Ib 1 Review Solutions - End of Year Exam 2012

8/2/2019 Ib 1 Review Solutions - End of Year Exam 2012

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1

SOLUTIONS

35. (a) For taking an appropriate ratio of consecutive terms (M1)

r= A1 N2

(b) For attempting to use the formula for the nth

term of a GP (M1)u15 = 1.39 A1 N2

(c) For attempting to use infinite sum formula for a GP (M1)

S = 1215 A1 N2[6]

36. (a) 5000(1.063)n A1 1

(b) Value = $5000(1.063)5 (= $6786.3511...)= $6790 to 3 sf (Accept $6786, or $6786.35) A1 1

(c) (i) 5000(1.063)n > 10000 or (1.063)n > 2 A1 1

(ii) Attempting to solve the inequality log (1.063) > log 2 (M1)n > 11.345... (A1)12 years A1 3

Note: Candidates are likely to use TABLE or LIST

on a GDC to find n. A good way of communicating

this is suggested below.

Lety = 1.063x (M1)Whenx = 11,y = 1.9582, whenx = 12,y = 2.0816 (A1)

x = 12 ie 12 years A1 3[6]

37. (a) (i) S4 = 20 A1 N1

(ii) u1 = 2, d= 2 (A1)

Attempting to use formula for Sn M1

S100 = 10100 A1 N2

3

2


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(b) (i) M2 = A2 N2

(ii) For writingM3 asM2MorMM2 M1

M3 = A2

M3 = AG N0

(c) (i) M4 = A1 N1

(ii) T4 = (M1)

= A1A1 N3

(d) T100 = (M1)

= A1A1 N3

[16]

38. (a) For taking three ratios of consecutive terms (M1)

A1

hence geometric AG N0

10

41

10

41

10

21or

1000

2401

10

61

10

81

10

81

10

61

10

41

10

21

40

204

10

2001...

10

41

10

21

1000

10100100

3162

486

54

162

18

54


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(b) (i) r= 3 (A1)

un = 18 3n 1 A1 N2

(ii) For a valid attempt to solve 18 3 n 1 = 1062882 (M1)

eg trial and error, logs

n = 11 A1 N2[6]

39. (a) 3, 6, 9 A1 N1

(b) (i) Evidence of using the sum of an AP M1

eg

A1 N1

(ii) METHOD 1

Correct calculation for (A1)

eg

Evidence of subtraction (M1)

eg 15150

630

A1 N2

METHOD 2

Recognising that first term is 63, the number of terms is 80 (A1)(A1)

eg

A1 N2

[6]

3120322

20

20

16303

nn

100

1

3n

n

15150,399322

100

145203100

21

n

n

3791262

80,30063

2

80

145203100

21

n

n


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40. (a) (i) logc 15 = logc 3 + logc 5 (A1)

=p + q A1 N2

(ii) logc 25 = 2 logc 5 (A1)

= 2q A1 N2

(b) METHOD 1

M1

d= 36 A1 N1

METHOD 2

For changing base M1

eg

d= 36 A1 N1

[6]

41. (a) (A1)

(A1) (C2)

Note:If candidates have an incorrect or no answer to part (a) award(A1)(A0)

if seen in part (b).

621

d

dd

1010

10

10 log6log2,2

1

log

6log

3 3 3log log ( 5) log5

xx x

x

5

xA

x

5log

x

x


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(b) EITHER

(M1)(A1)(A1)

(A1) (C4)

OR

(M1)(A1)

(A1)

(A1) (C4)[6]

42. (a) ln a3b = 3ln a + ln b (A1)(A1)

ln a3b = 3p + q A1 N3

(b) ln ln a ln b (A1)(A1)

ln pq A1 N3

[6]

3log 15

x

x

13 35

x

x

3 15x x

2 15x

15

2x

10

10

log5

1log 3

x

x

10 10log log 3

5

x

x

7.5x

2

1

b

a

2

1

b

a


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43. 9x1 =

32x2 = 32x (M1) (A1)2x2 =2x (A1)

x = (A1) (C4)

[4]

44. (a) (i) m = 3 A2 N2

(ii) p = 2 A2 N2

(b) Appropriate substitution M1

eg 0 = d(1 3)2 + 2, 0 = d(5 3)2 + 2, 2 = d(3 1)(3 5)

A1 N1[6]

45. (a) A1 N1

(b) (i) Attempt to form composite (f g) (x) =f(ln (1 + 2x)) (M1)

(f g) (x) = eln (1 + 2x) = (= 1 + 2x) A1 N2

(ii) Simplifyingy = eIn(1 + 2x) toy = 1 + 2x (may be seen in part(i) or later) (A1)

Interchangingx andy (may happen any time) M1

eg x = 1 + 2y x 1 = 2y

(f g)1 (x) = A1 N2

[6]

46. (a) 253250 (accept 253000) A1 N1

x2

3

1

2

1

2

1

d

xxf ln1

2

1x


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(b) 1972 2002 is 30 years, increase of 1.3% 1.013 (A1)(A1)

Evidence of any appropriate approach (M1)

Correct substitution 250000 1.01330 A1368000 (accept 368318) A1 N3

[6]

47. (a) (i) h = 3 A1 N1

(ii) k= 1 A1 N1

(b) g (x) =f(x 3) + 1, 5 (x 3)2 + 1, 6 (x 3)2, x2 + 6x 3 A2 N2

(c)

M1A1 N2

Note: Award M1 for attempt to reflect through

y-axis, A1 for vertex at approximately (3, 6).[6]

48. (a) (i) n = 5 (A1)

T= 280 1.125

T= 493 A1 N2

(ii) evidence of doubling (A1)eg 560setting up equation A1

eg 280 1.12n = 560, 1.12n = 2n = 6.116... (A1)in the year 2007 A1 N3

(b) (i) (A1)

P = 39 635.993... (A1)

P = 39 636 A1 N3

8 8

V

T

0

y

x

51.0e9010

2560000

P


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(ii)

P = 46 806.997... A1not doubled A1 N0valid reason for their answer R1egP < 51200

(c) (i) correct value A2 N2

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

finding the value 9.31.... (A1)

after 10 years A1 N2[17]

49. (a) METHOD 1

f(3) = (A1)

(gf) (3) = 7 A1 N2

METHOD 2

(gf) (x) = (=x + 4) (A1)

(gf) (3) = 7 A1 N2

(b) For interchangingx andy (seen anywhere) (M1)Evidence of correct manipulation A1

eg x =

f1(x) =x2 4 A1 N2

71.0e9010

2560000

P

7:640,4.91,280

25600..ge

70

12.1280e9010

2560000,70..

0.1n

nT

Pge

7

24x

4,4 2 yxy


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(c) x 0 A1 N1[6]

50. (a) when (may be implied by a sketch) (A1)

(A1) (C2)

(b) METHOD 1

Sketch of appropriate graph(s) (M1)Indicating correct points (A1)

(A1)(A1)(C2)(C2)

METHOD 2

, (A1)(A1)

,

, (A1)(A1)(C2)(C2)

[6]

51. a = 4, b = 2, c = A2A2A2 N6

[6]

52. (a) For using perimeter = r+ r+ arc length (M1)

20 = 2r + r A1

AG N0

0y

8or 2.79

9x

3.32 or 5.41x x

1sin

9 2x

7

9 6x

11

9 6x

7

6 9x

11

6 9x

19

18x

31

18x ( 3.32 , 5.41)x x

etc

2

3or

2

r

r220


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(b) FindingA = (A1)

For setting up equation in r M1Correct simplified equation, or sketch

eg 10rr2 = 25, r210r+ 25 = 0 (A1)r= 5 cm A1 N2

[6]

53. METHOD 1

Area sector OAB (M1)

(A1)

ON (A1)

AN (A1)

Area of

(A1)

Shaded area

(A1) (C6)

22 102202

1rr

r

rr

21 (5) (0.8)2

10

5cos0.8 3.483...

5sin 0.8 3.586.....

1AON ON AN

2

6.249... 2(cm )

10 6.249..

3.752(cm )


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METHOD 2

Area sector (M1)

(A1)

Area (M1)

(A1)

Twice the shaded area (M1)

Shaded area

(A1) (C6)

[6]

54.sinA =

cosA =

(A1)

ButA is obtuse cosA = (A1)

sin 2A = 2 sinA cosA (M1)

= 2

= (A1) (C4)

[4]

55. (a) 3 sin2x + 4 cosx = 3(1cos2x) + 4cosx

= 33 cos2 + 4 cosx (A1) (C1)

A

O N B

F

21ABF (5) (1.6)

2

20

21OAF (5) sin1.62

12.5

20 12.5 ( 7.5)

1(7.5)

2

3.752(cm )

13

5

13

12

13

12

13

12

13

5

169

120


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(b) 3 sin2x + 4 cosx4 = 0 33 cos2x + 4 cosx4 = 0

3 cos2x4 cosx + 1 = 0 (A1)(3 cosx1)(cosx1) = 0

cosx = or cosx = 1

x = 70.5 orx = 0 (A1)(A1) (C3)

Note:Award (C1) for each correct radian answer, ie x = 1.23 or x = 0.[4]

56. Note: Throughout this question, donot accept methods which involve

finding .

(a) Evidence of correct approach A1

eg sin =

sin

= AG N0

(b) Evidence of using sin 2= 2 sin cos (M1)

= A1

= AG N0

(c) Evidence of using an appropriate formula for cos 2 M1

eg

cos 2= A2 N2

[6]

57. (a) Evidence of choosing the double angle formula (M1)

f(x) = 15 sin (6x) A1 N2

3

1

523BC,AB

BC 22

3

5

3

2

3

52

9

54

81

801,

9

521,1

9

42,

9

5

9

4

9

1


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(b) Evidence of substituting forf(x) (M1)

eg 15 sin 6x = 0, sin 3x = 0 and cos 3x = 0

6x = 0, , 2

x = 0, A1A1A1 N4

[6]

58. (a) METHOD 1

Using the discriminant = 0 (M1)

k2 = 4 4 1

k= 4, k= 4 A1A1 N3

METHOD 2

Factorizing (M1)

(2x 1)2

k= 4, k= 4 A1A1 N3

(b) Evidence of using cos 2= 2 cos2 1 M1

eg 2(2 cos2 1) + 4 cos + 3

f() = 4 cos2 + 4 cos + 1 AG N0

3,

6


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(c) (i) 1 A1 N1

(ii) METHOD 1

Attempting to solve for cos M1

cos = (A1)

= 240, 120, 240, 120 (correct four values only) A2 N3

METHOD 2

Sketch ofy = 4 cos2 + 4 cos + 1 M1

Indicating 4 zeros (A1)

= 240, 120, 240, 120 (correct four values only) A2 N3

(d) Using sketch (M1)

c = 9 A1 N2[11]

59. (a) choosing sine rule (M1)

correct substitution A1

sinR = 0.676148...

= 42.5 A1 N2

2

1

y

x360 180 180 360

9

10

75sin

7

sin

R

QRP


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(b) P = 180 75 RP = 62.5 (A1)

substitution into any correct formula A1

eg area PQR = (their P)

= 31.0 (cm2) A1 N2[6]

60. (a) Attempting to multiply matrices (M1)

A1A1 N3

(b) Setting up equation M1

eg

(A1)

x = 3 A1 N2[6]

61. 2p2 + 12p = 14 (M1) (A1)

p2 + 6p7 = 0(p + 7)(p1) = 0 (A1)

p =7 orp = 1 (A1) (C4)

Note:Both answers are required for the final (A1).[4]

62. (a) (i) A1 = A2 N2

(ii) A2 = A2 N2

(b) (M1)

p = 2, q = 3 A1A1 N3

(c) Evidence of attempt to multiply (M1)

sin1072

1

x

x

x

xx

x

17

1

89

23

2

3

413

11 22

14

10

17

1,

28

20

234

22,

28

20

17

12

222

x

x

x

x

x

x

1417

101

28234

2022 22

x

x

x

x

02

1210

40

04

34

62

0

2

04

40

q

p


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egA1B =

A1B = A1 N2

(d) Evidence of correct approach (M1)

egX=A1B, setting up a system of equations

X= A1 N2

[11]

63. (a) A2 3

(b) For recognizing that the equations may be written as A (M1)

for attempting to calculate M1

x = 1.2,y = 0.6,z = 1.6 (Accept row or column vectors) A2 3[6]

02

12

10

30

22

112

30

12

12

10

accept

p

q

112

30

12

12

10

accept

p

q

8.0

3.0

1.0

2.0

2.0

4.0

2.1

7.0

1.0

3

2

1

z

y

x

6.1

6.0

2.1

3

2

11A

z

y

x


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64. (a)

Age range Frequency Mid - interval value

0 age < 20 40 10

20 age < 40 70 30

40 age < 60 100 50

60 age < 8050 70

80 age 100 10 90

A1A1 N2

(b) For attempting to find (M1)

Correct substitution (A1)

eg 40 10 + ... + 10 90 = 11900

For dividing by 270 (M1)

eg

Mean = 44.1 A1 N4[6]

65. (a) (i) m = 165 A1 N1

(ii) Lower quartile (1st quarter) = 160 (A1)

Upper quartile (3rd

quarter) = 170 (A1)IQR = 10 A1 N3

(b) Recognize the need to use the 40th percentile, or 48th student (M1)eg a horizontal line through (0, 48)a = 163 A1 N2

[6]

66. (a) (M1)

(Accept 50) (A1) (C2)

xf

270

11900

2230mean

45

x

n

49.6x


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(b) (may be implied) (M1)

(A1)

(A1)

(Accept 49) (A1) (C4)

[6]

67. (a) P = (= 0.957 (3 sf)) (A2) (C2)

(b)

(M1)OR

P = P (RRG) + P (RGR) + P (GRR) (M1)

(M1)(A1)

= (= 0.301 (3 sf)) (A1) (C4)

[6]

2

yy

n

2230 37 30y

2297

47y

48.9

23

22

325

2225

2124

323

R

R

G

G

etc

23

21

24

22

25

3

23

21

24

3

25

22

23

3

24

21

25

22

2300

693


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68. (a)

(A1)(A1)(A1)(A1) 4

Note: Award (A1) for the given probabilities in the correct

positions, and (A1) for eachboldvalue.

(b) Probability that Dumisani will be late is (A1)(A1)

= (0.294) (A1) (N2) 3

(c) P(WL) =

P(WL) = (A1)

P(L) = (A1)

P(WL) = (M1)

= (= 0.745) (A1) (N3) 4

[11]

L

L

W7

8

1

8

1

4

3

4

3

5

2

5

L '

L'

W'

5

3,

4

1,

8

7

5

3

8

1

4

1

8

7

160

47

)P(

)P(

L

LW

4

1

8

7

160

47

160

4732

7

47

35

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Ib 1 Review Solutions - End of Year Exam 2012