1. (a) choosing sine rule (m1) · substitution into any correct formula a1 e.g. area pqr = 7 10 sin...

IB Questionbank Maths SL 1

1. (a) choosing sine rule (M1)

correct substitution 10

75sin

7

sin

R A1

sin R = 0.676148...

QRP = 42.5 A1 N2

(b) P = 180 75 R

P = 62.5 (A1)

substitution into any correct formula A1

e.g. area PQR = sin1072

1 (their P)

= 31.0 (cm2) A1 N2

[6]

2. (a) evidence of choosing the formula cos2

A = 2 cos2

A 1 (M1)

Note: If they choose another correct formula, do

not award the M1 unless there is evidence

of finding sin2

A = 1 9

1.

correct substitution A1

e.g. cos 2A = 13

122cos,

9

8

3

122

A

9

72cos A A1 N2


(b) METHOD 1

evidence of using sin2

B + cos2

B = 1 (M1)

e.g. 9

5,1cos

3

2 2

2

B (seen anywhere),

cos B =

3

5

9

5 (A1)

cos B =

3

5

9

5 A1 N2

METHOD 2

diagram M1

e.g.

for finding third side equals 5 (A1)

cos B = 3

5 A1 N2

[6]

3. a = 4, b = 2, c =

etc

2

3or

2 A2A2A2 N6

[6]

4. (a) changing tan x into x

x

cos

sin A1

e.g. sin3 x + cos

3 x

x

x

cos

sin

simplifying A1

e.g. sin x (sin2 x + cos

2 x), sin

3 x + sin x – sin

3 x

f(x) = sin x AG N0


(b) recognizing f(2x) = sin 2x, seen anywhere (A1)

evidence of using double angle identity sin (2x) = 2 sin x cos x,

seen anywhere (M1)

evidence of using Pythagoras with sin x = 3

2 M1

e.g. sketch of right triangle, sin2 x + cos

2 x = 1

cos x =

3

5accept

3

5 (A1)

f(2x) = 2

3

5

3

2 A1

f(2x) = 9

54 AG N0

[7]

5. (a) period = A1 N1

(b)

4

3

2

1

–1

–2

–3

–4

y

x0 π2

π 32π 2π

A1A1A1 N3

Note: Award A1 for amplitude of 3, A1 for their

period, A1 for a sine curve passing through

(0, 0) and (0, 2).

(c) evidence of appropriate approach (M1)

e.g. line y = 2 on graph, discussion of number of solutions in

the domain

4 (solutions) A1 N2 [6]


6. (a) evidence of appropriate approach M1

e.g. 3 = 9

2r

r =13.5 (cm) A1 N1

(b) adding two radii plus 3 (M1)

perimeter = 27+3 (cm) (= 36.4) A1 N2

(c) evidence of appropriate approach M1

e.g. 9

25.13

2

1 2

area = 20.25 (cm2) (= 63.6) A1 N1

[6]

7. (a) For using perimeter = r + r + arc length (M1)

20 = 2r + r A1

r

r220 AG N0

(b) Finding A = 22 10220

2

1rr

r

rr

(A1)

For setting up equation in r M1

Correct simplified equation, or sketch

eg 10r – r2 = 25, r

2 – 10r + 25 = 0 (A1)

r = 5 cm A1 N2 [6]

h˜8. (a) (i) sin 140 = p A1 N1

(ii) cos 70 = q A1 N1


(b) METHOD 1

evidence of using sin2 + cos

2 = 1 (M1)

e.g. diagram, 21 p (seen anywhere)

cos 140 = 21 p (A1)

cos 140 = 21 p A1 N2

METHOD 2

evidence of using cos2 = 2 cos2 1 (M1)

cos 140 = 2 cos2 70 1 (A1)

cos 140 = 2( q)2 1 (= 2q

2 1) A1 N2

(c) METHOD 1

tan 140 = 21140cos

140sin

p

p

A1 N1

METHOD 2

tan 140 = 12 2 q

p A1 N1

[6]

9. (a) correct substitution into the formula for the area of a triangle A1

e.g. 2

1 × 5 × 13.6 × sin C = 20,

2

1 × 5 × h = 20

attempt to solve (M1)

e.g. sin C = 0.5882... , sin C = 6.13

8

C = 36.031...° (0.6288… radians) (A1)

BCA = 144° (2.51 radians) A1 N3

(b) evidence of choosing the cosine rule (M1)


e.g. (AB)2 = 5

2 + 13.6

2 – 2(5)(13.6)cos143.968...

AB = 17.9 A1 N2 [7]


10. (a) finding CBA = 110° (= 1.92 radians) (A1)

evidence of choosing cosine rule (M1)

e.g. AC2 = AB

2 + BC

2 – 2(AB)(BC) cos CBA


e.g. AC2 = 25

2 + 40

2 – 2(25)(40) cos 110°

AC = 53.9 (km) A1 N3

(b) METHOD 1

correct substitution into the sine rule A1

e.g. 9.53

110sin

40

CABsin

CAB = 44.2° A1

bearing = 074° A1 N1

METHOD 2

correct substitution into the cosine rule A1

e.g. )9.53)(25(2

9.532540CABcos

222

CAB = 44.3° A1

bearing = 074° A1 N1 [7]

11. (a) (i) evidence of finding the amplitude (M1)

e.g. 2

37 , amplitude = 5

p = –5 A1 N2

(ii) period = 8 (A1)

q = 0.785

4

π

8

π2 A1 N2

(iii) r = 2

37 (A1)

r = 2 A1 N2

(b) k = –3 (accept y = –3) A1 N1 [7]


12. (a) METHOD 1

evidence of choosing the cosine formula (M1)


e.g. 772

1377BCAcos

222

BCA = 2.38 radians (= 136°) A1 N2

METHOD 2

evidence of appropriate approach involving right-angled triangles (M1)


e.g. 7

5.6BCA

2

1sin

BCA = 2.38 radians (= 136°) A1 N2

(b) METHOD 1

DCA = π – 2.381 (180 – 136.4) (A1)

evidence of choosing the sine rule in triangle ACD (M1)


e.g. CDsinA

7

...760.0sin

5.6

CDA = 0.836... (= 47.9...°) A1

DAC = π – (0.760... + 0.836...) (180 – (43.5... + 47.9...))

= 1.54 (= 88.5°) A1 N3


METHOD 2

)4.136180(

2

1)381.2π(

2

1CBA (A1)

evidence of choosing the sine rule in triangle ABD (M1)


e.g. CDAsin

13

...380.0sin

5.6

CDA = 0.836... (= 47.9...°) A1

DAC = π – 0.836... – (π – 2.381...) (= 180 – 47.9... – (180 – 136.4))

= 1.54 (= 88.5°) A1 N3

Note: Two triangles are possible with the given information.

If candidate finds CDA ˆ = 2.31 (132°) leading to

DAC ˆ = 0.076 (4.35°), award marks as

per markscheme. [8]

13. evidence of substituting for cos2x (M1)

evidence of substituting into sin2 x + cos

2 x = 1 (M1)

correct equation in terms of cos x (seen anywhere) A1

e.g. 2cos2 x – 1 – 3 cos x – 3 = 1, 2 cos

2 x – 3 cos x – 5 = 0

evidence of appropriate approach to solve (M1)

e.g. factorizing, quadratic formula

appropriate working A1

e.g. (2 cos x – 5)(cos x + 1) = 0, (2x – 5)(x + 1), cos x = 4

493

correct solutions to the equation

e.g. cos x = 2

5, cos x = –1, x =

2

5, x = –1 (A1)

x = π A1 N4 [7]

14. (a) METHOD 1

choosing cosine rule (M1)

substituting correctly A1

e.g. AB = 8.1cos)9.3)(9.3(29.39.3 22

AB = 6.11(cm) A1 N2


METHOD 2

evidence of approach involving right-angled triangles (M1)


e.g. sin 0.9 = 2

1,

9.3

xAB = 3.9 sin 0.9

AB = 6.11 (cm) A1 N2

METHOD 3

choosing the sine rule (M1)


e.g. AB

8.1sin

9.3

...670.0sin

AB = 6.11 (cm) A1 N2

(b) METHOD 1

reflex BOA = 2π – 1.8 (= 4.4832) (A2)

correct substitution A = 2

1(3.9)

2(4.4832...) A1

area = 34.1 (cm2) A1 N2

METHOD 2

finding area of circle A = π(3.9)2 (= 47.78...) (A1)

finding area of (minor) sector A = 2

1(3.9)

2(1.8) (= 13.68...) (A1)

subtracting M1

e.g. π(3.9)2 – 0.5(3.9)

2(1.8), 47.8 – 13.7

area = 34.1 (cm2) A1 N2

METHOD 3

finding reflex BOA = 2π – 1.8 (= 4.4832) (A2)

finding proportion of total area of circle A1

e.g. 22 ππ2

,)9.3(ππ2

8.1π2r

area = 34.1 (cm) A1 N2 [7]

15. (a) (i) 7 A1 N1

(ii) 1 A1 N1

(iii) 10 A1 N1

(b) (i) evidence of appropriate approach M1


e.g. 2

218A

A = 8 AG N0

(ii) C = 10 A2 N2

(iii) METHOD 1

period = 12 (A1)

evidence of using B period = 2 (accept 360) (M1)

e.g. 12 = B

2

6

πB (accept 0.524 or 30) A1 N3

METHOD 2

evidence of substituting (M1)

e.g. 10 = 8 cos 3B + 10

simplifying (A1)

e.g. cos 3B = 0

23B

6

πB (accept 0.524 or 30) A1 N3

(c) correct answers A1A1

e.g. t = 3.52, t = 10.5, between 03:31 and 10:29 (accept 10:30) N2 [11]

16. (a) When t = 1, l = 33 + 5 cos 720 (M1)

l = 33 + 5 = 38 A1 N2

(b) Minimum when cos = –1 (M1)

lmin = 33 – 5 (M1)

= 28 A1 N3

(c) 33 = 33 + 5cos720t (0 = 5 cos 720t) M1

720t = 90 A1

t =

8

1

720

90 A1 N1

(d) Evidence of dividing into 360 (M1)

period =

2

1

720

360 A1 N2


[10]

17. (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]

18. (a)

A1A1A1 N3 3

Note: Award A1 for labelling 4° with horizontal, A1 for

labelling [AU] 25 metres, A1 for drawing [TU].

(b) TÂU = 86º (A1)



e.g. x2 = 25

2 + 36

2 – 2(25)(36) cos 86º

x = 42.4 A1 N3 4 [7]

19. (a) attempt to substitute 1 – 2 sin2 θ for cos 2θ (M1)


e.g. 4 – (1 – 2 sin2 θ) + 5 sin θ

4 – cos 2θ + 5 sin θ = 2 sin2 θ + 5 sin θ + 3 AG N0

(b) evidence of appropriate approach to solve (M1)

e.g. factorizing, quadratic formula

correct working A1


e.g. (2 sin θ + 3)(sin θ + 1), (2x + 3)(x + 1) = 0, sin x = 4

15

correct solution sin θ = –1

2

3–sin includingfor penalisenot do (A1)

θ = 2

π3 A2 N3

[7]

20. (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

(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 = 2

1 (A1)

= 240, 120, 240, 120 (correct four values only) A2 N3

METHOD 2

Sketch of y = 4 cos2 + 4 cos + 1 M1

y

x–360 –180 180 360

9

Indicating 4 zeros (A1)

= 240, 120, 240, 120 (correct four values only) A2 N3

(d) Using sketch (M1)

c = 9 A1 N2 [11]

21. (a) evidence of choosing cosine rule (M1)

e.g. a2 + b

2 – 2ab cos C


e.g. 72 + 9

2 – 2(7)(9) cos 120º

AC =13.9 (= 193 ) A1 N2 3


(b) METHOD 1

evidence of choosing sine rule (M1)

e.g. AC

ˆsin

BC

ˆsin BA


e.g. 9.13

120sin

9

ˆsin

A

1.34A A1 N2 3

METHOD 2


e.g. ACAB2

BCACABˆcos222

A


e.g. 9.1372

99.137ˆcos222

A

1.34A A1 N2 3 [6]

22. (a) choosing sine rule (M1)


e.g. 3.0sin

4

sin0.8

AD

AD = 9.71 (cm) A1 N2

(b) METHOD 1

finding angle OAD = π – 1.1 = (2.04) (seen anywhere) (A1)

choosing cosine rule (M1)


e.g. OD2 = 9.71

2 + 4

2 – 2 × 9.71 × 4 × cos(π – 1.1)

OD = 12.1 (cm) A1 N3


METHOD 2

finding angle OAD = π – 1.1 = (2.04) (seen anywhere) (A1)

choosing sine rule (M1)


e.g. 3.0sin

4

8.0sin

71.9

1.1)–sin(π

OD

OD = 12.1 (cm) A1 N3

(c) correct substitution into area of a sector formula (A1)

e.g. area = 0.5 × 42 × 0.8

area = 6.4 (cm2) A1 N2

(d) substitution into area of triangle formula OAD (M1)


e.g. A = 2

1 × 4 × 12.1 × sin 0.8, A =

2

1 × 4 × 9.71 × sin 2.04,

A = 2

1 × 12.1 × 9.71 × sin 0.3

subtracting area of sector OABC from area of triangle OAD (M1)

e.g. area ABCD = 17.3067 – 6.4

area ABCD = 10.9 (cm2) A1 N2

[13]

23. (a) tan θ =

x

4

3accept not do

4

3 A1 N1

(b) (i) sin θ = 5

3, cos θ =

5

4 (A1)(A1)


e.g. sin 2θ = 2

5

4

5

3

sin 2θ = 25

24 A1 N3


(ii) correct substitution A1

e.g. cos 2θ = 1 – 2

222

5

3

5

4,

5

3

cos 2θ = 25

7 A1 N1

[7]

24. (a)

A1A1A1 N3

Note: Award A1 for approximately sinusoidal shape,

A1 for end points approximately correct, (–2π, 4),

(2π, 4) A1 for approximately correct position of graph,

(y-intercept (0, 4) maximum to right of y-axis).

(b) (i) 5 A1 N1

(ii) 2π (6.28) A1 N1

(iii) –0.927 A1 N1

(c) f(x) = 5 sin (x + 0.927) (accept p = 5, q = 1, r = 0.927) A1A1A1 N3


(d) evidence of correct approach (M1)

e.g. max/min, sketch of f′(x) indicating roots

one 3 s.f. value which rounds to one of –5.6, –2.5, 0.64, 3.8 A1 N2

(e) k = –5, k = 5 A1A1 N2

(f) METHOD 1

graphical approach (but must involve derivative functions) M1

e.g.

each curve A1A1

x = 0.511 A2 N2

METHOD 2

g′(x) = 1

1

x A1

f′(x) = 3 cos x – 4 sin x (5 cos(x + 0.927)) A1

evidence of attempt to solve g′(x) = f′(x) M1

x = 0.511 A2 N2 [18]


25. (a) correct substitution in l = rθ (A1)

e.g. 10 × 6

1,

3

× 2π × 10

arc length =

3

π10

6

π20 A1 N2

(b) area of large sector =

6

π100

3

π10

2

1 2 (A1)

area of small sector =

6

π64

3

π8

2

1 2 (A1)

evidence of valid approach (seen anywhere) M1

e.g. subtracting areas of two sectors, )810(3

π

2

1 22

area shaded = 6π

.,

6

36πaccept etc A1 N3

[6]

26. METHOD 1

using double-angle identity (seen anywhere) A1

e.g. sin 2x = 2sin x cos x, 2cos x = 2sin x cos x

evidence of valid attempt to solve equation (M1)

e.g. 0 = 2sin x cos x – 2cos x, 2cos x (1– sin x) = 0

cos x = 0, sin x =1 A1A1

2

5,

2

3,

2

xxx A1A1A1 N4

[7]


METHOD 2

A1A1M1A1

Notes: Award A1 for sketch of sin 2x, A1 for a sketch of 2 cos x,

M1 for at least one intersection point seen, and A1 for 3

approximately correct intersection points. Accept sketches

drawn outside [0, 3π], even those with more than 3

intersections.

2

5,

2

3,

2

xxx A1A1A1 N4

[7]

27. (a) using the cosine rule a2 = b

2 + c

2 – 2bc Acos (M1)

substituting correctly BC2 = 65

2 + 104

2 – 2(65)(104)cos60° A1

= 4225 + 10 816 – 6760 = 8281

BC = 91m A1 N2

(b) finding the area, using Abc ˆsin2

1 (M1)

substituting correctly, area = 2

1(65)(104)sin60° A1

= 31690 (accept p = 1690) A1 N2

(c) (i) A1 =

2

1(65)(x)sin30° A1

= 4

65x AG N0


(ii) A2 =

2

1(104)(x)sin30° M1

= 26x A1 N1

(iii) stating A1 + A2 = A or substituting 4

65x + 26x = 31690 (M1)

simplifying 316904

169

x A1

x = 169

316904 A1

x = 340 (accept q = 40) A1 N2

(d) (i) Recognizing that supplementary angles have equal sines

e.g. CDA = 180° – BDsinACDsinABDA R1

(ii) using sin rule in ∆ADB and ∆ACD (M1)

substituting correctly BDsinA

sin30

65

BD

BDsinA

65

sin30

BD

A1

and CDsinA

sin30

104

DC

CDsinA

104

sin30

DC

M1

since CDsinABDsinA

104

65

DC

BD

104

DC

65

BD A1

8

5

DC

BD AG N0

[18]

28. (a) (i) attempt to substitute (M1)

e.g. a = 2

1529

a = 7 (accept a = –7) A1 N2

(ii) period = 12 (A1)

b = 12

π2 A1

b = 6

π AG N0


(iii) attempt to substitute (M1)

e.g. d = 2

1529

d = 22 A1 N2

(iv) c = 3 (accept c = 9 from a = –7) A1 N1

Note: Other correct values for c can be found,

c = 3 ± 12k, k .

(b) stretch takes 3 to 1.5 (A1)

translation maps (1.5, 29) to (4.5, 19) (so M′ is (4.5, 19)) A1 N2

(c) g(t) = 7 cos3

π(t – 4.5) + 12 A1A2A1 N4

Note: Award A1 for 3

π, A2 for 4.5, A1 for 12.

Other correct values for c can be found

c = 4.5 ± 6k, k .

(d) translation

10

3 (A1)

horizontal stretch of a scale factor of 2 (A1)

completely correct description, in correct order A1 N3

e.g. translation

10

3 then horizontal stretch of a scale factor of 2

[16]

29. (a) (i) range of f is [1, 1], (1 f (x) 1) A2 N2

(ii) sin3 x = 1 sin x = 1 A1

justification for one solution on [0, 2] R1

2..

xge , unit circle, sketch of sin x

1 solution (seen anywhere) A1 N1

(b) f (x) = 3 sin2

x cos x A2 N2


(c) using b

axyV d2 (M1)

xxxV dcossin3

2

2

0

2

1

(A1)

2

0

2 dcossin3 xxx A1

0sin2

sinsin 3320

3 xV A2

evidence of using sin 12

and sin 0 = 0 (A1)

e.g. (1 0)

V = A1 N1 [14]

30. (a) correct substitution A1

e.g. 25 + 16 – 40cos x, 52 + 4

2 – 2 × 4 × 5 cosx

AC = xcos4041 AG

(b) correct substitution A1

e.g. AC2

1,

30sin

4

sin

AC

x = 4 sin x

AC = 8 sin x

30sin

sin4accept

x A1 N1

(c) (i) evidence of appropriate approach using AC M1

e.g. 8 sin x = xcos4041 , sketch showing intersection

correct solution 8.682..., 111.317... (A1)

obtuse value 111.317... (A1)

x = 111.32 to 2 dp (do not accept the radian answer 1.94) A1 N2

(ii) substituting value of x into either expression for AC (M1)

e.g. AC = 8 sin 111.32

AC = 7.45 A1 N2


(d) (i) evidence of choosing cosine rule (M1)

e.g. cos B = ac

bca

2

222


e.g. 442

45.744 222

, 7.45

2 = 32 – 32 cos y, cos y = –0.734...

y = 137 A1 N2

(ii) correct substitution into area formula (A1)

e.g. 2

1 × 4 × 4 × sin 137, 8 sin 137

area = 5.42 A1 N2 [14]

31. (a)

A1A1A1 N3

Note: Award A1 for f being of sinusoidal shape, with

2 maxima and one minimum,

A1 for g being a parabola opening down,

A1 for two intersection points in approximately

correct position.

(b) (i) (2,0) (accept x = 2) A1 N1

(ii) period = 8 A2 N2

(iii) amplitude = 5 A1 N1


(c) (i) (2, 0), (8, 0) (accept x = 2, x = 8) A1A1 N1N1

(ii) x = 5 (must be an equation) A1 N1

(d) METHOD 1

intersect when x = 2 and x = 6.79 (may be seen as limits of integration) A1A1

evidence of approach (M1)

e.g.

79.6

2

2

4

πcos5855.0(,d)(d)(, xxxxxgxxffg

area = 27.6 A2 N3

METHOD 2

intersect when x = 2 and x = 6.79 (seen anywhere) A1A1

evidence of approach using a sketch of g and f, or g – f. (M1)

e.g. area A + B – C, 12.7324 + 16.0938 – 1.18129...

area = 27.6 A2 N3 [15]


32. (a) for using cosine rule 2 2 2 2 cosa b c ab C (M1)

2 2 2BC 15 17 2 15 17 cos29 (A1)

BC 8.24 m

(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)

29°

85°

A

B

C

17

ACB 180 (29 85) 66

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

AC 17

sin85 sin66 (A1)

17sin85AC

sin 66

AC (18.5380 ) 18.5 m (A1) (N2)

(ii) 1

Area 17 18.538... sin292

(A1)

276.4 m (Accept276.2 m ) (A1)(N1) 5


(c) BCA from previous triangle 66

Therefore alternative ˆACB 180 66 114 (or 29 85) (A1)

ˆABC 180 (29 114) 37

29° 114°

37°

A

B

C

17

AC 17

sin37 sin114 (M1)(A1)

AC (11.19906 ) 11.2 m

(A1) (N1) 4

(d)

29°

17

A

B

C

Minimum length for BC when BCA = 90°or diagram

showing right triangle (M1)

CBsin 29

17

CB 17sin29

CB (8.2417 ) 8.24 m

(A1) (N1) 2 [14]


33. (a) (i) 1

( ) 2cos2 sin2

f x x x

cos2 sinx x (A1)(A1) (N2)

Note: Award (A1)(A1) for 22sin sin 1x x only if work

shown, using product rule on sin cos cosx x x .

(ii) 22sin sin 1x x (2sin 1)(sin 1)x x or

2(sin 0.5)(sin 1)x x (A1) (N1)

(iii) 2sin 1 or sin 1x x

1sin

2x

π 5π 3π(0.524) (2.62) (4.71)

6 6 2x x x (A1)(A1)(A1)(N1)

(N1)(N1) 6

(b) π

0.5246

x (A1) (N1) 1

(c) (i) EITHER

curve crosses axis when π

2x (may be implied) (A1)

π 5π

2 6π π

6 2

Area ( )d ( )df x x f x x (M1)(A1) (N3)

OR

Area =5

6

6

( ) df x x

(M1)(A2) (N3)

(ii) Area 0.875 0.875 (M1)

1.75 (A1) (N2) 5 [12]


34. (a) appropriate approach (M1)

e.g. 6 = 8θ

COA = 0.75 A1 N2

(b) evidence of substitution into formula for area of triangle (M1)

e.g. area = 2

1 × 8 × 8 × sin(0.75)

area = 21.8… (A1)

evidence of substitution into formula for area of sector (M1)

e.g. area = 2

1 × 64 × 0.75

area of sector = 24 (A1)

evidence of substituting areas (M1)

e.g. Cabr sin2

1

2

1 2 , area of sector – area of triangle

area of shaded region = 2.19 cm2 A1 N4

(c) attempt to set up an equation for area of sector (M1)

e.g. 45 = 2

1 × 8

2 × θ

EOC = 1.40625 (1.41 to 3 sf) A1 N2

(d) METHOD 1

attempting to find angle EOF (M1)

e.g. π – 0.75 – 1.41

FOE = 0.985 (seen anywhere) A1



e.g. EF = 985.0cos88288 22

EF = 7.57 cm A1 N3

METHOD 2

attempting to find angles that are needed (M1)

e.g. angle EOF and angle OEF

FOE = 0.9853... and E)FO(or FEO = 1.078... A1

evidence of choosing sine rule (M1)

correct substitution (A1)

e.g. 08.1sin

8

sin0.985

EF

EF = 7.57 cm A1 N3

METHOD 3

attempting to find angle EOF (M1)

e.g. π – 0.75 – 1.41


FOE = 0.985 (seen anywhere) A1

evidence of using half of triangle EOF (M1)

e.g. x = 8 sin 2

985.0

correct calculation A1

e.g. x = 3.78

EF = 7.57 cm A1 N3 [15]

35. (a) (i) sin x = 0 A1

x = 0, x = π A1A1 N2

(ii) sin x = –1 A1

x = 2

π3 A1 N1

(b) 2

π3 A1 N1

(c) evidence of using anti-differentiation (M1)

e.g. xx d)sin66(2

π3

0

correct integral 6x – 6 cos x (seen anywhere) A1A1

correct substitution (A1)

e.g. ),0cos6(2

3π6cos

2

3π6

9π – 0 + 6

k = 9π + 6 A1A1 N3

(d) translation of

02

π

A1A1 N2

(e) recognizing that the area under g is the same as the shaded region in f (M1)

p = 2

π, p = 0 A1A1 N3

[17]

36. (a) evidence of finding height, h (A1)

e.g. sin θ = 2

h, 2 sin θ

evidence of finding base of triangle, b (A1)


e.g. cos θ = 2

b, 2 cos θ

attempt to substitute valid values into a formula for the area

of the window (M1)

e.g. two triangles plus rectangle, trapezium area formula

correct expression (must be in terms of θ) A1

e.g. cos422sin22

1,sin22sin2cos2

2

12

attempt to replace 2sinθ cosθ by sin 2θ M1

e.g. 4 sin θ + 2(2 sin θ cos θ)

y = 4 sin θ + 2 sin 2θ AG N0 5

(b) correct equation A1

e.g. y = 5, 4 sin θ + 2 sin 2θ = 5

evidence of attempt to solve (M1)

e.g. a sketch, 4 sin θ + 2 sin θ – 5 = 0

θ = 0.856 (49.0º), θ = 1.25 (71.4º) A1A1 N3 4

(c) recognition that lower area value occurs at θ = 2

(M1)

finding value of area at θ = 2

(M1)

e.g. 4 sin

22sin2

2, draw square

A = 4 (A1)

recognition that maximum value of y is needed (M1)

A = 5.19615… (A1)

4 < A < 5.20 (accept 4 < A < 5.19) A2 N5 7 [16]


37. (a) (i) OP = PQ (= 3cm) R1

So OPQ is isosceles AG N0

(ii) Using cos rule correctly eg cos QPO = 332

433 222

(M1)

cos QPO =

18

2

18

1699 A1

cos QPO = 9

1 AG N0

(iii) Evidence of using sin2

A + cos2

A = 1 M1

sin QPO =

81

80

81

11 A1

sin QPO = 9

80 AG N0

(iv) Evidence of using area triangle OPQ = PsinPQOP2

1 M1

eg 9938.02

9,

9

8033

2

1

Area triangle OPQ = 2

80 47.420 A1 N1

(b) (i) QPO = 1.4594...

QPO = 1.46 A1 N1

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

eg Area sector OPQ = 4594.132

1 2

= 6.57 A1 N2

(c) POQ = 841.02

4594.1

(A1)

Area sector QOS = 841.042

1 2 A1

= 6.73 A1 N2


(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]

1. (a) choosing sine rule (m1) · substitution into any correct formula a1 e.g. area pqr = 7 10 sin...

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