IB Math SL : Trig Practice Problems: MarkScheme Circular Functions and Trig - Practice Problems (to 07) MarkScheme. (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 A 4 + 6 5 eg, 5 4 + 6 4 6 cosq 4 6 (b) Area pr sin cos P Qˆ R 0. 565 48 P Qˆ R 55.8 (0.97 radians) AN P Qˆ R For substituting correctly 4 6 sin 55.8 A 9.9 (cm ) AN. Note: Throughout this question, do not accept methods which involve finding θ. (a) Evidence of correct approach A eg sin θ, BC AB 5 sin θ 5 AG (b) Evidence of using sin θ sin θ cos θ 5 A 4 5 9 AGN0 (c) Evidence of using an appropriate formula for cos θ M 4 5 4 5 80 eg,,, 9 9 9 9 8 cos θ 9 A N. (a) For using perimeter r + r + arc length 0 r + rθ A 0 r θ r AG N0 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 4. Notes: Candidates may have C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page of 9 N0
IB Math SL : Trig Practice Problems: MarkScheme 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 a + b c eg cos C, a b + c bc cos A ab Correct substitution + 4 eg cos A ÔP,4 + cos AÔP A cos A ÔP 0.5 6 A ÔP.8 (radians) 45 A N METHOD Area of AOBP 5.8 (from part (d)) Area of triangle AOP.905.9050 0.5 sin A ÔP A A ÔP. or.8 6 A ÔP.8 (radians) 45 A N (b) A ÔB (.8) (.64) 8.64 (radians) 45 AN (c) (i) Appropriate method of finding area eg area θr Area of sector PAEB 4. 6 A.0 (cm ) (accept the exact value.04) AN (ii) Area of sector OADB. 64 A.9 (cm ) AN (d) (i) Area AOBE Area PAEB Area AOBP (.0 5.8) M 7.9 (accept 7. from the exact answer for PAEB) AN (ii) Area shaded Area OADB Area AOBE (.9 7.9) M 4.7 (accept answers between 4.6 and 4.7) AN 5. (a) Evidence of choosing cosine rule eg a b + c bc cos A Correct substitution A eg (AD) 7. + 9. (7.) (9.) cos 60 (AD) 69.7 AD 8.5 (cm) AN (b) 80 6 8 Evidence of choosing sine rule Correct substitution A [4] C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page of 9
IB Math SL : Trig Practice Problems: MarkScheme 6. (a) DE 8.5 eg sin 8 sin 0 DE.75 (cm) A N (c) Setting up equation eg ab sin C 5.68, bh 5.68 Correct substitution eg 5.68 (.) (7.) sin sin D Bˆ C,. h 5.68, (h.55) D Bˆ C 0.5 D Bˆ C 0 and/or 50 A N (d) Finding A Bˆ C (60 + D Bˆ C) Using appropriate formula eg (AC) (AB) + (BC), (AC) (AB) + (BC) (AB) (BC) cos ABC Correct substitution (allow FT on their seen A Bˆ C ) eg (AC) 9. +. A AC 9.74 (cm) A N (e) For finding area of triangle ABD Correct substitution Area 9. 7. sin 60 8.8... A Area of ABCD 8.8... + 5.68 4.0 (cm ) AN y 0 A A [] 5 60 80 0 80 60 x 5 0 Correct asymptotes AA N (b) (i) Period 60 (accept ) A N (ii) f (90 ) A N (c) 70, 90 AA NN C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page of 9
IB Math SL : Trig Practice Problems: MarkScheme Notes: Penalize mark for any additional values. Penalize mark for correct answers given in radians,,or 4.7,.57. 7. (a) METHOD Using the discriminant 0 k 4 4 k 4, k 4 AA N METHOD Factorizing (x ± ) k 4, k 4 AA 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 (c) (i) A N (ii) METHOD Attempting to solve for cos θ M cos θ θ 40, 0, 40, 0 (correct four values only) AN METHOD Sketch of y 4 cos θ + 4 cos θ + y M 9 60 80 80 60 x Indicating 4 zeros θ 40, 0, 40, 0 (correct four values only) AN (d) Using sketch c 9 A N 8. Note: Accept exact answers given in terms of. (a) Evidence of using l rθ arc AB 7.85 (m) A N (b) Evidence of using A r θ Area of sector AOB 58.9 (m ) A N (c) METHOD [] C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 4 of 9
IB Math SL : Trig Practice Problems: MarkScheme 6 angle ( 0 ) 6 attempt to find 5 sin 6 M height 5 + 5 sin 6.5 (m) A N METHOD angle ( 60 ) attempt to find 5 cos M height 5 + 5 cos.5 (m) A N (d) (i) h 5 5cos + 4 4 5.6 (m) AN (ii) h(0) 5 5 cos 0 + 4 4.9(m) AN (iii) METHOD Highest point when h 0 R 0 5 5 cos t + 4 M cos t + 4 t.8 accept 8 AN METHOD C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 5 of 9
IB Math SL : Trig Practice Problems: MarkScheme h 0 t Sketch of graph of h M Correct maximum indicated t.8 AN METHOD Evidence of setting h (t) 0 M sin t + 0 4 Justification of maximum R eg reasoning from diagram, first derivative test, second derivative test t.8 accept 8 AN (e) h (t) 0 sin t + (may be seen in part (d)) 4 AA N (f) (i) h(t) 0 t 0 Notes: Award A for range 0 to 0, A for two zeros. Award A for approximate correct sinusoidal shape. (ii) METHOD Maximum on graph of h t 0.9 METHOD Minimum on graph of h t.96 METHOD Solving h (t) 0 One or both correct answers t 0.9, t.96 AAAN AN AN A N C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 6 of 9
IB Math SL : Trig Practice Problems: MarkScheme 9. (a) Vertex is (4, 8) AA N (b) Substituting 0 a(7 4) + 8 M a AN (c) For y-intercept, x 0 y 4 A N 0. METHOD Evidence of correctly substituting into A r θ A Evidence of correctly substituting into l rθ A For attempting to eliminate one variable leading to a correct equation in one variable A r 4 θ 6 ( 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 ) 6 A r 4 θ 6 ( 0.54, 0 ). a 4, b, c or etc. (a) (b) 5 PQ Using r a + tb x + 5 t y 6 AAA AA N AAA. METHOD Evidence of correctly substituting into l rθ A Evidence of correctly substituting into A r θ A For attempting to solve these equations eliminating one variable correctly A r 5 θ.6 ( 9.7 ) AA N METHOD Setting up and equating ratios 4 80 r r AA C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 7 of 9 N N6 N4 []
IB Math SL : Trig Practice Problems: MarkScheme Solving gives r 5 A rθ 4 or r θ 80 A θ.6 ( 9.7 ) A r 5 θ.6 ( 9.7 ) N 4. (a) For correct substitution into cosine rule A BD 4 + 8 4 8cosθ For factorizing 6, BD ( 5 4 cosθ) 6 A 4 5 4 cosθ AGN0 (b) (i) BD 5.565... sin CBˆ D sin 5 5.565 MA sin C Bˆ D 0.9 (accept 0.90, subject to AP) AN (ii) C Bˆ D 65.7 A N Or C Bˆ D 80 their acute angle 4 AN (iii) B Dˆ C 89. BC 5.565 BC or (or cosine rule) sin 89. sin 5 sin 89. sin 65.7 MA BC. (accept.7 ) A Perimeter 4 + 8 + +. 7. AN (c) Area 4 8 sin 40 A 0. A N 5. (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. (i) d y Using 0 dx x + 60 0 A x 5 AN (ii) y max 6(5 ) + 60(5) 56 y max 44 AN METHOD (i) Sketch of the correct parabola (may be seen in part (ii)) M x 5 AN (ii) y max 44 A N (b) (i) z 0 x (accept x + z 0) A N (ii) z x + 6 x 6 cos Z A N (iii) Substituting for z into the expression in part (ii) Expanding 00 0x + x x + 6 x cos Z A C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 8 of 9
IB Math SL : Trig Practice Problems: MarkScheme Simplifying x cos Z 0x 64 A 0x 64 Isolating cos Z x A 5x 6 cos Z x AGN0 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. (c) Evidence of using the formula for area of a triangle A 6 x sin Z M A x sin Z A 6 x sin Z 4 A A 9x sin Z AG N0 (d) Using sin Z cos Z 5x 6 Substituting x for cos Z A 5x 6 5x 60x + 56 for expanding to x 9x A for simplifying to an expression that clearly leads to the required answer A eg A 9x (5x 60x + 56) A 6x + 60x 56 AG (e) (i) 44 (is maximum value of A, from part (a)) A A max AN (ii) Isosceles A N 6. (a) Evidence of choosing the double angle formula f (x) 5 sin (6x) A N (b) Evidence of substituting for f (x) eg 5 sin 6x 0, sin x 0 and cos x 0 6x 0,, x 0,, 6 AAA 7. (a) (i) OP PQ ( cm) R So OPQ is isosceles AGN0 (ii) Using cos rule correctly eg cos O Pˆ Q 9+ 9 6 cos O Pˆ Q 8 8 cos O Pˆ Q 9 + 4 A AGN0 (iii) Evidence of using sin A + cos A M sin O Pˆ Q 80 8 A N4 [0] C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 9 of 9
IB Math SL : Trig Practice Problems: MarkScheme sin O Pˆ Q 80 9 AGN0 (iv) Evidence of using area triangle OPQ OP PQ sin P M 80 9 eg, 0.998 9 Area triangle OPQ (b) (i) O Pˆ Q.4594... 80 ( 0 ) ( 4.47) AN O PˆQ.46 AN (ii) Evidence of using formula for area of a sector eg Area sector OPQ.4594 6.57 AN.4594 (c) Q ÔP ( 0.84) Area sector QOS 4 0. 84 A 6.7 AN (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 M A eg 4.5 6.5675... 6.785... + 4.47..., 4.5 (6.785... +.095...), 4.5 (6.5675... +.56...) Area of the shaded region 5. A N 8. (a) p 0 A (b) METHOD Period q (M) q 4 A 4 METHOD Horizontal stretch of scale factor q (M) [7] scale factor 4 q 4 A 4 9. (a) using the cosine rule (A) b + c bc cos  substituting correctly BC 65 +04 (65) (04) cos 60 A 45 + 086 6760 88 BC 9 m A (b) finding the area, using bc sin  C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 0 of 9
IB Math SL : Trig Practice Problems: MarkScheme substituting correctly, area (65) (04) sin 60 A 690 (Accept p 690) A (c) (i) A (65) (x) sin 0 A 65x 4 AG (ii) A (04) (x) sin 0 M 6x A (iii) starting A + A A or substituting 65x + 6x 690 4 simplifying 69x 690 4 A 4 690 x 69 A 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 ADB and ACD substituting correctly sin 0 65 BD sin 0 sin ADˆ B 65 sin ADˆ B A and sin 0 04 DC sin 0 sin ADˆ B 04 sin ADˆ C M since sin A Dˆ B sin A Dˆ C BD DC BD 65 65 04 DC 04 A BD 5 DC 8 AG 5 0. (a) A r θ 7 (.5) r r 6 r 6 cm (C4) (b) Arc length rθ.5 6 Arc length 9 cm (C) Note: Penalize a total of ( mark) for missing units.. (a) when y 0 (may be implied by a sketch) [8] 8 x or.79 9 (C) (b) METHOD Sketch of appropriate graph(s) Indicating correct points x. or x 5.4 (C)(C) METHOD C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page of 9
IB Math SL : Trig Practice Problems: MarkScheme sin x + 9 7 x +, x + 9 6 9 6 7 x, x 6 9 6 9 9 x, x ( x., x 5.4) (C)(C) 8 8. (a) for using cosine rule ( a b c abcosc ) ( )( ) BC 5 + 7 5 7 cos 9 + 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). (i) A C 9 (c) 7 85 B ACB ɵ 80 (9 + 85) 66 for using sine rule (may be implied) AC 7 sin85 sin 66 7sin 85 AC sin 66 AC (8.580 ) 8.5 m (N) Area 7 8.58... sin 9 76.4 m (Accept 76. m ) (N) 5 (ii) ( )( ) A ĈB from previous triangle 66 Therefore alternative ACB ˆ 80 66 4 (or 9 + 85) ABC ˆ 80 (9 + 4) 7 A C 9 4 7 7 B C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page of 9
IB Math SL : Trig Practice Problems: MarkScheme AC 7 sin 7 sin4 AC (.9906 ). m (N) 4 (d) A 9 C 7 B Minimum length for BC when A ĈB 90 or diagram showing right triangle CB sin 9 7 CB 7sin 9 CB (8.47 ) 8.4 m (N). (a) (i) f ( x) cosx sin x cos x sin x (N) Note: Award for sin x sin x + only if work shown, using product rule on sin x cos x + cos x. (ii) sin x + sin x (sin x )(sin x + ) or (sin x 0.5)(sin x + ) (N) (iii) sin x or sin x sin x 5 x (0.54) x (.6) x (4.7) 6 6 (N) (N)(N)6 (b) x ( 0.54) 6 (N) (c) (i) EITHER curve crosses axis when x (may be implied) 5 6 6 (N) Area f ( x)d x + f ( x)dx 5 Area 6 f ( x ) dx (A)(N) 6 (ii) Area 0.875 + 0.875.75 (N) 5 4. Using area of a triangle ab sin C [4] [] C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page of 9
IB Math SL : Trig Practice Problems: MarkScheme 0 (0)(8)sin Q Note: Accept any letter for Q sin Q 0.5 P Qˆ R 0 or or 0.54 (C6) 6 5. (a) b 6 (C) (b) y B x (A) (C) (c) x.05 (accept (.05, 0.896) ) (correct answer only, no additional solutions) (A) (C) 6. (a) ( sin x) + sin x 6 sin x sin x 0 (p 6, q, r ) (C) (b) ( sin x )( sin x + ) (C) (c) 4 solutions (A) (C) 7. Area of large sector r θ 6.5 8. (a) Area of small sector r θ 0.5 9 75 Shaded area large area small area 9 75 7 (C6) C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 4 of 9
IB Math SL : Trig Practice Problems: MarkScheme y.5 0.5 0 0.5.5.5.5 x 0.5.5 (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. (b) (Maximum) x 0.85 4 x 0. ( dp) (C) (Minimum) x.856 4 x.9 ( dp) (C) 9. Area of a triangle 4 sin A 4 sin A 4.5 sin A 0.75 A 48.6 and A (or 0.848,.9 radians) (A) (C6) Note: Award (C4) for 48.6 only, (C5) for only. 0. METHOD 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 (C6) METHOD Graphical solutions EITHER for both graphs y cos x, y sin x, (M) for the graph of y cos x sin x. (M) THEN Points representing the solutions clearly indicated.57, 0.785 C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 5 of 9
IB Math SL : Trig Practice Problems: MarkScheme x, x 4 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.. (a) (i) 0 + 4 sin.4 (ii) At 00, t 0 + 4 sin 0.5 6.48 (N) Note: Award (A0) if candidates use t 00 leading to y.6. No other ft allowed. (b) (i) 4 metres t t (ii) 4 0 + 4 sin sin t (.4) (correct answer only) (N) (c) (i) 4 t (ii) 0 + 4 sin 7 t sin 0.75 t 7.98 (N) (iii) depth < 7 from 8 hours from 00 0 hours therefore, total 6 hours (N) 7. (a) Angle A 80 AB 5 sin 40 sin80 AB.6 cm (C) (b) Area ac sin B (5)(.6) sin 60 7.07 (accept 7.06) cm (C) Note: Penalize once in this question for absence of units.. METHOD (5) (0.8) Area sector OAB 0 5cos0.8.48... ON ( ) AN 5sin 0.8 (.586... ) Area of AON ON AN 6.49... (cm ) Shaded area 0 6.49...75 (cm ) (C6) METHOD (C6) [] C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 6 of 9
IB Math SL : Trig Practice Problems: MarkScheme A O N B F Area sector ABF (5) (.6) 0 Area OAF (5) sin.6.5 Twice the shaded area 0.5 ( 7.5) Shaded area (7.5).75 (cm ) (C6) 4. (a) (i) f ( x) 6sin x (ii) EITHER f ( x) sin xcos x 0 sin x 0 or cos x 0 sin x 0, for 0 x THEN x 0,, (N4) 6 (b) (i) translation in the y-direction of (ii). (.0 from TRACE is subject to AP) (A) 4 5. p + q cos 0 p + q p + q cos p q (a) p (C) (b) q (C) 6. Method [0] C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 7 of 9
IB Math SL : Trig Practice Problems: MarkScheme y 0.80.5 x 0 (C).80 [ sf] (G) (C).5 [ sf] (G) (C) 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) 7. (a) area of sector ΑΒDC 4 () area of segment BDCP area of ABC (C) (b) BP area of semicircle of radius BP ( ) area of shaded region ( ) 8. (a) PQ q p 0 7 (b) (C) PO PQ cos OPˆQ PO PQ C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 8 of 9
IB Math SL : Trig Practice Problems: MarkScheme ( 7) ( ) 58, PQ + ( ) PO + PO PQ + 6 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 OPˆ Q 754 (AG) (ii) sin P Qˆ R cos θ 5 754 5 754 59 754 754 (AG) 5 754 x P therefore x 754 5 59 x sin θ 754 Note: Award (A0) for the following solution. (AG) 5 cos θ θ 56.89 754 sin θ 0.876 0.876 sin θ 754 754 (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. C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 9 of 9
IB Math SL : Trig Practice Problems: MarkScheme Award final for the integer answer. 9. (a) Sine rule PR 9 sin5 sin0 9sin 5 PR sin0 5.96 km (b) EITHER Sine rule to find PQ 9sin 5 PQ sin0 4.9 km Cosine rule: PQ 5.96 + 9 ()(5.96)(9) cos 5 9.9 PQ 4.9 km 4.9 Time for Tom 8 5.96 Time for Alan a 4.9 5.96 Then 8 a a 0.9 7 (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 6.7 Therefore, Q ŜR 80 (5 + 6.7) 8. 9 QS SR sin8. sin6.7 sin 5 9sin6.7 9sin 5 QS sin8. sin8..9 6 40. (a) (i) cos 4, sin 4 [4] C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 0 of 9
IB Math SL : Trig Practice Problems: MarkScheme 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.6. x 4 (G) (b) y e x (cos x + sin x) d y dx e x (cos x + sin x) + e x ( sin x + cos x) 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 Note: Award (A0)(A0) for a.57, b 4.8. 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 4 y e 4 (cos 4 + sin 4 ) e 4 (AG) 5 (e) Required area 4 (cos x + sin x)dx 0 7.46 sq units (G) Αrea 7.46 sq units (G) Note: Award (G0) for the answer 9.8 obtained if the calculator is in degree mode. 4. (a) (i) 4 A is, 0 (C) (ii) B is (0, 4) [7] C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page of 9
IB Math SL : Trig Practice Problems: MarkScheme 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 (C) 8 (.67) 4. (a) ( sin x )(sin x ) (C) Note: Award A if x x + 6 correctly factorized to give (x )(x ) (or equivalent with another letter). (b) (i) ( sin x )(sin x ) 0 sin x sin x (C) (C) (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. 4. 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. (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 (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 C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page of 9
IB Math SL : Trig Practice Problems: MarkScheme 75 (d) Area of sector 75 or 80 60 94. (cm ) (accept 0 or 94. (cm )) (e) Shaded area area OPB area sector 6.4 (cm ) (accept 6. cm, 6. cm ) 44. Note: Do not penalize missing units in this question. (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 (b) (i) d s dt 0 sin t Note: Award for 0, and for sin t. (ii) d s v dt 0 sin t 0 sin t 0 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. (c) 48 +0 cos t 60 + 5 cos 4t t 0.6 secs t 0.6 secs (G) (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 0.6; for. 45. (a) l rθ or ACB OA 0 cm (C) (b) A ÔB (obtuse) Area θ r ( )(5) [] 46. 48 cm ( sf) (C4) C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page of 9
IB Math SL : Trig Practice Problems: MarkScheme B A d 50 80 70 P (A).5 0 50.5 80 d 50 + 80 50 80 cos 70 d 78.5 km (C6) 47. (a) (i) (C) (ii) 4 (accept 70 ) (A) (C) (b) y 48. number of solutions: 4 (G) (A) (C) Statement (a) Is the statement true for all (b) If not true, example real numbers x? (Yes/No) A No x l (log 0 0. ) (a) (A) (C) B No x 0 (cos 0 ) (b) (A) (C) C Yes N/A 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 C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 4 of 9
IB Math SL : Trig Practice Problems: MarkScheme 49. (a) 7 6 sin A sin 45 sin A 6 7 6 (AG) 7 (b) A D h B C (i) B Dˆ C + B ÂC 80 (ii) 6 sin A 7 (c) > A 59.0 or ( sf) > B ĈD 80 ( + 45 ) 4.0 ( sf) 7 (iii) BD sin 4 sin 45 >BD.69 6 BD h Area BDC Area BAC BA h BD BA (AG) 50. Using sine rule: BD 6sin 45 Area BCD Area BAC BA 6sin 45 BD BA (AG) sin B sin 48 5 7 sin B 7 5 sin 48 0.508 [0] B arcsin (0.508).06 (nearest degree) (C6) Note: Award a maximum of [5 marks] if candidates give the answer in radians (0.560). C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 5 of 9
IB Math SL : Trig Practice Problems: MarkScheme 5. (a) x is an acute angle > cos x is positive. cos x + sin x > cos x > cos x (b) cos x sin x 7 9 sin x 8 ( ) (C4) 9 Notes: (a) Award (M0)(A0) for cos sin 0.94. (b) Award (A0) for cos sin 0.778. 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) (c) cos x or cos x l > x 60, 80 or 00 (C) Note: Award (A0) if the correct answers are given in 5 radians (ie,,, or.05,.4, 5.4) 5. (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) 54. (a) sin x + 4 cos x ( cos x) + 4cos x cos + 4 cos x (C) [4] C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 6 of 9
IB Math SL : Trig Practice Problems: MarkScheme (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 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. 55. O Tˆ A 90 AT 6 6 T ÔA 60 Area area of triangle area of sector 6 6 6 6. cm (or 8 6) (C4) T ÔA 60 Area of 6 sin 60 [4] Area of sector 6 6 Shaded area 8 6. cm ( sf) (C4) 56. (a) (i) AP ( x 8) + (0 6) x 6x + 80 (AG) (b) (ii) OP ( x 0) + (0 0) x + 00 AP + OP OA cos OPˆA AP OP ( x 6x + 80) + ( x + 00) (8 + 6 ) x 6x + 80 x + 00 x x cos OPˆA 6x + 80 6x + 80 {( x x x + 00 8x + 40 6x + 80)( x + 00)} (AG) (c) For x 8, cos OPˆ A 0.780869 arccos 0.780869 8.7 ( sf) 8 tan OPˆA 0 O Pˆ A arctan (0.8) 8.7 ( sf) (d) O Pˆ A 60 cos OPˆ A 0.5 [4] C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 7 of 9
IB Math SL : Trig Practice Problems: MarkScheme 0.5 {( x x 8x + 40 6x + 80)( x + 00)} x 6x + 80 {( x 6x + 80)( x + 00)} 0 x 5.6 (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) (ii) x The line (OA) has equation y 4 When y 0, x x 40 ( ) 40 ( ) (G) 5 Note: Award (G) for.. 57. (a) Area r θ (5 )() 5 (cm ) (C) (b) Area OAB 5 sin 0. Area 5 0..7 (cm ) ( sf) (C) 58. (a) sin (AĈB) sin50 0 7 0sin 50 sin (AĈB) 0.90 7 A ĈB > 90 A ĈB 80 64. 5.7 A ĈB 6 ( sf) (C) (b) In Triangle, A ĈB 64. B ÂC 80 (64. + 50 ) 65.7 Area (0)(7) sin 65.7 55 (cm ) ( sf) (C) 59. METHOD The value of cosine varies between and +. Therefore: t 0 a + b 4. t 6 a b 0. a 4.6 a. (C) b 4.0 b (C) () Period hours k k (C) METHOD [4] [4] C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 8 of 9
IB Math SL : Trig Practice Problems: MarkScheme y 4. 60. 0. 6 8 4 t (h) From consideration of graph: Midpoint a. (C) Amplitude b (C) Period k k (C) km C [4] A 48km B 48 + 56 cos CÂB (48)() C ÂB arccos(0.065) 86 6. (a) cos x + sin x ( sin x) + sin x sin x + sin x (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 [4] 6. (a) sin x 0 x 0 or (0 or 80 ) Note: Award for both answers. 5 sin x x or (0 or 50 ) 6 6 Note: Award for both answers. [4] C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 9 of 9
IB Math SL : Trig Practice Problems: MarkScheme y 4 { 0.5< x<.5< y<4 MAXIMUM POINT integers on axis { 4 5 < x<.5.5< x<4 LEFT INTERCEPT {.< x<.6 0.< y <0 MINIMUM POINT x RIGHT INTERCEPT 5 (b) is a solution if and only if + cos 0. Now + cos + ( ) 0 (c) By using appropriate calculator functions x.696 7 9... x.6967 (6sf) (d) See graph: ( + x cos x)dx 0 (e) EITHER ( + x cos x)dx 7.86960 (6 sf) (A) 0 0 Note: This answer assumes appropriate use of a calculator eg fnint( Y, X,0, ) 7.86960440 fnint : withy + x cos x ( + x cos x)dx [x + x sin x + cos x] 0 ( 0) + ( sin 0 sin 0) + (cos cos 0) + 0 + 7.86960 (6 sf) 6. (a) (i) Q (4.6 8.). (ii) P (4.6 + 8.) (M0).4 (b) 0.4 +. cos t 6 7 so cos t 6 6 7 therefore arccos t 6 6 [5] C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 0 of 9
IB Math SL : Trig Practice Problems: MarkScheme which gives.06... 6 t or t.8648. t.86( sf) (c) (i) By symmetry, next time is.86... 8.5... t 8.4 ( sf) (ii) From above, first interval is.86 < t < 8.4 This will happen again, hours later, so 5.9 < t < 0. 4 64. cos x 5 sin x sin x cos x 5 tan x 0.6 x or x (to the nearest degree) (C)(C) Note: Deduct [ mark] if there are more than two answers. 5 65. sin A cos A ± [0] [4] But A is obtuse cos A sin A sin A cos A 5 0 69 66. (a) y sin x x y (C4) [4] (.5,.7) (., 0) (., 0) x (.5,.7) (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). C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page of 9
IB Math SL : Trig Practice Problems: MarkScheme 67. (a) Award for a smooth curve. Allow some flexibility, especially in the middle three marks here. (b) x. (c) x ( 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] 0º 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 (C) 68. (a) (b) PQ tan 6 40 PQ 9. m ( sf) 40m 0 B (C) (C) [4] Q 70 A A Qˆ B 80 AB 40 sin 80 sin 70 Note: Award for correctly substituting. C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page of 9
IB Math SL : Trig Practice Problems: MarkScheme AB 4 9. m ( sf) (C) 69. Perimeter 5( ) + 0 Note: Award for working in radians; for ; for +0. (0 + 5) cm ( 6.4, to sf) (C4) 70. From sketch of graph y 4 sin x + (M) or by observing sin θ. k > 4, k < 4 (C)(C) 4 [4] [4] 0 0 7. (a)(i) & (c)(i) 4 [4] C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page of 9
IB Math SL : Trig Practice Problems: MarkScheme y (., 0.55) 0 R (.5, 0) x (,.66) (A) Notes: The sketch does not need to be on graph paper. It should have the correct shape, and the points (0, 0), (., 0.55), (.57, 0) and (,.66) should be indicated in some way. Award for the correct shape. Award (A) for or 4 correctly indicated points, for or points. (ii) Approximate positions are positive x-intercept (.57, 0) maximum point (., 0.55) end points (0, 0) and (,.66) 7 (b) x cos x 0 x 0 cos x 0 x Note: Award (A) if answer correct. (c) (i) see graph (ii) 0 x cos x dx (A) Note: Award for limits, for rest of integral correct (do not penalize missing dx). (d) Integral 0.467 (G) C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 4 of 9
IB Math SL : Trig Practice Problems: MarkScheme / Integral [ ] x sin x + x cos x sin x 0 () + (0) () 4 [0 + 0 0] (exact) or 0.467 ( sf) [5] 7. (a) From graph, period (b) Range {y 0.4 < y < 0.4} d (c) (i) 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. (ii) f (x) 0 (d) 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 x (e) (i) (cos x )(sin x) dx sin x + c / (ii) Area (cosx)(sin x) dx sin (sin 0) 0 (f) At C f (x) 0 9 cos x 7 cos x 0 cos x(9 cos x 7) 0 4 x (reject) or x arccos 7 0.49 ( sf) 4 [0] cos α 5 7. Note: Award for identifying the largest angle. 4 + 5 7 4 5 α 0.5 C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 5 of 9
IB Math SL : Trig Practice Problems: MarkScheme Find other angles first β 44.4 γ 4.0 α 0.6 (C4) Note: Award (C) if not given to the correct accuracy. 74. AB rθ r θ r.6 5.4 8 cm (5.4) θ.6 [4] 4 θ.7 (.48 radians) AB rθ 4 5.4.7 8 cm (C4) 75. (a) OA 6 A is on the circle [4] OB 6 B is on the circle. (b) (c) OC 5 5 + 6 C is on the circle. AC OC OA 5 6 0 AO AC cos OAC ˆ AO AC 6. 0 6 + 6 6 6 C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 6 of 9
IB Math SL : Trig Practice Problems: MarkScheme (d) 76. tan x ˆ 6 + ( ) 6 cos O AC 6 as before 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 A number of possible methods here BC OC OB 5 6 0 BC ABC 6 ABC has base AB and height area 6 Given cos B AC ˆ 6 sin B AC ˆ ABC 6 6 6 4 [] tan x ± x 0 or x 50 (C)(C) 77. h r so r 00 r 50 l 0θ r 50 θ 0 5 0 [4] C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 7 of 9
IB Math SL : Trig Practice Problems: MarkScheme θ 4.44 (sf) (C4) Note: Accept either answer. 78. (a) f () f (5) (b) EITHER distance between successive maxima period 5 4 (AG) Period of sin kx ; k so period 4 (AG) (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. (d) f (x) sin x + f (x) cos x + 0 (A) Note: Award for the chain rule, for, for cos x. cos x 4 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 [4] C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 8 of 9
IB Math SL : Trig Practice Problems: MarkScheme x or or... x or 6... Since 0 x 5, we take x, so the point is (, ) (ii) Tangent line is: y (x ) + y ( + ) x k + 6 (f) f (x) sin x + sin x 5 x or or 6 6 6 5 x or or 5 [4] C:\Users\Bob\Documents\Dropbox\Desert\SL\Trig\LP_SLTrig-.doc on 0/9/0 at :05 AM Page 9 of 9