11.Note: Award (M1) for identifying the largest angle.cosa = (M1)= (A1) a = 101.5(A1)OR Find other angles firstb = 44.4g = 34.0(M1) a = 101.6(A1)(A1)Note: Award (C3) if not given to the correct accuracy[4]

2.AB = rq= (M1)(A1)= 21.6 (A1)= 8 cm(A1)OR (5.4)2q = 21.6 q = (= 1.481 radians)(M1)AB = rq(A1)= 5.4 (M1)= 8 cm(A1)[4]

3.(a) = 6A is on the circle(A1) = 6B is on the circle.(A1)

= = 6C is on the circle.(A1)3(b)= (M1)= (A1)2(c)(M1)= = (A1)= (A1)OR (M1)(A1)= as before(A1)OR using the triangle formed by and its horizontal andvertical components:(A1)(M1)(A1)3Note: The answer is 0.289 to 3 s.f.(d)A number of possible methods here

= (A1)= (A1)BC = DABC = (A1)= (A1)OR DABC has base AB = 12(A1)and height = (A1) area = (A1)= (A1)

OR Given (A1)(A1)(A1)= (A1)4[12]

4.tan2x = (M1) tanx = (M1) x = 30 or x = 150(A1)(A1)[4]

5.h = r so 2r2 = 100 r2 = 50(M1)l = 10q = 2pr(M1) q = (A1)= q = p = 4.44 (3s.f.)(A1)Note: Accept either answer.[4]

6.(a)f(1) = 3f(5) = 3(A1)(A1)2(b)EITHERdistance between successive maxima = period(M1) = 5 1(A1) = 4(AG)ORPeriod of sin kx = ;(M1)so period = (A1)= 4(AG)2

(c)EITHER Asin + B = 3 and Asin+ B = 1(M1) (M1) A + B = 3, A + B = 1(A1)(A1) A = 2, B = 1(AG)(A1)OR Amplitude = A(M1)A = (M1)A = 2(AG)Midpoint value = B(M1)B = (M1)B = 1(A1)5Note: 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) = 2sin + 1f(x) = + 0(M1)(A2)Note: Award (M1) for the chain rule, (A1) for , (A1) for 2 cos.= p cos(A1)4Notes: 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 px is a tangent p = pcos(M1) 1 = cos(A1) x = p or 3p 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 = p(x 2) + 1(M1)y = (2p + 1) pxk = 2p + 1(A1)6

(f)f(x) = 2 2sin + 1 = 2(A1) sin(A1) x = (A1)(A1)(A1)5[24]

7.3 cos x = 5 sin 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]

8.sin A = cos A = (A1)But A is obtuse cos A = (A1)sin 2A = 2 sin A cos A(M1)= 2 = (A1)(C4)[4]

9.(a)(A5)5Notes: 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).Award (A1) for a smooth curve.Allow some flexibility, especially in the middle three marks here.(b)x = 2.31(A1)1(c)(A1)(A1)Note: Do not penalise for the absence of C.Required area = (M1) = 0.944(G1)ORarea = 0.944(G2)4[10]

10.(a)(A2)(C2)(b)Period = (A1)(C1)[3]

11.(a)Using the formula for the area of a triangle givesA = x 3x sin (M1)sin = (A1)2(b)Using the cosine rule givescos = (M1)= (A1)2(c)(i)Substituting the answers from (a) and (b) into the identitycos2 = 1 sin2 gives(M1)(AG)(ii)(a)x = 1.24, 2.94(G1)(G1)(b) = arccos = 1.86 radians or = 0.171(accept 0.172) radians (3 s.f.)(G1)(G1)6Notes: Some calculators may not produce answers that are as accurate as required, especially if they use zoom and trace to find the answers. Allow 0.02 difference in the value of x, with appropriate ft for .Award (M1)(G1)(G0) for correct answers given in degrees (106 or 9.84).Award (M1)(G1)(G0) if the answers are not given to 3 s.f.Award (M0)(G2) for correct answers without working.[10]

12.(a)

Acute angle 30(M1)Note: Award the (M1) for 30 and/or quadrant diagram/graph seen2nd quadrant since sine positive and cosine negative q = 150(A1)(C2)(b)tan 150 = tan 30 or tan 150 = (M1)tan 150 = (A1)(C2)[4]

13.(a) = tan36 PQ 29.1 m (3 s.f.)(A1)(C1)(b)

= 80(A1)(M1)Note: Award (M1) for correctly substituting. AB = 41 9. m (3 s.f.)(A1)(C3)[4]

14.Perimeter = 5(2 1) + 10(M1)(A1)(A1)Note: Award (M1) for working in radians; (A1) for 2 1; (A1) for +10= (10 + 5)cm (= 36.4, to 3 s.f.)(A1)(C4)[4]

15.From sketch of graph y = 4sin(M2)or by observing |sin q| 1.k > 4, k < 4(A1)(A1)(C2)(C2)[4]

16.(a)(i)&(c)(i)

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 arepositive 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)x2cos x = 0x 0 cos x = 0(M1) x = (A1)2Note: Award (A2) if answer correct.(c)(i)see graph(A1)(ii)cos x dx(A2)3Note: Award (A1) for limits, (A1) for rest of integral correct (do not penalise missing dx).

(d)Integral = 0.467(G3)ORIntegral = (M1)= [0 + 0 0](M1)= 2 (exact) or 0.467 (3 s.f.)(A1)3[15]

17.(a)From graph, period = 2(A1)1(b)Range = {y |0.4 < y < 0.4}(A1)1(c)(i)f(x) = {cos x (sin x)2}= cos x (2sin x cos x) sin x (sin x)2 or 3sin3 x + 2sin 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{3cos x 1} = 0(A1) 3cos2 x 1 = 0 cos x = (A1)At A, f(x) > 0, hence cos x = (R1)(AG)(iii)f (x) = (M1) = (A1)9(d)x = (A1)1(e)(i)(M1)(A1)(ii)Area = (M1)= (A1)4

(f)At C f(x) = 0(M1) 9cos3 x 7 cos x = 0 cos x(9cos2 x 7) = 0(M1) x = (reject) or x = arccos = 0.491 (3 s.f.)(A1)(A1)4[20]

18.Note: Award full marks for exact answers or answers given to threesignificant figures.Method 1:Using the sine rule: sin C = C = 45, 135.(M1)Again, Thus, BC = sin 105 or sin 15BC = 8.20 cm or BC = 2.20 cm.(A1)(A1)(C3)Method 2:Using the cosine rule: AC2 = 62 + BC2 2(6)(BC)cos3018 = 36 + BC2 BC(M1)Therefore, BC2 ()BC + 18 = 0Therefore, (BC )2 = 27 18 = 9Therefore, BC = 3, i.e. BC = 8.20 cm or BC = 2.20 cm.(A1)(A1)(C3)Method 3:In DABD, AD = 3 cm,and BD = cm.(A1)In DAC1D, C1D = 3Also, C2D = 3.(A1)Therefore BC = ( 3) cm, i.e. BC = 8.20 cm or BC = 2.20 cm.(A1)(C3)Note: If only one answer is given, award a maximum of (M1)(A1).[3]

19.(a)(i)y = , 0 < a < b(M1)(C1)= (M1)(C1)= (AG)4(ii) = 0 cosx = 0 since b2 a2 0.This gives x = (+pk, k )(M1)(C1)When x = , y = = 1,and when x = , y = = 1.Therefore, maximum y = 1 and minimum y = 1.(A2)4(iii)A vertical asymptote at the point x exists if and only ifb + a sin x = 0.(R1)Then, since 0 < a < b, sin x = , which is impossible.(R1)Therefore, no vertical asymptote exists.(AG)2(b)(i)y-intercept = 0.8(A1)(ii)For x-intercepts, sin x = x = 4.069, 5.356.(A2)(iii)(C2)5(c)Area = (M1)(C1)ORArea = (M1)(C1)2[17]

20.

cos(M1)(A1) = arccos(0.0625)(A1) 86(A1)[4]

21.(a)2cos2 x + sin x = 2(1 sin2 x) + sin x= 2 2sin2 x +sin x(A1)(b)2cos2 x + sin x = 2 2 2sin2 x + sin x = 2sin x 2sin2 x = 0sin x(1 2sin x) = 0sin x = 0 or sin x = (M1)sin x = 0 x = 0 or p (0 or 180)(A1)Note: Award (A1) for both answerssin x = x = or (30 or 150)(A1)Note: Award (A1) for both answers[4]

22.(a)5(b)p is a solution if and only if p + p cosp = 0.(M1)Now p + p cosp = p + p(1)(A1)= 0(A1)3(c)By using appropriate calculator functions x = 3.696 722 9...(M1) x = 3.69672 (6s.f.)(A1)2(d)See graph:(A1)(A1)2(e)EITHER = 7.86960 (6 s.f.)(A3)3Note: This answer assumes appropriate use of a calculator e.g.fnInt:OR = p(p 0) + (p sinp 0 sin0) + (cosp cos0)(A1)= p2 + 0 + 2 = 7.86960 (6 s.f.)(A1)3[15]

23.(a)(i)Q = (14.6 8.2)(M1)= 3.2(A1)(ii)P = (14.6 + 8.2)(M0)= 11.4(A1)3(b)10 = 11.4 + 3.2cos(M1)so = costherefore arccos (A1)which gives 2.0236... = t or t = 3.8648. t = 3.86(3 s.f.)(A1)3(c)(i)By symmetry, next time is 12 3.86... = 8.135... t = 8.14 (3 s.f.)(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]

24.(a)

Notes: Award (A1) for end pointsAward (A1) for a maximum of 1.5Award (A1) for a local maximum of 0.5Award (A1) for a minimum of 0.75Award (A1) for the correct shape(b)C(x) = cos x + cos 2xC(x + 2p) = cos(x + 2p) + cos(2x + 4p) = cos x + cos2x = C(x)(M2)Therefore, C(x) is periodic with period 2p.

(c)C(x) is a maximum for x = 2p, 0, 2p(A2)Note: Do not penalise candidates who also write x = p, p(d)x0 = 1.2 (using a graphic display calculator)(A2)(e)(i)C(x) = cos(x) + cos(2x)= cos x + cos2x= C(x) for all x(AG)(ii)C(x0) = 0 so C(x0) = 0 [C(x) = C(x)]C(2p x0) = C(x0) = 0 [C(x) is periodic, period 2p](R1)Therefore, x1 = 2p x0(A1)[16]

25.(a)The smallest angle is opposite the smallest side.cos = (M1)= = 0.7857Therefore, = 38.2(A1)(C2)(b)Area = 8 7 sin38.2(M1) = 17.3 cm2(A1)(C2)[4]

26.(a)3sin2 x + 4cos x = 3(1 cos2 x) + 4cos x = 3 3cos2 + 4 cos x(A1)(C1)(b)3sin2 x + 4 cos x 4 = 0 3 3cos2 x + 4 cos x 4 = 0 3cos2 x 4cos x + 1 = 0(A1) (3cos x 1)(cos x 1) = 0 cos x = or cos x = 1x = 70.5 or x = 0(A1)(A1)(C3)Note: Award (C1) for each correct radian answer, i.e. x = 1.23 or x = 0[4]

27. = 90(A1)AT = = = 60 = (A1)Area = area of triangle area of sector= 6 6 6 (M1)= 12.3 cm2 (or 6p)(A1)(C4)OR = 60(A1)Area of D = 6 12 sin 60(A1)Area of sector = 6 6 (A1)Shaded area = 6p = 12.3 cm2 (3 s.f.)(A1)(C4)[4]

28.(a)(i)AP = (M1) (AG)(ii)OP = (A1)2(b)(M1)= (M1)= (M1)(AG)3(c)For x = 8, = 0.780869(M1)arccos 0.780869 = 38.7 (3 s.f.)(A1)OR(M1) = arctan (0.8) = 38.7 (3 s.f.)(A1)2

(d) = 60 = 0.50.5 = (M1)2x2 16x + 80 = 0(M1)x = 5.63(G2)4(e)(i)f(x) = 1 when = 1(R1)hence, when = 0.(R1)This occurs when the points O, A, P are collinear.(R1)(ii)The line (OA) has equation y = (M1)When y = 10, x = (= 13)(A1)ORx = (= 13)(G2)5Note: Award (G1) for 13.3.[16]

29.2sin x = tan x 2sin x cos x sin x = 0 sin x(2cos x 1) = 0(M1) sin x = 0, cos x = sin x = 0, x = or 1.05 (3 s. f.)(A1)(A1)(C3)ORx = 0, x = (or 1.05 (3 s. f.))(G1)(G1)(G1)(C3)Note: Award (G2) for x = 0, 60.[3]

30.(a)Area = (152)(2)(M1)= 225 (cm2)(A1)(C2)(b)Area OAB = 152 sin 2 = 102.3(A1)Area = 225 102.3 = 122.7 (cm2) = 123 (3 s.f.)(A1)(C2)[4]

31.(a)(M1) = 0.901 > 90 = 180 64.3 = 115.7 = 116 (3 s.f.)(A1)(C2)(b)In Triangle 1, = 64.3 = 180 (64.3 + 50)= 65.7(A1)Area = (20)(17)sin 65.7 = 155 (cm2) (3 s.f.)(A1)(C2)[4]

32.METHOD 1The value of cosine varies between 1 and +1. Therefore:t = 0 a + b = 14.3t = 6 a b = 10.3 2a = 24.6 a = 12.3(A1)(C1) 2b = 4.0 b = 2(A1)(C1)Period = 12 hours = 2(M1) k = 12(A1)(C2)METHOD 2From consideration of graph:Midpoint = a = 12.3(A1)(C1)Amplitude = b = 2(A1)(C1)Period = = 12(M1) k = 12(A1)(C2)[4]

33.METHOD 1(M1)a b=(A1)= = = 2sinq.(A1)(C3)METHOD 2In DOAM, AM = OAsinq.(M1)(A1)Therefore, a b= 2sinq..(A1)(C3)[3]

34.(a)l = rq or ACB = 2 OA(M1) = 30cm(A1)(C2)(b)(obtuse) = 2p 2(A1)Area = q r 2 = (2p 2) (15)2(M1)(A1) = 482 cm2 (3 s.f.)(A1)(C4)[6]

35.

(M1)(A2)OR2.5 20 = 50(M1)(A1)2.5 32 = 80(A1)d2 = 502 + 802 2 50 80 cos70(M1)(A1)d =78.5 km(A1)(C6)[6]

36.(a)(i)1(A1)(C1)(ii)4p (accept 720)(A2)(C2)

(b)

(G1)number of solutions: 4(A2)(C3)[6]

37.Statement(a) Is the statement true for all real numbers x? (YesfNo)(b) If not true, example

ANox = l (log100.1 = 1)(a) (A3) (C3)BNox = 0 (cos0 = 1)(b) (A3) (C3)CYesN/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 (e.g. arctan 1 = ).[6]

38.(a)(M1)sin A = 6 (A1) = (AG)2

(b)

(i) + = 180(A1)(ii)sin A = => A = 59.0 or 121 (3 s.f.)(A1)(A1)=> = 180 (121 + 45) = 14.0 (3 s.f.)(A1)(iii)(M1)=>BD = 1.69(A1)6(c)(M1)(A1) = (AG)2[10]

39.METHOD 1tanq + = 3 tan2q 3 tanq + 1 = 0(M1)tanq = (A1) = 0.382, 2.618(A1)(A1) q = 20.9, 69.1(A1)(A1)(C6)METHOD 2= 3 = 3(M1)(A1) = (A1) sin 2q = (A1) q = 20.9, 69.1(A1)(A1)(C6)[6]

40.(a)f(q ) = Rcosq cosa + Rsinq sina(M1)Rcosa = 4, Rsina = 3(M1)R = 5, a = arctan = 0.644(A1)(A1)f(q ) = 5cos(q 0.644)(C4)(b)f(q ) is maximum when q = a(M1) i.e. q = 0.644 radians(A1)(C2)[6]

41.Using sine rule: (M1)(A1) sin B = 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).[6]

42.(a)x is an acute angle => cosx is positive.(M1)cos2 x + sin2 x = 1 => cosx = (M1)=> cos x = (A1)= (= )(A1)(C4)(b)cos2x = 1 2sin2x = 1 2 (M1)= (A1)(C2)Notes: (a) Award (M1)(M0)(A1)(A0) for. cos = 0.943(b) Award (M1)(A0) for.cos = 0.778.

43.(a)2sin2x = 2(1 cos2x) = 2 2cos2x = l + cosx(M1)=> 2cos2x + cosx l = 0(A1)(C2)Note: Award the first (M1) for replacing sin2 x by 1 cos2 x.(b)2 cos2 x + cosx 1 = (2cos x 1) (cos x +1)(A1)(C1)(c)cos x = 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 (i.e.,p ,, or 1.05, 3.14, 5.24)[6]

44.METHOD 1

=> sin C = 0.4560(M1)(A1)(From diagram) smallest triangle when is obtuse,i.e. = 152.9 => = 7.13 (or 7.1)(A1)(A1)Area DABC = (8)(6)(sin7.13) (or sin7.1) (M1)Area DABC = 2.98(cm2) (accept 2.97)(A1)(C6)METHOD 2

Let AC = xBy the cosine rule 62 = 82 + x2 (2)(8)(x)cos 20(M1)(A1)=> 0 = x2 15.035x + 28 x = (A1) = 2.178(A1)Area = AB ACsin(20) = (8)(2.178)sin20 (M1)= 2.98(cm2)(A1)(C6)[6]

45.3 = p + q cos 0(M1)3 = p + q(A1)1 = p + q cos p(M1)1 = p q(A1)(a)p = 1(A1)(C3)(b)q = 2(A1)(C3)[6]

46.Method 1

0(C2)1.80 [3 s.f.](G2)(C2)2.51 [3 s.f.](G2)(C2)Method 23x = 0.5x + 2p (etc.)(M1)= > 3.5x = 0, 2p, 4p or 2.5x = 0, 2p, 4p(A1)7x = 0, 4p, (8p) or 5x = 0, 4p, (8p)(A1) x = 0, or x = 0, (A1)(A1)(A1) x = 0, , (C2)(C2)(C2)[6]

47.(a)area of sector DC = (2)2 = (A1)area of segment BDCP = area of DABC(M1) = 2(A1)(C3)

(b)BP = (A1)area of semicircle of radius BP = ()2 = (A1)area of shaded region = ( 2) = 2(A1)(C3)[6]

48.(a) = q p = (A1)(A1) = (A1)3(b)cos (A1)= , = (A1)(A1) = 21 + 6 = 15(A1)cos (AG)4(c)(i)Since + = 180(R1)cos = cos (AG)

(ii)sin = (M1) = (A1) = (AG)ORcos q =

(M1)therefore x2 = 754 225 = 529 = > x = 23(A1)= > sin q = (AG)Note: Award (A1)(A0) for the following solution.cos q = = > q = 56.89= > sin q = 0.8376 = 0.8376 = > sin q = (iii)Area of OPQR = 2 (area of triangle PQR)(M1) = 2 (A1) = 2 (A1) = 23 sq units.(A1)ORArea of OPQR = 2 (area of triangle OPQ)(M1) = 2 (A1)(A1) = 23 sq units.(A1)7Notes: Other valid methods can be used.Award final (A1) for the integer answer.[14]

49.(a)Sine rule (M1)(A1) PR = = 5.96 km(A1)3(b)EITHERSine rule to find PQ PQ = (M1)(A1) = 4.39 km(A1)ORCosine rule: PQ2 = 5.962 + 92 (2)(5.96)(9) cos 25(M1)(A1) = 19.29 PQ = 4.39km(A1)Time for Tom = (A1)Time for Alan = (A1) Then = (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(M1)= > sin sin 35(A1) = 16.7(A1)Therefore, = 180 (35 + 16.7)= 128.3(A1) (M1) QS = = 3.29(A1)6[16]

50.(a)(i)cos , sin (A1)therefore cos = 0(AG)(ii)cos x + sin x = 0 = > 1 + tan x = 0= > tan x = l(M1)x = (A1)Note: Award (A0) for 2.36.ORx = (G2)3(b)y = ex(cos x + sin x) = ex(cos x + sin x) + ex(sin x + cos x)(M1)(A1)(A1)3 = 2ex cos x(c) = 0 for a turning point = > 2excos x = 0(M1)= > cos x = 0(A1)= > x = = > a = (A1)y = e(cos + sin ) = eb = e(A1)4Note: Award (M1)(A1)(A0)(A0) for a = 1.57, b = 4.81.(d)At D, = 0(M1)2excos x 2exsin x = 0(A1)2ex(cos x sin x) = 0 = > cos x sin x = 0(A1) = > x = (A1) = > y = e(cos + sin )(A1) = e(AG)5

(e)Required area = (cos x + sin x) dx(M1) = 7.46 sq units(G1)ORrea = 7.46 sq units(G2)2Note: Award (M1)(G0) for the answer 9.81 obtained if the calculator is in degree mode.[17]

51.METHOD 12cos2 q l = l cos2q(or l 2sin2q = sin2q )(M1)(A1)= > 3cos2 q = 2(or 3sin2 q = 1)(M1) cos q = (or sin q = )(A1) q = 0.615, 2.53 (accept 0.196, 0.804)(A1)(A1)(C3)(C3)METHOD 2cos 2q = (1 cos 2q )(M1)(A1)cos 2 = (M1)(A1) q = 0.615, 2.53 (accept 0.196, 0.804)(A1)(A1)(C3)(C3)Notes: Do not penalize if the candidate has included extra solutions.Penalize [1 mark] if candidates give answers in degrees, i.e. award (A1) for 35.3, 145; (A0) for one correct answer in degrees.

52.sin C = (M1)(A1) = 56.4 or 123.6(A1)(A1) = 93.6 or 26.4(A1)(A1)(C6)Note: Award (C1) for one correct answer with no working.[6]

53.(a)cosx + sinx = R cosa cosx + R sina sinx(M1)= > R cosa = 1, R sina = = > R = 2, a = (A1)(A1)3Note: Award (M1)(A1)(A0) if degrees used instead of radians.

(b)(i)Since f(x) = 2cos ,fmax = 2 ; fmin = 1 (when x = 0)(A1)(A1)Range is [1, 2](A1)(ii)Inverse does not exist because f is not 1:1(R2)5Note: Award (R2) for a correct answer with a valid reason.Award (R1) for a correct answer with an attempt at a valid reason, e.g. horizontal line test.Award (R0) for just saying inverse does not exist, without any reason.(c)f(x) = = > cos = (M1) = (A1) x = (A1)ORf(x) =

(M1)x = 0.262(G1)= > x = (A1)3

(d)I = (M1) = (A1) = (A1)(A1) = = ln(3 + 2).(M1)(AG)5Note: Award zero marks for any work using GDC.[16]

54.(a)(i)A is (A1)(A1)(C2)(ii)B is (0, 4)(A1)(A1)(C2)Notes: In each of parts (i) and (ii), award C1 if A and B are interchanged, C1 if intercepts given instead of coordinates.(b)Area = 4 (M1) = (2.67)(A1)(C2)[6]

55.(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)(3sinx 2)(sinx 3) = 0sin x = 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.i.e. 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]

56.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(AG)2Note: The second (A1) is for transforming the initial expression to any simplified expression from which the given result can be clearly seen.(b) = 37.5(A1)BP = 12 tan 37.5(M1)= 9.21 cm(A1)OR = 105 = 37.5(A1)(M1)BP = = 9.21(cm)(A1)3(c)(i)Area OBP = (M1)= 55.3(cm2) (accept 55.2 cm2)(A1)(ii)Area ABP = (9.21)2 sin105(M1)= 41.0(cm2) (accept 40.9 cm2)(A1)4(d)Area of sector = (M1) = 94.2 (cm2) (accept 30 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]

57.Note: Do not penalize missing units in this question.(a)(i)At release(P), t = 0(M1)s = 48 + 10cos0= 58 cm below ceiling(A1)(ii)58 = 48 +10 cos 2t(M1)cos 2t = 1(A1)t = 1sec(A1)ORt = 1sec(G3)5

(b)(i) = 20 sin 2t(A1)(A1)Note: Award (A1) for 20, and (A1) for sin 2pt.(ii)v = = 20 sin 2t = 0(M1)sin 2 t = 0t = 0, ... (at least 2 values)(A1)s = 48 + 10 cos 0 or s = 48 +10 cos (M1) = 58 cm (at P)= 38 cm (20 cm above P)(A1)(A1)7Note: Accept these answers without working for full marks. May be deduced from recognizing that amplitude is 10.(c)48 +10 cos 2t = 60 + 15cos 4t(M1) t = 0.162 secs(A1)ORt = 0.162 secs(G2)2(d)12 times(G2)2Notes: 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]

58.(a)cos sin = r cos( + )where r = = 2(A1)and = arctan (or 30)(M1)(A1)3(b)Since cos sin = 2cos(M1)range will be [2, 2].(A1)2

(c)cos sin = 1 2cos = 1 cos(M1) q + (A1)(A1) = (A1)(A1)5Note: Answers must be multiples of .[10]

59.(M1)(M1)(A1)(A1)

(A1) tan(AG)[5]

60.Area of a triangle = 3 4sin A(A1) 3 4sin 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]

61.METHOD 12cos2 x = 2sin x cos x(M1)2cos2 x 2sin x cos x = 02 cos x(cos x sin x) = 0(M1)cos x = 0, (cos x sin x) = 0(A1)(A1)x = , x = (A1)(A1)(C6)METHOD 2Graphical solutionsEITHERfor both graphs y = 2cos2 x, y = sin 2 x,(M2)ORfor the graph of y = 2cos2 x sin 2 x.(M2)THENPoints representing the solutions clearly indicated(A1)1.57, 0.785(A1)x = , x = (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]

62.(a)d = dt(M1)(A1)(A1)(C3)Note: Award (M1) for , (A1) for both limits, (A1) for 4t + 5 5et(b)d = (A1)(A1)Note: Award (A1) for 2t2 + 5t, (A1) for 5et.= (32 + 20 + 5e4) (5)= 47 + 5e4 (47.1, 3sf )(A1)(C3)[6]

63.(a)(i)10 + 4sin1 = 13.4(A1)(ii)At 2100, t = 21(A1)10 + 4sin 10.5 = 6.48(A1)3Note: 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 + 4sin sin = 1(M1) t = (3.14) (correct answer only)(A1)3

(c)(i)4(A1)(ii)10 + 4sin = 7(M1) sin = 0.75(A1) t = 7.98(A1)(iii)depth < 7 from 8 11 = 3 hours(M1)from 2030 2330 = 3 hours(M1)therefore, total = 6 hours(A1)7[13]

64.(a)cos(A + B) = cos AcosB sin AsinB,cos(A B) = cos AcosB + sin AsinB(M1)(A1)Hence cos(A + B) + cos(A B) = 2cos AcosB(AG)2(b)(i)T1(x) = cos(arccosx)(M1)= x(A1)(ii)T2(x) = cos(2arccosx)(A1)= 2 cos (arccosx) 1(A2)= 2x2 1(AG)5(c)(i)Tn+1(x) + Tn1(x) = cos[(n + 1)arccosx] + cos[(n 1)arccosx](A1)Using part (a) with A = narccosx, B = arccosx(M1)Tn+1(x) + Tn1(x) = 2cos(narccosx)cos(arccosx)(A1)= 2xcos(arccosx)(A1)= 2xTn(x)(AG)(ii)Let Pn be the statement: Tn(x) is a polynomial of degree, n +T1(x) = x, a polynomial of degree one.(A1)So P1 is true.T2(x) = 2x2 1, is a polynomial of degree two.(A1)So P2 is true.Assume that Pk is true.(M1)From part (c)(i), Tk+1(x) = 2xTk(x) Tk1 (x)(M1)Assume Pk1 is true as well.(M1)Tk(x) has degree k 2xTk (x) has degree (k + 1)(A1)and as Tk+1(x) has degree (k 1) Tk+1(x) has degree (k + 1)(A1)By the principle of mathematical induction, Pn is true forall positive integers n.(R1)12Notes: These arguments may be in a different order.There is a maximum of 6 marks in part (ii) for candidates who do not consider a two stage process.[19]

65.(a)Angle (A1)(M1) cm(A1)(C3)(b)Area (M1)(A1) 7.07 (accept 7.06) (A1)(C3)Note: Penalize once in this question for absence of units.[6]

66.METHOD 1Area sector OAB (M1)(A1)ON(A1)AN(A1)Area of (A1)Shaded area(A1)(C6)

METHOD 2

Area sector (M1)(A1)Area (M1)(A1)Twice the shaded area (M1)Shaded area (A1)(C6)[6]

67.(a)(i)(A1)(A1)(ii)EITHER(M1)OR,for (M1)THEN(A1)(A1)(A1)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]

68.Area sector OAB (M1)(A1)Area of DOAB = (M1)(A1) Shaded area = area of sector OAB area of DOAB(M1)= 20.6 (cm)(A1)(C6)[6]

69.(a)p = 30A22(b)METHOD 1Period = (M2) = (A1) q = 4A14METHOD 2Horizontal stretch of scale factor = (M2)scale factor = (A1) q = 4A14[6]

70.(a)using the cosine rule (A2) = b2 + c2 2bccos(M1)substituting correctly BC2 = 652 +1042 2 (65) (104) cos 60 A1= 4225 + 10816 6760 = 8281 BC = 91 mA13(b)finding the area, using bc sin(M1)substituting correctly, area = (65) (104) sin 60A1= 1690 (Accept p = 1690)A13(c)(i)A1 = (65) (x) sin 30A1= AG1(ii)A2 = (104)(x)sin30M1= 26xA12

(iii)starting A1 + A2 = A or substituting + 26x = 1690(M1)simplifying = 1690A1x = A1 x = 40 (Accept q = 40)A14(d)(i)Recognizing that supplementary angles have equal sines e.g. = 180 sin = sin R1(ii)using sin rule in ADB and ACD(M1)substituting correctly A1and M1since sin = sin A1 AG5[18]

71.2tan2 5sec 10 = 0Using l + tan2 = sec2 , 2(sec2 l) 5sec 10 = 0(M1)2sec2 5sec 12 = 0A1Solving the equation eg (2sec + 3) (sec 4) = 0 (M1)sec = or sec = 4A1 in second quadrant sec is negative(R1)= sec = A1[6]

72.

(a)Using the cosine rule (a2 = b2 + c2 2bc cosA)(M1)Substituting correctlyBC2 = 652 + 1042 2(65)(104)cos60A1= 4225 + 10816 6760 = 8281 BC = 91 mA13(b)Finding the area using = bc sin A(M1)Substituting correctly, area = (65)(104)sin 60A1= 1690 (Accept p = 1690)A13(c)(i)Smaller area A1 = (65)(x) sin30(M1)A1= AGLarger area A2 = (104)(x)sin30M1 = 26xA1(ii)Using A1 + A2 = A(M1)Substituting + 26x = 1690A1Simplifying = 1690A1Solving x = 40 (Accept q = 40)A18

(d)using sin rule in ADB and ACD(M1)Substituting correctly A1and A1Since + = 180R1It follows that sin = sinR1A1AG6[20]

73.(a)z = z = e2i ei(M1)z = eiA12(b)|z| = (A2)|z| < 1AG2(c)Using S = (M1)S = A12(d)(i)S = (M1)(A1)Also S = ei +e2i + e2i + ..........(M1)S = A1

(ii)Taking real parts,cos + cos2 + cos3 +... = ReA1= ReM1= A1= A1= A1= A1AG10[16]

74.(a)(M1)(A1)(A1)(A1)(C4)(b)Arc length (M1)Arc length = 9 cm(A1)(C2)Note:Penalize a total of (1 mark) for missing units.[6]

75.(a)when (may be implied by a sketch)(A1)(A1)(C2)

(b)METHOD 1Sketch of appropriate graph(s)(M1)Indicating correct points(A1)(A1)(A1)(C2)(C2)METHOD 2

, (A1)(A1), , (A1)(A1)(C2)(C2)[6]

76.(a)for using cosine rule (M1)(A1)(A1)3Notes: 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 = 17sin29).(i)

for using sine rule (may be implied)(M1)(A1)

(A1)(ii)(A1) (Accept)(A1)5(c)from previous triangle Therefore alternative(A1)

(M1)(A1)(A1)4(d)

Minimum length for BC when or diagramshowing right triangle(M1)

(A1)(N1)2[14]

77.(a)(i)(A1)(A1)Note: Award (A1)(A1) for only if work shown, using product rule on .(ii)or(A1)(iii)

(A1)(A1)(A1)6(b)(A1)1(c)(i)EITHERcurve crosses axis when (may be implied)(A1)(M1)(A1)ORArea =(M1)(A2)(ii)Area (M1)(A1)5[12]

78.(M1)(A1)(A1)(A1)(M1)(A1)(C6)[6]

79.(a)(M1)(A1)(A1)(C3)

(b)Area (M1)(A1)(A1)(C3)[6]

80.

(a)(A1)(A1)Using (M1)(A1)(AG)4(b) (for max)(M1)(M1)(AG)(A1)(M1)

there is a maximum (when )(R1)5

(c)In triangle AOB: (M1)(A1)Perimeter OABC (M1)When , cm(A1)Area OABC (M1) (A1)6[15]