1. METHOD 1z = (2 i)(z + 2) M1 = 2z + 4 iz 2iz(1 i) = 4 + 2iz = i 1i 2 4+ A1z = i 1i 1i 1i 2 4+++ M1= 3 i A1 METHOD 2let z = a + ib2 ii+ ++b ab a = 2 i M1a + ib = (2 i)((a + 2) + ib)a + ib = 2(a + 2) + 2bi i(a + 2) + ba + ib = 2a + b + 4 + (2b a 2)iattempt to equate real and imaginary parts M1a = 2a + b + 4( a + b + 4 = 0)and b = 2b a 2( a + b 2 = 0) A1Note: Award A1 for two correct equations.b = 1;a = 3z = 3 i A1[4] 2. (a)AB = b a A1CB = a + b A1 (b)CB AB = (b a)(b + a) M1= b2a2A1= 0 since b=a R1Note: Only award the A1 and R1 if working indicates that theyunderstand that they are working with vectors.so AB is perpendicular to CB i.e. C BA is a right angle AG[5]IB Questionbank Mathematics Higher Level 3rd edition 13. (a)

,_

,_

134AC ,314ABA1A1Note: Accept row vectors. (b)

,_

161681 3 43 1 4 AC ABk j iM1A1normal n =

,_

221 so r

,_

,_

,_

221121221(M1)x + 2y + 2z = 7 A1Note: If attempt to solve by a system of equations:Award A1 for 3 correct equations, A1 for eliminating a variableand A2 for the correct answer. (c) r =

,_

+

,_

221735 (or equivalent) A11(5 + ) + 2(3 + 2) + 2(7 + 2) = 7 M19 = 18 = 2 A1Note: =

,_

16168if41 is used.distance = 2 2 22 2 1 2 + +(M1)= 6 A1 (d) (i) area = 2 2 216 16 821AC AB21+ + (M1)= 12 (accept 57621) A1IB Questionbank Mathematics Higher Level 3rd edition 2(ii) EITHERvolume = 31 area height (M1) = 31 12 6 = 24 A1ORvolume = ( ) ) AC AB ( AD61 M1= 24 A1 (e)2 2 216 16 8 AC AB + + 6 1 41 3 4 AD AC k j iM1= 19i 20j + 16k A1EITHER2 2 2 2 2 216 16 82116 20 1921+ + > + +M1therefore since area of ACD bigger than area ABC implies thatB is closer to opposite face than D R1ORcorrect calculation of second distance as 2 2 216 20 19144+ +A1which is smaller than 6 R1Note: Only award final R1 in each case if the calculations are correct.[19] 4. (a)CB = b c, AC = b + c A1A1Note: Condone absence of vector notation in (a).IB Questionbank Mathematics Higher Level 3rd edition 3(b)CB AC = (b + c) (b c) M1= b2 c2A1= 0 since b=c R1Note: Only award the A1 and R1 if working indicates that they understandthat they are working with vectors.so AC is perpendicular to CB i.e. B CA is a right angle AG[5] 5. METHOD 1equation of journey of ship S1r1 =

,_

2010tequation of journey of speedboat S2 ,setting off k minutes laterr2 =

,_

+

,_

3060) (3070k tM1A1A1Note: Award M1 for perpendicular direction, A1 for speed, A1 forchange in parameter (e.g. by using t k or T, k being the timedifference between the departure of the ships).solve

,_

+

,_

,_

3060) (30702010k t t(M1)Note: M mark is for equating their two expressions.10t = 70 60t + 60k20t = 30 + 30t 30k M1Note: M mark is for obtaining two equations involving two different parameters.7t 6k = 7t + 3k = 3k = 1528A1latest time is 11:52 A1IB Questionbank Mathematics Higher Level 3rd edition 4METHOD 2xyt i m e t a k e nt - kt i m et a k e n tS ( 2 6 , 5 2 )2 2 52 6 51 0 5 8( A )OB ( 7 0 , 3 0 )SB = 225M1A1(by perpendicular distance)SA = 265M1A1(by Pythagoras or coordinates)t = 5 105 26A1t k = 5 305 22A1k = 1528 leading to latest time 11:52 A1[7] 6. (a) x3 + 1 = 113+ x(1.26, 1) (= (32 , 1)) A1 (b) f(1.259...) = 4.762... (3 322) A1g(1.259...) = 4.762... (3 322) A1required angle = 2arctan

,_

... 762 . 41M1= 0.414 (accept 23.7) A1Note: Accept alternative methods including finding the obtuse angle first.[5]IB Questionbank Mathematics Higher Level 3rd edition 57. (a)

,_

,_

zyx150SR ,311PQ(M1)point S = (1, 6, 2) A1 (b)

,_

,_

142PS311PQA1

,_

2713PS PQm = 2 A1 (c) area of parallelogram PQRS = 2 2 2) 2 ( 7 ) 13 ( PS PQ + + M1= 222 = 14.9 A1 (d) equation of plane is 13x + 7y 2z = d M1A1substituting any of the points given gives d = 3313x + 7y 2z = 33 A1 (e) equation of line is r =

,_

+

,_

2713000A1Note: To get the A1 must have r = or equivalent. (f) 169 + 49 + 4 = 33 M1 = 22233 (= 0.149...) A1closest point is

,_

3711,7477,74143 (= (1.93, 1.04, 0.297)) A1 (g) angle between planes is the same as the angle between the normals (R1)cos = 6 2221 2 2 7 1 13 + M1A1 = 143 (accept = 37.4 or 2.49 radians or 0.652 radians) A1[17]IB Questionbank Mathematics Higher Level 3rd edition 68. (a) for using normal vectors (M1)

,_

,_

101121 = 1 1 = 0 M1A1hence the two planes are perpendicular AG (b) METHOD 1EITHER1 0 11 2 1 k j i= 2i 2j 2k M1A1ORif

,_

cba is normal to 3, thena + 2b c = 0 and a + c = 0 M1a solution is a = 1, b = 1, c = 1 A1THEN3 has equation x y z = d (M1)as it goes through the origin, d = 0 so 3 has equation x y z = 0 A1Note: The final (M1)A1 are independent of previous working. METHOD 2r =

,_

+

,_

+

,_

101121000t sA1(A1)A1A1[7]IB Questionbank Mathematics Higher Level 3rd edition 79. (a + b)(a b) = a a + b a a b b b M1= a a b b A1= |a|2 |b|2 = 0 since |a| = |b| A1the diagonals are perpendicular R1Note: Accept geometric proof, awarding M1 for recognizing OACB is arhombus, R1 for a clear indication that (a + b) and (a b) are thediagonals, A1 for stating that diagonals cross at right angles andA1 for hence dot product is zero.Accept solutions using components in 2 or 3 dimensions.[4] 10. (a) 2y + 8x = 4 M13x + 2y = 7 A12x + 6 2x = 6Note: Award M1 for attempt at components, A1 for two correct equations.No penalty for not checking the third equation.solving : x = 1, y = 2 A1 (b) a + 2b=

,_

+

,_

2242234=

,_

6742 2 26 ) 7 ( 4 2 + + + b a(M1)= 101A1[5] 11. (a) (i) use of a b = abcos (M1)a b = 1 (A1)a = 7, b = 5 (A1)cos = 351A1 (ii) the required cross product is4 3 02 3 6k j i = 18i 24j 18k M1A1 (iii) using r n = p n the equation of the plane is (M1)IB Questionbank Mathematics Higher Level 3rd edition 818x 24y 18z = 12 (3x 4y 3z = 2) A1 (iv) recognizing that z = 0 (M1)x-intercept = 32, y-intercept = 21(A1)area = 61212132 ,_

,_

,_

A1 (b) (i) p p = ppcos 0 M1A1= p2AG (ii) consider the LHS, and use of result from part (i)p + q2 = (p + q)(p + q) M1= p p + p q + q p + q q (A1)= p p + 2p q + q q A1= p2 + 2p q + q2AGIB Questionbank Mathematics Higher Level 3rd edition 9(iii) EITHERuse of p q pq M1so 0 |p + q|2 = p2 + 2p q + q2 p2 + 2pq+q2A1take square root (of these positive quantities) to establish A1p + qp+q AGORM1M1Note: Award M1 for correct diagram and M1 for correct labellingof vectors including arrows.since the sum of any two sides of a triangle is greater than the third side,p + q > p + q A1when p and q are collinear p + q = p + q p + q p + q AG[19]IB Questionbank Mathematics Higher Level 3rd edition 1012. (a) EITHERnormal to plane given by2 3 62 3 2k j iM1A1= 12i + 8j 24k A1equation of is 3x + 2y 6z = d (M1)as goes through (2, 3, 2) so d = 12 M1A1 :3x + 2y 6z = 12 AGORx = 2 + 2 + 6y = 3 + 3 3z = 2 + 2 + 2eliminating x + 2y = 4 + 82y + 3z = 12 M1A1A1eliminating 3(x + 2y) 2(2y + 3z) = 12 M1A1A1 : 3x + 2y 6z = 12 AG (b) therefore A(4, 0, 0), B(0, 6, 0) and C(0, 0, 2) A1A1A1Note: Award A1A1A0 if position vectors given instead of coordinates. (c) area of base OAB = 6 421 = 12 M1V = 2 1231 = 8 M1A1 (d)

,_

,_

001623= 3 = 7 1 cos M1A173arccos so = 90 arccos73 = 25.4 (accept 0.443 radians) M1A1 (e) d = 4 sin = 712 (= 1.71) (M1)A1IB Questionbank Mathematics Higher Level 3rd edition 11(f) 8 = area71231 area = 14 M1A1Note: If answer to part (f) is found in an earlier part, award M1A1,regardless of the fact that it has not come from their answersto part (c) and part (e).[20] 13. (a) use GDC or manual method to find a, b and c (M1)obtain a = 2, b = 1, c = 3 (in any identifiable form) A1 (b) use GDC or manual method to solve second set of equations (M1)obtain x = 27;211 4 tyt , z = t (or equivalent) (A1)r =

,_

+

,_

15 . 35 . 5002t (accept equivalent vector forms) M1A1Note: Final A1 requires r = or equivalent.[6] 14. (a) a =

,_

,_

k12plane tothe124e is parallel to the line (A1)(A1)Note: Award A1 for each correct vector written down, even if not identified.line plane e parallel to asince 2112124

,_

,_

kkt(M1)A1IB Questionbank Mathematics Higher Level 3rd edition 12(b) 4(3 2) 2

,_

+ 211 = 1 (M1)(A1)Note: FT their value of k as far as possible. = 78A1

,_

73,78,75PA1[8] 15. (a) cos =

,_

+21 3 sin2 21 2 cos sin cos 2 sin b aabM1A1 (b) ab cos = 0 M1sin 2 cos + sin cos 2 1 = 0 = 0.524

,_

6A1 (c) METHOD 11 sin cos1 2 cos 2 sin k j i(M1)assuming = 6 7Note: Allow substitution at any stage.1212312123 k j iA1=

,_

+

,_

+ ,_

2321212323232121k j i= 0 A1a and b are parallel R1Note: Accept decimal equivalents.IB Questionbank Mathematics Higher Level 3rd edition 13METHOD 2from (a) cos = 1 (and sin = 0) M1A1a b = 0 A1a and b are parallel R1[8] 16. (a)

,_

,_

,_

120OP and210ON ,221OMA1A1A1 (b) ,011MN and101MP

,_

,_

A1A1

,_

,_

1110 1 11 0 1 MN MPk j i(M1)A1 (c) (i) area of MNP = MN MP21M1 =

,_

11121 = 23A1 (ii)

,_

,_

220OG ,002OA

,_

222AGA1since ) MN MP ( 2 AG AG is perpendicular to MNP R1IB Questionbank Mathematics Higher Level 3rd edition 14(iii) r

,_

,_

,_

111221111M1A1r

,_

111 = 3 (accept x + y + z = 3) A1 (d) r =

,_

+

,_

222002A1

,_

,_

111222 2 = 3 M1A12 + 2 + 2 + 2 = 3 = 65A1r =

,_

+

,_

22265002M1coordinates of point

,_

35,35,31A1[20]IB Questionbank Mathematics Higher Level 3rd edition 1517. METHOD 1for finding two of the following three vectors (or their negatives)

,_

,_

,_

102BC ,222AC ,120AB(A1)(A1)and calculatingEITHERAC AB

,_

4222 2 21 2 0k j iM1A1area ABC = AC AB21M1ORBC BA

,_

4221 0 21 2 0k j iM1A1area ABC = BC BA21M1ORCB CA

,_

4221 0 22 2 2k j iM1A1area ABC = CB CA21M1THENarea ABC = 224A1= 6AG N0IB Questionbank Mathematics Higher Level 3rd edition 16METHOD 2for finding two of the following three vectors (or their negatives)

,_

,_

,_

102BC ,222AC ,120AB(A1)(A1)EITHERcos A = AC ABAC ABM1=

,_

153or60612 56sin A = 52A1area ABC = A sin AC AB21M1= 5212 521= 2421A1= 6AG N0 ORcos B = BC BABC BAM1= 515 51 sin B =

,_

524or2524A1area ABC = B sin BC BA21M1= 25245 521= 2421A1= 6AG N0IB Questionbank Mathematics Higher Level 3rd edition 17ORcos C = CB CACB CAM1=

,_

153or6065 126sin C = 52A1area ABC = C sin CB CA21M1= 525 1221= 2421A1= 6AG N0 METHOD 3for finding two of the following three vectors (or their negatives)

,_

,_

,_

102BC ,222AC ,120AB(A1)(A1)AB = a b c 5 BC , 3 2 12 AC , 5M1A1s = 5 325 3 2 5+ + +M1area ABC = ) )( )( ( c s b s a s s = ) 3 )( 3 5 )( 3 )( 5 3 ( += ) 3 5 ( 3 A1= 6AG N0IB Questionbank Mathematics Higher Level 3rd edition 18METHOD 4for finding two of the following three vectors (or their negatives)

,_

,_

,_

102BC ,222AC ,120AB(A1)(A1)AB = BC = 5 and AC = 3 2 12 M1A1ABC is isosceleslet M be the midpoint of [AC], the height BM = 2 3 5 M1area ABC = 22 3 2 A1= 6AG N0[6] 18. (a) identifies a direction vector e.g.

,_

,_

112BA or 112ABA1identifies the point (1, 1, 2) A1line l1: 121121 + z y xAG (b) r =

,_

+

,_

,_

+

,_

121321112211 r1 + 2 = 1 + , 1 + = 2 + 2, 2 + = 3 + (M1)equating two of the three equations gives = 1 and = 2 A1A1check in the third equationsatisfies third equation therefore the lines intersect R1therefore coordinates of intersection are (1, 2, 1) A1 (c) d1 = 2i + j + k, d2 = i + 2j + k A1IB Questionbank Mathematics Higher Level 3rd edition 19d1 d2 = 1 2 11 1 2k j i = i j + 3k M1A1Note: Accept scalar multiples of above vectors. (d) equation of plane is x y + 3z = k M1A1contains (1, 2, 3) (or (1, 2, 1) or (1, 1, 2)) k = 1 2 + 3 3 = 6 A1x y + 3z = 6 AG (e) direction vector of the perpendicular line is

,_

311(M1)r =

,_

+

,_

311413mA1Note: Award A0 if r omitted. (f) (i) find point where line meets plane(3 m) (1 m) + 3( 4 + 3m) = 6 M1m = 2 A1point of intersection is (1, 1, 2) A1 (ii) for T, m = 4 (M1)so T = (1, 3, 8) A1 (iii)2 2 28) 4 ( 3) (1 1) (3 T T + + + + (M1)= ) 11 4 ( 176 A1[22]IB Questionbank Mathematics Higher Level 3rd edition 2019. consider a vector parallel to each line,e.g. u =

,_

,_

133and124vA1A1let be the angle between the linescos = 19 211 6 12 + v uv uM1A1= 19 217 = 0.350... (A1)so 0 = 69.5

,_

,_

19 217arccos orrad 1.21 or A1 N4Note: Allow FT from incorrect reasonable vectors.[6] 20. (a) let A =

,_

,_

,_

522and ,4 1 53 1 22 1 1B Xzyx(M1)point of intersection is

,_

41,127,1211 (or (0.917, 0.583, 0.25)) A1 (b) METHOD 1(i) det01 53 1 22 1 1

,_

aM13a + 24 = 0 (A1)a = 8 A1 N1 (ii) consider the augmented matrix

,_

5228 1 53 1 22 1 1M1use row reduction to obtain

,_

,_

1000 0 01 00 1or1220 0 01 3 02 1 13135(or equivalent) A1any valid reason R1(e.g. as the last row is not all zeros, the planes do not meet)N0IB Questionbank Mathematics Higher Level 3rd edition 21METHOD 2use of row reduction (or equivalent manipulation of equations) M1e.g.

,_

,_

52210 6 01 3 02 1 15221 53 1 22 1 1a aA1A1Note: Award an A1 for each correctly reduced row. (i) a 10 = 2 a = 8 M1A1N1 (ii) when a = 8, row 3 2 row 2 R1N0[8] 21. (a)OP = i + 2j k (M1)the coordinates of P are (1, 2, 1) A1 (b) EITHERx = 1 + t, y = 2 2t, z = 3t 1 M1x 1 = t, tzty+31,22A1x 1 = 3122 + z yAGN0OR

,_

+

,_

,_

321121tzyxM1A1x 1 = 3122 + z yAG (c) (i) 2(1 + t) + (2 2t) + (3t 1) = 6 t = 1 M1A1N1 (ii) coordinates are (2, 0, 2) A1Note: Award A0 for position vector. (iii) distance travelled is the distance between the two points (M1)14 ) 1 2 ( ) 2 0 ( ) 1 2 (2 2 2 + + + (= 3.74) (M1)A1IB Questionbank Mathematics Higher Level 3rd edition 22(d) (i) distance from Q to the origin is given byd(t) = 2 2 2 4) 1 ( ) 1 ( t t t + + (or equivalent) M1A1e.g. for labelled sketch of graph of d or d2(M1)(A1)the minimum value is obtained for t = 0.761 A1N3 (ii) the coordinates are (0.579, 0.239, 0.421) A1Note: Accept answers given as a position vector. (e) (i)

,_

,_

,_

314and001,110c b a(M1)A1substituting in the equation a b = k(b c), we have (M1)

,_

,_

,_

,_

,_

,_

,_

313111314001001110k kA131and 1 k k which is impossibleso there is no solution for k R1 (ii)CB and BA are not parallel R2(hence A, B, and C cannot be collinear)Note: Only accept answers that follow from part (i).[23]IB Questionbank Mathematics Higher Level 3rd edition 2322. direction vector for line =

,_

11 or any multiple A1011sin 1sin 2

,_

,_

M12 sin 1 + sin = 0 A1Note: Allow FT on candidates direction vector just for line above only.3 sin = 1sin = 31A1 = 0.340 or 19.5 A1Note: A coordinate geometry method using perpendicular gradients is acceptable.[5] 23. EITHERl goes through the point (1, 3, 6), and the plane contains A(4, 2, 5)the vector containing these two points is on the plane, i.e.

,_

,_

,_

153524631(M1)A11 5 31 2 1153121

,_

,_

k j i = 7i + 4j + k M1A125147524

,_

,_

(M1)hence, Cartesian equation of the plane is 7x + 4y + z = 25 A1IB Questionbank Mathematics Higher Level 3rd edition 24ORfinding a third point M1e.g. (0, 5, 5) A1three points are (1, 3, 6), (4, 2, 5), (0, 5, 5)equation is ax + by + cz = 1system of equationsa + 3b + 6c = 1 M14a 2b + 5c = 15b + 5c = 1a = 251,254,257 c b, from GDC M1A1so 1251254257 + + z y xA1or 7x + 4y + z = 25[6] 24. (a) on l1 A(3 + 3, 4 + 2, 6 2) A1on l2l2 : r =

,_

+

,_

143374(M1) B(4 3, 7 + 4, 3 ) A1BA

,_

+ + + + 9 23 4 27 3 3 b a(M1)A1IB Questionbank Mathematics Higher Level 3rd edition 25EITHERBA0223BA1

,_

l 3(3 + 3 7) + 2(2 4 + 3) 2(2 + + 9) = 0 M117 = 33 A1BA0143BA2

,_

l 3(3 + 3 7) + 4(2 4 + 3) 2(2 + + 9) = 0 M1 26 = 24 A1solving both equations above simultaneously gives = 2; = 1 A(3, 0, 2), B(1, 3, 4) A1A1A1A1OR1 4 32 2 3 k j i = 6i + 9j + 18k M1A1so AB

,_

+ + + +

,_

9 23 4 27 3 3632 pM1A13 + 3 2p = 72 4 3p = 32 + 6p = 9 = 2, = 1, p = 1 A1A1A(3 + 6, 4 + 4, 6 4) = (3, 0, 2) A1B(4 3, 7 + 4, 3 1) = (1, 3, 4) A1 (b) AB =

,_

,_

,_

632203431(A1)|AB| = 49 ) 6 ( ) 3 ( ) 2 (2 2 2 + + = 7 M1A1IB Questionbank Mathematics Higher Level 3rd edition 26(c) from (b) 2i + 3j + 6k is normal to both linesl1 goes through (3, 4, 6)

,_

,_

632643 = 18 M1A1hence, the Cartesian equation of the plane through l1, but not l2,is 2x + 3y + 6z = 18 A1[19] 25. (a) (i) METHOD 1AB = b a =

,_

,_

,_

110211321(A1)AC = c a =

,_

,_

,_

112211103(A1)AB AC = 1 1 21 1 0 k j iM1= i (1 + 1) j(0 2) + k (0 2) (A1)= 2j 2k A1Area of triangle ABC = ( ) 2 8212 221 k j sq. units M1A1Note: Allow FT on final A1.IB Questionbank Mathematics Higher Level 3rd edition 27METHOD 26 AC , 12 BC , 2 AB A1A1A1Using cosine rule, e.g. on CM1cos C = 32 272 22 12 6 +A1C ab sin21ABC Area M1=

,_

32 2arccos sin 6 1221= ( ) 232 2arccos sin 2 3

,_

A1Note: Allow FT on final A1. (ii) AB = 2A12 = h h h , 221AB21 equals the shortest distance (M1)h = 2 A1 (iii) METHOD 1 has form r d

,_

220(M1) Since (1, 1, 2) is on the planed =

,_

2112 4 2221

,_

M1A1Hence r

,_

220= 22y 2z = 2 (or y z = 1) A1IB Questionbank Mathematics Higher Level 3rd edition 28METHOD 2r =

,_

+

,_

+

,_

112110211 (M1)x = 1 + 2 (i)y = 1 + (ii)z = 2 + (iii) A1Note: Award A1 for all three correct, A0 otherwise. From (i) = 21 xsubstitute in (ii) y = 1 +

,_

21 x = y 1 +

,_

21 xsubstitute and in (iii) M1z = 2 + y 1 +

,_

,_

2121 x xy z = 1 A1 (b) (i) The equation of OD isr =

,_

,_

,_

110or ,220 rM1This meets where2 + 2 = 1 (M1) = 41A1Coordinates of D are

,_

21,21, 0A1(ii)2121210 OD2 2 ,_

+ ,_

+ (M1)A1[20] IB Questionbank Mathematics Higher Level 3rd edition 2926. METHOD 1Use of | a b | = | a | | b | sin (M1)| a b |2 = | a |2 | b |2 sin2 (A1)Note: Only one of the first two marks can be implied.= | a |2 | b |2 (1 cos2) A1= | a |2 | b |2 | a |2 | b |2 cos2 (A1)= | a |2 | b |2 (| a | | b | cos)2(A1)Note: Only one of the above two A1 marks can be implied.= | a |2 | b |2 (a b)2A1Hence LHS = RHSAG N0 METHOD 2Use of a b = | a | | b | cos (M1)| a |2 | b |2 (a b)2 = | a |2 | b |2 (| a | | b | cos)2(A1)= | a |2 | b |2 | a |2 | b |2 cos2 (A1)Note: Only one of the above two A1 marks can be implied.= | a |2 | b |2 (1 cos2) A1= | a |2 | b |2 sin2 A1= | a b |2A1 Hence LHS = RHSAG N0Notes: Candidates who independently correctlysimplify both sides and show that LHS = RHS should be awarded full marks. If the candidate starts off with expression that they are trying to prove and concludes that sin2 = (1 cos2) award M1A1A1A1A0A0. If the candidate uses two general 3D vectors and explicitly finds the expressions correctly award full marks. Use of 2D vectors gains a maximum of 2 marks. If two specific vectors are used no marks are gained.[6]IB Questionbank Mathematics Higher Level 3rd edition 3027. (a) Use of cos = AB OAAB OA(M1)AB = i j + k A1AB = 3 and OA = 2 3A1 AB OA = 6 A1substituting gives cos =

,_

3662 or equivalent M1N1 (b) L1: r = OA + sABor equivalent (M1)L1: r = i j + 4k + s(i j + k) or equivalent A1Note: Award (M1)A0 for omitting r = in the final answer. (c) Equating components and forming equations involving s and t (M1)1 + s = 2 + 2t, 1 s = 4 + t, 4 + s = 7 + 3tHaving two of the above three equations A1A1Attempting to solve for s or t (M1)Finding either s = 3 or t = 2 A1Explicitly showing that these values satisfy the third equation R1Point of intersection is (2, 2, 1) A1N1Note: Position vector is not acceptable for final A1. (d) METHOD 1r =

,_

+

,_

+

,_

333312411 (A1)x = 1 + 2 3, y = 1 + + 3 and z = 4 + 3 3 M1A1Elimination of the parameters M1x + y = 3 so 4(x + y) = 12 and y + z = 4 + 3so 3(y + z) = 12 + 93(y + z) = 4(x + y) + 9 A1Cartesian equation of plane is 4x + y 3z = 9 (or equivalent) A1N1IB Questionbank Mathematics Higher Level 3rd edition 31METHOD 2EITHERThe point (2, 4, 7) lies on the plane.The vector joining (2, 4, 7) and (1, 1, 4) and 2i + j + 3kare parallel to the plane. So they are perpendicular to thenormal to the plane.(i j + 4k) (2i + 4j + 7k) = i 5j 3k (A1)3 1 23 5 1 k j inM1= 12i 3j + 9k or equivalent parallel vector A1 ORL1 and L2 intersect at D (2, 2,1)AD = (2i + 2j + k) (i j + 4k) = 3i + 3j 3k (A1)3 3 33 1 2 k j inM1= 12i 3j + 9k or equivalent parallel vector A1 THENr n = (i j + 4k) (12i 3j + 9k) M1= 27 A1Cartesian equation of plane is 4x + y 3z = 9 (or equivalent) A1N1[20]IB Questionbank Mathematics Higher Level 3rd edition 3228. The normal vector to the plane is .231

,_

(A1)EITHER is the angle between the line and the normal to the plane.

,_

,_

,_

6 7321 14321 14231214cos(M1)A1A1= 79.9 (= 1.394 ...) A1The required angle is 10.1 (= 0.176) A1 ORis the angle between the line and the plane.21 14321 14231214sin

,_

,_

(M1)A1A1= 10.1 (= 0.176) A2[6] 29. METHOD 1(from GDC)

,_

0611210 0 0321 0610 1(M1)12161 + xA16132 yA1r =

,_

+ + + ,_

k j i j i326161121A1A1A1N3IB Questionbank Mathematics Higher Level 3rd edition 33METHOD 2(Elimination method either for equations or row reduction of matrix)Eliminating one of the variables M1A1Finding a point on the line (M1)A1Finding the direction of the line M1The vector equation of the line A1N3[6] 30.BC = c bCA = a c a (c b) = 0 M1and b (a c) = 0 M1 a c = a b A1and a b = b c A1 a c = b c M1 b c a c = 0c (b a) = 0 A1OC is perpendicular to AB, as b a. AG[6] 31. a b = abcos (M1)a b =

,_

,_

m23321 = 7 + 3m A1213 14 m + b aA1 + 30 cos 13 14 cos2m b a7 + 3m = 213 14 m + cos 30 A1m = 2.27, m = 25.7 A1A1[6]IB Questionbank Mathematics Higher Level 3rd edition 3432. (a)

,_

5 7 0 54 2 1 33 2 1 kM1R1 2 R2

,_

5 7 0 54 2 1 38 7 0 5 k(A1)R1 + R3

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5 7 0 54 2 1 33 0 0 0 k(A1)Hence no solutions if k , k 3 A1 (b) Two planes meet in a line and the third plane is parallel to that line. A1[5] 33. (a) x = 3 + 2my = 2 mz = 7 + 2m A1x = 1 + 4ny = 4 nz = 2 + n A1 (b) 3 + 2m = 1 + 4n 2m 4n = 2(i)2 m = 4 n m n = 2(ii) M17 + 2m = 2 + n 2m n = 5(iii)(iii) (ii) m = 3 A1 n = 1 A1Substitute in (i), 6 + 4 = 2. Hence lines intersect. R1Point of intersection A is (3, 5, 1) A1IB Questionbank Mathematics Higher Level 3rd edition 35(c)

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2611 1 42 1 2k j iM1A1r

,_

,_

,_

261723261(M1)r

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261 = 29x + 6y + 2z = 29 A1Note: Award M1A0 if answer is not in Cartesian form. (d) x = 8 + 3y = 3 + 8 (M1)z = 2Substitute in equation of plane.8 + 3 18 + 48 + 4 = 29 M155 = 55 = 1 A1Coordinates of B are (5, 5, 2) A1 (e) Coordinates of C are

,_

23, 5 , 4(A1)r =

,_

+

,_

2612354M1A1Note: Award M1A0 unless candidate writes r = or

,_

zyx[18]IB Questionbank Mathematics Higher Level 3rd edition 3634. EITHERLet s be the distance from the origin to a point on the line, thens2 = (1 )2 + (2 3)2 + 4 (M1) = 102 14 + 9 A1 d) d(2s = 20 14 A1For minimum 107, 0d) d(2 sA1 ORThe position vector for the point nearest to the origin is perpendicular tothe direction of the line. At that point:

,_

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03123 21 = 0 (M1)A1Therefore, 10 7 = 0 A1Therefore, = 107A1 THENx = 101,103 y(A1)(A1)The point is

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2 ,101,103. N3[6]IB Questionbank Mathematics Higher Level 3rd edition 3735. (a)M1(M1)M1When a = 1 the augmented matrix isA1Hence the system is inconsistent a 1 R1 (b) When a 1, (a 1)z = 9 a2(a + 1)z = a2 9192+ aazM1A12y z = 0 ) 1 ( 29212+ aaz yM1A1x = 3y + z = ) 1 ( 29) 1 ( 2) 9 ( 2) 1 ( 2) 9 ( 32 2 2++++ aaaaaaM1A1The unique solution is

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+++19,) 1 ( 29,) 1 ( 292 2 2aaaaaa when a 1 (c) 2 a = 1 a = 1 M1

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28,48,48is solutionThe or (2, 2, 4) A1[13] 36. (a)AB = i 3j + k, BC = i + j A1A1IB Questionbank Mathematics Higher Level 3rd edition 38(b)0 1 11 3 1 BC AB k j iM1= i + j + 2k A1 (c) Area of ABC = 21i + j + 2k M1A1= 4 1 121+ += 26A1 (d) A normal to the plane is given by n = BC AB = i + j + 2k (M1)Therefore, the equation of the plane is of the form x + y + 2z = gand since the plane contains A, then 1 + 2 + 2 = g g = 3. M1Hence, an equation of the plane is x + y + 2z = 3. A1 (e) Vector n above is parallel to the required line.Therefore, x = 2 t A1y = 1 + t A1z = 6 + 2t A1 (f) x = 2 ty = 1 + tz = 6 + 2tx + y + 2z = 32 + t 1 + t 12 + 4t = 3 M1A115 + 6t = 36t = 18t = 3 A1Point of intersection (1, 2, 0) A1 (g) Distance = 54 6 3 32 2 2 + +(M1)A1 (h) Unit vector in the direction of n is e = nn 1(M1)= 61(i + j + 2k) A1Note: e is also acceptable.IB Questionbank Mathematics Higher Level 3rd edition 39(i) Point of intersection of L and P is (1, 2, 0).

,_

633DE(M1)A1

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633EFM1 coordinates of F are (4, 5, 6) A1[25] 37. (a) L1 : x = 2 + ; y = 2 + 3; z = 3 + (A1)L2 : x = 2 + ; y = 3 + 4; z = 4 + 2 (A1)At the point of intersection (M1)2 + = 2 + (1)2 + 3 = 3 + 4 (2)3 + = 4 + 2 (3)From (1), = A1Substituting in (2), 2 + 3 = 3 + 4 = = 1 A1We need to show that these values satisfy (3). (M1)They do because LHS = RHS = 2; therefore the lines intersect. R1So P is (1, 1, 2). A1N3 (b) The normal to is normal to both lines. It is therefore given bythe vector product of the two direction vectors.Therefore, normal vector is given by

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2 4 11 3 1k j iM1A1= 2i j + k A2The Cartesian equation of is 2x y + z = 2 + 1 + 2 (M1)i.e. 2x y + z = 5 A1N2IB Questionbank Mathematics Higher Level 3rd edition 40(c) The midpoint M of [PQ] is

,_

25,23, 2. M1A1The direction of MS is the same as the normal to , i.e. 2i j + k (R1)The coordinates of a general point R on MS are therefore

,_

+ + 25,23, 2 2(M1)It follows that PR = (1 + 2)i + k j ,_

+ + ,_

2125A1A1A1At S, length of PR is 3, i.e. (M1)(1 + 2)2 + 921252 2 ,_

+ + ,_

A11 + 4 + 42 + 425 5 + 2 + 41 + + 2 = 9 (A1)62 = 46A1 = 21tA1Substituting these values, (M1)the possible positions of S are (3, 1, 3) and (1, 2, 2) A1A1N2[29] 38. (a) Finding correct vectors

,_

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113AC134ABA1A1Substituting correctly in scalar product AC AB = 4(3) + 3(1) 1(1) A1 = 10AG N0 (b)11 AC 26 AB (A1)(A1)Attempting to use scalar product formula, 11 2610C AB cosM1= 0.591 (to 3 s.f.) A1C AB = 126 A1N3[8] 39. (a) A2 =

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+ 1 22 2a aa(M1)A1IB Questionbank Mathematics Higher Level 3rd edition 41 (b) METHOD 1det A2 = 4a2 + 2a 2a = 4a2M1a = 2 A1A1 N2METHOD 2det A = 2a M1det A = 4a = 2 A1A1 N2[5] 40. (a)

,_

,_

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k zyx132 1 23 1 11 2 02 1 23 1 11 2 0 = 0 2( 2 + 6) + ( 1 + 2) = 7 M1A1since determinant 0 unique solution to the system planes R1intersect in a point AGNote: For any method, including row reduction, leading to the explicitsolution

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+ 72 1,710,75 6 k k k, award M1 for an attempt ata correct method, A1 for two correct coordinates and A1 fora third correct coordinate. (b)2 1 23 1 11 2+ + aaa= a((a + 1)(a + 2) 3) 2(1(a + 2) + 6) + (1 + 2(a + 1)) M1(A1)planes not meeting in a point no unique solution i.e. determinant = 0 (M1)a(a2 + 3a 1) + (2a 8) + (2a + 1) = 0a3 + 3a2 + 3a 7 = 0 A1a = 1 A1[5]IB Questionbank Mathematics Higher Level 3rd edition 42(c)2 13 1 24 4 4 03 1 2 1r rk+

,_

M13 126 5 5 04 4 4 03 1 2 1r rk+

,_

+(A1)2 35 44 4 0 0 04 4 4 03 1 2 1r rk

,_

+(A1)for an infinite number of solutions to exist, 4 + 4k = 0 k = 1 A1x + 2y + z = 3y + z = 1 M1

,_

+

,_

,_

111011zyxA1Note: Accept methods involving elimination.Note: Accept any equivalent form e.g.

,_

+

,_

,_

,_

+

,_

,_

111120or 111102 zyxzyx.Award A0 if

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zyx or r = is absent.[14] 41. (a) det A = cos 2 cos + sin 2 sin M1A1= cos (2 ) A1Note: Allow use of double angle formulae if they lead to the correct answer= cos AG (b) cos2 = sin A1 = 0.666, 2.48 A1A1[6]IB Questionbank Mathematics Higher Level 3rd edition 4342. (a) BA =

,_

,_

,_

,_

4 414 184 52 32 23 1A2Note: Award A1 for one error, A0 for two or more errors. (b) det(BA) = (72 56) = 16 (M1)A1 (c) EITHERA(A1B + 2A1)A = BA + 2A (M1)(A1)=

,_

4 618 24A1ORA1 =

,_

3 52 421(A1)an attempt to evaluate (M1)A1B + 2A1 =

,_

,_

3 52 421 116 021=

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5 . 7 5 . 46 4A(A1B + 2A1)A =

,_

,_

,_

4 52 35 . 7 5 . 46 44 52 3=

,_

,_

,_

4 618 244 52 30 23 3A1[7]IB Questionbank Mathematics Higher Level 3rd edition 4443. EITHERusing row reduction (or attempting to eliminate a variable) M11 3 21 3 2 2 222 12 1 33 1 2R RR Rb a +

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5 / 22 21023 2 3 05 5 03 1 2Rb a

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++A1Note: For an algebraic solution award A1 for two correct equations in two variables.2 3 3 2 2223 2 3 01 1 03 1 2R R b a

,_

++

,_

++8 2226 2 0 01 1 03 1 2b aNote: Accept alternative correct row reductions.recognition of the need for 4 zeroes M1so for multiple solutions a = 3 and b = 4 A1A1 OR02 12 1 33 1 2aM1 2(a 4) + (3a + 2) + 3(6 + 1) = 0 5a + 15 = 0 a = 3 A102 12 1 32 1 2bM1 2(b + 4) + (3b 2) + 2(6 + 1) = 0 A1 5b + 20 = 0 b = 4 A1[5] 44. (a) (A + B)2 = A2 + AB + BA + B2A2(b) (A kI)3 = A3 3kA2 + 3k2A k3I A2 (c) CA = B C = BA1A2IB Questionbank Mathematics Higher Level 3rd edition 45Note: Award A1 in parts (a) to (c) if error is correctly identified, but not corrected.[6] 45. det02 021 02 01 22 01 2 01 1k k kkkkkk kkk+

,_

(M1)= 2k(k 2) + k(k 1) 2k A1Note: Allow expansion about any row or column.2k(k 2) + k(k 1) 2k = 0 M13k2 7k = 0k(3k 7) = 0k = 0 or k = 37A1A1 N2[5] 46. finding det A = ex ex(2 + ex) or equivalent A1A is singular det A = 0 (R1)ex ex(2 + ex) = 0e2x ex 2 = 0 A1solving for ex(M1)as ex > 0 (or equivalent explanation) (R1)ex = 2x = ln2 (only) A1 N0[6] 47. (a) det M = a2 + b2A1a2 + b2 > 0, therefore M is non-singular or equivalent statement R1 (b) M2 =

,_

,_

,_

2 22 222b a abab b aa bb aa bb aM1A1IB Questionbank Mathematics Higher Level 3rd edition 46(c) EITHERdet(M2) = (a2 b2)(a2 b2) + (2ab)(2ab) A1det(M2) = (a2 b2)2 + (2ab)2(= (a2 + b2)2)since the first term is non-negative and the second is positive R1therefore det(M2) > 0Note: Do not penalise first term stated as positive.ORdet(M2) = (det M)2A1since det M is positive so too is det (M2) R1[6] 48. (a) a = 16A1 (b) A1 =

,_

1 3 00 1 31 2 1161(M1)A1 (c) AX = C X = A1C (M1)=

,_

,_

4081 3 00 1 31 2 1161=

,_

,_

,_

25 . 05 . 175 . 042412161A1[5] 49. If M is singular, then det M = 0 R1| M | = 11 0211(M1)= (2 + 1) 2 (= 3 ) A1(2 1) = 0 = 0, = 1, = 1 A1A1A1[6] IB Questionbank Mathematics Higher Level 3rd edition 4750. det A = 4 75 926 78 916 48 53 + M1= 3(30 32) 1(54 56) + 2(36 35) (A1)(A1)(A1)= 3 (2) 1 (2) + 2(1)= 6 + 2 + 2 (= 2) A1[5] 51. A2 =

,_

,_

12 112 1k kM1=

,_

++1 2 00 2 1kkA2Note: Award A2 for 4 correct, A1 for 2 or 3 correct.1 + 2k = 0 M1k = 21A1[5] 52. M2 =

,_

,_

,_

19 186 74 31 24 31 2M1A1

,_

,_

24 186 1219 186 7 + kI = 0 (M1)

,_

5 00 5 + kI = 0 (A1)5 kA1[5] 53. For multiplying (I X)(I + X + X2) M1= I2 + IX + IX2 XI X2 X3 = I + X + X2 X X2 X3(A1)(A1)= I X3A1= I A1AB = I A1 = B (R1)(I X) (I + X + X2) = I (I X)1 = I + X + X2AG N0[6] 54. (a) A1 =

,_

8 . 0 2 . 0 2 . 13 . 0 2 . 0 7 . 01 . 0 4 . 0 1 . 0A2N2IB Questionbank Mathematics Higher Level 3rd edition 48 (b) For attempting to calculate

,_

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3211AzyxM1x = 1.2, y = 0.6, z = 1.6 (so the point is (1.2, 0.6, 1.6)) A2N2 (c) (1.2, 0.6, 1.6) lies on x + y + z = d d = 3.4 A1 N1[6] 55. (a) u1 = 27r 127281M1r = 31A1 (b) v2 = 9v4 = 12d = 8 d = 4 (A1)v1 = 13 (A1)2N(2 13 4(N 1)) > 0 (accept equality) M12N(30 4N) > 0N(15 2N) > 0N < 7.5 (M1)N = 7 A1Note: 13 + 9 + 5 + 1 3 7 11 > 0 N = 7 or equivalentreceives full marks.[7]IB Questionbank Mathematics Higher Level 3rd edition 4956. (a)1 cos 2 1cos sin 22 cos 12 sin2 ++ M1Note: Award M1 for use of double angle formulae.= 2cos 2cos sin 2A1= cossin= tan AG (b)4cos 14sin8tan+(M1)4sin4cos 18cot+M1= 22221+= 1 + 2A1[5] 57. (a) area of AOP = 21r2 sin A1 (b) TP = r tan (M1)area of POT = 21r(r tan ) = 21r2 tan A1IB Questionbank Mathematics Higher Level 3rd edition 50(c) area of sector OAP = 21r2 A1area of triangle OAP < area of sector OAP < area of triangle POT R1 tan2121sin212 2 2r r r < 0 , arctan x + arctan y = arctan

,_

+xyy x1 if xy < 1 M1so, arctan4312113121arctan31arctan21

,_

+ +A1AG METHOD 3an appropriate sketch M1e.g.correct reasoning leading to 4R1AGIB Questionbank Mathematics Higher Level 3rd edition 64(b) METHOD 1arctan(2) + arctan(3) =

,_

+ ,_

31arctan221arctan2(M1)=

,_

,_

+ ,_

31arctan21arctan (A1)Note: Only one of the previous two marks may be implied.= 434 A1N1 METHOD 2let x = arctan 2 tan x = 2 and y = arctan 3 tan y = 3tan (x + y) = 13 2 13 2tan tan 1tan tan ++y xy x(M1)as

,_

< < < 1 (M1)so, arctan 2 + arctan 3 = arctan

,_

+3 2 13 2 + (A1)Note: Only one of the previous two marks may be implied.= 4 3A1 N1IB Questionbank Mathematics Higher Level 3rd edition 65METHOD 4an appropriate sketch M1e.g.correct reasoning leading to 4 3R1A1[5] 77. A = 2(R2 r2) A1B = 22rA1from A: B = 2:1, we have R2 r2 = 2r2M1R = r 3(A1)hence exact value of the ratio R : r is 3 :1 A1 N0[5] 78. (a) a reasonable attempt to show either that n2 + n + 1 > 2n + 1 orn2 + n + 1 > n2 1 M1complete solution to each inequality A1A1IB Questionbank Mathematics Higher Level 3rd edition 66(b) cos = ) 1 )( 1 2 ( 2) 1 ( ) 1 ( ) 1 2 (22 2 2 2 2 ++ + + +n nn n n nM1A1= ) 1 )( 1 2 ( 21 2 222 3 ++ + n nn n nM1= ) 1 )( 1 2 ( 2) 1 2 )( 1 )( 1 (2 ++ + n nn n nA1= 21A1 = 120 AG[8] 79. (a) PQ = 50 and non-intersecting R1 (b) a construction QT (where T is on the radius MP), parallel to MN,so that M TQ = 90 (angle between tangent and radius = 90) M1lengths 50, x 10 and angle marked on a diagram, or equivalent R1Note: Other construction lines are possible. (c) (i) MN = 2 2) 10 ( 50 xA1 (ii) maximum for MN occurs when x = 10 A1 (d) (i) = 2 2 M1= 2 2 arccos

,_

5010 xA1 (ii) = 2 (= 2) A1=

,_

,_

5010cos 21xA1 (e) (i) b(x) = x + 10 + 2 2) 10 ( 50 2 xA1A1A1= 2 2 1 1) 10 ( 50 25010cos 205010cos 2 2 +

,_

,_

+

,_

,_

,_

xx xxM1A1IB Questionbank Mathematics Higher Level 3rd edition 67(ii) maximum value of perimeter = 276 A2 (iii) perimeter of 200 cm b (x) = 200 (M1)when x = 21.2 A1[18] 80. (a) shaded area = area of triangle area of sector, i.e. (M1)

,_

,_

x x2 2221sin 421 = 8 sin x 2x A1A1AG (b) EITHERany method from GDC gaining x 1.32 (M1)(A1)maximum value for given domain is 5.11 A2ORxAdd = 8cosx 2 A1set xAdd = 0, hence 8 cos x 2 = 0 M1cos x = 41x 1.32 A1hence Amax = 5.11 A1[7] 81.B C sin12sin9(M1)C C 2 sin12sin9A1Using double angle formula C C C cos sin 212sin9M19(2 sin C cos C) = 12 sin C6 sin C (3 cos C 2) = 0 or equivalent (A1)(sin C 0)cos C = 32A1[5] 82. METHOD 1IB Questionbank Mathematics Higher Level 3rd edition 68AC = 5 and AB = 13(may be seen on diagram) (A1)53cos and 54sin (A1)133cos and 132sin (A1)Note: If only the two cosines are correctly given award (A1)(A1)(A0).Use of cos ( ) = cos cos + sin sin (M1)= 1325413353 + (substituting) M1=

,_

6513 1713 517A1N1 METHOD 2AC = 5 and AB = 13(may be seen on diagram) (A1)Use of cos ( + ) = ( ) ( ) AB AC 2BC AB AC2 2 2 +(M1)= 13 5 236 13 25 +

,_

13 51A1Use of cos ( + ) + cos ( ) = 2 cos cos (M1)53cos and 133cos (A1)( )

,_

,_

6513 1713 5113353213 517cos A1N1[6]IB Questionbank Mathematics Higher Level 3rd edition 6983. 10 cm water depth corresponds to 16 sec 6 3236 ,_

x(A1)Rearranging to obtain an equation of the form sec kx ,_

36 orequivalentie making a trignometrical function the subject of the equation. M113836cos ,_

x(A1)36x = tarccos 138M1x = t36 arccos 138A1Note: Do not penalize the omission of t .Width of water surface is 72 arccos 138 (cm) R1N1Note: Candidate who starts with 10 instead of 6 has the potentialto gain the two M1 marks and the R1 mark.[6] 84. (a) CD = AC AD = b c cos A R1AG (b) METHOD 1BC2 = BD2 + CD2(M1)a2 = (c sin A)2 + (b c cos A)2(A1)= c2 sin2 A + b2 2bc cos A + c2 cos2 A A1= b2 + c2 2bc cos A A1METHOD 2BD2 = AB2 AD2 = BC2 CD2(M1)(A1)c2 c2 cos2 A = a2 b2 + 2bc cos A c2 cos2 A A1a2 = b2 + c2 2bc cos A A1IB Questionbank Mathematics Higher Level 3rd edition 70(c) METHOD 1b2 = a2 + c2 2ac cos 60 b2 = a2 + c2 ac (M1)A1c2 ac + a2 b2 = 0 M1c = ( ) ( )242 2 2b a a a t(M1)A1= 43 42 23 42 22 2a b aa b at t(M1)A1= 2 24321a b a tAGNote: Candidates can only obtain a maximum of the firstthree marks if they verify that the answer given inthe question satisfies the equation. METHOD 2b2 = a2 + c2 2ac cos 60 b2 = a2 + c2 ac (M1)A1c2 ac = b2 a2(M1)c2 ac + 22 222 2

,_

+ ,_

aa baM1A12 22432a bac ,_

(A1)2 2432a bac t A12 24321a b a c t AG[12] 85. PR = h tan 55, QR = h tan 50 where RS = h M1A1A1Use the cosine rule in triangle PQR. (M1)202 = h2 tan2 55 + h2 tan2 50 2h tan 55 h tan 50 cos 45 A1 + 45 cos 50 tan 55 tan 2 50 tan 55 tan4002 22h(A1)= 379.9... (A1)h = 19.5 (m) A1[8] IB Questionbank Mathematics Higher Level 3rd edition 7186. (a) Either finding depths graphically, using 16sin t tor solving h (t) = 0 for t (M1)h (t)max = 12 (m), h (t)min = 4 (m) A1A1N3 (b) Attempting to solve 8 + 486sin t algebraically or graphically (M1)t [0, 6] [12, 18] {24} A1A1N3[6] 87. METHOD 1Attempting to use the cosine rule i.e.BC2 = AB2 + AC2 2 AB AC cos C AB(M1)62 = 8.752 + AC2 2 8.75 AC cos 37.8 (or equivalent) A1Attempting to solve the quadratic in AC e.g. graphically, numerically orwith quadratic formula M1A1Evidence from a sketch graph or their quadratic formula (AC =)that there are two values of AC to determine. (A1)AC = 9.60 or AC = 4.22 A1A1 N4Note: Award (M1)A1M1A1(A0)A1A0 for one correct value of AC.IB Questionbank Mathematics Higher Level 3rd edition 72METHOD 2Attempting to use the sine rule i.e. B CA sinABC AB sinBC(M1)( ) ... 8938 . 068 . 37 sin 75 . 8sin C(A1)C = 63.3576... A1C = 116.6423... and B = 78.842... or B = 25.5576... A1EITHERAttempting to solve or37.8 sin 678.842... sin AC 37.8 sin 625.5576... sin ACM1ORAttempting to solve AC2 = 8.752 + 62 2 8.75 6 cos 25.5576... orAC2 = 8.752 + 62 2 8.75 6 cos 78.842... M1AC = 9.60 or AC = 4.22 A1A1N4Note: Award (M1)(A1)A1A0M1A1A0 for one correct value of AC.[7] 88. (a)A2Note: Award A1 for shape.A1 for scales given on each axis.IB Questionbank Mathematics Higher Level 3rd edition 73(b)A5Asymptotes x = 0, x = 2, x =

,_

,_

1 ,4Min , 1 ,43MaxNote: Award A1 for shape A2 for asymptotes, A1 for one error, A0 otherwise. A1 for max. A1 for min. (c) tanx + cot x xxxxsincoscossin+M1 x xx xcos sincos sin2 2+A1 x 2 sin211A1 2 csc 2x AG (d) tan 2x + cot 2x 2 csc 4x (M1)Max is at

,_

2 ,2 3A1A1Min is at

,_

2 ,8A1A1IB Questionbank Mathematics Higher Level 3rd edition 74(e) csc 2x = 1.5 tan x 0.521tan23cot21tan21 + x x xM1tan x + cot x = 3 tan x 12 tan x x tan1 1 = 0 M12 tan2 x tan x 1 = 0 A1(2 tan x + 1)(tan x 1) = 0 M1tan x = 21 or 1 A1x = 4A1Note: Award A0 for answer in degrees or if more than one value given for x.[21] 89.435 sin5 . 6sin BM1B = 68.8 or 111 A1A1C = 76.2 or 33.8(accept 34) A1A sinBCsinCAB 35 sin4sin76.2AB(M1)AB = 6.77 cm 35 sin4sin33.8ABA1AB = 3.88cm(accept 3.90) A1[7] 90. 2 sin x cos x 2cos x = 0 (M1)cos x (2 sin x 2) = 0 (A1)cos x = 0 sin x = 22A1x = 2x = 4 3,4A1A1A1[6] 91. (a) sin B = 135A1IB Questionbank Mathematics Higher Level 3rd edition 75 (b) cos B = 1312A1 (c) sin 2B = 2sin B cos B (M1)= 2 1312135 = 169120A1 (d) cos2B = 2cos2 B 1 (M1)= 21169144 ,_

= 169119A1[6] 92. Using tan 2 = 2tan 1tan 2(M1)43tan 1tan 22 3 tan2 + 8 tan 3 = 0 A1Using factorisation or the quadratic formula (M1)tan = 31 or 3 A1A1[5] 93. (a) 4(1 2s2) 3s31s + 6 = 0 M1A14s2 8s4 + 6s2 3 = 0 A18s4 10s2 + 3 = 0 AGIB Questionbank Mathematics Higher Level 3rd edition 76(b) Attempt to factorise or use the quadratic formula (M1)sin2 x = 21 or sin2 x = 43(A1)sin x = 4 3or 422 x xA1A1sin x = 3 2or 323 x xA1A1Note: Penalise A1 if extraneous solutions given.[9] 94. (a) sinx cos cosx sin = k sin x cos + k cos x sin (M1) tan x cos sin = k tan x cos + k sin M1 tan x =

,_

+ + tan) 1 () 1 (cos ) 1 (sin ) 1 (kkkkA1 (b) tan x = 210 cos21210 sin23(M1)Now sin 210 = sin 30 = 21 and cos 210 = cos 30 = 23A1A1tan x = 333322323213 A1 x = 60, 240 A1A1[9] 95. 2 tan2 5 sec 10 = 0Using 1 + tan2 = sec2 , 2(sec2 1) 5 sec 10 = 0 (M1)2 sec2 5 sec 12 = 0 A1Solving the equation e.g. (2 sec + 3)(sec 4) = 0 (M1)sec = 23 or sec = 4 A1 in second quadrant sec is negative (R1) sec = 23A1 N3[6] 96. (a)IB Questionbank Mathematics Higher Level 3rd edition 77(M1)(x + 2)2 = (x 2)2 + x2 2(x 2) xcos120 M1A1x2 + 4x + 4 = x2 4x + 4 + x2 + x2 2x (M1)0 = 2x2 10x A10 = x(x 5)x = 5 A1 (b) Area = 21 5 3 sin 120 M1A1= 231521 A1= 43 15AG (c) sin A = 232143 15 5 7 sin B 143 3sin BM1A1Similarly sin C = 143 5A1sin A + sin B + sin C = 143 15A1[13]IB Questionbank Mathematics Higher Level 3rd edition 7897.(a) Using the cosine rule (a2 = b2 + c2 2bc cos A) (M1)Substituting correctlyBC2 = 652 + 1042 2 (65) (104) cos 60 A1= 4225 + 10 816 6760 = 8281 BC = 91m A1N2 (b) Finding the area using = 21bc sin A (M1)Substituting correctly, area = 21(65) (104) sin 60 A1= 3 1690 (accept p = 1690) A1N2 (c) (i) Smaller area A1 =

,_

21(65) (x)sin 30 (M1)A1= 465xAG N0Larger area A2 =

,_

21(104) (x) sin 30 M1= 26x A1 N1IB Questionbank Mathematics Higher Level 3rd edition 79(ii) Using A1 + A2 = A (M1)Substituting 3 1690 26465 + xxA1Simplifying 3 16904169xA1Solving x = 1693 1690 43 40 x (accept q = 40) A1N1 (d) Using sin rule in ADB and ACD (M1)Substituting correctly B DA sin30 sin65BDB DsinA65sin30BD A1and C DA sin30 sin104DCC DA sin104sin30DC A1Since C DA B DA + = 180 R1It follows that C DsinA B DA sin R110465DCBD104DC65BD A185DCBD AG N0[20] 98. (a) (i) (g f)(x) = 23,3 21 +xx (or equivalent) A1 (ii) (f g)(x) = x2 + 3, x 0 (or equivalent) A1IB Questionbank Mathematics Higher Level 3rd edition 80(b) EITHERf(x) = (g1 f g)(x)(g f)(x) = (f g)(x) (M1)323 21+ + x xA1OR(g1 f g)(x) = 321+xA12x + 3 = 321+xM1THEN6x2 + 12x + 6 = 0 (or equivalent) A1x = 1, y = 1 (coordinates are (1, 1)) A1[6] 99. (a) f (x a) b (M1)x 0 and x 2a (or equivalent) A1IB Questionbank Mathematics Higher Level 3rd edition 81(b) vertical asymptotes x = 0, x = 2a A1horizontal asymptote y = 0 A1Note: Equations must be seen to award these marks.maximum

,_

ba1,A1A1Note: Award A1 for correct x-coordinate and A1 for correct y-coordinate.one branch correct shape A1other 2 branches correct shape A1[8] 100. (a) METHOD 1f(x) = q 2x = 0 M1f(3) = q 6 = 0q = 6 A1f(3) = p + 18 9 = 5 M1p = 4 A1METHOD 2f(x) = (x 3)2 + 5 M1A1 = x2 + 6x 4q = 6, p = 4A1A1IB Questionbank Mathematics Higher Level 3rd edition 82(b) g(x) = 4 + 6(x 3) (x 3)2 (= 31 + 12x x2) M1A1Note: Accept any alternative form that is correct.Award M1A0 for a substitution of (x + 3).[6] 101. (a)A3Note: Award A1 for each correct branch with position of asymptotesclearly indicated. If x = 2 is not indicated, only penalise once. (b)A3Note: Award A1 for behaviour at x = 0, A1 for intercept at x = 2,A1 for behaviour for large x.[6] IB Questionbank Mathematics Higher Level 3rd edition 83102. x = 2ey ye1M1Note: The M1 is for switching the variables and may be awarded at anystage in the process and is awarded independently. Further marksdo not rely on this mark being gained.xey = 2e2y 12e2y xey 1 = 0 A148e2+ tx xyM1A1y = ln

,_

+ t482x xtherefore h1(x) = ln

,_

+ +482x xA1since ln is undefined for the second solution R1Note: Accept y = ln

,_

+ +482x x.Note: The R1 may be gained by an appropriate comment earlier.[6] 103. (a) f(a) = 4a3 + 2a2 7a = 10 M14a3 + 2a2 7a + 10 = 0(a + 2) (4a2 6a + 5) = 0 or sketch or GDC (M1)a = 2A1 (b) substituting a = 2 into f(x)f(x) = 4x3 4x + 14 = 0 A1EITHERgraph showing unique solution which is indicated (must includemax and min) R1ORconvincing argument that only one of the solutions isreal (1.74, 0.8681.12i) R1[5]IB Questionbank Mathematics Higher Level 3rd edition 84104.M1A1A1A1A1A1A1Note: Award A1 for both vertical asymptotes correct, M1 for recognizing that there are two turning points near the origin, A1 for both turning points near the origin correct, (only this A mark is dependent on the M mark) A1 for the other pair of turning points correct, A1 for correct positioning of the oblique asymptote, A1 for correct equation of the oblique asymptote, A1 for correct asymptotic behaviour in all sections.[7] 105. using the factor theorem or long division M1A + B 1 + 6 = 0 A B = 5 A18A + 4B + 2 + 6 = 0 2A + B = 2 A13A = 3 A = 1 A1B = 4 A1 N3Note: Award M1A0A0A1A1 for using (x 3) as the third factor, withoutjustification that the leading coefficient is 1.[5]IB Questionbank Mathematics Higher Level 3rd edition 85106. g(x) = 0 or 3 (M1)(A1)x = 1 or 4 or 1 or 2 A1A1Notes: Award A1A1 for all four correct values,A1A0 for two or three correct values,A0A0 for less than two correct values. Award M1 and corresponding A marks for correct attempt to find expressions for f and g.[4] 107. 22x2 = 2x + 8 (M1)4122x = 2x + 8 (A1)22x 4 2x 32 = 0 A1(2x 8)(2x + 4) = 0 (M1)2x = 8 x = 3 A1Notes: Do not award final A1 if more than 1 solution is given.[5] 108. (a) an attempt to use either asymptotes or intercepts (M1)a = 2, b = 1, c = 21A1A1A1 (b)A4Note: Award A1 for both asymptotes, A1 for both intercepts, A1, A1 for the shape of each branch, ignoring shape at (x = 2).[8] 109. (a) attempt at completing the square (M1)IB Questionbank Mathematics Higher Level 3rd edition 863x2 6x + 5 = 3(x2 2x) + 5 = 3 (x 1)2 1 + 5 (A1)= 3(x 1)2 + 2 A1(a = 3, b = 1, c = 2) (b) definition of suitable basic transformations:T1 = stretch in y direction scale factor 3 A1T2 = translation

,_

01A1T3 = translation

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20A1[6] 110. (a) Note: Interchange of variables may take place at any stage.for the inverse, solve for x iny = 13 2xxy(x 1) = 2x 3 M1yx 2x = y 3x(y 2) = y 3 (A1)x = 23yy23) (1 xxx f (x 2) A1Note: Do not award final A1 unless written in the form f1(x) = ... (b) f1(x) = 1 + f1(x) leads to232xx = 1(M1)A1x = 38A1[6] 111. (a) x 0 and x 16 A1A1IB Questionbank Mathematics Higher Level 3rd edition 87(b)graph not to scale (M1)finding crossing pointse.g. 4 x2 = 4 xx = 0 or x = 1 (A1)0 x 1 or x > 16 A1A1Note: Award M1A1A1A0 for solving the inequality only for the case x < 16[6] 112. (a) rewrite the equation as (4x 1)ln 2 = (x + 5)ln 8 + (1 2x) log216 (M1)(4x 1)ln 2 = (3x + 15)ln 2 + 4 8x (M1)(A1)x = 2 ln 82 ln 16 4++A1 (b) x = a2(M1)a = 1.318 A1Note: Treat 1.32 as an AP.Award A0 for .[6] 113. (a) solving to obtain one root: 1, 2 or 5 A1obtain other roots A1 (b) D = x [5, 2] [1, ) (or equivalent) M1A1Note: M1 is for 1 finite and 1 infinite interval. (c) coordinates of local maximum 3.73 ( 2 3), 3.22(3 6) A1A1IB Questionbank Mathematics Higher Level 3rd edition 88(d) use GDC to obtain one root: 1.41, 3.18 or 4.23 A1obtain other roots A1 (e)A1A1A1Note: Award A1 for shape, A1 for max and for min clearly in correctplaces, A1 for all intercepts.Award A1A0A0 if only the complete top half is shown. (f) required area is twice that of y = f(x) between 5 and 2 M1A1answer 14.9 A1 N3Note: Award M1A0A0 for 25d ) ( x x f = 7.47... or N1 for 7.47.[14] 114. (a) g f(x) = xe 11+A11 < 1 + ex < (M1)range g f is ]0, 1[ A1N3IB Questionbank Mathematics Higher Level 3rd edition 89(b) Note: Interchange of variables and rearranging can be done in either order.attempt at solving y = xe 11+M1rearrangingex = yy 1M1(g f)1(x) = ln

,_

xx 1A1Note: The A1 is for RHS.domain is ]0, 1[ A1Note: Final A1 is independent of the M marks. (c) (i) y = f g h = 1 + ecos xM1A1xydd = sin xecos xM1A1= (1 y)sin x AGNote: Second M1A1 could also be obtained by solving the differentialequation. (ii) EITHERrearrangingy sin x = sin x xyddA1xxyx x x x y dddd sin d sin M1= cos x y(+ c) A1= cos x ecos x(+ d) A1OR + x x x x yxd sin ) e 1 ( d sincosA1= + x x x xxd e sin d sincosNote: Either the first or second line gains the A1.= cos x ecos x(+ d) A1M1A1IB Questionbank Mathematics Higher Level 3rd edition 90(iii) use of definition of y and the differential equation orGDC to identify first minimum at x = (3.14...) (M1)A1EITHERthe required integral isx xyyd maxmin2M1A1Note: ymax = 1 + e and ymin = 1 + e1 but these do not need to be specified.... 32 . 4 d e sin cos02 x x xx = 13.6 (M1)A1ORthe required integral is++e 1e 121d y xM1A1= ++e 1e 121) 1 ln( arccos ydy = 4.32... = 13.6 M1A1Note: 1 + e = 3.7182... and 1 + e1 = 1.3678...[21] 115. EITHERx 1 > 2x 1 (x 1)2 > (2x 1)2M1x2 2x + 1 > 4x2 4x + 13x2 2x < 0 A10 < x < 32A1A1 N2Note: Award A1A0 for incorrect inequality signs.ORx 1 > 2x 1x 1 = 2x 1 x 1 = 1 2x M1A1x = 03x = 2x = 0 x = 32Note: Award M1 for any attempt to find a critical value. If graphicalmethods are used, award M1 for correct graphs, A1 for correctvalues of x.0 < x < 32A1A1 N2Note: Award A1A0 for incorrect inequality signs[4] 116.IB Questionbank Mathematics Higher Level 3rd edition 91correct concavities A1A1Note: Award A1 for concavity of each branch of the curve.correct x-intercept of gf (which is EXACTLY the x-intercept of f) A1correct vertical asymptotes of gf (which ONLY occur when x equals thex-intercepts of g) A1A1[5] 117. g(x) = 0log5|2log3x| = 0 (M1)|2log3x| = 1 A1log3x = 21(A1)x = 213tA1so the product of the zeros of g is 21213 3 = 1 A1 N0[5]IB Questionbank Mathematics Higher Level 3rd edition 92118. (a)A1A1Note: Award A1 for the correct x-intercept,A1 for completely correct graph. (b) METHOD 1the area under the graph of y = 2ax for a x a, can be dividedinto ten congruent triangles; M1A1the area of eight of these triangles is given by 0d2axaxand the areas of the other two by axax0d2M1A1so, xax xaxaad24 d200 k = 4 A1N0IB Questionbank Mathematics Higher Level 3rd edition 93METHOD 2use area of trapezium to calculate M1202 2321d2aa aa xaxa ,_

+ A1and area of two triangles to obtain M14 2 212 d2220a axaxa ,_

A1so, k = 4 A1 N0 METHOD 3use integration to find the area under the curvexax xaxa ad2d20 0 + M1= 22 2022 2 2 2aa axa xa + 1]1

+ A1andxax xax xaxaaaad2d2d2220 0 + + M1= 4 4 8 2 2 4 8 2 2 2 22 2 2 2 2 2 222202a a a a a a axa xxa xaaa + + + 1]1

+1]1

+ A1so, k = 4 A1 N0[7] 119. (a) f(1) = 1 arctan1 = 1 4A133 ) 3 arctan( 3 ) 3 ( + fA1 (b) f(x) = x arctan(x) M1= x + arctan x A1= (x arctan x)= f(x) AG N0IB Questionbank Mathematics Higher Level 3rd edition 94(c) as 2arctan2< < x, for any x A12arctan2< < x, for any x then by adding x (or equivalent) R1we have x 2arctan2+ < < x x xAGN0 (d) f(x) = 1 2221or 11xxx + +A1A1f(x) = 2 2 2 23 2) 1 (2or) 1 (2 ) 1 ( 2xxxx x x+ + +M1A1f(0) = f(0) = 0 A1A1EITHERas f (x) 0 for all values of x ((0,0) is not an extreme of the graph of f (or equivalent)) R1ORas f(x) > 0 for positive values of x and f(x) < 0 fornegative values of x R1THEN(0, 0) is a point of inflexion of the graph of f (with zero gradient) A1N2IB Questionbank Mathematics Higher Level 3rd edition 95(e)A1A1A1Note: Award A1 for both asymptotes.A1 for correct shape (concavities) x < 0.A1 for correct shape (concavities) x > 0. (f) (see sketch above)as f is increasing (and therefore one-to-one) and its range is,f1 is defined for all x R1use the result that the graph of y = f1(x) is the reflectionin the line y = x of the graph of y = f(x) to draw the graph of f1(M1)A1[20] 120. q(1) = k + 9 M1A1q(2) = 4k + 9 A1k + 9 = 7(4k + 9) M1k = 2A1Notes: The first M1 is for one substitution and the consequent equations.Accept expressions for q(1) and q(2) that are not simplified.[5]IB Questionbank Mathematics Higher Level 3rd edition 96121. (a)A1Note: Award A1 for correct concavity, passing through (0, 0) and increasing.Scales need not be there. (b) a statement involving the application of the Horizontal Line Test or equivalent A1 (c) y = x kfor either x = 22or kyx y k A1f1(x) = 22kxA1dom(f1(x)) = [0, [ A1 (d)x kkx22 or equivalent method M1k = xk = 2 A1 (e) (i) A = bax y y d ) (2 1(M1)A = x x x d41240221

,_

A1= 403231213411]1

x xA1= 316A1IB Questionbank Mathematics Higher Level 3rd edition 97(ii) attempt to find either f(x) or (f1)(x) M1f(x) =

,_

2) ( ) ( ,11xx fxA1A121 ccM1c = 322A1[16] 122.f(x) = 2) 1 (2x +M1A1Note: Alternatively, award M1A1 for correct sketch of the derivative.find at least one point of intersection of graphs (M1)y = f(x) and y = f(x) for x = 3 or 1.73 (A1)y = f(x) and y = g (x) for x = 0 (A1)forming inequality 0 x 3 (or 0 x 1.73) A1A1 N4Note: Award A1 for correct limits and A1 for correct inequalities.[7]IB Questionbank Mathematics Higher Level 3rd edition 98123. (a) METHOD 1using GDCa = 1, b = 5, c = 3 A1A2A1METHOD 2x = x + 2 cos x cos x = 0 M1...2 3,2 xa = 1, c = 3 A11 2 sin x = 0 M165or 621sin x xb = 5 A1Note: Final M1A1 is independent of previous work. (b)36 56 5 ,_

f (or 0.886) (M1)f(2) = 2 + 2 (or 8.28) (M1)the range is 1]1

+ 2 2 , 36 5 (or [0.886, 8.28]) A1 (c) f(x) = 1 2 sin x (M1)f

,_

2 3 = 3 A1gradient of normal = 31(M1)equation of the normal is y

,_

2 3312 3x(M1)y = 31x + 2 (or equivalent decimal values) A1N4 (d) (i) V = + 2 322 2d ) ) cos 2 ( ( x x x x (or equivalent) A1A1Note: Award A1 for limits and A1 for and integrand.IB Questionbank Mathematics Higher Level 3rd edition 99(ii) V = x x x x d ) ) cos 2 ( ( 2 322 2+ = x x x x d ) cos 4 cos 4 ( 2 322+ using integration by parts M1and the identity 4cos2 x = 2cos 2x + 2, M1V = 2 32)] 2 2 (sin ) cos 4 sin 4 [( x x x x x + + +A1A1Note: Award A1 for 4x sin x + 4 cos x and A1 for sin 2x + 2x.= 1]1

,_

+ + + ,_

+ + + sin2cos 42sin 2 3 3 sin2 3cos 42 3sin 6 A1= (6 + 3 2 )= 62AG N0Note: Do not accept numerical answers.[19] 124. (a) EITHERtranslation of 21 parallel to the x-axisstretch of a scale factor of 21 parallel to the x-axis A1A1ORstretch of a scale factor of 21 parallel to the x-axistranslation of 1 parallel to the x-axis A1A1Note: Accept clear alternative terminologies for either transformation. (b) EITHER1.16 < x < 5.71 6.75 < x 10 A1A1A1A1OR]1.16, 5.71[ ]6.75,10] A1A1A1A1Note: Award A1 for 1 intersection value, A1 for the other 2,A1A1 for the intervals.[6] IB Questionbank Mathematics Higher Level 3rd edition 100125. (a)xydd = 24x2 + 2bx + c (A1)24x2 + 2bx + c = 0 (M1) = (2b)2 96(c) (A1)4b2 96c > 0 A1b2 > 24c AG (b) 1 + c b2141+ + d = 126 + b + c = 027 + c b2349 + d = 2054 3b + c = 0 A1A1A1Note: Award A1 for each correct equation, up to 3, not necessarily simplified.b = 12, c = 18, d = 7 A1[8] 126. (a) f(1) = 3 a + b (A1)f(1) = 3 + a + b (A1)3 a + b = 3 + a + b M12a = 6a = 3 A1 N4 (b) b is any real number A1[5] 127. (a) l ln x 1 (M1)e1 x e A1A1 (b) y = arcsin (ln x) ln x = sin y (M1)ln y = sin x y = esinx(M1) f1(x) = esinxA1[6]IB Questionbank Mathematics Higher Level 3rd edition 101128. (a) EITHERgraph of the cubic is shifted horizontally one unit to the right (M1) x = 0.796 A1OR(x 1) = 1.796 (M1)x = 0.796 A1 (b) EITHERstretch factor of 0.5 in the x-direction (M1) 2x = 1.796 (M1)Note: At least one of the above lines must be seen to award the M marks. x = 0.898 A1OR8x3 2x + 4 = (2x)3 2x + 4 = 0 (M1) 2x = 1.796 (M1)Note: At least one of the above lines must be seen to award the M marks. x = 0.898 A1[5]IB Questionbank Mathematics Higher Level 3rd edition 102129. (a)A3Note: Award A1 for each graph A1 for the point of tangency.point on curve and line is (a, ln a) (M1)y = ln (x)a xyx xy 1dd 1dd (when x = a) (M1)A1EITHERgradient of line, m, through (0, 0) and (a, ln a) is aa ln(M1)A1e1e 1 ln1 ln m a aa aaM1A1ORy ln a = a1(x a)(M1)A1passes through 0 ifln a 1 = 0 M1a = e e1 mA1THENx ye1 A1 (b) the graph of ln x never goes above the graph of y = xe1, hence ln x exR1 (c) lnx x x x xx eln ln eeM1A1exponentiate both sides of lnxe x xe exR1AG (d) equality holds when x = e R1IB Questionbank Mathematics Higher Level 3rd edition 103letting x = e < eA1N0[17] 130. from GDC, sketch a relevant graph A1maximum: y = 3 or (1, 3) A1minimum: y = 1.81 or (0.333, 1.81)

,_

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2749,31or2749or yA1hence, 1.81 < k < 3 A1A1 N3Note: Award A1 for 1.81 k 3.[5] 131. (a) a = 2.24 ) 5 (A1 (b) (i)A2Note: Award A1 for end point A1 for its asymptote. (ii) sketch of g1 (see above) A2Note: Award A1 for end point A1 for its asymptote.IB Questionbank Mathematics Higher Level 3rd edition 104(c) y = +253xxyx2 3x + 5y = 0 M1yyx220 9 32 t A1xxx g220 9 3) (21 + A1[8] 132. (a) h (x) = g f (x) = ( ) 0 ,3 e12+xx(M1)A1 (b) 0 < x 41A1A1Note: Award A1 for limits and A1 for correct inequality signs. (c)3 e12+xy1 3 e2 + y yxM1yyx3 1e2A1x2 = lnyy 3 1M1x = tyy 3 1lnh1 (x) =

,_

,_

31ln3 1lnx xxA1[8]IB Questionbank Mathematics Higher Level 3rd edition 105133. METHOD 1As (x + 1) is a factor of P(x), then P(1) = 0 (M1)a b + 1 = 0 (or equivalent) A1As (x 2) is a factor of P(x), then P(2) = 0 (M1)4a + 2b + 10 = 0 (or equivalent) A1Attempting to solve for a and b M1a = 2 and b = 1 A1 N1 METHOD 2By inspection third factor must be x 1. (M1)A1(x + 1)(x 2)(x 1) = x3 2x2 x + 2 (M1)A1Equating coefficients a = 2, b = 1 (M1)A1N1 METHOD 3Considering ( )22 x xx P or equivalent (M1)( )( )( ) ( )22 2 3122 2 + + + ++ + + x xa x b aa xx xx PA1A1Recognizing that (a + b + 3) x + 2 (a + 2) = 0 (M1)Attempting to solve for a and b M1a = 2 and b = 1 A1 N1[6] 134. (a)( )

,_

+24xg x h(M1)=

,_

+ + xxx 22124A1 (b) METHOD 1124+yx(interchanging x and y) M1Attempting to solve for y M1(y + 2)(x + 1) = 4

,_

+ +142xy(A1)( ) ( ) 1 2141 +xxx hA1 N1 IB Questionbank Mathematics Higher Level 3rd edition 106METHOD 2yyx+22(interchanging x and y) M1Attempting to solve for y M1xy + y = 2 2x (y(x + 1) = 2(1 x)) (A1)( )( )( ) 111 21 +xxxx hA1 N1Note: In either METHOD 1 or METHOD 2 rearranging firstand interchanging afterwards is equally acceptable.[6] 135. (a)A1A1A1A1Note: Award A1 for y-interceptA1A1 for x-interceptsA1 for shape. (b) correct line A15 solutions A1[6] 136. (a) y = arccos (1.2 cos x) A1y = arcsin (1.4 sin x) A1 (b) The solutions areIB Questionbank Mathematics Higher Level 3rd edition 107x = 1.26, y = 0.464 A1A1x = 0.464, y = 1.26 A1A1[6] 137.A1A1A1A1A1 Notes: Award A1 for vertical asymptotes at x = 1, x = 2 and x = 5.A1 for x 2, ( )+01x fA1 for x 8, ( )11 x fA1 for local maximum at

,_

21, 0 (branch containinglocal max. must be present)A1 for local minimum at (3, 1) (branch containing localmin. must be present)In each branch, correct asymptotic behaviour must bedisplayed to obtain the A1.Disregard any stated horizontal asymptotes such as y = 0 or y = 1.[5]IB Questionbank Mathematics Higher Level 3rd edition 108138. Attempting to solve |0.1x2 2x + 3| = log10 x numerically or graphically. (M1)x = 1.52, 1.79 (A1)(A1)x = 17.6, 19.1 (A1)(1.52 < x < 1.79) (17.6 < x < 19.1) A1A1N2[6] 139. f(2) = 16 + 24 + 4p 4 + q = 15 M1 4p + q = 21 A1f(3) = 81 81 + 9p + 6 + q = 0 M1 9p + q = 6 A1 p = 3 and q = 33 A1A1 N0[6] 140. (a)A1Note: Award A1 for shape.x intercepts 0.354, 1.36, 2.59, 2.95 A2Note: Award A1 for three correct, A0 otherwise.maximum = (1.57, 0.352) =

,_

352 . 0 ,2A1minimum = (1, 0.640) and (2.77, 0.0129) A1IB Questionbank Mathematics Higher Level 3rd edition 109(b) 0 < x < 0.354, 1.36 < x < 2.59, 2.95 < x < 4 A2Note: Award A1 if two correct regions given.[7] 141. (a) For 29 x x , 3 x 3 and for 2arcsin

,_

3x, 3 x 3 A1D is 3 x 3 A1 (b) V = xxx x d3arcsin 2 9 28 . 202

,_

+ M1A1= 181 A1 (c)9132) 9 () 9 (dd22122212xxxxxy+ M1A1= 2122122212) 9 (2) 9 () 9 (x xxx+ A1= 2122 2) 9 (2 9xx x+ A1= 2292 11xxA1 (d)ppppxx x xxx1]1

+ 3arcsin 2 9 d92 11222M1= 3arcsin 2 93arcsin 2 92 2pp ppp p + + + A1=

,_

+ 3arcsin 4 9 22pp pAG (e) 11 2p2 = 0 M1p = 2.35

,_

211A1Note: Award A0 for p = 2.35. IB Questionbank Mathematics Higher Level 3rd edition 110(f) (i) f(x) = 2212 22129) 9 )( 2 11 ( ) 4 ( ) 9 (xx x x x x + M1A1= 2322 2) 9 () 2 11 ( ) 9 ( 4xx x x x + A1= 2323 3) 9 (2 11 4 36xx x x x + + A1= 2322) 9 () 25 2 (xx xAG (ii) EITHERWhen 0 < x < 3, f(x) < 0. When 3 < x < 0, f(x) > 0. A1ORf(0) = 0 A1THENHence f(x) changes sign through x = 0, giving a point of inflexion. R1EITHERx = 225 is outside the domain of f. R1ORx = 225 is not a root of f(x) = 0. R1[21]IB Questionbank Mathematics Higher Level 3rd edition 111142.A1A1Note: Award A1 for each graph.2x = 1 x 31 xM1A131< xA1[5] 143. Attempting to find f (2) = 8 + 12 + 2a + b (M1)= 2a + b + 20 A1Attempting to find f (1) = 1 + 3 a + b (M1)= 2 a + b A1Equating 2a + 20 = 2 a A1a = 6 A1 N2[6] 144. (a) f : x ex f1 : x ln x f1(3) = ln 3 A1g : x x + 2 g1 : x x 2 g1(3) = 1 A1f1(3) g1(3) = ln3 A1N1IB Questionbank Mathematics Higher Level 3rd edition 112(b) EITHERf g(x) = f(x + 2) = ex+2A1ex+2 = 3 x + 2 = ln3 M1x = ln 3 2 A1 N0ORf g(x) = ex+2f g1(x) = ln (x) 2 A1f g1(3) = ln (3) 2 M1x = ln 3 2 A1 N0[6] 145. METHOD 1Graph of f (x) g(x) M1A1A1A1Note: Award A1 for each branch.x < 1 or 4 < x 14 A1A1 N3Note: Each value and inequality sign must be correct.IB Questionbank Mathematics Higher Level 3rd edition 113METHOD 204214++xxxxM10) 4 )( 1 (2 162 2 ++ + x xx x x0) 4 )( 1 (14 +x xxA1Critical value of x = 14 A1Other critical values x = 1 and x = 4 A1x < 1 or 4 < x 14 A1A1 N3Note: Each value and inequality sign must be correct.[6] 146. (a) (cos + i sin )3 = cos3 + 3cos2 (i sin ) + 3 cos (isin )2 + (isin )3(M1)= cos3 3 cos sin2 + i(3 cos2 sin sin3 ) A1 (b) from De Moivres theorem(cos + i sin )3 = cos 3 + i sin 3 (M1)cos 3 + i sin 3 = (cos3 3 cos sin2 ) + i(3 cos2 sin sin3 )equating real partscos 3 = cos3 3 cos sin2 M1 = cos3 3 cos (1 cos2 ) A1 = cos3 3 cos + 3 cos3 = 4 cos3 3 cos AGNote: Do not award marks if part (a) is not used. (c) (cos + i sin )5 =cos5 + 5 cos4 (i sin ) + 10 cos3 (i sin )2 + 10 cos2 (i sin )3+ 5cos (i sin )4 + (i sin )5(A1)from De Moivres theoremcos 5 = cos5 10 cos3 sin2 + 5 cos sin4 M1 = cos5 10 cos3 (1 cos2 ) + 5cos (1 cos2 )2A1 = cos5 10 cos3 + 10 cos5 + 5 cos 10 cos3 + 5 cos5 cos 5 = 16 cos5 20 cos3 + 5 cos AGNote: If compound angles used in (b) and (c), then marks can beallocated in (c) only.IB Questionbank Mathematics Higher Level 3rd edition 114(d) cos 5 + cos 3 + cos = (16 cos5 20 cos3 + 5 cos ) + (4 cos3 3 cos ) + cos = 0 M116 cos5 16 cos3 + 3 cos = 0 A1cos (16 cos4 16 cos2 + 3) = 0cos (4 cos2 3)(4 cos2 1) = 0 A121;23; 0 cos t t A12;3;6t t t A2 (e) cos 5 = 05 = ...... ;27;25;23;2

,_

(M1) = ...... ;107;105;103;10

,_

(M1)Note: These marks can be awarded for verifications later in the question.now consider 16 cos5 20 cos3 + 5 cos = 0 M1cos (16 cos4 20 cos2 + 5) = 0cos2 = 32) 5 )( 16 ( 4 400 20 t; cos = 0 A1cos = 32) 5 )( 16 ( 4 400 20 tt32) 5 )( 16 ( 4 400 2010cos +t since max value of cosine angleclosest to zero R185 58 . 4) 5 ( 4 25 4 5 . 410cos+ +A185 5107cos A1A1[22] 147. (a) AB = 2 2) 3 2 ( 1 +M1= 3 4 88 A1= 3 2 2 A1IB Questionbank Mathematics Higher Level 3rd edition 115(b) METHOD 13arg ,4z arg2 1 zA1A1Note: Allow 3and4.Note: Allow degrees at this stage.43B OA = )12(accept12A1Note: Allow FT for final A1.METHOD 2attempt to use scalar product or cosine rule M12 23 1B OA cos+A112B OA A1[6] 148. (a) let the first three terms of the geometric sequence be given by u1, u1 r, u1r2 u1 = a + 2d, u1r = a + 3d and u1r2 = a + 6d (M1)d ad ad ad a2336++++A1a2 + 8ad + 12d2 = a2 + 6ad + 9d2A12a + 3d = 0a = 23d AGIB Questionbank Mathematics Higher Level 3rd edition 116(b) u1 =

,_

29,23,221 1dr udr udM1r = 3 A1geometric 4th term u1r3 = 227dA1arithmetic 16th term a + 15d = d d 1523+ M1= 227dA1Note: Accept alternative methods.[8] 149. (a) using the factor theorem z + 1 is a factor (M1)z3 + 1 = (z + 1)(z2 z + 1) A1 (b) (i) METHOD 1z3 = 1 z3 + 1 = (z + 1)(z2 z + 1) = 0 (M1)solving z2 z + 1 = 0 M1z = 23 i 124 1 1 t tA1therefore one cube root of 1 is AGMETHOD 22 = 23 i 123 i 12+

,_

+M1A13 = 43 123 i 123 i 1 + + A1= 1 AG METHOD 3 = 3i23 i 1e + A13 = ei 1 A1IB Questionbank Mathematics Higher Level 3rd edition 117(ii) METHOD 1as is a root of z2 z + 1 = 0 then 2 + 1 = 0 M1R1 2 = 1 AGNote: Award M1 for the use of z2 z + 1 = 0 in any way.Award R1 for a correct reasoned approach. METHOD 22 = 23 i 1+ M1 1 = 23 i 1123 i 1 + +A1 (iii) METHOD 1(1 )6 = (2)6(M1) = ()12A1 = (3)4(M1) = (1)4 = 1 A1METHOD 2(1 )6= 1 6 + 152 203 + 154 65 + 6M1A1Note: Award M1 for attempt at binomial expansion.use of any previous result e.g. = 1 6 + 15 2 + 20 15 + 6 2 + 1 M1= 1 A1Note: As the question uses the word hence, other methods thatdo not use previous results are awarded no marks.IB Questionbank Mathematics Higher Level 3rd edition 118(c) METHOD 1A2 =

,_

+

,_

,_

22101101101 A1A2 A + I =

,_

+ + + 11 1011122 M1from part (b)2 + 1 = 0) 1 (1112+ + = 0 A1) 1 (111 122 2+ + + = 0 A1hence A2 A + I = 0 AG METHOD 2A2 =

,_

+ 23 i 10123 i 1A1A1A1Note: Award 1 mark for each of the non-zero elements expressed in this form.verifying A2 A + I = 0 (d) (i) A2 = A I A3 = A2 A M1A1 = A I A A1 = I AGNote: Allow other valid methods. (ii) I = A A2A1 = A1A A1A2M1A1 A1 = I A AGNote: Allow other valid methods.[20]IB Questionbank Mathematics Higher Level 3rd edition 119150. (a) 2x2 + x 3 = (2x + 3)(x 1) A1Note: Accept 2

,_

+23x(x 1).Note: Either of these may be seen in (b) and if so A1 should be awarded. (b) EITHER(2x2 + x 3)8 = (2x + 3)8(x 1)8M1 = (38 + 8(37)(2x) + ...)((1)8 + 8(1)7(x) + ...) (A1)coefficient of x = 38 8 (1)7 + 37 8 2 (1)8M1 = 17 496 A1Note: Under FT, final A1 can only be achieved for an integer answer.OR(2x2 + x 3)8 = (3 (x 2x2))8M1= 38 + 8((x 2x2)(37) + ...) (A1)coefficient of x = 8 ( 1) 37M1 = 17 496 A1Note: Under FT, final A1 can only be achieved for an integer answer.[5] 151. logx+1 y = 2logy+1 x = 41so (x + 1)2 = y A141) 1 ( + y = x A1 EITHERx4 1 = (x + 1)2M1x = 1, not possible R1x = 1.70, y = 7.27 A1A1 OR412) 2 2 ( + + x x x = 0 M1attempt to solve or graph of LHS M1x = 1.70, y = 7.27 A1A1[6] IB Questionbank Mathematics Higher Level 3rd edition 120152. (a) A2 =

,_

,_

cos sinsin coscos sinsin cos=

,_

+ + 2 22 2cos sin sin cos cos sincos sin sin cos sin cosM1(A1)=

,_

2 22 2sin cos cos sin 2cos sin 2 sin cosA1=

,_

2 cos 2 sin2 sin 2 cosAG (b) let P(n) be the proposition that

,_

,_

n nn nncos sinsin coscos sinsin cosfor all n +P(1) is true A1

,_

,_

cos sinsin coscos sinsin cos1assume P(k) to be true A1Note: Must see the word true or equivalent, that makes clear anassumption is being made that P(k) is true.

,_

,_

k kk kkcos sinsin coscos sinsin cosconsider P(k + 1)

,_

,_

,_

+ cos sinsin coscos sinsin coscos sinsin cos1 k k(M1)

,_

,_

cos sinsin coscos sinsin cosk kk kA1

,_

+ + cos cos sin sin sin cos cos sincos sin sin cos sin sin cos cosk k k kk k k kA1

,_

+ + + + ) 1 cos( ) 1 sin() 1 sin( ) 1 cos(k kk kA1if P(k) is true then P(k + 1) is true and since P(1) is true then P(n) is truefor all n +R1Note: The final R1 can only be gained if the M1 has been gained.IB Questionbank Mathematics Higher Level 3rd edition 121(c) EITHERA1 =

,_

) cos( ) sin() sin( ) cos( from formula =

,_

cos sinsin cosA1A1A = AA1 =

,_

,_

,_

,_

cos sinsin coscos sinsin coscos sinsin coscos sinsin cosM1Note: Accept either just A1A or just AA1.=

,_

1 00 1A1A1 is inverse of AORA1 =

,_

+ cos sinsin cossin cos12 2M1A1 =

,_

cos sinsin cosA1putting n = 1 in formula gives inverse A1[13] 153. u4 = u1 + 3d = 7, u9 = u1 + 8d = 22 A1A1Note: 5d = 15 gains both above marksu1 = 2, d = 3 A1Sn = 2n(4 + (n 1)3) > 10000 M1n = 83A1[5]IB Questionbank Mathematics Higher Level 3rd edition 122154. prove that 1 + 211 3 222421...21421321+ ,_

+ + ,_

+ ,_

+ ,_

nnnnfor n = 1LHS = 1, RHS = 4 022 1+ = 4 3 = 1so true for n = 1 R1assume true for n= k M1so 1 + 211 3 222421...21421321+ ,_

+ + ,_

+ ,_

+ ,_

kkkknow for n = k + 1LHS: 1 + 2k kk k ,_

+ + ,_

+ + ,_

+ ,_

+ ,_

21) 1 (21...214213211 3 2A1= 4 kkkk

,_

+ ++21) 1 (221M1A1= 4 k kk k212) 2 ( 2 +++ (or equivalent) A1= 4 1 ) 1 (22 ) 1 ( ++ +kk (accept 4 kk23 +) A1Therefore if it is true for n = k it is true for n = k + 1. It has been shownto be true for n = 1 so it is true for all n ( +). R1Note: To obtain the final R mark, a reasonable attempt at inductionmust have been made.[8]IB Questionbank Mathematics Higher Level 3rd edition 123155. P (six in first throw) = 61(A1)P (six in third throw) = 613625(M1)(A1)P (six in fifth throw) = 6136252 ,_

P(A obtains first six) = ...613625613625612+ ,_

+ +(M1)recognizing that the common ratio is 3625(A1)P(A obtains first six) = 3625161 (by summing the infinite GP) M1= 116A1[7] 156. (a) sin (2n + 1)x cosx cos (2n + 1)x sinx = sin (2n + 1)x x M1A1= sin 2nx AGIB Questionbank Mathematics Higher Level 3rd edition 124(b) if n = 1 M1LHS = cos xRHS = xx xxxsin 2cos sin 2sin 22 sin = cos x M1so LHS = RHS and the statement is true for n = 1 R1assume true for n = k M1Note: Only award M1 if the word true appears.Do not award M1 for let n = k only.Subsequent marks are independent of this M1.so cos x + cos 3x + cos 5x + ... + cos(2k 1)x = xkxsin 22 sinif n = k + 1 thencos x + cos 3x + cos 5x + ... + cos(2k 1)x + cos(2k + 1)x M1= xkxsin 22 sin cos (2k + 1)x A1= xx x k kxsin 2sin ) 1 2 cos( 2 2 sin + +M1= xx x k x x k x x ksin 2sin ) 1 2 cos( 2 sin ) 1 2 cos( cos ) 1 2 sin( + + + +M1= xx x k x x ksin 2sin ) 1 2 cos( cos ) 1 2 sin( + + +A1= xx ksin 2) 2 2 sin( +M1= xx ksin 2) 1 ( 2 sin +A1so if true for n = k, then also true for n = k + 1as true for n = 1 then true for all n +R1Note: Final R1 is independent of previous work.IB Questionbank Mathematics Higher Level 3rd edition 125(c)21sin 24 sinxxM1A1sin 4x = sin x4x = x x = 0 but this is impossible4x = x 5 xA14x = 2 + x 3 2 xA14x = 3 x 5 3 xA1for not including any answers outside the domain R1Note: Award the first M1A1 for correctly obtaining 8 cos3 x 4 cos x 1 = 0or equivalent and subsequent marks as appropriate including theanswers arccos

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t45 1,21.[20] 157. (a) (i) 1 2 + 2 3 + ... + n(n + 1) = 31n(n + 1)(n + 2) R1 (ii) LHS = 40; RHS = 40 A1 (b) the sequence of values are:5, 7, 11, 19, 35 or an example A135 is not prime, so Bills conjecture is false R1AGIB Questionbank Mathematics Higher Level 3rd edition 126(c) P(n) : 5 7n + 1 is divisible by 6P(1): 36 is divisible by 6 P(1) true A1assume P(k) is true (5 7k + 1 = 6r) M1Note: Do not award M1 for statement starting let n = k.Subsequent marks are independent of this M1.consider 5 7k+1 + 1 M1= 7(6r 1) + 1 (A1)= 6(7r 1) P(k + 1) is true A1P(1) true and P(k) true P(k + 1) true, so by MI P(n) is true for all n +R1Note: Only award R1 if there is consideration of P(1), P(k) and P (k + 1)in the final statement.Only award R1 if at least one of the two preceding A marks hasbeen awarded.[10] 158. (a) (i) 3 = 332isin3 2cos

,_

,_

+ ,_

=

,_

+ ,_

323 isin3 23 cos(M1)= cos 2 + i sin 2 A1= 1 AG (ii) 1 + + 2 = 1 +

,_

+ ,_

+ ,_

+ ,_

34sin i34cos32sin i3 2cosM1A1= 1 + 23i2123i21 + A1= 0 AG (b) (i)

,_

+ ,_

++ +34i32iie e e =

,_

,_

+ +34ii 32ii ie e e e e (M1)=

,_

,_

+ +

,_

,_

34i32iie e 1 e= ei(1 + + 2) A1= 0 AGIB Questionbank Mathematics Higher Level 3rd edition 127(ii)A1A1Note: Award A1 for one point on the imaginary axis and anotherpoint marked with approximately correct modulus and argument.Award A1 for third point marked to form an equilateral trianglecentred on the origin. (c) (i) attempt at the expansion of at least two linear factors (M1)(z 1)z2 z( + 2) + 3 or equivalent (A1)use of earlier result(M1)F(z) = (z 1)(z2 + z + 1) = z3 1 A1 (ii) equation to solve is z3 = 8 (M1)z = 2, 2, 22A2Note: Award A1 for 2 correct solutions.[16]IB Questionbank Mathematics Higher Level 3rd edition 128159. (a) z3 = 4i 3e 2 2(M1)(A1)z1 = 4ie 2A1adding or subtracting 3i 2M1z2 = 12i 113i 24ie 2 e 2 +A1z3 = 12i 53i 24ie 2 e 2 A1Notes: Accept equivalent solutions e.g. z3 = 12i 19e 2 Award marks as appropriate for solving (a + bi)3 = 2 + 2i. Accept answers in degrees. (b)

,_

,_

+ 2i212 e 24iA1= 1 + i AGNote: Accept geometrical reasoning.[7] 160. (100 + 101 + 102 + ... + 999) (102 + 105 + ... + 999) (M1)= ) 999 102 (2300) 999 100 (2900+ +M1A1A1= 329 400 A1 N5Note: A variety of other acceptable methods may be seen includingfor example ) 998 100 (2600or) 1995 201 (2300+ +.[5] 161. (a) There are 3! ways of arranging the Mathematics books, 5! waysof arranging the English books and 4! ways of arranging theScience books. (A1)Then we have 4 types of books which can be arranged in 4! ways. (A1)3! 5! 4 4! = 414 720 (M1)A1IB Questionbank Mathematics Higher Level 3rd edition 129(b) There are 3! ways of arranging the subject books, and for eachof these there are 2 ways of putting the dictionary next to theMathematics books. (M1)(A1)3! 5! 4! 3! 2 = 207 360 A1[7] 162. (a) 18n 10 (or equivalent) A1 (b)nr1) 10 18 ( (or equivalent) A1 (c) by use of GDC or algebraic summation or sum of an AP (M1)151) 10 18 ( r = 2010 A1[4] 163. (a) the expression is! 2 )! 2 2 ()! 2 (! 3 )! 3 (! nnnn(A1)2) 1 2 ( 26) 2 )( 1 ( n n n n nM1A1=

,_

+ + 6) 8 156) 8 15 (3 2n n n n n nA1 (b) the inequality is68 152 3n n n + > 32nattempt to solve cubic inequality or equation (M1)n3 15n2 184n > 0n(n 23)(n + 8) > 0n > 23(n 24) A1[6]IB Questionbank Mathematics Higher Level 3rd edition 130164. (a) using de Moivres theoremzn + nz1 = cos n + i sin n + cos n i sin n (= 2 cos n), imaginarypart of which is 0 M1A1so Im

,_

+nnzz1 = 0 AG (b)1 sin i cos1 sin i cos11+ + ++ zz= ) sin i 1 )(cos sin i 1 (cos) sin i 1 )(cos sin i 1 (cos + + + + + M1A1Note: Award M1 for an attempt to multiply numerator and denominatorby the complex conjugate of their denominator.r denominato realsin ) 1 )(cos 1 (cos11Re2 + + ,_

+zzM1A1Note: Award M1 for multiplying out the numerator.r denominato real1 sin cos2 2 + A1= 0 AG[7] 165. (a) the area of the first sector is 2221(A1)the sequence of areas is 2, 2k, 2k2... (A1)the sum of these areas is 2(1 + k + k2 + ...) (M1)= k 12 = 4M1A1hence = 2(1 k) AGNote: Accept solutions where candidates deal with angles instead of area. (b) the perimeter of the first sector is 4 + 2 (A1)the perimeter of the third sector is 4 + 2k2 (A1)the given condition is 4 + 2k2 = 2 + M1which simplifies to 2 = (1 2k2) A1eliminating , obtain cubic in k: (1 k) (1 2k2) 1 = 0 A1or equivalentsolve for k = 0.456 and then = 3.42 A1A1[12] IB Questionbank Mathematics Higher Level 3rd edition 131166.4 3222 2 3 2 4 2422 2) ( 42) ( 62) ( 4 ) (2

,_

+ ,_

+ ,_

+ ,_

+ ,_

x xxxxxx xxx(M1)= x8 8x5 + 24x2 416 32xx+A3Note: Deduct one A mark for each incorrect or omitted term.[4] 167. METHOD 15(2a + 9d) = 60 (or 2a + 9d = 12) M1A110(2a + 19d) = 320 (or 2a + 19d = 32) A1solve simultaneously to obtain M1a = 3, d = 2 A1the 15th term is 3 + 14 2 = 25 A1Note: FT the final A1 on the values found in the penultimate line. METHOD 2with an AP the mean of an even number of consecutive terms equalsthe mean of the middle terms (M1)211 10a a + = 16 (or a10 + a11 = 32) A126 5a a + = 6 (or a5 + a6 = 12) A1a10 a5 + a11 a6 = 20 M15d + 5d = 20d = 2 and a = 3 (or a5 = 5 or a10 = 15) A1the 15th term is 3 + 14 2 = 25 (or 5 + 10 2 = 25 or 15 + 5 2 = 25) A1Note: FT the final A1 on the values found in the penultimate line.[6] 168. METHOD 1(a) un = Sn Sn1(M1)= 11 17777 nn nnn na aA1IB Questionbank Mathematics Higher Level 3rd edition 132(b) EITHERu1 = 1 7aA1u2 = 1

,_

71722a aM1=

,_

717a aA1common ratio = 7aA1 ORun = 1 1717

,_

+ ,_

n na aM1=

,_

,_

7171a an= 17 77

,_

na aA1u1 = 77 a , common ratio = 7aA1A1 (c) (i) 0 < a < 7 (accept a < 7) A1(ii) 1 A1 METHOD 2(a) un = brn1 = 17 77

,_

,_

na aA1A1IB Questionbank Mathematics Higher Level 3rd edition 133(b) for a GP with first term b and common ratio rSn = nnrrbrbrr b

,_

,_

1 1 1) 1 (M1as Sn = nnn na a

,_

7177comparing both expressions M1rb 1 = 1 and r = 7ab = 1 777a a u1 = b = 77 a , common ratio = r = 7aA1A1Note: Award method marks if the expressions for b and r are deducedin part (a). (c) (i) 0 < a < 7 (accept a < 7) A1(ii) 1 A1[8] 169. (a) METHOD 12i++zz = iz + i = iz + 2i M1(1 i)z = i A1z = i 1iA1EITHERz =

,_

,_

43cis 22cisM1z =

,_

,_

,_

43cis21or43cis22A1A1ORz =

,_

+ + i21212i 1M1z =

,_

,_

,_

43cis21or43cis22A1A1 IB Questionbank Mathematics Higher Level 3rd edition 134METHOD 2i = y xy xi 2) 1 i(+ ++ +M1x + i(y + 1) = y + i(x + 2) A1x = y; x + 2 = y + 1 A1solving, x = 21;21 yA1z = i2121+ z =

,_

,_

,_

43cis21or43cis22A1A1Note: Award A1 fort the correct modulus and A1 for the correct argument,but the final answer must be in the form r cis .Accept 135 for the argument. (b) substituting z = x + iy to obtain w = i ) 2 (i ) 1 (y xy x+ ++ +(A1)use of (x + 2) yi to rationalize the denominator M1 = 2 2) 2 ()) 2 )( 1 ( i( ) 1 ( ) 2 (y xx y xy y y x x+ ++ + + + + + +A1= 2 22 2) 2 () 2 2 i( ) 2 (y xy x y y x x+ ++ + + + + +AG (c) Re = 2 22 2) 2 (2y xy y x x+ ++ + + = 1 M1 x2 + 2x + y2 + y = x2 + 4x + 4 + y2A1 y = 2x + 4 A1which has gradient m = 2 A1IB Questionbank Mathematics Higher Level 3rd edition 135(d) EITHERarg (z) = 4x = y (and x, y > 0) (A1) = 2 2 2 22) 2 () 2 i(3) 2 (3 2x xxx xx x+ ++++ ++if arg() = x xx3 22 3tan2++ (M1)13 22 32++x xxM1A1ORarg (z) = 4 x = y (and x, y > 0) A1arg (w) = 4 x2 + 2x + y2 + y = x + 2y + 2 M1solve simultaneously M1x2 + 2x + x2 + x = x + 2x + 2 (or equivalent) A1THENx2 = 1x = 1 (as x > 0) A1Note: Award A0 for x = 1.z = 2A1Note: A1low FT from incorrect values of x.[19] 170. (a) (i) the period is 2 A1 (ii) v = tsdd = 2 cos (t) + 2 cos (2t) (M1)A1a = tvdd = 22 sin (t) 42 sin (2t) (M1)A1IB Questionbank Mathematics Higher Level 3rd edition 136(iii) v = 02 (cos (t) + cos (2t)) = 0EITHERcos (t) + 2 cos2 (t) 1 = 0 M1(2 cos (t) 1) (cos (t) + 1) = 0 (A1)cos (t) = 21 or cos (t) = 1 A1t = 31, t = 1 A1t = 311,37,35 t t t = 3 A1OR2 cos

,_

,_

23cos2 t t = 0 M1cos023cos or02 ,_

,_

t tA1A1t = 31, 1 A1t = 311, 3 ,37,35A1 (b) P(n) :f(2n)(x) = (1)n(Aa2n sin (ax) + Bb2n sin (bx))P(1): f(x) = (Aa cos (ax) + Bb cos (bx)) M1= Aa2 sin (ax) Bb2 sin (bx)= 1(Aa2 sin (ax) + Bb2 sin(bx)) A1P(1) trueassume thatP(k) :f(2k)(x) = (1)k [(Aa2k sin (ax) + Bb2k sin (bx)) is true M1consider P(k + 1)f(2k+1)(x) = (1)k (Aa2k+1 cos(ax) + Bb2k+1 cos(bx)) M1A1f(2k+2) (x) = (1)k (Aa2k+2 sin (ax) + Bb2k+2 sin (bx)) A1= (1)k+1 (Aa2k+2 sin (ax) + Bb2k+2 sin (bx)) A1P(k) true implies P(k + 1) true, P(1) true so P(n) true n +R1Note: Award the final R1 only if the previous three M marks have been awarded.[18]IB Questionbank Mathematics Higher Level 3rd edition 137171. EITHER'' + +5 ln 2 ln 31 ln ln5 ln ln1 ln2 3y xy xy xyxM1A1solve simultaneously'52ln57lnyxM1x = 57e (= 4.06) and y = 52e (= 1.49) A1A1 ORlnyx = 1 x = ey A1ln x3 + ln y2 = 5ln x3y2 = 5x3y2 = e5M1e3y5 = e5y5 = e2M1y = 5752e , e xA1A1[5] 172. METHOD 11 + i is a zero 1 i is a zero (A1)1 2i is a zero 1 + 2i is a zero (A1)(x (1 i))(x (1 + i)) = (x2 2x + 2) (M1)A1(x (1 2i))(x (1 + 2i)) = (x2 2x + 5) A1p(x) = (x2 2x + 2) (x2 2x + 5) M1= x4 4x3 + 11x2 14x + 10 A1a = 4, b = 11, c = 14, d = 10IB Questionbank Mathematics Higher Level 3rd edition 138METHOD 2p(1 + i) = 4 + (2 + 2i)a + (2i)b + (1 + i)c + d M1p(1 + i) = 0 ' + + + + 0 2 20 2 4c b ad c aM1A1A1p(1 2i) = 7 + 24i + (11 + 2i)a + (3 4i)b + (1 2i)c + dp(1 2i) = 0 ' + + + 0 2 4 2 240 3 11 7c b ad c b aA1

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1014114247040 2 4 21 1 3 110 1 2 21 1 0 21dcbaM1A1a = 4, b = 11, c = 14, d = 10[7] 173. (a) z = 5 and w= 24 a +w = 2z5 2 42 + aattempt to solve equation M1Note: Award M0 if modulus is not used.a = 4 A1A1 N0 (b) zw = (2 2a) + (4 + a)i A1forming equation 2 2a = 2 (4 + a) M1a = 23A1 N0[6]IB Questionbank Mathematics Higher Level 3rd edition 139174. (a) METHOD 1V = a3 31aA1x3 = 31

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aaM1= a3 3a + 31 3aa = a3

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aaa1313 (or equivalent) (A1)x xaa 31333+ V = x3 + 3x A1 N0 METHOD 2V = a3 31aA1attempt to use difference of cubes formula, x3 y3 = (x y) (x2 + xy + y2) M1V =

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+ + ,_

22111aaaa=

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+ ,_

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31 12aaaa(A1)= x(x2 + 3) or x3 + 3x A1 N0IB Questionbank Mathematics Higher Level 3rd edition 140METHOD 3diagram showing that the solid can be decomposed M1into three congruent x a a1 cuboids with volume x A1and a cube with edge x with volume x3A1so, V = x3 + 3x A1 N0 (b) Note: Do not accept any method where candidate substitutesthe given value of a into x = a a1.METHOD 1V = 4x x3 + 3x = 4x x3 x = 0 x(x 1)(x + 1) = 0 M1 x = 1 as x > 0 A1so, a 11a a2 a 1 = 0 25 1t aM1A1as a > 1, a = 25 1+AGN0IB Questionbank Mathematics Higher Level 3rd edition 141METHOD 2a3 ,_

aaa1413 = a6 4a4 + 4a2 1 = 0 (a2 1)(a4 3a2 + 1) = 0 M1A1as a > 1 22 2 225 125 3, 1

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+ + > a a aM1A125 1+ aAGN0[8] 175. (a) (i) a, 2a, 3a, ..., na are n consecutive terms of an APwith first term a and common difference aso their mean is nn nanna a a a2) 1 (... 3 2++ + + +M1A1 = 2) 1 ( + n aAG N0 (ii) 4 + 2 4 + 3 4 + ... + 4n > 2) 1 ( 4 + n + 100 M12) 1 ( 4 + n n > 2(n + 1) + 100 A12n2 + 2n > 2n + 102attempt to solve (M1)n2 > 51so the minimum value of n that satisfies the condition is 8 A1N0Note: Award M1A1(M1)A1 for use of equations if there is aclear conversion to an inequality. (b) (i) M = n my y x xn m++ + + + + ... ...1 1M1= n mn m+ + 1 0A1= n mn+AG N0IB Questionbank Mathematics Higher Level 3rd edition 142EITHERS = n mnn mnmn mn+ ,_

+ + ,_

+2 21 0M1A1attempt to simplifyS = 322 2) () ( ) (n mn m mnn mn mm n n m+++++= ) (2n mmn+A1= n mmn+AG N0 ORVar(x) = 2 1212Mn my xniimii++ M1A1attempt to simplifyM1Var(x) = 22) ( n mnn mn++=

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++ n mnn mn1= n mmn mn++= 2) ( n mmn+A1n mmnS+ AG N0IB Questionbank Mathematics Higher Level 3rd edition 143(ii) M = S n mmnn mn++A1attempt to solve M1mn n n = m, as n > 0 A1so, then the set has 2n numbers, x1, ..., xn, y1, ..., ynfrom which the first n are all 0 and the last n are all 1 (M1)hence the value of the median is 2121+ y xnA1N0[17] 176. (a) z = z, arg(z) = 0 A1A1so L(z) = ln z AGN0 (b) (i) L(1) = ln 1 + i = i A1A1N2 (ii) L(1 i) = 47i 2 ln +A1A1N2 (iii) L(1 + i) = 43i 2 ln +A1N1 (c) for comparing the product of two of the above results with the third M1for stating the result 1 + i = 1 (1 i) and L (1 + i) L (1) + L (1 i) R1hence, the property L(z1z2) = L(z1) + L(z2)does not hold for all values of z1 and z2AGN0[9] 177. (a) 1 3 iA1IB Questionbank Mathematics Higher Level 3rd edition 144(b) EITHER(z (1 + 3 i))(z (1 3 i)) = z2 2z + 4 (M1)A1p(z) = (z 2)(z2 2z + 4) (M1)= z3 4z2 + 8z 8 A1therefore b = 4, c = 8, d = 8ORrelating coefficients of cubic equations to rootsb = 2 + 1 + 3 i + 1 3 i = 4 M1c = 2 (1 + 3 i) + 2 (1 3 i) + (1 + 3 i) (1 3 i) = 8d = 2(1 + 3 i)(1 3 i) = 8b = 4, c = 8, d = 8 A1A1A1 (c) z2 = 3i33ie 2 , e 2 zA1A1A1Note: Award A1 for modulus, A1 for each argument.[8] 178. let n = 1LHS = 1 1! = 1RHS = (1 + 1)! 1 = 2 1 = 1hence true for n = 1 R1assume true for n = kkrr r1) ! ( = (k + 1)! 1 M1+11) ! (krr r = (k + 1)! 1 + (k + 1) (k + 1)! M1A1= (k + 1)!(1 + k + 1) 1= (k + 1)!(k + 2) 1 A1= (k + 2)! 1 A1hence if true for n = k, true for n = k + 1 R1since the result is true for n = 1 and P (k) P(k + 1) the result is provedby mathematical induction n +R1[8] 179. (a) any appropriate form, e.g. (cos + i sin )n = cos (n) + i sin (n) A1IB Questionbank Mathematics Higher Level 3rd edition 145(b) zn = cos n + i sin n A1nz1 = cos(n) + i sin(n) (M1)= cos n i sin (n) A1therefore zn nz1 = 2i sin (n) AG (c)5 4 3223 4 551 1451351251151

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+ ,_

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+ ,_

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+ ,_

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+ ,_

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+ ,_

z zzzzzzzz zzz(M1)(A1)= z5 5z3 + 10z 5 31 5 10z zz +A1 (d)

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+ ,_

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zzzzzzzz110151 133555M1A1(2i sin )5 = 2i sin 5 10i sin 3 + 20i sin M1A116 sin5 = sin 5 5 sin 3 + 10 sin AG (e) 16 sin5 = sin 5 5 sin 3 + 10 sin LHS = 54sin 16 ,_

= 52216

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=

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242 2A1RHS =

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+ ,_

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4sin 104 3sin 54 5sin=

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+

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221022522M1A1Note: Award M1 for attempted substitution.=

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242 2A1hence this is true for = 4AGIB Questionbank Mathematics Higher Level 3rd edition 146(f) + 20205d ) sin 10 3 sin 5 5 (sin161d sin M1= 20cos 1033 cos 555 cos1611]1

+ A1= 1]1

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+ 1035510161A1= 158A1 (g)158d cos205 , with appropriate reference to symmetry and graphs. A1R1R1Note: Award first R1 for partially correct reasoning e.g. sketchesof graphs of sin and cos.Award second R1 for fully correct reasoning involving sin5 and cos5.[22] 180. (a) i4 5i3 + 7i2 5i + 6 = 1 + 5i 7 5i + 6 M1A1= 0 AG N0 (b) i root i is second root (M1)A1moreover, x4 5x3 + 7x2 5x + 6 = (x i) (x + i) q(x)where q(x) = x2 5x + 6finding roots of q(x)the other two roots are 2 and 3 A1A1Note: Final A1A1 is independent of previous work.[6] 181. (a) (i) X = B A1 =

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0 1 11 0 10 1 01 0 01 1 00 1 11 1 10 1 10 0 1A1Y = B1 A =

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0 1 01 0 11 1 01 0 01 1 01 1 11 1 00 1 10 0 1A1IB Questionbank Mathematics Higher Level 3rd edition 147(ii) X1 + Y1 =

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0 1 01 0 10 1 0(A1)X1 + Y1 has no inverse A1as det(X1 + Y1) = 0 R1 (b) if P(n): An =

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+1 0 01 02) 1 (1nn nnfor n = 1, P(1) : A =

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+1 0 01 1 02) 1 1 ( 11 1P(1) is true A1assume P(k) is true i.e. Ak =

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+1 0 01 02) 1 (1kk kkM1for n = k + 1,Ak+1 = Ak A or AAkM1=

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+1 0 01 1 01 1 11 0 01 02) 1 (1kk kk=

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+++ + +1 0 01 1 02) 1 (1 1 1kk kk kM1A1=

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++ ++1 0 01 1 02) 2 )( 1 (1 1kk kkA1hence P(k) P(k + 1) and P(1) is true, so P(n) is true for all n +. R1N0IB Questionbank Mathematics Higher Level 3rd edition 148(c) (i) An(An)1 = I

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+1 0 00 1 00 0 11 0 01 011 0 01 02) 1 (1xy xnn nnM1

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+++ + +1 0 00 1 00 0 11 0 01 02) 1 (1n xn nnx y n xA1solve simultaneous equations to obtainx + n = 0 and y + nx + 02) 1 (+ n nM1x = n and y = 2) 1 ( n nA1A1N2 (ii) An + (An)1 =

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+

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+2 0 00 2 00 21 0 01 02) 1 (11 0 01 02) 1 (12nnn nnnn nnA1[18] 182. EITHERwith no restrictions six people can be seated in 5! = 120 ways A1we now count the number of ways in which the two restricted people will besitting next to each othercall the two restricted people p1 and p2they sit next to each other in two ways A1the remaining people can then be seated in 4! ways A1the six may be seated (p1 and p2 next to each other) in 2 4! = 48 ways M1 with p1 and p2 not next to each other the number of ways = 120 48 = 72 A1 N3ORperson p 1 seated at table in 1 way A1p2 then sits in any of 3 seats (not next to p1) M1A1the remaining 4 people can then be seated in 4! ways A1 number ways with p1 not next to p2 = 3 4! = 72 ways