4 + 5 7 cos α 4 5 5 α 0.5. Note: Award for identifying the largest angle. Find other angles first β 44.4 γ 4.0 α 0. (C4) Note: Award (C) if not given to the correct accuracy.. (a) p (C) 4. (a) OA A is on the circle (b) OB B is on the circle. & 5 # OC! % " 5 + C is on the circle. AC OC OA & 5 # &! #! % " % 0" &! # % " (b) Area 0 cos x dx [sin x ] 0 square units (C). AB rθ r θ r. 5.4 8 cm (5.4) θ. (c) AO AC cos OAC ˆ AO AC & # & #. 0!! % " % " + ˆ + ( ) coso AC as before 4 θ.7 (.48 radians) AB rθ 4 5.4.7 8 cm (C4) using the triangle formed by AC and its horizontal and vertical components: AC cos O AC ˆ Note: The answer is 0.89 to sf
(d) A number of possible methods here BC OC OB & 5 # &! #! % " % 0 " &! # % " %BC % %ΔABC % ΔABC has base AB and height area. (a) I (b) (c) III IV Note: Award (C4) for correct, (C) for correct, (C) for correct. 7. h r so r 00 r 50 l 0θ r 50 θ 0 5 0 θ 4.44 (sf) (C4) Note: Accept either answer. 5. tan x Given cos B AC ˆ sin B AC ˆ ΔABC 4 tan x ± x 0 or x 50 (C)(C) [] 8. (a) f () f (5) (b) EITHER distance between successive maxima period 5 4 (AG) Period of sin kx ; k so period 4 (AG) 4
(c) & # & # EITHER A sin! + B and A sin!" + B % " % A + B, A + B A, B (AG) Amplitude A ( ) 4 A A (AG) Midpoint value B + ( ) B B 5 Note: As the values of A and B are likely to be quite obvious to a bright student, do not insist on too detailed a proof. (ii) Tangent line is: y (x ) + y ( + ) x k + (f) & # f (x) sin x! + % " & # sin x! % " 5 x or or 5 x or or 5 & # (d) f (x) sin x! + % " & # & # f )(x)! cos x! + 0 (A) % " % " & # cos x! % " & # Note: Award for the chain rule, for!, for % " & # cos x!. % " Notes: Since the result is given, make sure that reasoning is valid. In particular, the final 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 x! % " & # cos x! % " x or or... 4 9. cos x 5 sin x sin x cos x 5 tan x 0. x or x (to the nearest degree) (C)(C) Note: Deduct [ mark] if there are more than two answers. 5 0. sin A cos A ± But A is obtuse cos A sin A sin A cos A 5 & #! % " 0 9 (C4) x or... Since 0 x 5, we take x, so the point is (, ) 5
. (a) y sin x x. (a) y (.5,.7) 0º (., 0) (.5,.7) (., 0) (A5) 5 Notes: Award for appropriate scales marked on the axes. Award for the x-intercepts at (±., 0). Award for the maximum and minimum points at (±.5, ±.7). Award for the end points at (±, ±.55). Award for a smooth curve. Allow some flexibility, especially in the middle three marks here. (b) x. x. (a) Acute angle 0 Note: Award the for 0 and/or quadrant diagram/graph seen. nd quadrant since sine positive and cosine negative θ 50 (b) tan 50 tan 0 or tan 50 tan 50 PQ tan 40 PQ 9. m ( sf) (C) (C) (C) x (c) ( sin x x)dx cos x + C Note: Do not penalize for the absence of C. Required area ( sin x x)dx 0 0.944 (G) area 0.944 (G) 4 [0] 7 8
(b) Q 40m 0 70 A B A Qˆ B 80 AB 40 sin 80 sin 70 AB 4 9. m ( sf) Note: Award for correctly substituting. (C) 5. & # From sketch of graph y 4 sin x +! % " (M) or by observing sin θ. k > 4, k < 4 (C)(C) 4 0 0 4. Perimeter 5( ) + 0 Note: Award for working in radians; for ; for +0. (0 + 5) cm (.4, to sf) (C4) 4. (a) From graph, period (b) Range {y 0.4 < y < 0.4} (c) (i) d f )(x) {cos x (sin x) } dx cos x ( sin x cos x) sin x (sin x) or sin x + sin x Note: Award for using the product rule and for each part. 9 0
(ii) f )(x) 0 sin x{ cos x sin x} 0 or sin x{ cos x } 0 cos x 0 cos x ± & #! % " At A, f (x) > 0, hence cos x & #! % " (R)(AG) (iii) f (x) & # & # & # & #!!!!! % "! % % % " " " 9 9 7. C km A 48km B 48 + 5 cos CÂB (48)() C ÂB arccos(0.05) 8 (d) x 8. (a) cos x + sin x ( sin x) + sin x sin x + sin x (e) (i) (cos x )(sin x) dx sin x + c (ii) / "( "% /, Area (cosx)(sin x) dx '-sin * (sin 0) 0 "&. + " # 4 (b) cos x + sin x sin x + sin x sin x sin x 0 sin x( sin x) 0 sin x 0 or sin x sin x 0 x 0 or (0 or 80 ) Note: Award for both answers. 5 sin x x or (0 or 50 ) (f) At C f,(x) 0 9 cos x 7 cos x 0 cos x(9 cos x 7) 0 x (reject) or x arccos 7 0.49 ( sf) 4 [0] Note: Award for both answers.
{ 9. (a) y (e) EITHER ( + x cos x)dx 7.890 ( sf) (A) 0 Note: This answer assumes appropriate use of a calculator eg # fnint( Y, X,0, ) 7.890440 fnint : "! withy + x cos x 4 { 0.5< x<.5< y<4 MAXIMUM POINT 0 ( + x cos x)dx [x + x sin x + cos x] ( 0) + ( sin 0 sin 0) + (cos cos 0) + 0 + 7.890 ( sf) 0 [5] integers on axis 0. (a) (i) Q (4. 8.). 4 5 < x<.5.5< x<4 LEFT INTERCEPT {.< x<. 0.< y <0 MINIMUM POINT x RIGHT INTERCEPT (b) is a solution if and only if + cos 0. Now + cos + ( ) 0 (c) By using appropriate calculator functions x.9 7 9... x.97 (sf) (d) See graph: ( + x cos x)dx 0 5 (ii) P (4. + 8.) (M0).4 & # (b) 0.4 +. cos t! % " 7 & # so cos t! % " & 7 # therefore arccos! t % " which gives.0... t or t.848. t.8( sf) (c) (i) By symmetry, next time is.8... 8.5... t 8.4 ( sf) (ii) From above, first interval is.8 < t < 8.4 This will happen again, hours later, so 5.9 < t < 0. 4 [0] 4
. (a) The smallest angle is opposite the smallest side. 8 + 7 5 cos θ 8 7 88 0.7857 4 Therefore, θ 8. (b) Area 8 7 sin 8. (C) 7. cm (C). O Tˆ A 90 AT T ÔA 0 Area area of triangle area of sector. cm (or 8 ) (C4) T ÔA 0. (a) sin x + 4 cos x ( cos x) + 4cos x cos + 4 cos x (C) Area of Δ sin 0 Area of sector (b) sin x + 4 cos x 4 0 cos x + 4 cos x 4 0 cos x 4 cos x + 0 ( cos x )(cos x ) 0 Shaded area 8. cm ( sf) (C4) cos x or cos x x 70.5 or x 0 (C) Note: Award (C) for each correct radian answer, ie x. or x 0. 4. (a) (i) AP ( x 8) + (0 ) x x + 80 (AG) (b) (ii) OP ( x 0) + (0 0) x + 00 AP + OP OA cos OPˆA AP OP ( x x + 80) + ( x + 00) (8 + ) x x + 80 x + 00 x x cos OPˆA x + 80 x + 80 {( x x + 00 x 8x + 40 x + 80)( x + 00)} (AG) 5
(c) For x 8, cos OPˆ A 0.78089 arccos 0.78089 8.7 ( sf) (d) 8 tan OPˆA 0 O PˆA arctan (0.8) 8.7 ( sf) O Pˆ A 0 cos OPˆ A 0.5 0.5 {( x x 8x + 40 x + 80)( x + 00)} x x + 80 {( x x + 80)( x + 00)} 0 x 5. (G) 4 (e) (i) f (x) when cos OPˆ A (R) hence, when O Pˆ A 0. (R) This occurs when the points O, A, P are collinear. (R) x (ii) The line (OA) has equation y 4 0sin 50 sin (AĈB) 0.90 7 A ĈB > 90 A ĈB 80 4. 5.7 A ĈB ( sf) (C) (b) In Triangle, A ĈB 4. B ÂC 80 (4. + 50 ) 5.7 7. METHOD Area (0)(7) sin 5.7 55 (cm ) ( sf) (C) The value of cosine varies between and +. Therefore: t 0 a + b 4. t a b 0. a 4. a. (C) b 4.0 b (C) () Period hours k k (C) When y 0, x 40 ( ) METHOD y x 40 ( ) (G) 5 Note: Award (G) for.. 4. 0. 8 4 t (h) 5. (a) Area r θ (5 )() 5 (cm ) (C) (b) Area OAB 5 sin 0. Area 5 0..7 (cm ) ( sf) (C) From consideration of graph: Midpoint a. (C) Amplitude b (C) Period k k (C) 8. (a) l rθ or ACB OA 0 cm (C). (a) sin (AĈB) sin 50 0 7 (b) A ÔB (obtuse) 7 8
Area θ r ( )(5) 48 cm ( sf) (C4) (b) y 9. A d B 50 80 70 P (A).5 0 50.5 80 d 50 + 80 50 80 cos 70 d 78.5 km (C). number of solutions: 4 Statement (a) Is the statement true for all real numbers x? (Yes/No) (b) If not true, example (G) (A) (C) A No x l (log 0 0. ) (a) (A) (C) B No x 0 (cos 0 ) (b) (A) (C) C Yes N/A 0. (a) (i) (C) (ii) 4 (accept 70 ) (A) (C) Notes: (a) Award for each correct answer. (b) Award (A) marks for statements A and B only if NO in column (a). Award (A) for a correct counter example to statement A, for a correct counter example to statement B (ignore other incorrect examples). Special Case for statement C: Award if candidates write NO, and give a valid reason (eg 5 arctan ). 4. (a) 7 sin A sin 45 sin A 7 (AG) 7 (b) 9 0
(i) B D B Dˆ C + B ÂC 80 (ii) sin A 7 h A C. Using sine rule: sin B sin 48 5 7 sin B 7 5 sin 48 0.508 B arcsin (0.508).0 (nearest degree) (C) Note: Award a maximum of [5 marks] if candidates give the answer in radians (0.50). (iii) > A 59.0 or ( sf) > B ĈD 80 ( + 45 ) 4.0 ( sf) 7 BD sin 4 sin 45 >BD.9 4. (a) x is an acute angle > cos x is positive. cos x + sin x > cos x > cos x & #! % " sin x 8 ( ) (C4) 9 (c) BD h Area ΔBDC Area ΔBAC BA h BD BA (AG) BD sin 45 Area ΔBCD Area ΔBAC BA sin 45 BD BA (AG) (b) cos x sin & # x!" % 7 9 Notes: (a) Award (M0)(A0) for & & ## cos sin!! 0.94. % % " " & & ## (b) Award (A0) for cos sin!! 0.778. % % " " (C) [0] 5. (a) sin x ( cos x) cos x l + cos x > cos x + cos x l 0 (C) Note: Award the first for replacing sin x by cos x. (b) cos x + cos x ( cos x )(cos x +) (C)
(c) cos x or cos x l > x 0, 80 or 00 (C) Note: Award (A0) if the correct answers are given in 5 radians (ie,,, or.05,.4, 5.4) Method x ±0.5x + (etc.).5x 0,, 4 or.5x 0,, 4 7x 0, 4, (8) or 5x 0, 4, (8) 4 4 x 0, or x 0, 7 5 4 4 x 0,, 7 5 (C)(C)(C). p + q cos 0 p + q p + q cos p q (a) p (C) (b) q (C) 8. (a) area of sector ΑΒDC () 4 area of segment BDCP area of ΔABC (C) (b) BP 7. Method area of semicircle of radius BP ( ) y area of shaded region ( ) (C) 0.80.5 x 9. (a) PQ q p & 0# & 7#!! % " % " & #! % " 0 (C).80 [ sf] (G) (C).5 [ sf] (G) (C) (b) PO PQ cos OPˆQ PO PQ ( 7) ( ) 58, PQ + ( ) PO + PO PQ + 5 5 5 cos O PˆQ (AG) 4 58 754 (c) (i) Since O Pˆ Q + P Qˆ R 80 (R) & 5 # cos P Qˆ R cos O Pˆ Q! % 754 " (AG) 4
(ii) sin P Qˆ R cos θ 5 754 & 5 #! % 754 " 59 754 754 5 754 (AG) (iii) Area of OPQR (area of triangle PQR) PQ QR sin PQˆ R 58 754 sq units. Area of OPQR (area of triangle OPQ) ' % "( 7 0) & # sq units. 7 Notes: Other valid methods can be used. Award final for the integer answer. therefore x 754 5 59 x sin θ 754 Note: Award (A0) for the following solution. cos θ x 5 θ 5.89 754 sin θ 0.87 0.87 sin θ 754 754 P (AG) 40. (a) Sine rule (b) EITHER PR 9 sin5 sin0 9sin5 PR sin0 5.9 km Sine rule to find PQ 9sin 5 PQ sin0 4.9 km Cosine rule: PQ 5.9 + 9 ()(5.9)(9) cos 5 9.9 PQ 4.9 km Time for Tom 4.9 8 5.9 Time for Alan a 4.9 5.9 Then 8 a a 0.9 7 5
(c) RS 4QS 4QS QS + 8 8 QS cos 5 QS + 4.74QS 8 0 (or x + 4.74x 8 0) QS 8.0 or QS.9 (G) therefore QS.9 QS QS sinsrˆ Q sin5 sin S Rˆ Q sin 5 S Rˆ Q.7 Therefore, Q ŜR 80 (5 +.7) 8. 9 QS & SR #! sin8. sin.7 % sin 5 " 9sin.7 & 9sin5 # QS! sin8. % sin8. ".9 4. (a) (i) & # & # cos!, sin! % 4 " % 4 " & # & # therefore cos! + sin! 0 % 4 " % 4 " (AG) (ii) cos x + sin x 0 + tan x 0 tan x l x 4 Note: Award (A0) for.. x 4 (G) (b) y e x (cos x + sin x) d y e x (cos x + sin x) + e x ( sin x + cos x) dx e x cos x (c) d y 0 for a turning point e x cos x 0 dx cos x 0 x a y e (cos + sin ) e b e d y (d) At D, 0 dx e x cos x e x sin x 0 e x (cos x sin x) 0 cos x sin x 0 x 4 Note: Award (A0)(A0) for a.57, b 4.8. y e 4 (cos 4 + sin 4 ) e 4 4 (AG) 5 (e) Required area 4 (cos x + sin x)dx 0 7.4 sq units (G) Αrea 7.4 sq units (G) Note: Award (G0) for the answer 9.8 obtained if the calculator is in degree mode. [7] 7 8
& 4 # 4. (a) (i) A is, 0! % " (C) (ii) B is (0, 4) (C) Note: In each of parts (i) and (ii), award C if A and B are interchanged, C if intercepts given instead of coordinates. (b) Area 4 4 8 (.7) (C) (b) P ÔB 7.5 BP tan 7.5 9. cm B PˆA 05 B ÂP 7.5 AB BP sin05 sin 7. 5 ABsin 7.5 BP 9.(cm) sin 05 4. (a) ( sin x )(sin x ) (C) Note: Award A if x x + correctly factorized to give (x )(x ) (or equivalent with another letter). (c) (i) & # Area OBP 9. or tan 7. 5! % " 55. (cm ) (accept 55. cm ) (ii) Area ABP (9.) sin05 4.0 (cm ) (accept 40.9 cm ) 4 (b) (i) ( sin x )(sin x ) 0 sin x sin x (C) & 75 # (d) Area of sector 75 or! 80 % 0 " 94. (cm ) (accept 0 or 94. (cm )) (ii) x 4.8, 8 (C) Notes: Penalize [ mark] for any extra answers and [ mark] for answers in radians. ie Award A A0 for 4.8, 8 and any extra answers. Award A A0 for 0.70,.4. Award A0 A0 for 0.70,.4 and any extra answers. (e) Shaded area area OPB area sector.4 (cm ) (accept. cm,. cm ) 45. Note: Do not penalize missing units in this question. [] 44. Note: Do not penalize missing units in this question. (a) AB + cos 75 ( cos 75 ) ( cos 75 ) AB ( cos 75 ) (AG) Note: The second is for transforming the initial expression to any simplified expression from which the given result can be clearly seen. (a) (i) At release(p), t 0 s 48 + 0 cos 0 58 cm below ceiling (ii) 58 48 +0 cos t cos t t sec t sec (G) 5 9 0
(b) (i) d s 0 sin t dt Note: Award for 0, and for sin t. d s (ii) v 0 sin t 0 dt sin t 0 47. (a) y.5 0.5 t 0,... (at least values) s 48 + 0 cos 0 or s 48 +0 cos 58 cm (at P) 8 cm (0 cm above P) 7 Note: Accept these answers without working for full marks. May be deduced from recognizing that amplitude is 0. 0 0.5.5 0.5.5.5.5 x (c) 48 +0 cos t 0 + 5 cos 4t t 0. secs t 0. secs (G) (C) Note: Award for the graph crossing the y-axis between 0.5 and, and for an approximate sine curve crossing the x- axis twice. Do not penalize for x >.4. (d) times (G) Note: If either of the correct answers to parts (c) and (d) are missing and suitable graphs have been sketched, award (G) for sketch of suitable graph(s); for t 0.; for. & # (b) (Maximum) x 0.85! % 4 " x 0. ( dp) & # (Minimum) x.85! % 4 " x.9 ( dp) (C) (C) 4. Area of large sector r θ.5 9 48. Area of a triangle 4 sin A Area of small sector r θ 0.5 75 Shaded area large area small area 9 75 7 (C) 4 sin A 4.5 sin A 0.75 A 48. and A (or 0.848,.9 radians) (A) (C) Note: Award (C4) for 48. only, (C5) for only.
49. METHOD (c) (i) 4 cos x sin x cos x cos x sin x cos x 0 cos x(cos x sin x) 0 cos x 0, (cos x sin x) 0 x, x 4 (C) (ii) & t # 0 + 4 sin! 7 % " & t # sin! 0.75 % " t 7.98 (N) METHOD Graphical solutions EITHER (iii) depth < 7 from 8 hours from 00 0 hours therefore, total hours (N) 7 [] for both graphs y cos x, y sin x, (M) THEN for the graph of y cos x sin x. (M) Points representing the solutions clearly indicated.57, 0.785 x, x 4 (C) Notes: If no working shown, award (C4) for one correct answer. Award (C)(C) for each correct decimal answer.57, 0.785. Award (C)(C) for each correct degree answer 90, 45. Penalize a total of [ mark] for any additional answers. o 5. (a) Angle A 80 (b) AB 5 o o sin 40 sin80 AB. cm (C) Area ac sin B (5)(.)sin 0 o 7.07 (accept 7.0) cm Note: Penalize once in this question for absence of units. (C) 50. (a) (i) 0 + 4 sin.4 (ii) At 00, t 0 + 4 sin 0.5.48 (N) Note: Award (A0) if candidates use t 00 leading to y.. No other ft allowed. (b) (i) 4 metres & t # & t # (ii) 4 0 + 4 sin! sin! % " % " t (.4) (correct answer only) (N) 4
5. METHOD 5. (a) (i) f! ( x) sin x Area sector OAB (5) (0.8) 0 ON 5cos0.8 (.48... ) AN 5sin 0.8 (.58... ) Area of Δ AON ON AN.49... (cm ) Shaded area 0.49.. (ii) THEN EITHER f! ( x) sin xcos x 0 sin x 0 or cos x 0 sin x 0, for 0 x x 0,, (N4).75 (cm ) (C) (b) (i) translation in the y-direction of METHOD A (ii). (.0 from TRACE is subject to AP) (A) 4 [0] O N B 54. (a) p 0 A (b) METHOD Area sector ABF (5) (.) F Period q (M) q 4 A 4 0 Area Δ OAF (5) sin..5 Twice the shaded area 0.5 ( 7.5) Shaded area (7.5) METHOD Horizontal stretch of scale factor q scale factor 4 (M) q 4 A 4.75 (cm ) (C) 55. (a) using the cosine rule (A) b + c bc cos  substituting correctly BC 5 +04 (5) (04) cos 0 A 45 + 08 70 88 BC 9 m A (b) finding the area, using bc sin  5
substituting correctly, area (5) (04) sin 0 A 90 (Accept p 90) A (c) (i) A &! # (5) (x) sin 0 A % " 5x AG 4 (ii) A &! # (04) (x) sin 0 M % " x A (b) Arc length rθ.5 Arc length 9 cm Note: Penalize a total of ( mark) for missing units. 57. (a) when y 0 (may be implied by a sketch) (C) 8 x or.79 (C) 9 (iii) starting A + A A or substituting 5x + x 90 4 simplifying 9x 90 A 4 4 90 x 9 x 40 (Accept q 40) A 4 (d) (i) Recognizing that supplementary angles have equal sines eg A Dˆ C 80 A Dˆ B sin A Dˆ C sin A Dˆ B R (ii) using sin rule in ΔADB and ΔACD substituting correctly BD sin 0 5 BD sin 0 sin ADˆ B 5 sin ADˆ B and DC sin 0 04 DC sin 0 sin ADˆ B 04 sin ADˆ C since sin A Dˆ B sin A Dˆ C BD DC BD 5 A 5 04 DC 04 BD 5 AG 5 DC 8 A A M [8] (b) METHOD Sketch of appropriate graph(s) Indicating correct points x. or x 5.4 (C)(C) METHOD! " sin x + % & 9' 7 x +, x + 9 9 7 x, x 9 9 9 x, 8 x ( x., x 5.4) 8 (C)(C) 5. (a) A r θ 7 (.5) r r r cm (C4) 7 8
58. (a) for using cosine rule ( a b c abcosc) + (c) o A ĈB from previous triangle ( )( ) o BC 5 + 7 5 7 cos9 BC 8.4 m (N0) Notes: Either the first or the second line may be implied, but not both. Award no marks if 8.4 is obtained by assuming a right (angled) triangle (BC 7 sin 9). Therefore alternative ACB ˆ 80 4 o (or 9 + 85) ABC ˆ 80 (9 + 4) 7 A o C 9 4 (i) A 9 7 85 B ACB o 80 (9 + 85) for using sine rule (may be implied) C 7 AC 7 o o sin7 sin4 AC (.990 K ). m (N) 4 7 B AC 7 o o sin85 sin 7sin85 AC o sin o AC (8.580 K ) 8.5 m (N) (d) A 9 7 C o Area 7 8.58... sin 9 (ii) ( )( ) 7.4 m (Accept 7. m ) (N) 5 Minimum length for BC when A ĈB 90 or diagram showing right triangle o CB sin 9 7 o CB 7sin 9 B CB (8.47 K ) 8.4 m (N) 9 40
59. (a) (i) (ii) f! ( x) cosx sin x cosx sin x (N) Note: Award for sin x sin x+ only if work shown, using product rule on sin xcos x+ cos x. sin x+ sin x (sin x )(sin x+ ) or (sin x 0.5)(sin x+ ) (N) 0. Using area of a triangle ab sin C 0 (0)(8)sin Q Note: Accept any letter for Q sin Q 0.5 P Qˆ R 0 or or 0.54 (C) (iii) sin x or sin x sin x 5 x (0.54) x (.) x (4.7) (N) (N)(N). (a) ( sin x) + sin x sin x sin x 0 (p, q, r ) (C) (b) ( sin x )( sin x + ) (C) x (N) (b) ( 0.54) (c) 4 solutions (A) (C) (c) (i) EITHER curve crosses axis when 5 Area f( x)d x+ f( x)dx x (may be implied) (N) 5 Area f ( x ) dx (A) (N) (ii) Area 0.875 + 0.875.75 (N) 5 [] 4 4
. METHOD 5. METHOD Evidence of correctly substituting into A Evidence of correctly substituting into l rθ r θ A A Evidence of correctly substituting into l rθ A Evidence of correctly substituting into A r θ A For attempting to eliminate one variable leading to a correct equation in one variable A For attempting to solve these equations eliminating one variable correctly A r 4 θ ( 0.54, 0 ) AA N METHOD Setting up and equating ratios 4 r r AA Solving gives r 4 A rθ & 4 # or r θ! % " A θ ( 0.54,0 ) A r 5 θ. ( 9.7 ) AA N METHOD Setting up and equating ratios 4 80 r r AA Solving gives r 5 A rθ 4 & # or r θ 80! % " A θ. ( 9.7 ) A r 5 θ. ( 9.7 ) N r 4 θ ( 0.54,0 ) N. (a) For correct substitution into cosine rule A BD 4 + 8 4 8cosθ & #. a 4, b, c or etc! % " AAA N For factorizing, BD ( 5 4 cos θ) A 4 5 4 cos θ AG N0 4. (a) PQ & 5! # % " AA N (b) Using r a + tb & x # & # & #! + 5! t! % y" % " % " AAA N4 4 44
(ii) z x + x cos Z A N (b) (i) BD 5.55... (iii) Substituting for z into the expression in part (ii) (ii) (iii) sin CBˆ D sin 5 MA 5.55 sin C Bˆ D 0.9 (accept 0.90, subject to AP) A N C Bˆ D 5.7 A N Or C Bˆ D 80 their acute angle 4 A N B Dˆ C 89. BC 5.55 BC or (or cosine rule) MA sin 89. sin 5 sin 89. sin 5.7 Expanding 00 0x + x x + x cos Z Simplifying x cos Z 0x 4 0x 4 Isolating cos Z x 5x cos Z x Note: Expanding, simplifying and isolating may be done in any order, with the final A being awarded for an expression that clearly leads to the required answer. A A A AG N0 BC. (accept.7 ) A Perimeter 4 + 8 + +. 7. A N (c) Area 4 8 sin 40 A (c) Evidence of using the formula for area of a triangle & # A x sin Z! % " & # A x sin Z A x sin Z! A % 4 " A 9x sin Z AG N0 M 0. A N (d) Using sin Z cos Z 7. (a) METHOD Note: There are many valid algebraic approaches to this problem (eg completing the square, b using x ). Use the following mark a allocation as a guide. 5x Substituting for cos Z x & 5x # & 5x 0x + 5 # for expanding! to! % " x % 9x " for simplifying to an expression that clearly leads to the required answer A eg A 9x (5x 0x + 5) A A d y (i) Using 0 dx A x + 0x 5 AG x + 0 0 x 5 A N (ii) y max (5 ) + 0(5) 5 y max 44 A N A (e) (i) 44 (is maximum value of A, from part (a)) A A max A N (ii) Isosceles A N [0] METHOD (i) Sketch of the correct parabola (may be seen in part (ii)) M x 5 A N (ii) y max 44 A N (b) (i) z 0 x (accept x + z 0) A N 45 4
8. (a) Evidence of choosing the double angle formula f (x) 5 sin (x) A N (b) Evidence of substituting for f (x) eg 5 sin x 0, sin x 0 and cos x 0 x 0,, x 0,, AAA N4 (b) (i) O Pˆ Q.4594... (c) O PˆQ.4 A N (ii) Evidence of using formula for area of a sector eg Area sector OPQ.4594.4594 Q ÔP ( 0.84).57 A N 9. (a) (i) OP PQ ( cm) R So Δ OPQ is isosceles AG N0 (ii) Using cos rule correctly eg cos O Pˆ Q 9+ 9 & # cos O Pˆ Q! 8 % 8 " cos O Pˆ Q 9 + 4 (iii) Evidence of using sin A + cos A M sin sin O Pˆ Q O Pˆ Q A AG & # 80! 8 % 8 A " 80 9 (iv) Evidence of using area triangle OPQ OP PQ sin P M eg 80 9, 0.998 9 Area triangle OPQ 80 ( 0 ) ( 4.47) AG N0 N0 A N Area sector QOS 4 0. 84 A.7 A N (d) Area of small semi-circle is 4.5 ( 4.7...) A Evidence of correct approach eg Area area of semi-circle area sector OPQ area sector QOS + area triangle POQ Correct expression eg 4.5.575....785... + 4.47..., 4.5 (.785... +.095...), 4.5 (.575... +.5...) Area of the shaded region 5. A N 70. (a) Evidence of using the cosine rule p + r q eg cos P Qˆ R, q p + r pr cos P Qˆ R pr Correct substitution eg 4 + 5 4, 5 4 + 4 cosq M cos P Qˆ R 7 0. 55 48 P Qˆ R 55.8 (0.97 radians) A N A A [7] 47 48
(b) Area pr sin P Qˆ R For substituting correctly 4 sin 55.8 A 9.9 (cm ) A N ' 0 r (b) Finding A r % " ( 0r r ) & r # For setting up equation in r M Correct simplified equation, or sketch eg 0r r 5, r 0r + 5 0 r 5 cm A N 7. Note: Throughout this question, do not accept methods which involve finding θ. (a) Evidence of correct approach A eg sin θ BC, BC 5 AB 7. 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. (a) METHOD Evidence of using the cosine rule sin θ 5 AG N0 a eg cos C + b c ab, a b + c bc cos A Correct substitution (b) Evidence of using sin θ sin θ cos θ & # # 5! &! A % "% " + 4 eg cos A ÔP,4 + cos AÔP cos A ÔP 0.5 A 4 5 9 AG N0 & # A ÔP.8! (radians) A N % 45 " METHOD (c) Evidence of using an appropriate formula for cos θ M Area of AOBP 5.8 (from part (d)) 4 5 4 5 & 80 # eg,,,! 9 9 9 9 % 8 " Area of triangle AOP.905.9050 0.5 sin A ÔP A cos θ A N 9 A ÔP. or.8 & # A ÔP.8! (radians) A N % 45 " 7. (a) For using perimeter r + r + arc length 0 r + rθ A 0 r θ AG N0 r (b) A ÔB (.8) (.4) & 8 #.4! (radians) A N % 45 " 49 50
(c) (i) Appropriate method of finding area eg area θr Area of sector PAEB 4. A.0 (cm ) (accept the exact value.04) A N (ii) Area of sector OADB. 4 A.9 (cm ) A N (d) (i) Area AOBE Area PAEB Area AOBP (.0 5.8) M 7.9 (accept 7. from the exact answer for PAEB) A N (ii) Area shaded Area OADB Area AOBE (.9 7.9) M 4.7 (accept answers between 4. and 4.7) A N (c) Setting up equation eg ab sin C 5.8, bh 5.8 Correct substitution eg 5.8 (.) (7.) sin sin D Bˆ C,. h 5.8, (h.55) D Bˆ C 0.5 D Bˆ C 0 and/or 50 A N (d) Finding A Bˆ C (0 + D Bˆ C) Using appropriate formula eg (AC) (AB) + (BC), (AC) (AB) + (BC) (AB) (BC) cos ABC Correct substitution (allow FT on their seen eg (AC) 9. +. A Bˆ C ) A AC 9.74 (cm) A N A 74. (a) Evidence of choosing cosine rule eg a b + c bc cos A Correct substitution A eg (AD) 7. + 9. (7.) (9.) cos 0 (AD) 9.7 AD 8.5 (cm) A N (e) For finding area of triangle ABD Correct substitution Area 9. 7. sin 0 A 8.8... A Area of ABCD 8.8... + 5.8 4.0 (cm ) A N [] (b) 80 8 Evidence of choosing sine rule Correct substitution A DE 8.5 eg sin 8 sin 0 DE.75 (cm) A N 5 5
75. (a) y 0 (c) y V T 5 8 0 8 x 0 80 0 80 0 x 5 0 Correct asymptotes AA N Note: Award M for attempt to reflect through y-axis, A for vertex at approximately (, ). MA N (b) (i) Period 0 (accept ) A N (ii) f (90 ) A N (c) 70, 90 AA NN Notes: Penalize mark for any additional values. Penalize mark for correct answers given & # in radians,,or 4.7,.57!. % " 77. (a) METHOD Using the discriminant Δ 0 k 4 4 k 4, k 4 AA N METHOD Factorizing (x ± ) k 4, k 4 AA N 7. (a) (i) h A N (ii) k A N (b) Evidence of using cos θ cos θ M eg ( cos θ ) + 4 cos θ + f (θ) 4 cos θ + 4 cos θ + AG N0 (b) g (x) f (x ) +, 5 (x ) +, (x ), x + x A N 5 54
(c) (i) A N (ii) METHOD Attempting to solve for cos θ cos θ M θ 40, 0, 40, 0 (correct four values only) A N METHOD Sketch of y 4 cos θ + 4 cos θ + y 9 M (c) METHOD angle ( 0 ) attempt to find 5 sin height 5 + 5 sin M 0 80 80 0 x.5 (m) A N METHOD Indicating 4 zeros θ 40, 0, 40, 0 (correct four values only) A N (d) Using sketch c 9 A N [] angle ( 0 ) attempt to find 5 cos M 78. Note: Accept exact answers given in terms of. (a) Evidence of using l rθ arc AB 7.85 (m) A N height 5 + 5 cos.5 (m) A N (b) Evidence of using A r θ Area of sector AOB 58.9 (m ) A N 55 5
& # & # (d) (i) h! 5 5cos +! % 4 " % 4 " 5. (m) A N & # (ii) h(0) 5 5 cos 0 +! % 4 " (iii) METHOD 4.9(m) A N Highest point when h 0 & # 0 5 5 cos t +! M % 4 " & # cos t +! % 4 " & # t.8 accept! A N % 8 " METHOD h 0 Sketch of graph of h M Correct maximum indicated t.8 A N METHOD Evidence of setting h)(t) 0 & # sin t +! 0 % 4 " Justification of maximum eg reasoning from diagram, first derivative test, second derivative test & # t.8 accept! A N % 8 " R t M R (f) (i) h(t) 0 0 (ii) METHOD Notes: Award A for range 0 to 0, A for two zeros. Award A for approximate correct sinusoidal shape. t AAA Maximum on graph of h) t 0.9 A N METHOD Minimum on graph of h) t.9 A N METHOD Solving h))(t) 0 One or both correct answers t 0.9, t.9 79. (a) (i) sin 40 p A N (ii) cos 70 q A N A N N [] & # (e) h)(t) 0 sin t +! (may be seen in part (d)) AA N % 4 " 57 58
(b) METHOD evidence of using sin θ + cos θ eg diagram, cos 40 ± cos 40 METHOD p (seen anywhere) p p A N evidence of using cos θ cos θ cos 40 cos 70 cos 40 ( q) ( q ) A N (c) evidence of appropriate approach eg line y on graph, discussion of number of solutions in the domain 4 (solutions) A N 8. (a) evidence of choosing the formula cos A cos A Note: If they choose another correct formula, do not award the M unless there is evidence of finding sin A 9. (c) METHOD sin 40 tan 40 cos 40 METHOD tan 40 p q p p A A N N correct substitution ' 8 ' eg cos A % ", cos A % " & # 9 & # 7 cos A A N 9 A 80. (a) period A N (b) y 4 0 4 x Note: Award A for amplitude of, A for their period, A for a sine curve passing through (0, 0) and (0, ). AAA N 59 0
(b) METHOD evidence of using sin B + cos B & # eg! % " 5 + cos B, (seen anywhere), 9 (c) using V y dx b a ( % V sin x cos & x# dx 0 & # ' cos B ± 5 9 & ± % 5 #! " 0 sin x cos x dx A cos B 5 9 METHOD diagram eg & % 5 #! " A M N [ ]! # & & & # # sin x 0 sin! sin 0! % % % " "" V A evidence of using sin and sin 0 0 eg ( 0) V A N for finding third side equals 5 cos B 5 A N 8. (a) evidence of using area of a triangle eg A sinθ A sin θ A N 8. (a) (i) range of f is [, ], ( f (x) ) A N (ii) sin x sin x A justification for one solution on [0, ] R e. g. x, unit circle, sketch of sin x solution (seen anywhere) A N (b) f )(x) sin x cos x A N (b) METHOD P ÔA θ area ΔOPA sin ( θ) ( sin ( θ)) A since sin ( θ) sin θ R then both triangles have the same area AG N0 METHOD triangle OPA has the same height and the same base as triangle OPB R then both triangles have the same area AG N0 (c) area semi-circle ( ) ( ) A area Δ APB sin θ + sin θ ( 4 sin θ) A S area of semicircle area ΔAPB ( 4 sin θ) M S ( sin θ) AG N0
(d) METHOD attempt to differentiate ds eg 4cos θ dθ setting derivative equal to 0 correct equation eg 4 cos θ 0, cos θ 0, 4 cos θ 0 θ EITHER evidence of using second derivative S))(θ) 4 sin θ & S)) 4! # % " & it is a minimum because S)) 0! # % " > evidence of using first derivative A A A A R for θ <, S )(θ) < 0 (may use diagram) A for θ >, S )(θ) > 0 (may use diagram) A it is a minimum since the derivative goes from negative to positive R N0 METHOD 4 sin θ is minimum when 4 sin θ is a maximum 4 sin θ is a maximum when sin θ (A) θ R A N N0 N 84. (a) choosing sine rule (b) sin R sin 75 correct substitution 7 0 sin R 0.748... P Rˆ Q 4.5 A N P 80 75 R P.5 substitution into any correct formula eg area Δ PQR 7 0 sin (their P) A.0 (cm ) A N 85. (a) evidence of appropriate approach M eg r 9 r.5 (cm) A N (b) adding two radii plus perimeter 7+ (cm) (.4) A N (c) evidence of appropriate approach M eg.5 9 area 0.5 (cm ) (.) A N A (e) S is greatest when 4 sin θ is smallest (or equivalent) (R) θ 0 (or ) A N [8] 4
8. (a) y 5 (b) (i) evidence of appropriate approach M eg 8 A A 8 AG N0 0 (ii) C 0 A N (iii) METHOD period 5 evidence of using B period (accept 0 ) eg B 0 x B ( accept 0.54or 0) A N Note: 5 Award A for passing through (0, 0), A for correct shape, A for a range of approximately to 5. AAA N METHOD evidence of substituting eg 0 8 cos B + 0 simplifying & # eg cos B 0 B! % " (b) evidence of attempt to solve f (x) sin x eg line on sketch, using tan x cos x x 0.07 x 0.77 AA N B ( accept 0.54or 0) A N (c) correct answers AA eg t.5, t 0.5, between 0: and 0:9 (accept 0:0) N [] 87. (a) (i) 7 A N (ii) A N (iii) 0 A N 5