IB Math – SL 1: Trig Practice Problems: MarkScheme Alei ... · PDF file9 5 1, 1 2 9 4, 2...
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Circular Functions and Trig - Practice Problems (to 07) MarkScheme 1. (a) Evidence of using the cosine rule (M1)
eg cos RQP = prrpqpr
qrp2,
2222
222
−+=−+
cos RQP
Correct substitution A1
eg Qcos642645,642
564 222222
××−+=××−+
cos RQP = 5625.048
27 = (A1)
RQP = 55.8° (0.973 radians) A1N2
(b) Area = pr2
1sin RQP
For substituting correctly 642
1 ×× sin 55.8 A1
= 9.92 (cm2) A1N1 [6]
2. Note: Throughout this question, do not accept methods which involve finding θ .
(a) Evidence of correct approach A1
eg sin θ = 523BC,AB
BC 22 =−=
sin θ = 3
5 AG N0
(b) Evidence of using sin 2θ = 2 sin θ cos θ (M1)
=
3
2
3
52 A1
= 9
54 AGN0
(c) Evidence of using an appropriate formula for cos 2θ M1
eg
−×−−×−81
801,
9
521,1
9
42,
9
5
9
4
cos 2θ = 9
1− A2 N2
[6] 3. (a) For using perimeter = r + r + arc length (M1)
20 = 2r + rθ A1
r
r220−=θ 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, r2 – 10r + 25 = 0 (A1) r = 5 cm A1 N2
[6] 4. Notes: Candidates may have
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differing answers due to using approximate answers from previous parts or using answers from the GDC. Some leeway is provided to accommodate this.
(a) METHOD 1 Evidence of using the cosine rule (M1)
eg cos C = Abccbaab
cbacos2,
2222
222
−+=−+
Correct substitution
eg cos POA = POAcos232234,232
423 222222
××−+=××−+
A1
cos POA = −0.25
POA = 1.82
π=45
26 (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 POA A1
POA = 1.32 or 1.82
POA = 1.82
π=45
26 (radians) A1 N2
(b) BOA = 2(π − 1.82) (= 2π − 3.64) (A1)
= 2.64
π=45
38 (radians) A1N2
(c) (i) Appropriate method of finding area (M1)
eg area = 2
2
1θr
Area of sector PAEB = 63.142
1 2 ×× A1
= 13.0 (cm2) (accept the exact value 13.04) A1N2
(ii) Area of sector OADB = 64.232
1 2 ×× 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
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eg °18sin
DE =
°110sin
35.8
DE = 2.75 (cm) A1 N2 (c) Setting up equation (M1)
eg 2
1 ab sin C = 5.68,
2
1 bh = 5.68
Correct substitution A1
eg 5.68 = 2
1(3.2) (7.1) sin CBD ,
2
1 × 3.2 × h = 5.68, (h = 3.55)
sin CBD = 0.5 (A1)
CBD 30° and/or 150° A1 N2
(d) Finding AB C (60° + DB 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 CBA ) 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 = 2
1 × 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)
360°180°–180°–360° 0
10
5
–5
–10
y
x
Correct asymptotes A1A1 N2
(b) (i) Period = 360° (accept 2π) A1 N1 (ii) f (90°) = 2 A1 N1
(c) 270°, −90° A1A1 N1N1
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Notes: Penalize 1 mark for any additional values. Penalize 1 mark for correct answers given
in radians .57.1,71.4or,2
,2
3
−π−π
[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
(b) Evidence of using cos 2θ = 2 cos2 θ − 1 M1 eg 2(2 cos2 θ − 1) + 4 cos θ + 3 f (θ) = 4 cos2 θ + 4 cos θ + 1 AG N0
(c) (i) 1 A1 N1 (ii) METHOD 1
Attempting to solve for cos θ M1
cos θ = 2
1− (A1)
θ = 240, 120, − 240, −120 (correct four values only) A2N3 METHOD 2 Sketch of y = 4 cos2 θ + 4 cos θ + 1 M1
y
x–360 –180 180 360
9
Indicating 4 zeros (A1) θ = 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 π.
(a) Evidence of using l = rθ (M1) arc AB = 7.85 (m) A1 N2
(b) Evidence of using θrA 2
2
1= (M1)
Area of sector AOB = 58.9 (m2) A1 N2 (c) METHOD 1
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π 62
3π
angle = ( )°π30
6 (A1)
attempt to find 15 sin 6
π M1
height = 15 + 15 sin 6
π
= 22.5 (m) A1 N2 METHOD 2
π3
angle = ( )°π60
3 (A1)
attempt to find 15 cos 3
π M1
height = 15 + 15 cos 3
π
= 22.5 (m) A1 N2
(d) (i)
π+π−=
π42
cos15154
h (M1)
= 25.6 (m) A1N2
(ii) h(0) = 15 − 15 cos
π+4
0 (M1)
= 4.39(m) A1N2 (iii) METHOD 1
Highest point when h = 30 R1
30 = 15 − 15 cos
π+4
2t M1
cos
π+4
2t = −1 (A1)
t = 1.18
π8
3accept A1N2
METHOD 2
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h30
t2π Sketch of graph of h M2 Correct maximum indicated (A1) t = 1.18 A1N2 METHOD 3 Evidence of setting h′(t) = 0 M1
sin 04
2 =
π+t (A1)
Justification of maximum R1 eg reasoning from diagram, first derivative test, second
derivative test
t = 1.18
π
8
3accept A1N2
(e) h′(t) = 30 sin
π+4
2t (may be seen in part (d)) A1A1 N2
(f) (i) h(t)30
–30
tππ2
A1A1A1N3
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
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[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 = θr 2
2
1 A1
Evidence of correctly substituting into l = rθ A1 For attempting to eliminate one variable … (M1) leading to a correct equation in one variable A1
r = 4 θ = 6
π (= 0.524, 30°) A1A1 N3
METHOD 2 Setting up and equating ratios (M1)
rr π
π=
π
π
23
2
3
4
2 A1A1
Solving gives r = 4 A1
rθ =
π=θπ3
4
2
1or
3
2 2r A1
θ = ( )°=π30,524.0
6 A1
r = 4 θ = ( )°=π30,524.0
6 N3
[6]
11. a = 4, b = 2, c =
ππetc
2
3or
2 A2A2A2 N6
[6]
12. (a) →
PQ =
− 3
5 A1A1 N2
(b) Using r = a + tb
−+
=
3
5
6
1t
y
x A2A1A1 N4
[6] 13. METHOD 1
Evidence of correctly substituting into l = rθ A1
Evidence of correctly substituting into A = θ2
2
1r A1
For attempting to solve these equations (M1) eliminating one variable correctly A1 r = 15 θ = 1.6 (= 91.7°) A1A1 N3 METHOD 2 Setting up and equating ratios (M1)
2
180
2
24
rr π=
π A1A1
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Solving gives r = 15 A1
rθ = 24
=1802
1or 2
θr A1
θ = 1.6 (= 91.7°) A1 r = 15 θ = 1.6 (= 91.7°) N3
[6] 14. (a) For correct substitution into cosine rule A1
BD = θcos84284 22 ××−+
For factorizing 16, BD = ( )θcos4516 − A1
= θcos454 − AGN0
(b) (i) BD = 5.5653 ... (A1)
5653.5
25sin
12
DBCsin = M1A1
sin DBC = 0.911 (accept 0.910, subject to AP) A1N2
(ii) DBC = 65.7° A1 N1
Or DBC = 180 − their acute angle (M1) = 114° A1N2
(iii) CDB = 89.3° (A1)
65.7sin
12
89.3sin
BCor
25sin
5.5653
89.3sin
BC == (or cosine rule) M1A1
BC = 13.2 (accept 13.17…) A1 Perimeter = 4 + 8 + 12 + 13.2 = 37.2 A1N2
(c) Area = 2
1 × 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 )2a
bx
−= . Use the following mark
allocation as a guide.
(i) Using 0d
d =x
y (M1)
−32x + 160 = 0 A1 x = 5 A1N2
(ii) ymax = −16(52) + 160(5) − 256 ymax = 144 A1N1
METHOD 2 (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
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Simplifying 12x cos Z = 20x − 64 A1
Isolating cos Z = x
x
12
6420 − A1
cos Z = x
x
3
165 − 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
×××= ZxA sin62
1 M1
××== ZxAZxA 222 sin634
1sin3 A1
A2 = 9x2 sin2 Z AG N0 (d) Using sin2 Z = 1 − cos2 Z (A1)
Substituting x
x
3
165 − for cos Z A1
for expanding
+−
−2
22
9
25616025to
3
165
x
xx
x
x 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, π, 2π
x = 0, 3
,6
ππ A1A1A1 N4
[6] 17. (a) (i) OP = PQ (= 3cm) R1
So ∆ OPQ is isosceles AGN0
(ii) Using cos rule correctly eg cos QPO = 332
433 222
××−+
(M1)
cos QPO =
=−+18
2
18
1699 A1
cos QPO = 9
1 AGN0
(iii) Evidence of using sin2 A + cos2 A = 1 M1
sin QPO =
=−
81
80
81
11 A1
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sin QPO = 9
80 AGN0
(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 == A1N1
(b) (i) QPO = 1.4594...
QPO = 1.46 A1N1 (ii) Evidence of using formula for area of a sector (M1)
eg Area sector OPQ = …4594.132
1 2 ××
= 6.57 A1N2
(c) POQ = ( )841.02
4594.1 =−π … (A1)
Area sector QOS = 841.042
1 2 ×× A1
= 6.73 A1N2 (d) Area of small semi-circle is 4.5π (= 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 4.5π − 6.5675... − 6.7285... + 4.472..., 4.5π − (6.7285... + 2.095...), 4.5π −(6.5675... + 2.256...) Area of the shaded region = 5.31 A1 N1
[17] 18. (a) p = 30 A22
(b) METHOD 1
Period = qπ2 (M2)
= 2π (A1)
⇒ q = 4 A1 4 METHOD 2
Horizontal stretch of scale factor = q1 (M2)
scale factor = 4
1 (A1)
⇒ q = 4 A1 4 [6]
19. (a) using the cosine rule (A2) = b2 + c2 –2bc cosA (M1) substituting correctly BC2 = 652 +1042 –2 (65) (104) cos 60° A1 = 4225 + 10816 – 6760 = 8281 ⇒ BC = 91 m A13
(b) finding the area, using 2
1bc sin A (M1)
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substituting correctly, area = 21 (65) (104) sin 60° A1
= 1690 3 (Accept p = 1690) A1 3
(c) (i) A1 =
21 (65) (x) sin 30° A1
= 4
65x AG 1
(ii) A2 =
21 (104) (x) sin 30° M1
= 26x A1 2
(iii) starting A1 + A2 = A or substituting 4
65x + 26x = 1690 3 (M1)
simplifying 4
169x = 1690 3 A1
x = 169
316904× A1
⇒ x = 40 3 (Accept q = 40) A1 4 (d) (i) Recognizing that supplementary angles have equal sines
eg CDA = 180 – BDA ⇒ sin CDA = sin BDA R1 (ii) using sin rule in ∆ADB and ∆ACD (M1)
substituting correctly BDAsin
30sin65BD
BDAsin65
30sinBD °=⇒=
° A1
and CDAsin
30sin104DC
BDAsin104
30sinDC °
=⇒=°
M1
since sin BDA = sin CDA
10465
DCBD
104DC
65BD =⇒= A1
⇒ 85
DCBD = AG 5
[18]
20. (a) 21
2A r θ=
2127 (1.5)
2r= (M1)(A1)
2 36r = (A1) 6 cmr = (A1) (C4)
(b) Arc length 1.5 6rθ= = × (M1) Arc length = 9 cm (A1) (C2)
Note: Penalize a total of (1 mark) for missing units. [6]
21. (a) when 0y = (may be implied by a sketch) (A1)
8πor 2.79
9x = (A1) (C2)
(b) METHOD 1 Sketch of appropriate graph(s) (M1)
Indicating correct points (A1) 3.32 or 5.41x x= = (A1)(A1)(C2)(C2)
METHOD 2
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π 1sin
9 2x + = −
π 7π
9 6x+ = ,
π 11π
9 6x + = (A1)(A1)
7π π
6 9x = − ,
11π π
6 9x = −
19π
18x = ,
31π
18x = ( 3.32, 5.41)x x= = (A1)(A1)(C2)(C2)
[6] 22. (a) for using cosine rule ( )2 2 2 2 cosa b c ab C= + − (M1)
( )( )2 2 2BC 15 17 2 15 17 cos 29= + − � (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= − + = �ɵ 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) ( )( )1Area 17 18.538... sin 29
2= � (A1)
276.4 m= (Accept 276.2 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
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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) CB
sin 2917
=�
CB 17sin29= � CB (8.2417 ) 8.24 m= =… (A1) (N1) 2
[14]
23. (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) ( )π0.524
6x = = (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 xπ
π∫ (M1)(A2)(N3)
(ii) Area 0.875 0.875= + (M1) 1.75= (A1) (N2) 5
[12]
24. Using area of a triangle = 2
1 ab sin C (M1)
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Qsin(10)(8)2
120= (A1)(A1)(A1)
Note: Accept any letter for Q sin Q = 0.5 (A1)
RQP = 30° or 6
π or 0.524 (A1) (C6)
[6] 25. (a) b = 6 (A1) (C1)
(b)
1 2
By
x
(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 2
1r2θ =
2
1162 × 1.5 (M1)
= 192 (A1)
Area of small sector 2
1r2θ =
2
1 × 102 × 1.5 (M1)
= 75 (A1) Shaded area = large area – small area = 192 – 75 (M1)
= 117 (A1) (C6) [6]
28. (a)
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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
(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…
−2
1
4
π (A1)
x = 0.3 (1 dp) (A1) (C2)
(Minimum) x = 1.856…
−2
1
4
3π (A1)
x = 1.9 (1 dp) (A1) (C2) [6]
29. Area of a triangle = 2
1 × 3 × 4 sin A (A1)
2
1 × 3 × 4 sin A = 4.5 (A1)
sin A = 0.75 (A1) 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 = 2
π, x =
4
π (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)
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x = 2
π, x =
4
π (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
2
t ⇒ sin
2
t = 1 (M1)
⇒ t = π (3.14) (correct answer only) (A1) (N2) 3 (c) (i) 4 (A1)
(ii) 10 + 4 sin
2
t = 7 (M1)
⇒ sin
2
t = –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 80A = � (A1)
AB 5
sin 40 sin80=
� � (M1)
AB 3.26= cm (A1) (C3)
(b) Area 1 1
sin (5)(3.26)sin602 2
ac B= = � (M1)(A1)
= 7.07 (accept 7.06) 2cm (A1) (C3) Note: Penalize once in this question for absence of units.
[6] 33. METHOD 1
Area sector OAB 21(5) (0.8)
2= (M1)
10= (A1) ON ( )5cos0.8 3.483...= = (A1)
AN ( )5sin 0.8 3.586.....= = (A1)
Area of 1
AON ON AN2
∆ = ×
6.249...= 2(cm ) (A1) Shaded area 10 6.249..= −
3.75= 2(cm ) (A1) (C6) METHOD 2
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A
O N B
F
Area sector 21ABF (5) (1.6)
2= (M1)
20= (A1)
Area 21OAF (5) sin1.6
2∆ = (M1)
12.5= (A1) Twice the shaded area 20 12.5 ( 7.5)= − = (M1)
Shaded area 1
(7.5)2
=
3.75= 2(cm ) (A1) (C6)
[6] 34. (a) (i) ( ) 6sin 2f x x′ = − (A1)(A1)
(ii) EITHER ( ) 12sin cos 0f x x x′ = − =
sin 0 or cos 0x x⇒ = = (M1) OR
sin 2 0x = , for 0 2 2x≤ ≤ π (M1)
THEN π
0, ,π2
x = (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 π (M1) –1 = p – q (A1) (a) p = 1 (A1)
(C3) (b) q = 2 (A1)
(C3) [6]
36. Method 1
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y
x1.80 2.510
0 (C2)
1.80 [3 sf] (G2) (C2) 2.51 [3 sf] (G2) (C2)
Method 2 3x = ±0.5x + 2π (etc.) (M1)
⇒ 3.5x = 0, 2π, 4π or 2.5x = 0, 2π, 4π (A1) 7x = 0, 4π, (8π) or 5x = 0, 4π, (8π) (A1)
x = 0, 7π4
or x = 0, 5π4
(A1)(A1)(A1)
x = 0, 7π4
, 5π4
(C2)(C2)(C2)
[6]
37. (a) area of sector ΑΒDC = 41π(2)2 = π (A1)
area of segment BDCP = π – area of ∆ABC (M1) = π – 2 (A1) (C3)
(b) BP = 2 (A1)
area of semicircle of radius BP = 21π( 2 )2 = π (A1)
area of shaded region = π – (π – 2) = 2 (A1) (C3)
[6] 38. (a) PQOR=
= q – p
=
3
7–
1
10 (A1)(A1)
=
2–
3 (A1) 3
(b) cos PQPO
PQPO QPO
×⋅= (A1)
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( ) ( )22 3–7–PO += = 58 , ( )22 2–3PQ += = 13 (A1)(A1)
PQPO⋅ = –21 + 6 = –15 (A1)
cos 754
15–
1358
15–QPO == (AG) 4
(c) (i) Since QPO + RQP = 180° (R1)
cos RQP = –cos QPO
=754
15 (AG)
(ii) sin RQP = 2
754
15–1
(M1)
= 754529
(A1)
= 754
23 (AG)
OR
cos θ = 754
15
15754
Px (M1) therefore x2 = 754 – 225 = 529 ⇒ x = 23 (A1)
⇒ sin θ = 754
23 (AG)
Note: Award (A1)(A0) for the following solution.
cos θ = 754
15⇒ θ = 56.89°
⇒ sin θ = 0.8376
754
23 = 0.8376 ⇒ sin θ =
754
23
(iii) Area of OPQR = 2 (area of triangle PQR) (M1)
= 2 × RQPsinQRPQ21 ×× (A1)
= 2 × 754
235813
21
(A1)
= 23 sq units. (A1) OR Area of OPQR = 2 (area of triangle OPQ) (M1)
= 2 ( )103–1721 ××
(A1)(A1)
= 23 sq units. (A1) 7 Notes: Other valid methods can be used.
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Award final (A1) for the integer answer. [14]
39. (a) Sine rule 120sin9
sin35PR = (M1)(A1)
PR = 120sin
35sin9
= 5.96 km (A1) 3 (b) EITHER Sine rule to find PQ
PQ = 120sin
25sin9 (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 = 839.4
(A1)
Time for Alan = a
96.5 (A1)
Then 839.4
= a
96.5 (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
sin35QS2
QRsinS
QS = (M1)
⇒ sin 21
QRS = sin 35 (A1)
QRS = 16.7° (A1)
Therefore, RSQ = 180 – (35 + 16.7) = 128.3° (A1)
==35sin
SR
sin16.7
QS
3.128sin
9 (M1)
QS =
=3.128sin2
35sin93.128sin7.16sin9
= 3.29 (A1) 6 [16]
40. (a) (i) cos 2
1
4
π– =
, sin
2
1–
4
π– =
(A1)
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therefore cos
+
4π
–sin4π
– = 0 (AG)
(ii) cos x + sin x = 0 ⇒ 1 + tan x = 0 ⇒ tan x = –l (M1)
x = 4π3
(A1)
Note: Award (A0) for 2.36. OR
x = 4π3
(G2) 3
(b) y = ex(cos x + sin x)
x
y
dd
= ex(cos x + sin x) + ex(–sin x + cos x) (M1)(A1)(A1) 3
= 2ex cos x
(c) x
y
d
d = 0 for a turning point ⇒ 2ex cos x = 0 (M1)
⇒ cos x = 0 (A1)
⇒ x = 2π
⇒ a = 2π
(A1)
y = e2
π
(cos 2π
+ sin 2π
) = e2
π
b = e2
π
(A1) 4 Note: Award (M1)(A1)(A0)(A0) for a = 1.57, b = 4.81.
(d) At D, 2
2
dd
x
y= 0 (M1)
2ex cos x – 2exsin x = 0 (A1) 2ex (cos x – sin x) = 0 ⇒ cos x – sin x = 0 (A1)
⇒ x = 4π
(A1)
⇒ y = e4
π
(cos 4π
+ sin 4π
) (A1)
= 2 e4
π
(AG) 5
(e) Required area = ∫ 4
3
0ex (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
0,
3
4 (A1)(A1)(C2)
(ii) B is (0, –4) (A1)(A1)
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(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 = 21
× 4 × 34
(M1)
= 38
(= 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
sin x = 3
2 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 )75cos1(2 °− (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) BOP = 37.5° (A1) BP = 12 tan 37.5° (M1) = 9.21 cm (A1)
OR APB = 105° PAB = 37.5° (A1)
°=
° 5.37sin
BP
105sin
AB (M1)
BP = °
°105sin
37.5sinAB = 9.21(cm) (A1) 3
(c) (i) Area ∆OBP =
°×××× 5.37tan12122
1or 21.912
2
1 (M1)
= 55.3 (cm2) (accept 55.2 cm2) (A1)
(ii) Area ∆ABP = 2
1(9.21)2 sin105° (M1)
= 41.0 (cm2) (accept 40.9 cm2) (A1) 4
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(d) Area of sector =
××××× 22 12π360
75or
180
π7512
2
1 (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) t
s
d
d = –20π sin 2πt (A1)(A1)
Note: Award (A1) for –20π, and (A1) for sin 2πt.
(ii) v = t
s
d
d = –20π sin 2πt = 0 (M1)
sin 2 πt = 0
t = 0, 2
1... (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 = rθ or ACB = 2 × OA (M1)
= 30 cm (A1) (C2)
(b) BOA (obtuse) = 2π – 2 (A1)
Area = 2
1 θ r 2 = 2
1(2π – 2)(15)2 (M1)(A1)
= 482 cm2 (3 sf) (A1) (C4)
[6] 46.
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AB
50 8070°
d
P (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) 4π (accept 720°) (A2) (C2)
(b)
π π32
y
(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
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 = 4
π5).
[6]
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49. (a) °
=45sin2
27
Asin
6 (M1)
sin A = 27
222
6 ×× (A1)
= 7
6 (AG) 2
(b)
B
D
C
A
h
(i) CDB + CAB = 180° (A1)
(ii) sin A = 7
6
=> A = 59.0° or 121° (3 sf) (A1)(A1)
=> DCB = 180° – (121° + 45°) = 14.0° (3 sf) (A1)
(iii) °
=° 45sin
2
27
14sin
BD (M1)
=>BD = 1.69 (A1) 6
(c)
h
h
××
××=
∆∆
BA2
1
BD2
1
BAC Area
BDC Area (M1)(A1)
= BA
BD (AG) 2
OR
45sin6BA2
1
45sin6BD2
1
∆BACArea∆BCDArea
×
×= (M1)(A1)
=BABD
(AG) 2
[10]
50. Using sine rule: 748sin
5sin °=B
(M1)(A1)
⇒ sin B = 75
sin 48° = 0.5308… (M1)
⇒ B = arcsin (0.5308) = 32.06° (M1)(A1) = 32° (nearest degree) (A1) (C6)
Note: Award a maximum of [5 marks] if candidates give the answer in radians (0.560).
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[6] 51. (a) x is an acute angle => cos x is positive. (M1)
cos2 x + sin2 x = 1 => cos x = x2sin–1 (M1)
=> cos x = 2
3
1–1
(A1)
= 98
(= 3
22) (A1) (C4)
(b) cos 2x = 1 – 2 sin2 x = 1 – 2 2
31
(M1)
= 9
7 (A1)
(C2) Notes: (a) Award (M1)(M0)(A1)(A0) for
cos
31
sin 1– = 0.943.
(b) Award (M1)(A0) for cos
31
sin2 1– = 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 = 21
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 3
π,π ,
3
5π, or 1.05, 3.14, 5.24)
[6] 53. (a) The smallest angle is opposite the smallest side.
cos θ = 782
578 222
××−+
(M1)
= 14
11
112
88 = = 0.7857
Therefore, θ = 38.2° (A1) (C2)
(b) Area = 2
1 × 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)
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(b) 3 sin2 x + 4 cos x – 4 = 0 ⇒3 – 3 cos2 x + 4 cos x – 4 = 0 ⇒ 3 cos2 x – 4 cos x + 1 = 0 (A1) (3 cos x – 1)(cos x – 1) = 0
cos x = 3
1 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. ATO = 90° (A1)
AT = 22 612 −
= 36
AOT = 60° = 3
π (A1)
Area = area of triangle – area of sector
= 2
1 × 6 × 36 –
2
1 × 6 × 6 ×
3
π (M1)
= 12.3 cm2 (or 318 – 6π) (A1) (C4) OR AOT = 60° (A1)
Area of ∆ = 2
1× 6 × 12 × sin 60 (A1)
Area of sector = 2
1 × 6 × 6 ×
3
π (A1)
Shaded area = 318 – 6π = 12.3 cm2 (3 sf) (A1) (C4) [4]
56. (a) (i) AP = 8016)610()8( 222 +−=−+− xxx (M1) (AG)
(ii) OP = 100)010()0( 222 +=−+− xx (A1) 2
(b) OPAP2
OAOPAPAPOcos
222
×−+= (M1)
= 10080162
)68()100()8016(22
2222
++−
+−+++−
xxx
xxx (M1)
= 10080162
8016222
2
++−
+−
xxx
xx (M1)
)}100)(8016{(
408APOcos
22
2
++−
+−=xxx
xx (AG) 3
(c) For x = 8, APOcos = 0.780869 (M1) arccos 0.780869 = 38.7° (3 sf) (A1)
OR
10
8APOtan = (M1)
APO = arctan (0.8) = 38.7° (3 sf) (A1) 2
(d) APO = 60° ⇒ APOcos = 0.5
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0.5 = )}100)(8016{(
40822
2
++−
+−
xxx
xx (M1)
2x2 – 16x + 80 – )}100)(8016{( 22 ++− xxx = 0 (M1)
x = 5.63 (G2) 4
(e) (i) f (x) = 1 when APOcos = 1 (R1)
hence, when APO = 0. (R1) This occurs when the points O, A, P are collinear. (R1)
(ii) The line (OA) has equation y = 4
3x (M1)
When y = 10, x = 3
40 (= 133
1 ) (A1)
OR
x = 3
40 (= 13
31 ) (G2) 5
Note: Award (G1) for 13.3. [16]
57. (a) Area = 2
1
2
1 2 =θr (152)(2) (M1)
= 225 (cm2) (A1) (C2)
(b) Area ∆OAB = 2
1152 sin 2 = 102.3 (A1)
Area = 225 – 102.3 = 122.7 (cm2) = 123 (3 sf) (A1) (C2)
[4]
58. (a) 17
50sin
20
B)CA(sin °= (M1)
⇒ 17
50sin20B)CA(sin
°= = 0.901
BCA > 90° ⇒ BCA = 180° – 64.3° = 115.7°
BCA = 116 (3 sf) (A1) (C2) (b) In Triangle 1, BCA = 64.3°
⇒ CAB = 180° – (64.3° + 50°) = 65.7° (A1)
Area = 2
1(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 = 14.3 t = 6 ⇒ a – b = 10.3 ⇒ 2a = 24.6 ⇒ a = 12.3 (A1) (C1) ⇒ 2b = 4.0 ⇒ b = 2 (A1) (C1)
Period = 12 hours ⇒ k
)12(π2 = 2π (M1)
⇒ k = 12 (A1) (C2) METHOD 2
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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 =
k
π2π2
= 12 (M1)
⇒ k = 12 (A1) (C2) [4]
60.
AB
C
32km
48km
cos)32)(48(2
563248BAC
222 −+= (M1)(A1)
BAC = 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 = 2
1 (M1)
sin x = 0 ⇒ x = 0 or π (0° or 180°) (A1) Note: Award (A1) for both answers.
sin x = 2
1 ⇒ x =
6
π or
6
5π (30° or 150°) (A1)
Note: Award (A1) for both answers. [4]
62. (a)
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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
5 (b) π is a solution if and only if π + π cos π = 0. (M1)
Now π + π cos π = π + π(–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)
∫ +π
0d)cosπ( xxx (A1) 2
(e) EITHER ∫ +π
0d)cosπ( xxx = 7.86960 (6 sf) (A3) 3
Note: This answer assumes appropriate use of a calculator eg
‘fnInt’:
+==xxYwith
XYfnInt
cosπ
869604401.7)π,0,,(
1
1
OR π
0
π
0]cossinπ[d)cosπ( xxxxxxx ++=+∫
= π(π – 0) + (π sin π – 0 × sin 0) + (cos π – cos 0) (A1) = π2 + 0 + –2 = 7.86960 (6 sf) (A1) 3
[15]
63. (a) (i) Q = 2
1(14.6 – 8.2) (M1)
= 3.2 (A1)
(ii) P = 2
1(14.6 + 8.2) (M0)
= 11.4 (A1) 3
(b) 10 = 11.4 + 3.2 cos
t
6
π (M1)
so 16
7− = cos
t
6
π
therefore arccos t6
π
16
7 =
− (A1)
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which gives 2.0236... = 6
πt 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
⇒ 5
3
cos
sin =x
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 = 13
5 ⇒ cos A = ±
13
12 (A1)
But A is obtuse ⇒ cos A = –13
12 (A1)
sin 2A = 2 sin A cos A (M1)
= 2 ×
−×13
12
13
5
= –169
120 (A1)
(C4) [4]
66. (a) y = π sin x – x
(–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
(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).
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Award (A1) for a smooth curve. Allow some flexibility, especially in the middle three marks here.
(b) x = 2.31 (A1) 1
(c) ∫ +−−=− Cx
xxxx2
cosπd)sinπ(2
(A1)(A1)
Note: Do not penalize for the absence of C.
Required area = ∫ −1
0d)sinπ( xxx (M1)
= 0.944 (G1) OR area = 0.944 (G2) 4
[10] 67. (a)
30º
Acute angle 30° (M1)
Note: Award the (M1) for 30° and/or quadrant diagram/graph seen. 2nd quadrant since sine positive and cosine negative
⇒ θ = 150° (A1) (C2)
(b) tan 150° = –tan 30° or tan 150° =
2
32
1
− (M1)
tan 150° = –3
1 (A1)
(C2) [4]
68. (a) 40
PQ = tan 36°
⇒ PQ ≈ 29.1 m (3 sf) (A1) (C1) (b)
40m
A
B
Q
30
70
BQA = 80° (A1)
°=
° 70sin
40
80sin
AB (M1)
Note: Award (M1) for correctly substituting.
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⇒ 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
+2
π3x (M2)
or by observing |sin θ| ≤ 1. k > 4, k < –4 (A1)(A1) (C2)(C2)
4
3
2
1
–1
–2
–3
–4
00–2π –π 2ππ
[4]
71. (a)(i) & (c)(i)
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1
0
–1
–2
1 2
(1.1, 0.55)
R
(1.51, 0)
(2, –1.66)
y
x
(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
(b) x2 cos x = 0 x ≠ 0 ⇒ cos x = 0 (M1)
⇒ x = 2
π (A1) 2
Note: Award (A2) if answer correct.
(c) (i) see graph (A1)
(ii) ∫π2
0
2x 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
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Integral = [ ] 2/π
02 sin2cos2sin xxxxx −+ (M1)
=
−
+ )1(2)0(2
π2)1(
4
π2
– [0 + 0 – 0] (M1)
= 2
π – 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) = xd
d{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 = ±
3
1 (A1)
At A, f (x) > 0, hence cos x =
3
1 (R1)(AG)
(iii) f (x) =
2
3
1–1
3
1 (M1)
= 39
2
3
1
3
2 =× (A1) 9
(d) x = 2
π (A1) 1
(e) (i) cxxxx +=∫32 sin
3
1d))(sin(cos (M1)(A1)
(ii) Area = ∫
−
=2/π
0
33
2 )0(sin2
πsin
3
1d))(sin(cos xxx (M1)
= 3
1 (A1) 4
(f) At C f ″(x) = 0 (M1) ⇔ 9 cos3 x – 7 cos x = 0 ⇔ cos x(9 cos2 x – 7) = 0 (M1)
⇒ x = 2
π (reject) or x = arccos
3
7 = 0.491 (3 sf) (A1)(A1) 4
[20] 73. Note: Award (M1) for identifying the largest angle.
cos α = 542
754 222
××−+
(M1)
= –5
1 (A1)
⇒ α = 101.5° (A1)
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OR Find other angles first β = 44.4° γ = 34.0° (M1) ⇒ α = 101.6° (A1)(A1)
(C4) Note: Award (C3) if not given to the correct accuracy.
[4] 74. AB = rθ
= r
r2
2
1 2 ×θ (M1)(A1)
= 21.6 × 4.5
2 (A1)
= 8 cm (A1)
OR 2
1 × (5.4)2θ = 21.6
⇒ θ = 7.2
4 (= 1.481 radians) (M1)
AB = rθ (A1)
= 5.4 × 7.2
4 (M1)
= 8 cm (A1) (C4)
[4]
75. (a) OA = 6 ⇒ A is on the circle (A1)
OB = 6 ⇒ B is on the circle. (A1)
=
11
5OC
= 1125+ = 6 ⇒ C is on the circle. (A1) 3
(b) OAOCAC −=
=
−
0
6
11
5 (M1)
=
−11
1 (A1) 2
(c) ACAO
ACAOCAO
⋅=ˆcos (M1)
= 1116
11
1.
0
6
+
−
−
= 126
6 (A1)
= 6
3
32
1 = (A1)
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OR 1262
6)12(6ˆcos222
××−+=CAO (M1)(A1)
= 12
1 as before (A1)
OR using the triangle formed by AC and its horizontal and vertical components:
12=AC (A1)
12
1ˆcos =CAO (M1)(A1) 3
Note: The answer is 0.289 to 3 sf (d) A number of possible methods here
OBOCBC −=
=
−−
0
6
11
5 (A1)
=
11
11 (A1)
BC = 132
∆ABC = 121322
1 ×× (A1)
= 116 (A1) OR ∆ABC has base AB = 12 (A1)
and height = 11 (A1)
⇒ area = 11122
1 ×× (A1)
= 116 (A1)
OR Given 6
3ˆcos =CAB
6
331212
2
1
6
33ˆsin ×××=∆⇒= ABCCAB (A1)(A1)(A1)
= 116 (A1) 4 [12]
76. tan2 x = 3
1 (M1)
⇒ tan x = ±3
1 (M1)
⇒ x = 30° or x = 150° (A1)(A1) (C2)(C2)
[4] 77. h = r so 2r2 = 100 ⇒ r2 = 50 (M1)
l = 10θ = 2πr (M1)
⇒ θ = 10
50π2 (A1)
= 10
25π2
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θ = π 2 = 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 = k
π2; (M1)
so period =
2
π
π2 (A1)
= 4 (AG) 2
(c) EITHER A sin
2
π + B = 3 and A sin
2
π3+ B = –1 (M1) (M1)
⇔ A + B = 3, – A + B = –1 (A1)(A1) ⇔ A = 2, B = 1 (AG)(A1) OR Amplitude = A (M1)
A = 2
4
2
)1(3 =−− (M1)
A = 2 (AG) Midpoint value = B (M1)
B = 2
2
2
)1(3 =−+ (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
x
2
π + 1
f ′(x) =
x
2
πcos2
2
π+ 0 (M1)(A2)
Note: Award (M1) for the chain rule, (A1) for
2
π, (A1) for
2 cos
x
2
π.
= π cos
x
2
π (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 – πx is a tangent ⇒ –π = π cos
x
2
π (M1)
⇒ –1 = cos
x
2
π (A1)
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⇒ 2
πx = π or 3π 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 = –π(x – 2) + 1 (M1)
y = (2π + 1) – πx k = 2π + 1 (A1) 6
(f) f (x) = 2 ⇒ 2 sin
x
2
π + 1 = 2 (A1)
⇒ sin2
1
2
π =
x (A1)
⇒ 6
13πor
6
5πor
6
π
2
π =x
x = 3
13or
3
5or
3
1 (A1)(A1)(A1) 5
[24]