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
IB Questionbank Maths SL 2
(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
IB Questionbank Maths SL 3
(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]
IB Questionbank Maths SL 4
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
IB Questionbank Maths SL 5
(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)
correct substitution A1
e.g. (AB)2 = 5
2 + 13.6
2 – 2(5)(13.6)cos143.968...
AB = 17.9 A1 N2 [7]
IB Questionbank Maths SL 6
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
correct substitution A1
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]
IB Questionbank Maths SL 7
12. (a) METHOD 1
evidence of choosing the cosine formula (M1)
correct substitution A1
e.g. 772
1377BCAcos
222
BCA = 2.38 radians (= 136°) A1 N2
METHOD 2
evidence of appropriate approach involving right-angled triangles (M1)
correct substitution A1
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)
correct substitution A1
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
IB Questionbank Maths SL 8
METHOD 2
)4.136180(
2
1)381.2π(
2
1CBA (A1)
evidence of choosing the sine rule in triangle ABD (M1)
correct substitution A1
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
IB Questionbank Maths SL 9
METHOD 2
evidence of approach involving right-angled triangles (M1)
substituting correctly A1
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)
substituting correctly A1
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
IB Questionbank Maths SL 10
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
IB Questionbank Maths SL 11
[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)
evidence of choosing cosine rule (M1)
correct substitution 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)
correct substitution A1
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
IB Questionbank Maths SL 12
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
IB Questionbank Maths SL 13
(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
correct substitution A1
e.g. 72 + 9
2 – 2(7)(9) cos 120º
AC =13.9 (= 193 ) A1 N2 3
IB Questionbank Maths SL 14
(b) METHOD 1
evidence of choosing sine rule (M1)
e.g. AC
ˆsin
BC
ˆsin BA
correct substitution A1
e.g. 9.13
120sin
9
ˆsin
A
1.34A A1 N2 3
METHOD 2
evidence of choosing cosine rule (M1)
e.g. ACAB2
BCACABˆcos222
A
correct substitution A1
e.g. 9.1372
99.137ˆcos222
A
1.34A A1 N2 3 [6]
22. (a) choosing sine rule (M1)
correct substitution A1
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)
correct substitution A1
e.g. OD2 = 9.71
2 + 4
2 – 2 × 9.71 × 4 × cos(π – 1.1)
OD = 12.1 (cm) A1 N3
IB Questionbank Maths SL 15
METHOD 2
finding angle OAD = π – 1.1 = (2.04) (seen anywhere) (A1)
choosing sine rule (M1)
correct substitution A1
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)
correct substitution A1
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)
correct substitution A1
e.g. sin 2θ = 2
5
4
5
3
sin 2θ = 25
24 A1 N3
IB Questionbank Maths SL 16
(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
IB Questionbank Maths SL 17
(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]
IB Questionbank Maths SL 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]
IB Questionbank Maths SL 19
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
IB Questionbank Maths SL 20
(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
IB Questionbank Maths SL 21
(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
IB Questionbank Maths SL 22
(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
IB Questionbank Maths SL 23
(d) (i) evidence of choosing cosine rule (M1)
e.g. cos B = ac
bca
2
222
correct substitution A1
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
IB Questionbank Maths SL 24
(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]
IB Questionbank Maths SL 25
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
IB Questionbank Maths SL 26
(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]
IB Questionbank Maths SL 27
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]
IB Questionbank Maths SL 28
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
evidence of choosing cosine rule (M1)
correct substitution 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
IB Questionbank Maths SL 29
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)
IB Questionbank Maths SL 30
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]
IB Questionbank Maths SL 31
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
IB Questionbank Maths SL 32
(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]