Math Sl End of Year Review Answers

8/22/2019 Math Sl End of Year Review Answers

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IBSL- Review Ex. Mark Schemes

1. (a) evidence of setting function to zero (M1)e.g. f(x) = 0, 8x = 2x2

evidence of correct working A1

e.g. 0 = 2x(4 x), 4648

x-intercepts are at 4 and 0 (accept (4, 0) and (0, 0), orx = 4,x = 0) A1A1 N1N1

(b) (i) x = 2(must be equation) A1 N1

(ii) substitutingx = 2intof(x) (M1)y = 8 A1 N2

[7]

2. (a) WP=

65

13

A1A1A1 N3

Note:Award A1for each correct element.

(b) Note: The first two steps may be done in any order.subtracting (A1)

e.g.

10

12

26

2WP

multiplying WPby 2 (A1)

e.g.

12

10

26

S=

22

0

A1 N2[6]

3. (a) evidence of expanding M1e.g. 24 + 4(23)x + 6(22)x2 + 4(2)x3 +x4, (4 + 4x +x2)(4 + 4x +x2)

(2 +x)4 = 16 + 32x + 24x2 + 8x3 +x4 A2 N2

(b) finding coefficients 24 and 1 (A1)(A1)

term is 25x2 A1 N3

IB Questionbank Maths SL 1


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[6]

4. (a) tan =

x4

3acceptnotdo

4

3

A1 N1

(b) (i) sin = 53

, cos = 54

(A1)(A1)correct substitution A1

e.g. sin 2 = 2

5

4

5

3

sin 2 = 2524

A1 N3

(ii) correct substitution A1

e.g. cos 2 = 1 2

222

5

3

5

4,

5

3

cos 2 = 257

A1 N1[7]

5. (a) interchangingx andy (seen anywhere) (M1)

e.g. x = ylog (accept any base)evidence of correct manipulation A1

e.g. 3x = 21

,3, 21

== xxy ylog

3y, 2y = log

3x

f1(x) = 32x AG N0

(b) y > 0,f1(x) > 0 A1 N1

(c) METHOD 1

findingg(2) = log3

2(seen anywhere) A1

attempt to substitute (M1)

e.g.

(f1

g

)(2) =

2log33

evidence of using log or index rule (A1)

e.g. (f1g)(2) =

22log4log 33 3,3

(f1g)(2) = 4 A1 N1

METHOD 2

attempt to form composite (in any order) (M1)

e.g. (f1

g)(x) =

x3log2

3

evidence of using log or index rule (A1)



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e.g.(f1g)(x) =

2loglog 32

3 3,3xx

(f1g)(x) =x2 A1

(f1g)(2) = 4 A1 N1

[7]

6. (a) (i) p = 0.2 A1 N1

(ii) q = 0.4 A1 N1

(iii) r= 0.1 A1 N1

(b) P(AB) = 32

A2 N2

Note: Award A1 for an unfinished answer such as 3.0

2.0

.

(c) valid reason R1

e.g. 32

0.5, 0.35 0.3thus,A andB are not independent AG N0

[6]

7. (a) f(x) =x2 2x 3 A1A1A1

evidence of solvingf(x) = 0 (M1)e.g. x

2 2x 3 = 0


e.g. (x + 1)(x 3), 2162

x = 1(ignorex = 3) (A1)evidence of substituting their negativex-value intof(x) (M1)

e.g.31

3

1),1(3)1()1(

3

1 23 +

y = 35

A1

coordinates are

3

5,1

N3

(b) (i) (3, 9) A1 N1



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(ii) (1, 4) A1A1 N2

(iii) reflection gives (3, 9) (A1)

stretch gives

9,

2

3

A1A1 N3[14]

8. (a) any correct equation in the form r= a + tb (accept any parameter) A2 N2

e.g.r=

+

8

1

2

25

5

8

t

Note:Award A1 fora + tb, A1 for L = a + tb,A0 forr= b + ta.

(b) recognizing scalar product must be zero (seen anywhere) R1

e.g.a b = 0

evidence of choosing direction vectors

k2

7

,

8

1

2

(A1)(A1)

correct calculation of scalar product (A1)e.g. 2(7) + 1(2) 8k

simplification that clearly leads to solution A1e.g. 16 8k, 16 8k=0k= 2 AG N0

(c) evidence of equating vectors (M1)

e.g. L1

=L3,

+

=

+

227

305

812

2533

qp

any two correct equations A1A1

e.g.3 + 2p = 5 7q, 1 +p = 2q, 25 8p = 3 2qattempting to solve equations (M1)finding one correct parameter (p = 3, q = 2) A1

the coordinates of A are (9, 4, 1) A1 N3

(d) (i) evidence of appropriate approach (M1)

e.g.

==+149

2558

AB,OBABOA

=26

1

1

AB

A1 N2



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(ii) finding

=

2

2

7

AC

A1

evidence of finding magnitude (M1)

e.g.222 227AC ++=

57AC =A1 N3

[18]

9. (a) A1 =

=

3

10

3

23

11

3

53

52

3

13

333.00667.0

333.0167.1

67.1233.4

A2 N2

(b) evidence of attempting to solve equation (M1)e.g. multiply byA1(on left or right), setting up system of equations

X=

1

0

1

(acceptx = 1,y = 0,z= 1) A2 N3[5]

10. (a) common difference is 6 A1 N1

(b) evidence of appropriate approach (M1)e.g. u

n= 1353

correct working A1

e.g. 1353 = 3 + (n 1)6, 631353+

n = 226 A1 N2

(c) evidence of correct substitution A1

e.g.S226

= 2226

,2

)13533(226 +

(2 3 + 225 6)

S226

= 153 228(accept 153 000) A1 N1[6]



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11. (a) fx =1(2) + 2(4) + ... + 7(4), fx = 146 + 5x (seen anywhere) A1

evidence of substituting into mean =

f

fx

(M1)

correct equation A1

e.g. xx

+

+34

5146

= 4.5, 146 + 5x = 4.5(34 +x)x = 14 A1 N2

(b) = 1.54 A2 N2[6]

12. (a) (i) evidence of finding the amplitude (M1)

e.g. 2

37 +

, amplitude = 5p = 5 A1 N2

(ii) period = 8 (A1)

q = 0.785

==

4

8

2

A1 N2

(iii) r= 237

(A1)r

= 2 A1 N2

(b) k= 3(accepty = 3) A1 N1[7]

13. (a) correct substitution A1e.g. 25 + 16 40cosx, 52 + 42 2 4 5 cosx

AC = xcos4041 AG

(b) correct substitution A1

e.g.AC

2

1,

30sin

4

sin

AC=

x = 4 sinx

AC = 8 sinx

30sin

sin4accept

x

A1 N1

(c) (i) evidence of appropriate approach using AC M1

e.g. 8 sinx = xcos4041 , sketch showing intersection

correct solution 8.682..., 111.317... (A1)obtuse value 111.317... (A1)

x = 111.32 to 2 dp (do not accept the radian answer 1.94) A1 N2



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(ii) substituting value ofx into either expression for AC (M1)e.g. AC = 8 sin 111.32AC = 7.45 A1 N2

(d) (i) evidence of choosing cosine rule (M1)

e.g. cosB = acbca

2

222 +

correct substitution A1

e.g. 44245.744 222

+

, 7.452 = 32 32 cosy, cosy = 0.734...y = 137 A1 N2

(ii) correct substitution into area formula (A1)

e.g. 21

4 4 sin 137, 8 sin 137area = 5.42 A1 N2

[14]

14. (a) q = 2, r= 4 orq = 4, r= 2 A1A1 N2

(b) x = 1 (must be an equation) A1 N1

(c) substituting (0, 4) into the equation (M1)e.g.4 =p(0 (2))(0 4), 4 =p(4)(2)

correct working towards solution (A1)e.g.4 = 8p

p =

=

2

1

8

4

A1 N2[6]

15. (a) evidence of appropriate approach (M1)

e.g.

+=32

6

232

,ACBABC

=1

1

8

BC

A1 N2

(b) attempt to find the length ofAB (M1)

)7499436(3)2(6AB222 ==++=++=

(A1)



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unit vector is

=

7

37

27

6

3

2

6

7

1

A1 N2

(c) recognizing that the dot product or cos being 0 implies perpendicular (M1)

correct substitution in a scalar product formula A1

e.g. (6) (2) + (2) (3) + (3) (2), cos = 177

6612

++

correct calculation A1

e.g. ACAB = 0, cos = 0

therefore, they are perpendicular AG N0 [8]

16. (a) evidence of multiplying (M1)e.g. one correct element

AB =

5

15

A1A1 N3

(b) METHOD 1

evidence of multiplying byA (on left or right) (M1)

e.g.AA1

X= AB,X= AB

X =

5

15

(acceptx = 15,y = 5) A1 N2

METHOD 2

attempt to set up a system of equations (M1)

e.g.5

10

3,5

10

24=

+=

+ yxyx

X =5

15

(acceptx = 15,y = 5) A1 N2[5]

17. (a)cos

2

=

f

(A1)

= 1 A1 N2

(b) (gf)

2

=g(1) (= 2(1)2 1) (A1)

= 1 A1 N2



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(c) (gf)(x) = 2(cos (2x))2 1 (= 2 cos2(2x) 1) A1

evidence of 2 cos2 1 = cos 2 (seen anywhere) (M1)

(gf)(x) = cos 4x

k= 4 A1 N2

[7]18. recognizing log a + log b = log ab (seen anywhere) (A1)

e.g. log2(x(x 2)),x2 2x

recognizing logab =x ax = b (seen anywhere) (A1)

e.g. 23 = 8

correct simplification A1e.g. x(x 2) = 23,x2 2x 8

evidence of correct approach to solve (M1)e.g. factorizing, quadratic formula

correct working A1

e.g. (x 4)(x + 2), 2362

x = 4 A2 N3[7]

19. (a) (i) evidence of appropriate approach (M1)e.g. 9 + 25 + 35, 34 + 35

p = 69 A1 N2

(ii) evidence of valid approach (M1)e.g. 109 their value ofp, 120 (9 + 25 + 35 + 11)q = 40 A1 N2

(b) evidence of appropriate approach (M1)

e.g. substituting into n

fx, division by 120

mean = 3.16 A1 N2

(c) 1.09 A1 N1[7]

20. (a) evidence of equation foru27

M1

e.g. 263 = u1

+ 26 11, u27

= u1

+ (n 1) 11, 263 (11 26)

u1

= 23 A1 N1

(b) (i) correct equation A1e.g. 516 = 23 + (n 1) 11, 539 = (n 1) 11n = 50 A1 N1



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(ii) correct substitution into sum formula A1

e.g.S50

= 2)1149)23(2(50

,2

)51623(5050

+=

+S

S50

= 12325 (accept 12300) A1 N1

[6]

21. (a) 36 outcomes (seen anywhere, even in denominator) (A1)

valid approach of listing ways to get sum of 5, showing at least two pairs (M1)e.g. (1, 4)(2, 3), (1, 4)(4, 1), (1, 4)(4, 1), (2, 3)(3, 2) , lattice diagram

P(prize) =

=

9

1

36

4

A1 N3

(b) recognizing binomial probability (M1)

e.g. B

9

1,8

, binomial pdf,

53

9

8

9

1

3

8

P(3 prizes) = 0.0426 A1 N2[5]

22. evidence of substituting into binomial expansion (M1)

e.g.a5 +...2

5

1

5234 +

+

baba

identifying correct term forx4 (M1)evidence of calculating the factors, in any order A1A1A1

e.g.

2

32

2

6 2)3(10;4

,27,2

5

x

xx

x

Note:AwardA1for each correct factor.

term = 1080x4 A1 N2

Note:Award M1M1A1A1A1A0 for 1080 with working shown.

[6]

23. (a)



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A1A1A1 N3

(b) x = 1.32,x = 1.68 (acceptx = 1.41,x = 1.39 if working in degrees) A1A1 N2

(c) 1.32


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EF = 7.57 cm A1 N3

METHOD 2

attempting to find angles that are needed (M1)e.g. angle EOF and angle OEF

FOE = 0.9853... and E)FO(orFEO = 1.078... A1evidence of choosing sine rule (M1)correct substitution (A1)

e.g. 08.1sin8

sin0.985

EF=

EF = 7.57 cm A1 N3

METHOD 3

attempting to find angle EOF (M1)

e.g. 0.75 1.41FOE = 0.985 (seen anywhere) A1

evidence of using half of triangle EOF (M1)

e.g. x = 8 sin 2985.0

correct calculation A1e.g.x = 3.78

EF = 7.57 cm A1 N3[15]

25. (a) (i) (3, 4, 0) A1 N1

(ii) choosing velocity vector

1

3

2

(M1)finding magnitude of velocity vector (A1)

e.g. 194,13)2(222 ++++

speed = 3.74 ( 14 ) A1 N2

(b) (i) substitutingp = 7 (M1)B = (11, 17, 7) A1 N2

(ii) METHOD 1

appropriate method to find BAorAB (M1)

e.g. OBAO + , A B



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=

=

7

21

14

BAor

7

21

14

AB

(A1)

distance = 26.2 )147( A1 N3

METHOD 2

evidence of applying distance is speed time (M2)e.g. 3.74 7

distance = 26.2 )147( A1 N3

METHOD 3

attempt to find AB2, AB (M1)

e.g. (3 (11))2

+ (4 17)2

+ (0 7)2

,

222 )70()174())11(3( ++

AB2 = 686, AB = 686 (A1)

distance AB = 26.2 ( 147 ) A1 N3

(c) correct direction vectors

a

2

1

and

1

3

2

(A1)(A1)

+=

aa

a2

1

13

2

,52

12

= a + 8 (A1)(A1)substituting M1

e.g. cos 40 = 514

8

2 +

+

a

a

a = 3.21, a = 0.990 A1A1 N3[16]

26. (a) (i) g(0) = e0 2 (A1)= 1 A1 N2

(ii) METHOD 1

substituting answer from (i) (M1)e.g. (f

g)(0) =f(1)

correct substitution f(1) = 2(1)3 + 3 (A1)f(1) = 1 A1 N3

METHOD 2



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attempt to find (fg)(x) (M1)

e.g. (fg)(x) =f(e3x 2) = 2(e3x 2)3 + 3

correct expression for (fg)(x) (A1)

e.g. 2(e3x 2)3+ 3

(fg)(0) = 1 A1 N3

(b) interchangingx andy (seen anywhere) (M1)e.g. x = 2y3 + 3

attempt to solve (M1)

e.g. y3 = 2

3x

f1(x) =

3

2

3x

A1 N3[8]

27. (a) evidence of equating scalar product to 0 (M1)2 3 + 3 (1) + (1) p = 0 (6 3 p = 0, 3 p = 0) A1

p = 3 A1 N2

(b) evidence of substituting into magnitude formula (M1)

e.g. 2512 ++ q , 1 + q2 + 25

setting up a correct equation A1

e.g. 422512 =++ q , 1 + q2 + 25 = 42, q2 = 16

q = 4 A1 N2[6]

28. (a) A2 N2

(b) evidence of appropriate approach (M1)



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e.g. reference to any horizontal shift and/or stretch factor,x = 3 + 1,y =2

2

1

P is (4, 1) (acceptx = 4,y = 1) A1A1 N3[5]

29. evidence of substituting for cos2x (M1)evidence of substituting into sin2x + cos2x = 1 (M1)correct equation in terms of cosx (seen anywhere) A1

e.g. 2cos2x 1 3 cosx 3 = 1, 2 cos2x 3 cosx 5 = 0

evidence of appropriate approach to solve (M1)e.g. factorizing, quadratic formula

appropriate working A1

e.g. (2 cosx 5)(cosx + 1) = 0, (2x 5)(x + 1), cosx = 4493

correct solutions to the equation

e.g. cosx = 25

, cosx = 1,x = 25

,x = 1 (A1)

x = A1 N4[7]

30. (a) METHOD 1

recognizing thatf(8) = 1 (M1)

e.g. 1 = klog2 8recognizing that log

28 = 3 (A1)

e.g. 1 = 3k

k= 31

A1 N2

METHOD 2

attempt to find the inverse off(x) = klog2x (M1)

e.g. x = klog2y,y = k

x

2

substituting 1 and 8 (M1)

e.g. 1 = klog2

8, k1

2 = 8

k=

=

3

1

8log

1

2

k

A1 N2



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(b) METHOD 1

recognizing thatf(x) = 32

(M1)

e.g.

x2log3

1

3

2=

log2x = 2 (A1)

f1

3

2

= 4 (acceptx = 4) A2 N3

METHOD 2

attempt to find inverse off(x) = 31

log2x (M1)

e.g. interchangingx andy , substituting k= 31

intoy = kx

2

correct inverse (A1)e.g. f

1(x) = 23x, 23x

f1

3

2

= 4 A2 N3[7]

31. (a) (i) evidence of substituting into n(A B) = n(A) + n(B) n(A B) (M1)

e.g. 75 + 55 100, Venn diagram30 A1 N2

(ii) 45 A1 N1

(b) (i) METHOD 1

evidence of using complement, Venn diagram (M1)e.g. 1 p, 100 30

= 107

100

70

A1 N2

METHOD 2

attempt to find P(only one sport), Venn diagram (M1)

e.g. 10045

100

25+

=

10

7

100

70

A1 N2



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(ii)

=

14

9

70

45

A2 N2

(c) valid reason in words or symbols (R1)e. g. P(A B) = 0 if mutually exclusive, P(A B) if not mutually exclusive

correct statement in words or symbols A1 N2e.g. P(A B) = 0.3, P(A B) P(A) + P(B), P(A) + P(B) > 1, somestudents play both sports, sets intersect

(d) valid reason for independence (R1)e.g. P(A B) = P(A) P(B), P(BA) = P(B)

correct substitution A1A1 N3

e.g. 10075

55

30,

100

55

100

75

100

30

[12]

32. METHOD 1

substituting into formula forS40

(M1)


e.g. 1900 = 2)106(40 1 +u

u1

= 11 A1 N2

substituting into formula foru40

orS40

(M1)


e.g. 106 = 11 + 39d, 1900 = 20(22 + 39d)d= 3 A1 N2

METHOD 2

substituting into formula forS40

(M1)

correct substitution A1e.g. 20(2u

1+ 39d) = 1900

substituting into formula foru40

(M1)

correct substitution A1e.g. 106 = u1 + 39du

1= 11, d= 3 A1A1 N2N2

[6]

33. (a) A1 =

968

212

323

A2 N2

(b) evidence of subtracting matrices (M1)



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e.g.

15510

1221

8410

,

1029

435

123

571

856

767

,D C

evidence of multiplying on leftbyA1 (M1)

e.g.A1AB,A1(D C),

15510

1221

8410

968

212

323

B =

114

201

312

A2 N36]

34. (a) choosing sine rule (M1)correct substitution A1

e.g. 3.0sin4

sin0.8

AD=

AD = 9.71 (cm) A1 N2

(b) METHOD 1

finding angle OAD = 1.1 = (2.04) (seen anywhere) (A1)

choosing cosine rule (M1)correct substitution A1e.g. OD2 = 9.712 + 42 2 9.71 4 cos( 1.1)

OD = 12.1 (cm) A1 N3

METHOD 2

finding angle OAD = 1.1 = (2.04) (seen anywhere) (A1)

choosing sine rule (M1)correct substitution A1

e.g. 3.0sin

4

8.0sin

71.9

1.1)sin(

OD==

OD = 12.1 (cm) A1 N3

(c) correct substitution into area of a sector formula (A1)e.g. area = 0.5 42 0.8

area = 6.4 (cm2) A1 N2

(d) substitution into area of triangle formula OAD (M1)correct substitution A1

e.g. A = 21

4 12.1 sin 0.8,A = 21

4 9.71 sin 2.04,



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A = 21

12.1 9.71 sin 0.3

subtracting area of sector OABC from area of triangle OAD (M1)e.g. area ABCD = 17.3067 6.4

area ABCD = 10.9 (cm2) A1 N2[13]

35. (a) evidence of correct approach A1

e.g.

=

51

2

83

3,OPOQPQ

=3

2

1

PQ

AG N0

(b) (i) correct description R1 N1

e.g. reference to

83

3

being the position vector of a point on the line,a vector to the line, a point on the line.

(ii) any correct expression in the form r= a + tb A2 N2

where a is

83

3

, and b is a scalar multiple of

32

1

e.g.r=

++

=

+

s

s

s

t

6843

23,

321

83

3r

(c) one correct equation (A1)e.g. 3 +s = 1, 3 2s = 5

s = 4 A1p = 4 A1 N2

(d) one correct equation A1e.g. 3 + t= 1, 9 2t= 5

t= 2 A1substituting t= 2e.g. 2 + 2q = 4, 2q = 6 A1

q = 3 AG N0



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(e) choosing correct direction vectors

32

1and

32

1

(A1)(A1)finding correct scalar product and magnitudes (A1)(A1)(A1)

scalar product (1)(1) + (2)(2) + (3)(3) (= 4)

magnitudes 14)3()2(1,143)2(1232222 =++=++

evidence of substituting into scalar product M1

e.g. cos = ...741.3...741.34

= 1.86 radians (or 107) A1 N4[17]

36. (a) evidence of addition (M1)

e.g. at least two correct elements

A +B =

01

24

A1 N2

(b) evidence of multiplication (M1)

e.g. at least two correct elements

3A =

39

63

A1 N2

(c) evidence of matrix multiplication (in correct order) (M1)

e.g.AB =

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

++++

11032133

12012231

AB =

111

21

A2 N3[7]

37. (a) (i) sin 140 =p A1 N1

(ii) cos 70 = q A1 N1

(b) METHOD 1

evidence of using sin2 + cos2 = 1 (M1)

e.g. diagram,21 p (seen anywhere)

cos 140 = 21 p (A1)

cos 140 = 2

1 p A1 N2

METHOD 2



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evidence of using cos2 = 2 cos2 1 (M1)

cos 140 = 2 cos2 70 1 (A1)

cos 140 = 2(q)21 (= 2q2 1) A1 N2

(c) METHOD 1

tan 140 =21140cos

140sin

p

p

=

A1 N1

METHOD 2

tan 140 = 122 qp

A1 N1[6]

38. (a) d= 3 (A1)

evidence of substitution into un

= a + (n 1) d (M1)

e.g.u101

= 2 + 100 3

u101

= 302 A1 N3

(b) correct approach (M1)

e.g. 152 = 2 + (n 1) 3

correct simplification (A1)

e.g. 150 = (n 1) 3, 50 = n 1, 152 = 1 + 3n

n = 51 A1 N2[6]

39. (a) period = A1 N1

(b)

43

2

1

1

2

3

4

y

x0 2

32 2

A1A1A1 N3Note: Award A1 for amplitude of 3, A1 fortheirperiod, A1 for a sine curve passing through

(0, 0) and (0, 2).



22/51

(c) evidence of appropriate approach (M1)

e.g. liney = 2 on graph, discussion of number of solutions inthe domain

4 (solutions) A1 N2

[6]40. (a) METHOD 1

ln (x + 5) + ln 2 = ln (2(x + 5)) (= ln (2x + 10)) (A1)

interchangingx andy (seen anywhere) (M1)

e.g.x = ln (2y + 10)

evidence of correct manipulation (A1)

e.g. ex = 2y + 10

( )2

101 =xe

xf

A1 N2METHOD 2

y = ln (x + 5) + ln 2

y ln 2 = ln (x + 5) (A1)

evidence of correct manipulation (A1)

e.g. ey ln 2 =x + 5

interchangingx andy (seen anywhere) (M1)

e.g. ex ln 2 =y + 5

f1 (x) = ex ln 2 5 A1 N2

(b) METHOD 1

evidence of composition in correct order (M1)

e.g. (gf) (x) =g(ln (x + 5) + ln 2)

= eln (2(x + 5)) = 2(x + 5)

(g f) (x) = 2x + 10 A1A1 N2

METHOD 2

evidence of composition in correct order (M1)

e.g. (gf) (x) = eln(x + 5) + ln 2

= eln (x + 5) eln 2 = (x + 5) 2

(g f) (x) = 2x + 10 A1A1 N2[7]

41. (a) f(x) = 3(x2 + 2x + 1) 12 A1

= 3x2 + 6x + 3 12 A1

= 3x2 + 6x 9 AG N0



23/51

(b) (i) vertex is (1, 12) A1A1 N2

(ii) x = 1 (must be an equation) A1 N1

(iii) (0, 9) A1 N1

(iv) evidence of solvingf(x) = 0 (M1)e.g. factorizing, formula,

correct working A1

e.g. 3(x + 3)(x 1) = 0, 6108366 +

=x

(3, 0), (1, 0) A1A1 N1N1

(c)

y

x

3

9

1 2

1

A1A1 N2Notes: Award A1 for a parabola opening upward,

A1 for vertex and intercepts in

approximately correct positions.

(d),

12

1

=

q

p

t= 3 (acceptp = 1, q = 12, t= 3) A1A1A1 N3[15]

42. (a) (i) P(B) = 43

A1 N1

(ii) P(R) = 41

A1 N1

(b) 43

=pA1 N1

4

3,

4

1== ts

A1 N1

(c) (i) P(X= 3)

= P (getting 1 and 2) = 41

43

A1



24/51

= 163

AG N0

(ii) P(X= 2) = 41

41

+ 43

16

31or

(A1)

= 1613

A1 N2

(d) (i)

X 2 3

P(X=x)16

13

16

3

A2 N2

(ii) evidence of using E(X) = xP(X=x) (M1)

E(X) =

+

16

33

16

132

(A1)

=

=

16

32

16

35

A1 N2

(e) win $10 scores 3 one time, 2 other time (M1)

P(3) P(2) = 1613

163

(seen anywhere) A1

evidence of recognizing there are different ways of winning $10 (M1)

e.g. P(3) P(2) + P(2) P(3),,

16

3

16

132

256

3

256

36

256

3

256

36+++

P(win $10) =

=

128

39

256

78

A1 N3[16]

43. (a) (i) p = 65 A1 N1

(ii) for evidence of using sum is 125 (or 99 p) (M1)

q = 34 A1 N2

(b) evidence of median position (M1)

e.g. 63rd student, 2125

median is 17 (sit-ups) A1 N2



25/51

(c) evidence of substituting into 125

)( xf(M1)

( ) ( ) ( ) ( ) ( ) ( )

125

2176,

125

82018193418331721161115..

+++++ge

mean = 17.4 A1 N2[7]

44. (a) choosing sine rule (M1)

correct substitution 1075sin

7

sin =

R

A1

sinR = 0.676148...

QRP

= 42.5

A1 N2(b) P= 180 75 R

P= 62.5 (A1)

substitution into any correct formula A1

e.g. area PQR =sin107

2

1

(theirP)

= 31.0 (cm2) A1 N2[6]

45. (a) evidence of appropriate approach M1

e.g. 3 = 92

r

r=13.5 (cm) A1 N1

(b) adding two radii plus 3 (M1)

perimeter = 27+3 (cm) (= 36.4) A1 N2

(c) evidence of appropriate approach M1

e.g. 92

5.132

1 2

area = 20.25 (cm2) (= 63.6) A1 N1

[6]

46. (a)



26/51

1 5

1 0

5

0

5

3

3

y

x

A1A1A1 N3Note: Award A1 for passing through (0, 0), A1

for correct shape, A1 for a range of

approximately 1 to 15.

(b) evidence of attempt to solvef(x) = 1 (M1)

e.g. line on sketch, using xx

xcos

sintan =

x = 0.207 x = 0.772 A1A1 N3[6]

47. (a) evidence of binomial distribution (may be seen in parts (b) or (c)) (M1)

e.g. np, 100 0.04

mean = 4 A1 N2

(b) P(X= 6) =( ) ( ) 946 96.004.0

6

100

(A1)

= 0.105 A1 N2

(c) for evidence of appropriate approach (M1)

e.g. complement, 1 P(X= 0)

P(X= 0) = (0.96)100 = 0.01687... (A1)

P(X 1) = 0.983 A1 N2[7]

48. pw =pi+ 2pj 3pk(seen anywhere) (A1)

attempt to find v +pw (M1)

e.g.

3i

+ 4j

+k

+p

(i

+ 2j

3k

)collecting terms (3 +p)i+ (4 + 2p)j+ (1 3p) k A1

attempt to find the dot product (M1)



27/51

e.g. 1(3 +p) + 2(4 + 2p) 3(1 3p)

setting their dot product equal to 0 (M1)

e.g. 1(3 +p) + 2(4 + 2p) 3(1 3p) = 0

simplifying A1

e.g.

3 +p

+ 8 + 4p

3 + 9p

= 0, 14p

+ 8 = 0

P= 0.571

14

8

A1 N3[7]

49. (a) (i) evidence of approach M1

e.g.AO +

OB =

AB, B A

AB =

1

6

4

AG N0

(ii) for choosing correct vectors, (

AOwithAB , or

OA with

BA ) (A1)(A1)

Note: Using AO with BA will lead to

0.799. If they then say OAB

= 0.799, this is a correct solution.

calculatingAO AB ,

AB,AO (A1)(A1)(A1)

e.g.d1d

2= (1)(4) + (2)(6) + (3)(1) (= 19)

( ) ( ) ( ) ,14321 2221 =++=d

( ) ( ) ( )53164 2222 =++=d

evidence of using the formula to find the angle M1

e.g. cos =

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ),

164321

136241

222222 ++++

++

...69751.0,5314

19

OAB = 0.799 radians (accept 45.8) A1 N3

(b) two correct answers A1A1

e.g. (1, 2, 3), (3, 4, 2), (7, 10, 1), (11, 16, 0) N2



28/51

(c) (i) r=

+

2

4

3

3

2

1

t

A2 N2

(ii) C onL2, so

+

=

2

4

3

3

2

1

5

tk

k

(M1)

evidence of equating components (A1)

e.g. 1 3t= k, 2 + 4t= k, 5 = 3 + 2t

one correct value t= 1, k= 2 (seen anywhere) (A1)

coordinates of C are (2, 2, 5) A1 N3

(d) for setting up one (or more) correct equation using

+

=

12

1

08

3

52

2

p

(M1)

e.g. 3 +p = 2, 8 2p = 2, p = 5

p = 5 A1 N2[18]

50. (a) evidence of using fi= 100 (M1)

k= 4 A1 N2

(b) (i) evidence of median position (M1)

e.g. 50th item, 26 + 10 + 20 = 56

median = 3 A1 N2

(ii) Q1

= and Q3

= 5 (A1)(A1)

interquartile range = 4 (accept 1 to 5 or 51, etc.) A1 N3[7]

51. (a) evidence of attempting to solvef(x) = 0 (M1)


e.g.( ) ( )

2

91,21

+ xx

intercepts are (1, 0) and (2, 0) (acceptx = 1,x = 2) A1A1 N1N1

(b) evidence of appropriate method (M1)

e.g.,

2,

221

abxxxx vv =+= reference to symmetry

xv

= 0.5 A1 N2



29/51

[6]

52. (a) detM= 4 A1 N1

(b) M1 =

=

2

1

2

14

1

4

1

22

11

4

1

A1A1 N2

Note: Award A1 for 4

1

and A1 for the correct

matrix.

(c) X=M1

=

8

4

22

11

4

1

8

4X

M1

X=( )2,3

2

3==

yx

A1A1 N0Note: Award no marks for an algebraic solution

of the system 2x + y = 4, 2x y = 8.

[6]

53. (a) evidence of choosing the formula cos

2

A = 2 cos

2

A 1 (M1)Note: If they choose another correct formula, donot award the M1 unless there is evidence

of finding sin2A = 1 9

1

.


e.g. cos 2A =1

3

122cos,

9

8

3

122

=

A

9

7

2cos =A A1 N2

(b) METHOD 1

evidence of using sin2B + cos2B = 1 (M1)

e.g. 95

,1cos3

2 22

=+

B

(seen anywhere),

cosB =

=3

5

9

5

(A1)



30/51

cosB =

=3

5

9

5

A1 N2

METHOD 2

diagram M1

e.g.

for finding third side equals 5 (A1)

cosB = 35

A1 N2

[6]

54. (a) (i) correct calculation (A1)

e.g. 203324

,20

2

20

5

20

9 ++++

P(male or tennis) =

=

5

3

20

12

A1 N2

(ii) correct calculation (A1)

e.g. 1133

,20

11

20

6 +

P(not football | female) = 116

A1 N2

(b) P(first not football) = 2011

, P(second not football) = 1910

A1

P(neither football) = 2011

1910

A1

P(neither football) =

=

38

11

380

110

A1 N1 [7]

55. (a)



31/51

4

3

2

1

1

2

3

4

4321 1 2 0

y

x

M1A1 N2

Note: Award M1 for evidence of reflection inx-axis, A1 for correct vertex andall

intercepts approximately correct.

(b) (i) g(3) =f(0) (A1)

f(0) = 1.5 A1 N2

(ii) translation (accept shift, slide, etc.) of

0

3

A1A1 N2[6]

56. (a) (i) evidence of combining vectors (M1)

e.g.AB =

OB

OA (or

AD =

AO +

OD in part (ii))

AB =

2

4

2

A1 N2

(ii)AD =

2

5

2

k

A1 N1

(b) evidence of using perpendicularity scalar product = 0 (M1)

0

2

5

2

2

4

2

.. =

kge

4 4(k

5) + 4 = 0 A14k+ 28 = 0 (accept any correct equation clearly leading to k= 7) A1

k= 7 AG N0



32/51

(c)AD =

22

2

(A1)

BC =

111

A1

evidence of correct approach (M1)

e.g.

=

+

+=

1

1

1

2

1

3

,

1

1

1

2

1

3

,BCOBOC

z

y

x

OC =

12

4

A1 N3

(d) METHOD 1

choosing appropriate vectors,BC,BA (A1)

finding the scalar product M1

e.g.2(1) + 4(1) + 2(1), 2(1) + (4)(1) + (2)(1)

cos CBA = 0 A1 N1

METHOD 2

BC parallel to

AD (may show this on a diagram with points labelled) R1

BC

AB (may show this on a diagram with points labelled) R1

CBA = 90

cos CBA = 0 A1 N1[13]

57. (a) evidence of using area of a triangle (M1)

e.g.A sin22

2

1=

A = 2 sin A1 N2



33/51

(b) METHOD 1

AOP = (A1)

area OPA =( ) sin22

2

1

(= 2 sin ()) A1

since sin () = sin R1then both triangles have the same area AG N0

METHOD 2

triangle OPA has the same height and the same base as triangle OPB R3

then both triangles have the same area AG N0

(c) area semi-circle =( ) ( )= 22

2

1 2

A1

area APB = 2 sin + 2 sin (= 4 sin ) A1S= area of semicircle area APB (= 2 4 sin ) M1

S= 2( 2 sin ) AG N0

(d) METHOD 1

attempt to differentiate (M1)

e.g.

Scos4

d

d=

setting derivative equal to 0 (M1)

correct equation A1

e.g.4 cos = 0, cos = 0, 4 cos = 0

= 2

A1 N3

EITHER

evidence of using second derivative (M1)

S() = 4 sin A1

S4

2=

A1

it is a minimum because S0

2>

R1 N0

OR

evidence of using first derivative (M1)



34/51

for,

2

S() > 0 (may use diagram) A1

it is a minimum since the derivative goes from negativeto positive R1 N0

METHOD 2

2 4 sin is minimum when 4 sin is a maximum R3

4 sin is a maximum when sin = 1 (A2)

= 2

A3 N3

(e) Sis greatest when 4 sin is smallest (or equivalent) (R1)

= 0 (or) A1 N2[18]

58. (a) evidence of dividing two terms (M1)

e.g. 10801800

,3000

1800

r = 0.6 A1 N2

(b) evidence of substituting into the formula for the 10th term (M1)

e.g.u10

= 3000( 0.6)9

u10

= 30.2 (accept the exact value 30.233088) A1 N2

(c) evidence of substituting into the formula for the infinite sum (M1)

6.1

3000.. =Sge

S= 1875 A1 N2

[6]

59. evidence of using binomial expansion (M1)

e.g. selecting correct term,...

2

8

1

826708 +

+

+ bababa

evidence of calculating the factors, in any order A1A1A1

e.g. 56,( ) 5

3

5

3

3

33

2

5

8,3,

3

2

x

4032x3 (accept = 4030x 3 to 3 s.f.) A1 N2[5]

60. (a) intercepts whenf(x) = 0 (M1)



35/51

(1.54, 0) (4.13, 0) (acceptx = 1.54 x = 4.13) A1A1 N3

(b)

3

2

1

1

2

3

4

5

6

7

8

9

1 0

6543210 1 2

y

x

A1A1A1 N3Note: Award A1 for passing through

approximately (0, 4), A1 for correct

shape, A1 for a range of approximately

9 to 2.3.

(c) gradient is 2 A1 N1[7]

61. (a) evidence of binomial distribution (seen anywhere) (M1)

e.g.X~ B

4

1,3

mean =( )75.0

4

3=

A1 N2

(b) P(X= 2) =

4

3

4

1

2

3 2

(A1)

P(X= 2) = 0.141

=

64

9

A1 N2

(c) evidence of appropriate approach M1

e.g. complement, 1 P(X= 0), adding probabilities

P(X= 0) = (0.75)3= 64

27,422.0

(A1)



36/51

P(X 1) = 0.578

=

64

37

A1 N2[7]

62. evidence of equating vectors (M1)

e.g.L1

=L2

for any two correct equations A1A1

e.g. 2 +s = 3 t, 5 + 2s = 3 + 3t, 3 + 3s = 8 4t

attempting to solve the equations (M1)

finding one correct parameter (s = 1, t= 2) A1

the coordinates of T are (1, 3, 0) A1 N3[6]

63. (a) (i) 7 A1 N1

(ii) 1 A1 N1

(iii) 10 A1 N1

(b) (i) evidence of appropriate approach M1

e.g. 2218

=A

A = 8 AG N0

(ii) C= 10 A2 N2

(iii) METHOD 1

period = 12 (A1)

evidence of usingB period = 2 (accept 360) (M1)

e.g. 12 = B2

6

=B (accept 0.524 or 30) A1 N3

METHOD 2

evidence of substituting (M1)

e.g. 10 = 8 cos 3B + 10

simplifying (A1)

e.g. cos 3B = 0

=

23B

6

=B

(accept 0.524 or 30) A1 N3



37/51

(c) correct answers A1A1

e.g.t= 3.52, t= 10.5, between 03:31 and 10:29 (accept 10:30) N2[11]

64. (a) (i) n = 5 (A1)T= 280 1.125

T= 493 A1 N2

(ii) evidence of doubling (A1)

e.g. 560

setting up equation A1

e.g. 280 1.12n = 560, 1.12n = 2

n = 6.116... (A1)

in the year 2007 A1 N3

(b) (i)( )51.0e9010

0005602+

=P(A1)

P= 39 635.993... (A1)

P= 39 636 A1 N3

(ii)

( )71.0

e9010

0005602

+

=P

P= 46 806.997... A1

not doubled A1 N0

valid reason fortheir answer R1

e.g.P< 51200

(c) (i) correct value A2 N2

7:640,4.91,

280

25600..ge

(ii) setting up an inequality (accept an equation, or reversedinequality) M1

( )70

12.1280e9010

0005602,70..

0.1

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Math Sl End of Year Review Answers