing the square,

using . Use the following mark

allocation as a guide.

(i) Using (M1)

32x + 160 = 0 A1 x = 5 A1N2

(ii) ymax = 16(52) + 160(5) 256 ymax = 144 A1N1

METHOD 2

2

21 r

2180

224

rr

180

21or 2θr

θcos84284 22

θcos4516

θcos454

5653.525sin

12DBCsin

DBCDBC

DBC

CDB

65.7sin 12

89.3sinBCor

25sin 5.5653

89.3sinBC

21

)2a

bx

0dd

xy



(i) Sketch of the correct parabola (may be seen in part (ii)) M1 x = 5 A2N2

(ii) ymax = 144 A1 N1 (b) (i) z = 10 x (accept x + z = 10) A1 N1

(ii) z2 = x2 + 62 2 x 6 cos Z A2 N2 (iii) Substituting for z into the expression in part (ii) (M1)

Expanding 100 20x + x2 = x2 + 36 12x cos Z A1 Simplifying 12x cos Z = 20x 64 A1

Isolating cos Z = A1

cos Z = AGN0

Note: Expanding, simplifying and isolating may be done in any order, with the final A1 being awarded for an expression that clearly leads to the required answer.

(c) Evidence of using the formula for area of a triangle

M1

A1

A2 = 9x2 sin2 Z AG N0

(d) Using sin2 Z = 1 cos2 Z (A1)

Substituting for cos Z A1

for expanding A1

for simplifying to an expression that clearly leads to the required answer A1

eg A2 = 9x2 (25x2 160x + 256)

A2 = 16x2 + 160x 256 AG

(e) (i) 144 (is maximum value of A2, from part (a)) A1 Amax = 12 A1N1

(ii) Isosceles A1 N1 [20]

16. (a) Evidence of choosing the double angle formula (M1) f (x) = 15 sin (6x) A1 N2

(b) Evidence of substituting for f (x) (M1) eg 15 sin 6x = 0, sin 3x = 0 and cos 3x = 0 6x = 0, , 2

x = 0, A1A1A1 N4

[6] 17. (a) (i) OP = PQ (= 3cm) R1

So OPQ is isosceles AGN0

(ii) Using cos rule correctly eg cos = (M1)

cos = A1

xx12

6420

xx3

165

ZxA sin6

21

ZxAZxA 222 sin63

41sin3

xx3

165

2

22

925616025to

3165

xxx

xx

3,

6

QPO332433 222

QPO

182

181699



cos = AGN0

(iii) Evidence of using sin2 A + cos2 A = 1 M1

sin = A1

sin = AGN0

(iv) Evidence of using area triangle OPQ = M1

eg

Area triangle OPQ = A1N1

(b) (i) = 1.4594...

= 1.46 A1N1 (ii) Evidence of using formula for area of a sector (M1)

eg Area sector OPQ =

= 6.57 A1N2

(c) = (A1)

Area sector QOS = A1

= 6.73 A1N2 (d) Area of small semi-circle is 4.5 (= 14.137...) A1

Evidence of correct approach M1 eg Area = area of semi-circle area sector OPQ area sector QOS +

area triangle POQ Correct expression A1 eg 4.5 6.5675... 6.7285... + 4.472..., 4.5 (6.7285... + 2.095...), 4.5 (6.5675... + 2.256...) Area of the shaded region = 5.31 A1 N1

[17] 18. (a) p = 30 A22

(b) METHOD 1

Period = (M2)

= (A1)

q = 4 A1 4 METHOD 2

Horizontal stretch of scale factor = (M2)

scale factor = (A1)

q = 4 A1 4 [6]

19. (a) using the cosine rule (A2) = b2 + c2 –2bc cos (M1)

QPO91

QPO

8180

8111

QPO980

PsinPQOP21

9938.029,

98033

21

280 47.420

QPOQPO

4594.1321 2

POQ 841.024594.1

841.0421 2

q2

2

q1

41

A



substituting correctly BC2 = 652 +1042 –2 (65) (104) cos 60° A1 = 4225 + 10816 – 6760 = 8281 BC = 91 m A13

(b) finding the area, using bc sin (M1)

substituting correctly, area = (65) (104) sin 60° A1

= 1690 (Accept p = 1690) A1 3

(c) (i) A1 = (65) (x) sin 30° A1

= AG 1

(ii) A2 = (104) (x) sin 30° M1

= 26x A1 2

(iii) starting A1 + A2 = A or substituting + 26x = 1690 (M1)

simplifying = 1690 A1

x = A1

x = 40 (Accept q = 40) A1 4 (d) (i) Recognizing that supplementary angles have equal sines

eg = 180 – sin = sin R1 (ii) using sin rule in ΔADB and ΔACD (M1)

substituting correctly A1

and M1

since sin = sin

A1

AG 5

[18]

20. (a)

(M1)(A1)

(A1) (A1) (C4)

(b) Arc length (M1) Arc length = 9 cm (A1) (C2)

Note: Penalize a total of (1 mark) for missing units. [6]

21. (a) when (may be implied by a sketch) (A1)

(A1) (C2)

21 A

21

3

21

465x

21

465x 3

4169x 3

169316904

3

CDA BDA CDA BDA

BDAsin30sin

65BD

BDAsin65

30sinBD

CDAsin30sin

104DC

BDAsin104

30sinDC

BDA CDA

10465

DCBD

104DC

65BD

85

DCBD

212

A r

2127 (1.5)2

r

2 36r

6 cmr 1.5 6r

0y 8π or 2.799

x



(b) METHOD 1 Sketch of appropriate graph(s) (M1)

Indicating correct points (A1) (A1)(A1) (C2)(C2)

METHOD 2

, (A1)(A1)

,

, (A1)(A1) (C2)(C2)

[6] 22. (a) for using cosine rule (M1)

(A1)

(A1) (N0) 3 Notes: Either the first or the second line may be implied, but not both. Award no marks if 8.24 is obtained by assuming a right (angled) triangle (BC = 17 sin 29).

(i)

for using sine rule (may be implied) (M1)

(A1)

(A1) (N2)

(ii) (A1)

(Accept ) (A1) (N1) 5

(c) from previous triangle Therefore alternative (A1)

(M1)(A1)

(A1) (N1) 4

3.32 or 5.41x x

π 1sin9 2

x

π 7π9 6

x π 11π9 6

x

7π π6 9

x 11π π

6 9x

19π18

x 31π18

x ( 3.32, 5.41)x x

2 2 2 2 cosa b c ab C

2 2 2BC 15 17 2 15 17 cos29

BC 8.24 m

29°

85°

A

B

C

17

ACB 180 (29 85) 66

AC 17sin85 sin66

17sin85ACsin66

AC (18.5380 ) 18.5 m

1Area 17 18.538... sin 292

276.4 m 276.2 mBCA 66

ˆACB 180 66 114 (or 29 85)

ˆABC 180 (29 114) 37

29° 114°

37°

A

B

C

17

AC 17sin37 sin114

AC (11.19906 ) 11.2 m



(d)

Minimum length for BC when = 90°or diagram

showing right triangle (M1)

(A1) (N1) 2

[14]

23. (a) (i)

(A1)(A1) (N2) Note: Award (A1)(A1) for only if work shown, using product rule on .

(ii) or (A1) (N1)

(iii)

(A1)(A1)(A1) (N1)

(N1)(N1) 6

(b) (A1) (N1) 1

(c) (i) EITHER

curve crosses axis when (may be implied) (A1)

(M1)(A1) (N3)

OR

Area = (M1)(A2) (N3)

(ii) Area (M1) (A1) (N2) 5

[12]

24. Using area of a triangle = ab sin C (M1)

(A1)(A1)(A1)

Note: Accept any letter for Q sin Q = 0.5 (A1)

= 30 or or 0.524 (A1) (C6)

[6] 25. (a) b = 6 (A1) (C1)

29°

17

A

B

C

BCA

CBsin 2917

CB 17sin29

CB (8.2417 ) 8.24 m

1( ) 2cos2 sin2

f x x x

cos2 sinx x 22sin sin 1x x

sin cos cosx x x22sin sin 1x x (2sin 1)(sin 1)x x

2(sin 0.5)(sin 1)x x

2sin 1 or sin 1x x

1sin2

x

π 5π 3π(0.524) (2.62) (4.71)6 6 2

x x x

π 0.5246

x

π2

x

π 5π2 6

π π6 2

Area ( )d ( )df x x f x x

56

6

( ) df x x

0.875 0.875

1.75

21

Qsin(10)(8)2120

RQP6



(b)

(A3) (C3)

(c) x = 1.05 (accept (1.05, −0.896) ) (correct answer only, no additional solutions) (A2) (C2)

[6] 26. (a) 3(1 2 sin2 x) + sin x = 1 (A1)

6 sin2 x sin x 2 = 0 (p = 6, q = 1, r = 2) (A1) (C2) (b) (3 sin x 2)(2 sin x + 1) (A1)(A1) (C2) (c) 4 solutions (A2) (C2)

[6]

27. Area of large sector r2θ = 162 × 1.5 (M1)

= 192 (A1)

Area of small sector r2θ = × 102 × 1.5 (M1)

= 75 (A1) Shaded area = large area – small area = 192 – 75 (M1)

= 117 (A1) (C6) [6]

28. (a)

(A1)(A1) (C2) Note: Award (A1) for the graph crossing the y-axis between 0.5 and 1, and (A1) for an approximate sine curve crossing the x-axis twice. Do not penalize for x >3.14.

(b) (Maximum) x = 0.285… (A1)

x = 0.3 (1 dp) (A1) (C2)

(Minimum) x = 1.856… (A1)

x = 1.9 (1 dp) (A1) (C2) [6]

29. Area of a triangle = × 3 × 4 sin A (A1)

× 3 × 4 sin A = 4.5 (A1)

sin A = 0.75 (A1)

1 2

By

x

21

21

21

21

2

1.5

1

0.5

0

–0.5

–1

–1.5

–2

0.5 1 1.5 2 2.5 3 3.5 x

y

21

4π

21

43π

21

21



A = 48.6° and A = 131° (or 0.848, 2.29 radians) (A1)(A2) (C6) Note: Award (C4) for 48.6° only, (C5) for 131° only.

[6] 30. METHOD 1 2 cos2 x = 2 sin x cos x (M1)

2 cos2 x – 2 sin x cos x = 0 2 cos x(cos x – sin x) = 0 (M1) cos x = 0, (cos x – sin x) = 0 (A1)(A1)

x = , x = (A1)(A1) (C6)

METHOD 2 Graphical solutions

EITHER for both graphs y = 2 cos2 x, y = sin 2 x, (M2) OR for the graph of y = 2 cos2 x – sin 2 x. (M2)

THEN Points representing the solutions clearly indicated (A1)

1.57, 0.785 (A1)

x = , x = (A1)(A1) (C6)

Notes: If no working shown, award (C4) for one correct answer. Award (C2)(C2) for each correct decimal answer 1.57, 0.785. Award (C2)(C2) for each correct degree answer 90°, 45°. Penalize a total of [1 mark] for any additional answers.

[6] 31. (a) (i) 10 + 4 sin 1 = 13.4 (A1)

(ii) At 2100, t = 21 (A1) 10 + 4 sin 10.5 = 6.48 (A1) (N2) 3

Note: Award (A0)(A1) if candidates use t = 2100 leading to y = 12.6. No other ft allowed.

(b) (i) 14 metres (A1)

(ii) 14 = 10 + 4 sin sin = 1 (M1)

t = π (3.14) (correct answer only) (A1) (N2) 3 (c) (i) 4 (A1)

(ii) 10 + 4 sin = 7 (M1)

sin = –0.75 (A1)

t = 7.98 (A1) (N3) (iii) depth < 7 from 8 –11 = 3 hours (M1)

from 2030 – 2330 = 3 hours (M1) therefore, total = 6 hours (A1) (N3) 7

[13] 32. (a) Angle (A1)

(M1)

cm (A1) (C3)

(b) Area (M1)(A1)

7.07 (accept 7.06) (A1) (C3)

2π

4π

2π

4π

2t

2t

2t

2t

80A

AB 5sin 40 sin80

AB 3.26

1 1sin (5)(3.26)sin602 2

ac B

2cm



Note: Penalize once in this question for absence of units. [6]

33. METHOD 1

Area sector OAB (M1)

(A1) ON (A1)

AN (A1)

Area of

(A1) Shaded area

(A1) (C6) METHOD 2

Area sector (M1)

(A1)

Area (M1)

(A1) Twice the shaded area (M1)

Shaded area

(A1) (C6) [6]

34. (a) (i) (A1)(A1) (ii) EITHER

(M1) OR

, for (M1)

THEN

(A1)(A1)(A1) (N4) 6

(b) (i) translation (A1) in the y-direction of –1 (A1)

(ii) 1.11 (1.10 from TRACE is subject to AP) (A2) 4 [10]

35. 3 = p + q cos 0 (M1) 3 = p + q (A1) –1 = p + q cos (M1) –1 = p – q (A1) (a) p = 1 (A1) (C3) (b) q = 2 (A1) (C3)

21 (5) (0.8)2

10

5cos0.8 3.483...

5sin0.8 3.586.....

1AON ON AN2

6.249... 2(cm )10 6.249..

3.75 2(cm )

A

O N B

F

21ABF (5) (1.6)2

20

21OAF (5) sin1.62

12.5

20 12.5 ( 7.5)

1 (7.5)2

3.75 2(cm )

( ) 6sin2f x x

( ) 12sin cos 0f x x x

sin 0 or cos 0x x

sin2 0x 0 2 2x

π0, , π2

x



[6] 36. Method 1

0 (C2)

1.80 [3 sf] (G2) (C2) 2.51 [3 sf] (G2) (C2)

Method 2 3x = ±0.5x + 2 (etc.) (M1)

3.5x = 0, 2, 4 or 2.5x = 0, 2, 4 (A1) 7x = 0, 4, (8) or 5x = 0, 4, (8) (A1)

x = 0, or x = 0, (A1)(A1)(A1)

x = 0, , (C2)(C2)(C2)

[6]

37. (a) area of sector ΑΒDC = π(2)2 = π (A1)

area of segment BDCP = π – area of ABC (M1) = π – 2 (A1) (C3)

(b) BP = (A1)

area of semicircle of radius BP = π( )2 = π (A1)

area of shaded region = π – (π – 2) = 2 (A1) (C3) [6]

38. (a) = q – p

= (A1)(A1)

= (A1) 3

(b) cos (A1)

= , = (A1)(A1)

= –21 + 6 = –15 (A1)

cos (AG) 4

(c) (i) Since + = 180° (R1)

y

x1.80 2.510

7π4

5π4

7π4

5π4

41

2

21

2

PQOR

37

–1

10

2–3

PQPOPQPO QPO

22 3–7–PO 58 22 2–3PQ 13

PQPO

75415–

135815–QPO

QPO RQP



cos = –cos (AG)

(ii) sin = (M1)

= (A1)

= (AG)

OR

cos =

(M1)

therefore x2 = 754 – 225 = 529 x = 23 (A1)

sin = (AG)

Note: Award (A1)(A0) for the following solution.

cos = = 56.89°

sin = 0.8376

= 0.8376 sin =

(iii) Area of OPQR = 2 (area of triangle PQR) (M1)

= 2 × (A1)

= 2 × (A1)

= 23 sq units. (A1) OR Area of OPQR = 2 (area of triangle OPQ) (M1)

= 2 (A1)(A1)

= 23 sq units. (A1) 7 Notes: Other valid methods can be used. Award final (A1) for the integer answer.

[14]

39. (a) Sine rule (M1)(A1)

PR =

= 5.96 km (A1) 3 (b) EITHER

RQP QPO

75415

RQP2

75415–1

754529

75423

75415

15754

Px

75423

75415

75423

75423

RQPsinQRPQ21

754235813

21

103–1721

120sin9

sin35PR

120sin35sin9



Sine rule to find PQ

PQ = (M1)(A1)

= 4.39 km (A1) OR Cosine rule: PQ2 = 5.962 + 92 – (2)(5.96)(9) cos 25 (M1)(A1)

= 19.29 PQ = 4.39 km (A1)

Time for Tom = (A1)

Time for Alan = (A1)

Then = (M1)

a = 10.9 (A1) 7

(c) RS2 = 4QS2 (A1)

4QS2 = QS2 + 81 – 18 × QS × cos 35 (M1)(A1)

3QS2 + 14.74QS – 81 = 0 (or 3x2 + 14.74x – 81 = 0) (A1) QS = –8.20 or QS = 3.29 (G1) therefore QS = 3.29 (A1) OR

(M1)

sin sin 35 (A1)

= 16.7° (A1)

Therefore, = 180 – (35 + 16.7) = 128.3° (A1)

(M1)

QS =

= 3.29 (A1) 6 [16]

40. (a) (i) cos , sin (A1)

therefore cos = 0 (AG)

(ii) cos x + sin x = 0 1 + tan x = 0 tan x = –l (M1)

x = (A1)

Note: Award (A0) for 2.36. OR

x = (G2) 3

120sin25sin9

839.4

a96.5

839.4

a96.5

sin35QS2

QRsinSQS

21 QRS

QRSRSQ

35sin SR

sin16.7QS

3.128sin9

3.128sin235sin9

3.128sin7.16sin9

21

4π–

21–

4π–

4π–sin

4π–

4π3

4π3



(b) y = ex(cos x + sin x)

= ex(cos x + sin x) + ex(–sin x + cos x) (M1)(A1)(A1) 3

= 2ex cos x

(c) = 0 for a turning point 2ex cos x = 0 (M1)

cos x = 0 (A1)

x = a = (A1)

y = e (cos + sin ) = e

b = e (A1) 4 Note: Award (M1)(A1)(A0)(A0) for a = 1.57, b = 4.81.

(d) At D, = 0 (M1)

2ex cos x – 2exsin x = 0 (A1)

2ex (cos x – sin x) = 0 cos x – sin x = 0 (A1)

x = (A1)

y = e (cos + sin ) (A1)

= e (AG) 5

(e) Required area = (cos x + sin x)dx (M1)

= 7.46 sq units (G1) OR Αrea = 7.46 sq units (G2) 2

Note: Award (M1)(G0) for the answer 9.81 obtained if the calculator is in degree mode.

[17]

41. (a) (i) A is (A1)(A1) (C2)

(ii) B is (0, –4) (A1)(A1) (C2) Note: In each of parts (i) and (ii), award C1 if A and B are interchanged, C1 if intercepts given instead of coordinates.

(b) Area = × 4 × (M1)

= (= 2.67) (A1) (C2)

[6] 42. (a) (3 sin x – 2)(sin x – 3) (A1)(A1) (C2)

Note: Award A1 if 3x2 – 11x + 6 correctly factorized to give (3x – 2)(x – 3) (or equivalent with another letter).

(b) (i) (3 sin x – 2)(sin x – 3) = 0

xy

dd

xy

dd

2π

2π

2π

2π

2π 2

π

2π

2

2

dd

xy

4π

4π

4π

4π

2 4π

43

0e x

0,34

21

34

38



sin x = sin x = 3 (A1)(A1) (C2)

(ii) x = 41.8°, 138° (A1)(A1) (C2) Notes: Penalize [1 mark] for any extra answers and [1 mark] for answers in radians. ie Award A1 A0 for 41.8°, 138° and any extra answers. Award A1 A0 for 0.730, 2.41. Award A0 A0 for 0.730, 2.41 and any extra answers.

[6] 43. Note: Do not penalize missing units in this question.

(a) AB2 = 122 + 122 – 2 × 12 × 12 × cos 75° (A1)

= 122(2 – 2 cos 75°) (A1)

= 122 × 2(1 cos 75°) AB = 12 (AG) 2

Note: The second (A1) is for transforming the initial expression to any simplified expression from which the given result can be clearly seen.

(b) = 37.5° (A1) BP = 12 tan 37.5° (M1) = 9.21 cm (A1)

OR = 105° = 37.5° (A1)

(M1)

BP = = 9.21(cm) (A1) 3

(c) (i) Area ∆OBP = (M1)

= 55.3 (cm2) (accept 55.2 cm2) (A1)

(ii) Area ∆ABP = (9.21)2 sin105° (M1)

= 41.0 (cm2) (accept 40.9 cm2) (A1) 4

(d) Area of sector = (M1)

= 94.2 (cm2) (accept 30π or 94.3 (cm2)) (A1) 2 (e) Shaded area = 2 × area ∆OPB – area sector (M1)

= 16.4 (cm2) (accept 16.2 cm2, 16.3 cm2) (A1) 2 [13]

44. Note: Do not penalize missing units in this question. (a) (i) At release(P), t = 0 (M1)

s = 48 + 10 cos 0 = 58 cm below ceiling (A1)

(ii) 58 = 48 +10 cos 2πt (M1) cos 2πt = 1 (A1) t = 1sec (A1)

OR t = 1sec (G3) 5

(b) (i) = –20π sin 2πt (A1)(A1)

Note: Award (A1) for –20π, and (A1) for sin 2t.

32

)75cos1(2

BOP

APB PAB

5.37sinBP

105sinAB

105sin37.5sinAB

5.37tan1212

21or 21.912

21

21

22 12π

36075or

180π7512

21

ts

dd



(ii) v = = –20π sin 2πt = 0 (M1)

sin 2 πt = 0

t = 0, ... (at least 2 values) (A1)

s = 48 + 10 cos 0 or s = 48 +10 cos π (M1) = 58 cm (at P) = 38 cm (20 cm above P) (A1)(A1) 7

Note: Accept these answers without working for full marks. May be deduced from recognizing that amplitude is 10.

(c) 48 +10 cos 2πt = 60 + 15 cos 4πt (M1) t = 0.162 secs (A1)

OR t = 0.162 secs (G2) 2 (d) 12 times (G2) 2

Note: If either of the correct answers to parts (c) and (d) are missing and suitable graphs have been sketched, award (G2) for sketch of suitable graph(s); (A1) for t = 0.162; (A1) for 12.

[16] 45. (a) l = r or ACB = 2 × OA (M1)

= 30 cm (A1) (C2)

(b) (obtuse) = 2 – 2 (A1)

Area = r 2 = (2 – 2)(15)2 (M1)(A1)

= 482 cm2 (3 sf) (A1) (C4) [6]

46.

(M1)(A2)

OR 2.5 × 20 = 50 (M1)(A1) 2.5 × 32 = 80 (A1)

d2 = 502 + 802 – 2 × 50 × 80 × cos 70° (M1)(A1) d = 78.5 km (A1) (C6)

[6] 47. (a) (i) –1 (A1) (C1)

(ii) 4 (accept 720°) (A2) (C2) (b)

(G1)

number of solutions: 4 (A2) (C3) [6]

48. Statement (a) Is the statement true for all

real numbers x? (Yes/No) (b) If not true, example

A No x = –l (log10 0.1 = –1) (a) (A3) (C3) B No x = 0 (cos 0 = 1) (b) (A3) (C3) C Yes N/A

ts

dd

21

BOA

21

21

AB

50 8070°

d

P

32

y



Notes: (a) Award (A1) for each correct answer. (b) Award (A) marks for statements A and B only if NO in column (a). Award (A2) for a correct counter example to statement A, (A1) for a correct counter example to statement B (ignore other incorrect examples). Special Case for statement C: Award (A1) if candidates write NO, and give a

valid reason (eg arctan 1 = ).

[6]

49. (a) (M1)

sin A = (A1)

= (AG) 2

(b)

(i) + = 180° (A1)

(ii) sin A =

=> A = 59.0° or 121° (3 sf) (A1)(A1)

=> = 180° – (121° + 45°) = 14.0° (3 sf) (A1)

(iii) (M1)

=>BD = 1.69 (A1) 6

(c) (M1)(A1)

= (AG) 2

OR

(M1)(A1)

= (AG) 2

[10]

50. Using sine rule: (M1)(A1)

sin B = sin 48° = 0.5308… (M1)

B = arcsin (0.5308) = 32.06° (M1)(A1)

4π5

45sin2

27

Asin6

272

22 6

76

B

D

C

A

h

CDB CAB

76

DCB

45sin2

27

14sin BD

h

h

BA21

BD21

BAC AreaBDC Area

BABD

45sin6BA21

45sin6BD21

ΔBACAreaΔBCDArea

BABD

748sin

5sin

B

75



= 32° (nearest degree) (A1) (C6) Note: Award a maximum of [5 marks] if candidates give the answer in radians (0.560).

[6] 51. (a) x is an acute angle => cos x is positive. (M1)

cos2 x + sin2 x = 1 => cos x = (M1)

=> cos x = (A1)

= (= ) (A1) (C4)

(b) cos 2x = 1 – 2 sin2 x = 1 – 2 (M1)

= (A1) (C2)

Notes: (a) Award (M1)(M0)(A1)(A0) for cos = 0.943.

(b) Award (M1)(A0) for cos = 0.778.

[6] 52. (a) 2 sin2 x = 2(1 – cos2 x) = 2 – 2 cos2 x = l + cos x (M1)

=> 2 cos2 x + cos x – l = 0 (A1) (C2)

Note: Award the first (M1) for replacing sin2 x by 1 – cos2 x.

(b) 2 cos2 x + cos x – 1 = (2 cos x – 1)(cos x +1) (A1) (C1)

(c) cos x = or cos x = –l

=> x = 60°, 180° or 300° (A1)(A1)(A1) (C3)

Note: Award (A1)(A1)(A0) if the correct answers are given in radians (ie

, , , or 1.05, 3.14, 5.24)

[6] 53. (a) The smallest angle is opposite the smallest side.

cos θ = (M1)

= = 0.7857

Therefore, θ = 38.2° (A1) (C2)

(b) Area = × 8 × 7 × sin 38.2° (M1)

= 17.3 cm2 (A1) (C2) [4]

54. (a) 3 sin2 x + 4 cos x = 3(1 – cos2 x) + 4cos x

= 3 – 3 cos2 + 4 cos x (A1) (C1)

(b) 3 sin2 x + 4 cos x – 4 = 0 3 – 3 cos2 x + 4 cos x – 4 = 0

x2sin–12

31–1

98

322

2

31

97

31sin 1–

31sin2 1–

21

3

35

782578 222

1411

11288

21



3 cos2 x – 4 cos x + 1 = 0 (A1) (3 cos x – 1)(cos x – 1) = 0

cos x = or cos x = 1

x = 70.5° or x = 0° (A1)(A1) (C3) Note: Award (C1) for each correct radian answer, ie x = 1.23 or x = 0.

[4] 55. = 90° (A1)

AT =

=

= 60° = (A1)

Area = area of triangle – area of sector

= × 6 × – × 6 × 6 × (M1)

= 12.3 cm2 (or – 6) (A1) (C4) OR = 60° (A1)

Area of = × 6 × 12 × sin 60 (A1)

Area of sector = × 6 × 6 × (A1)

Shaded area = – 6 = 12.3 cm2 (3 sf) (A1) (C4) [4]

56. (a) (i) AP = (M1) (AG)

(ii) OP = (A1) 2

(b) (M1)

= (M1)

= (M1)

(AG) 3

(c) For x = 8, = 0.780869 (M1) arccos 0.780869 = 38.7° (3 sf) (A1)

OR

(M1)

= arctan (0.8) = 38.7° (3 sf) (A1) 2 (d) = 60° = 0.5

0.5 = (M1)

31

ATO22 612

36

AOT3π

21 36

21

3π

318

AOT

21

21

3π

318

8016)610()8( 222 xxx

100)010()0( 222 xx

OPAP2OAOPAPAPOcos

222

10080162

)68()100()8016(22

2222

xxx

xxx

10080162

8016222

2

xxx

xx

)}100)(8016{(

408APOcos22

2

xxx

xx

APOcos

108APOtan

APOAPO APOcos

)}100)(8016{(

40822

2

xxx

xx



2x2 – 16x + 80 – = 0 (M1) x = 5.63 (G2) 4

(e) (i) f (x) = 1 when = 1 (R1) hence, when = 0. (R1) This occurs when the points O, A, P are collinear. (R1)

(ii) The line (OA) has equation y = (M1)

When y = 10, x = (= 13 ) (A1)

OR

x = (= 13 ) (G2) 5

Note: Award (G1) for 13.3. [16]

57. (a) Area = (152)(2) (M1)

= 225 (cm2) (A1) (C2)

(b) Area ∆OAB = 152 sin 2 = 102.3 (A1)

Area = 225 – 102.3 = 122.7 (cm2) = 123 (3 sf) (A1) (C2)

[4]

58. (a) (M1)

= 0.901

> 90° = 180° – 64.3° = 115.7° = 116 (3 sf) (A1) (C2)

(b) In Triangle 1, = 64.3° = 180° – (64.3° + 50°) = 65.7° (A1)

Area = (20)(17) sin 65.7° = 155 (cm2) (3 sf) (A1) (C2)

[4] 59. METHOD 1

The value of cosine varies between –1 and +1. Therefore:t = 0 a + b = 14.3t = 6 a – b = 10.3

2a = 24.6 a = 12.3 (A1) (C1) 2b = 4.0 b = 2 (A1) (C1)

Period = 12 hours = 2π (M1)

k = 12 (A1) (C2) METHOD 2

)}100)(8016{( 22 xxx

APOcosAPO

43x

340

31

340

31

21

21 2 r

21

1750sin

20B)CA(sin

1750sin20B)CA(sin

BCA BCABCA

BCACAB

21

k)12(π2

y

t (h)6 12 18 24

10.3

14.3



From consideration of graph: Midpoint = a = 12.3 (A1) (C1) Amplitude = b = 2 (A1) (C1)

Period = = 12 (M1)

k = 12 (A1) (C2) [4]

60.

cos (M1)(A1)

= arccos(0.0625) (A1) 86° (A1)

[4] 61. (a) 2 cos2 x + sin x = 2(1 – sin2 x) + sin x

= 2 – 2 sin2 x + sin x (A1)

(b) 2 cos2 x + sin x = 2

2 – 2 sin2 x + sin x = 2

sin x – 2 sin2 x = 0 sin x(1 – 2 sin x) = 0

sin x = 0 or sin x = (M1)

sin x = 0 x = 0 or (0° or 180°) (A1) Note: Award (A1) for both answers.

sin x = x = or (30° or 150°) (A1)

Note: Award (A1) for both answers. [4]

62. (a)

5 (b) is a solution if and only if + cos = 0. (M1)

Now + cos = + (–1) (A1) = 0 (A1) 3

(c) By using appropriate calculator functions x = 3.696 722 9... (M1) x = 3.69672 (6sf) (A1) 2

(d) See graph: (A1)

kπ2π2

AB

C

32km

48km

)32)(48(2563248BAC

222

BAC

21

21

6π

65π

4

3

2

1

1 2 3 4 5

–1

integerson axis

0.5< <13.5< <4

xy

3.2< <3.6–0.2< <0

xy

{

{

{

(A1)

(A1) (A1)

(A1)

(A1)

LEFTINTERCEPT

RIGHTINTERCEPT3< <3.5x 3.5< <4x

MAXIMUMPOINT

MINIMUMPOINT

y

x



(A1) 2

(e) EITHER = 7.86960 (6 sf) (A3) 3

Note: This answer assumes appropriate use of a calculator eg

‘fnInt’:

OR

= ( – 0) + ( sin – 0 × sin 0) + (cos – cos 0) (A1)

= 2 + 0 + –2 = 7.86960 (6 sf) (A1) 3 [15]

63. (a) (i) Q = (14.6 – 8.2) (M1)

= 3.2 (A1)

(ii) P = (14.6 + 8.2) (M0)

= 11.4 (A1) 3

(b) 10 = 11.4 + 3.2 cos (M1)

so = cos

therefore arccos (A1)

which gives 2.0236... = t or t = 3.8648. t = 3.86(3 sf) (A1) 3

(c) (i) By symmetry, next time is 12 – 3.86... = 8.135... t = 8.14 (3 sf) (A1) (ii) From above, first interval is 3.86 < t < 8.14 (A1)

This will happen again, 12 hours later, so (M1) 15.9 < t < 20.1 (A1) 4

[10] 64. 3 cos x = 5 sin x

(M1)

tan x = 0.6 (A1) x = 31° or x = 211° (to the nearest degree) (A1)(A1) (C2)(C2)

Note: Deduct [1 mark] if there are more than two answers. [4]

65. sin A = cos A = (A1)

But A is obtuse cos A = – (A1)

sin 2A = 2 sin A cos A (M1)

= 2 ×

= – (A1) (C4)

[4] 66. (a) y = sin x – x

π

0d)cosπ( xxx

π

0d)cosπ( xxx

xxYwithXYfnInt

cosπ869604401.7)π,0,,(

1

1

π0

π

0]cossinπ[d)cosπ( xxxxxxx

21

21

t6π

167

t6π

t6π

167

6π

53

cossin

xx

135

1312

1312

1312

135

169120



(A5) 5 Notes: Award (A1) for appropriate scales marked on the axes. Award (A1) for the x-intercepts at (2.3, 0). Award (A1) for the maximum and minimum points at (1.25, 1.73). Award (A1) for the end points at (3, 2.55). Award (A1) for a smooth curve. Allow some flexibility, especially in the middle three marks here.

(b) x = 2.31 (A1) 1

(c) (A1)(A1)

Note: Do not penalize for the absence of C.

Required area = (M1)

= 0.944 (G1) OR area = 0.944 (G2) 4

[10] 67. (a)

Acute angle 30° (M1)

Note: Award the (M1) for 30° and/or quadrant diagram/graph seen. 2nd quadrant since sine positive and cosine negative

= 150° (A1) (C2)

(b) tan 150° = –tan 30° or tan 150° = (M1)

tan 150° = – (A1) (C2)

[4]

68. (a) = tan 36°

PQ 29.1 m (3 sf) (A1) (C1) (b)

(–2.3, 0)

(–1.25, –1.73)

(1.25, 1.73)

(2.3, 0)

–3

–3

–2

–2

–1

–1

1

1

2

2

3

3

y

x

Cxxxxx2

cosπd)sinπ(2

1

0d)sinπ( xxx

30º

23

21

31

40PQ

40m

A

B

Q

30

70



= 80° (A1)

(M1)

Note: Award (M1) for correctly substituting. AB = 41 9. m (3 sf) (A1) (C3)

[4] 69. Perimeter = 5(2π – 1) + 10 (M1)(A1)(A1)

Note: Award (M1) for working in radians; (A1) for 2π – 1; (A1) for +10. = (10π + 5) cm (= 36.4, to 3 sf) (A1) (C4)

[4]

70. From sketch of graph y = 4 sin (M2)

or by observing sin 1.k > 4, k < –4 (A1)(A1) (C2)(C2)

[4] 71. (a)(i) & (c)(i)

(A3) Notes: The sketch does not need to be on graph paper. It should have the correct shape, and the points (0, 0), (1.1, 0.55), (1.57, 0) and (2, –1.66) should be indicated in some way. Award (A1) for the correct shape. Award (A2) for 3 or 4 correctly indicated points, (A1) for 1 or 2 points.

(ii) Approximate positions are positive x-intercept (1.57, 0) (A1) maximum point (1.1, 0.55) (A1) end points (0, 0) and (2, –1.66) (A1)(A1) 7

BQA

70sin40

80sinAB

2π3x

4

3

2

1

–1

–2

–3

–4

00–2 – 2

1

0

–1

–2

1 2

(1.1, 0.55)

R

(1.51, 0)

(2, –1.66)

y

x



(b) x2 cos x = 0 x ≠ 0 ⇒ cos x = 0 (M1)

x = (A1) 2

Note: Award (A2) if answer correct. (c) (i) see graph (A1)

(ii) cos x dx (A2) 3

Note: Award (A1) for limits, (A1) for rest of integral correct (do not penalize missing dx).

(d) Integral = 0.467 (G3) OR

Integral = (M1)

= – [0 + 0 – 0] (M1)

= – 2 (exact) or 0.467 (3 sf) (A1) 3

[15] 72. (a) From graph, period = 2π (A1) 1

(b) Range = {y –0.4 < y < 0.4} (A1) 1

(c) (i) f (x) = {cos x (sin x)2}

= cos x (2 sin x cos x) – sin x (sin x)2 or –3 sin3 x + 2 sin x (M1)(A1)(A1) Note: Award (M1) for using the product rule and (A1) for each part.

(ii) f (x) = 0 (M1)

sin x{2 cos x – sin2 x} = 0 or sin x{3 cos x – 1} = 0 (A1)

3 cos2 x – 1 = 0

cos x = ± (A1)

At A, f (x) > 0, hence cos x = (R1)(AG)

(iii) f (x) = (M1)

= (A1) 9

(d) x = (A1) 1

(e) (i) (M1)(A1)

(ii) Area = (M1)

= (A1) 4

(f) At C f (x) = 0 (M1)

9 cos3 x – 7 cos x = 0

2π

20

2x

2/π0

2 sin2cos2sin xxxxx

)1(2)0(

2π2)1(

4π2

2π

xdd

31

31

2

31–1

31

392

31

32

2π

cxxxx 32 sin

31d))(sin(cos

2/π

0

33

2 )0(sin2πsin

31d))(sin(cos xxx

31



cos x(9 cos2 x – 7) = 0 (M1)

x = (reject) or x = arccos = 0.491 (3 sf) (A1)(A1) 4

[20] 73. Note: Award (M1) for identifying the largest angle.

cos = (M1)

= – (A1)

= 101.5° (A1) OR Find other angles first = 44.4° = 34.0° (M1) = 101.6° (A1)(A1) (C4)

Note: Award (C3) if not given to the correct accuracy. [4]

74. AB = r

= (M1)(A1)

= 21.6 × (A1)

= 8 cm (A1)

OR × (5.4)2 = 21.6

= (= 1.481 radians) (M1)

AB = r (A1)

= 5.4 × (M1)

= 8 cm (A1) (C4) [4]

75. (a) = 6 A is on the circle (A1)

= 6 B is on the circle. (A1)

= = 6 C is on the circle. (A1) 3

(b)

= (M1)

= (A1) 2

(c) (M1)

2π

37

542754 222

51

rr 2

21 2

4.52

21

7.24

7.24

OA

OB

115

OC

1125

OAOCAC

06

115

111

ACAOACAOCAO

ˆcos



=

= (A1)

= (A1)

OR (M1)(A1)

= as before (A1)

OR using the triangle formed by and its horizontal and vertical components:

(A1)

(M1)(A1) 3

Note: The answer is 0.289 to 3 sf (d) A number of possible methods here

= (A1)

= (A1)

BC =

ABC = (A1)

= (A1) OR ABC has base AB = 12 (A1) and height = (A1)

area = (A1)

= (A1)

OR Given

(A1)(A1)(A1)

= (A1) 4 [12]

76. tan2 x = (M1)

tan x = (M1)

x = 30° or x = 150° (A1)(A1) (C2)(C2)

1116

111

.06

1266

63

321

12626)12(6ˆcos

222

CAO

121

AC

12AC

121ˆcos CAO

OBOCBC

06

115

1111

132

1213221

116

11

111221

116

63ˆcos CAB

6331212

21

633ˆsin ABCCAB

116

31

31



[4] 77. h = r so 2r2 = 100 r2 = 50 (M1)

l = 10 = 2r (M1)

= (A1)

=

= = 4.44 (3sf) (A1) (C4) Note: Accept either answer.

[4] 78. (a) f (1) = 3 f (5) = 3 (A1)(A1) 2

(b) EITHER distance between successive maxima = period (M1) = 5 – 1 (A1) = 4 (AG)

OR Period of sin kx = ; (M1)

so period = (A1)

= 4 (AG) 2

(c) EITHER A sin + B = 3 and A sin + B = –1 (M1) (M1)

A + B = 3, – A + B = –1 (A1)(A1) A = 2, B = 1 (AG)(A1) OR Amplitude = A (M1)

A = (M1)

A = 2 (AG) Midpoint value = B (M1)

B = (M1)

B = 1 (A1) 5 Note: As the values of A = 2 and B = 1 are likely to be quite obvious to a bright student, do not insist on too detailed a proof.

(d) f (x) = 2 sin + 1

f (x) = + 0 (M1)(A2)

Note: Award (M1) for the chain rule, (A1) for , (A1) for 2 cos .

= cos (A1) 4

Notes: Since the result is given, make sure that reasoning is valid. In particular, the final (A1) is for simplifying the result of the chain rule calculation. If the preceding steps are not valid, this final mark should not be given. Beware of “fudged” results.

(e) (i) y = k – x is a tangent – = cos (M1)

1050π2

1025π2

2

kπ2

2ππ2

2π

2π3

24

2)1(3

22

2)1(3

x2π

x2πcos2

2π

2π

x2π

x2π

x2π



–1 = cos (A1)

x = or 3 or ...

x = 2 or 6 ... (A1) Since 0 x 5, we take x = 2, so the point is (2, 1) (A1)

(ii) Tangent line is: y = –(x – 2) + 1 (M1) y = (2 + 1) – x k = 2 + 1 (A1) 6

(f) f (x) = 2 2 sin + 1 = 2 (A1)

sin (A1)

x = (A1)(A1)(A1) 5

[24]

x2π

2π

x2π

21

2π

x

613πor

65πor

6π

2π

x

313or

35or

31