1

IB Math HL: Cumulative review 01 Markscheme

1. Let a be the first term and d be the common difference, then

a + d = 7 and S4 = (2a + 3d) = 12 (M1)

(M1)

a = 15, d = –8 (A2)(C2)(C2) [4]

2. The sum to infinity of a geometric series is

S = , r = < 1 (from formulae and statistical tables)

In this case, u1 = –12 and r = , therefore

S = (M1)(A1)

= or – 7.2 (A1)

[3]

3. and u1 + u1r + u1r2 = 13 (M1)

(1 – r)(1 + r + r2) = 13 (M1)

1 – r3 = giving r =

Therefore, u1 = 9. (A1) (C3)

[3]

2

4

1264

7

da

da

r

u

1

1

3

2

3

21

12

5

36

2

27

1

1 r

u

2

27

27

26

3

1

2

4. METHOD 1

Consider the group as two groups – one group of the two oldest

and one group of the rest (M2)

Either one of the two oldest is chosen or neither is chosen (M1)

Then the number of ways to choose the committee is

(M1)(M1)

= 40 + 15

= 55 ways (A1) (C6)

METHOD 2

The number of ways to choose a committee of 4 minus the number

of ways to have both the oldest (M3)

(M1)(M1)

= 70 – 15

= 55 ways (A1) (C6) [6]

5. The first student can receive x coins in ways, 1 x 5. (M1)

[The second student then receives the rest.]

Therefore, the number of ways = (A1)

= 26 – 2

= 62. (A1) (C3) [3]

6. (a) (2 + x)5 = 2

5 + 5(2)

4(x) + 10(2)

3x

2 + 10(2)

2x

3 + 5(2)x

4 + x

5 (M1)(A1)

= 32 + 80x + 80x2 + 40x

3 + 10x

4 + x

5 (A1) (C3)

Note: Award (C2) for 5 correct terms, (C1) for 4 correct terms.

(b) Let x = 0.01 = 10–2

(2.01)5 = 32 + 0.8 + 0.008 + 0.000 04 + 0.000 0001 + 0.000 000 0001(M1)(A1)

= 32.808 040 1001 (A1) (C3) [6]

7. The required term is 210–7

37 (M2)

= 2 099 520 (A1) (C3) [3]

0

2

4

6

1

2

3

6

2

6

4

8

x

6

5

6

4

6

3

6

2

6

1

6

7

10

3

8. (a) METHOD 1

f (g (x)) = f (x2 + x) = (M1) (A1)

=>x2

+ x – 2 0 (M1)

=>x –2, x l

a = –2, b = 1 (A1)(A1) (C5)

METHOD 2

f (g (x)) = (M1) (A1)

=

(x + 2) (x – 1) 0 (M1)

=> a = –2, b = 1 (A1)(A1) (C5)

(b) range is y 0 (A1) (C1) [6]

9. (f ° g) : x x3 + 1 (M1)

(f ° g)–1

: x (x – 1)1/3

(M1)(A1) (C3)

[3]

10. Let f (x) = ax2 + bx + c where a = 1, b = (2 – k) and c = k

2.

Then for a > 0, f (x) > 0 for all real values of x if and only if

b2 – 4ac < 0 (M1)

(2 – k)2 – 4k

2 < 0 (A1)

4 – 4k + k2 – 4k

2 < 0

3k2 – 4k – 4 > 0

(3k – 2)(k + 2) > 0 (M1)

k > , k < –2 (A1)(C2)(C2)

[4]

y x x = + –22

–2 1

2–3 xx

2–2 xx

1–2 xx

3

2

–2 0 k23

4

11. For kx2 – 3x + (k + 2) = 0 to have two distinct real roots then

k 0 (A1)

and 9 – 4k(k + 2) > 0 (M1)

4k2 + 8k – 9 < 0

–2.803 < k <0.803 (A1)

Set of values of k is –2.80 < k < 0.803, k 0 (C2)(C1) [3]

12. Given = 3 (1)

y = sin x

xy – y = x2 + (2)

y = (M1)

Substituting into (2) gives

32x4 – 32x

3 – 12x

2 – 27 = 0 (or equivalent) (A1)

x = , y = 9 (A1) (C3)

Notes: Award final (A1) only if both values are correct.

If no working is shown, and only 1 answer is correct, award

(C1).

Some candidates may be using calculators that cannot find these

exact answers ie 3/2 and 9. Award marks as appropriate, where

answers seem incorrect. Candidates should sketch graphs as

part of their answers, and this should help identify why answers

may be incorrect.

GDC example: finding solutions from a graph. [3]

y

x38

4

9

3

8 3x

2

3

5

13. METHOD 1

(M2)

x2 – 4 + < 0

=> –2.30 < x < 0 or l < x < 1.30 (G2)(G2) (C6)

METHOD 2

x2 – 4 + < 0

=> < 0 (M1)

=> < 0 (M1)

Critical values: l, (–l ± ), 0 (A2)

=> – ( + 1) < x < 0 or 1 < x < ( – 1) (A1)(A1) (C6)

[6]

14. (a) g (x) = –(x – 3)2 – 4, therefore the maximum point is (3, –4) (A1)

Note: Other methods are possible, including the use of a graphic

display calculator

(b) f (x) is mapped onto g (x) by a reflection in the x-axis followed by

the translation . (A2)

Note: Award (A2) for other correct answers

Award (A1) for a correct single transformation or a correct

combination that is not a reflection followed by a translation. [3]

–2.3 1 1.3

x

3

x

3

x

xx 34–3

x

xxx 3–1– 2

2

113

+ – + – +

12

– (1+ 13)12 (1+ 13)0 1

2

113

2

113

4

3

6

15.

(A2)(A2) (C4)

Notes: The graph of y2 is y1 shifted k units to the right.

Award (A2) for the correct graph.

Award (A1) for indicating each point of intersection with the

x-axis ie (m + k, 0) and (n + k, 0).

Award (C4) if the graph of y2 is drawn correctly and correctly

labelled with m + k and n + k. [4]

16. y = 1 – (M1) (A1)

= 1 – (A1)

Asymptotes are y = 1, (A1) (C2)

x = 4, x = 1. (A1)(A1)(C2)(C2) [6]

17. Since range goes from 4 to 2 a = 3 (M1)A1

Since curve is shifted right by (M1)A1

Since curve has been shifted in vertical by one unit down c = 1 (M1)A1

N2N2N2

[6]

x

y

m + m k

0 n n k +

y f x k = ( – )

y f x = ( )

k

1

2

45–

82 xx

1–4–

8

xx

4,

4

b

14

3

cba

7

18. METHOD 1

(a) The equation of the tangent is y = –4x – 8. (G2) (C2)

(b) The point where the tangent meets the curve again is (–2, 0). (G1) (C1)

METHOD 2

(a) y = –4 and = 3x2 + 8x + 1 = –4 at x = –1. (M1)

Therefore, the tangent equation is y = –4x – 8. (A1) (C2)

(b) This tangent meets the curve when –4x – 8 = x3 + 4x

2 + x – 6 which gives

x3 + 4x

2 + 5x + 2 = 0 (x + 1)

2(x + 2) = 0.

The required point of intersection is (–2, 0). (A1) (C1) [3]

19. (a) M1A1 N2

(b) Hence when x = 2, gradient of tangent (A1)

gradient of normal is (A1)

M1

A1 N4

(accept y = 2.33x + 6.61) [6]

20. (a) Given f (x) = esin

x

Then f (x) = cos x × esin x

(A1) (C1)

(b) f (x) = cos2 x × e

sinx – sin x × e

sin x

= esin x

(cos2 x – sin x) (M1)

For the point of inflexion, put f (x) = 0

esin

x (cos

2 x – sin x) = 0 (or equivalent) (A1) (C2)

Note: Award (C1) if the candidate only writes

f (x) = 0 [3]

x

y

d

d

13

33

13

1

xxxf

7

3

3

7

23

77ln xy

7ln3

14

3

7 xy

8

21. y y = f (x)

x

(a) Minimum points (A1) (C1)

(b) Maximum point (A1) (C1)

(c) Points of inflexion (A1) (C1)

Note: There is no scale on the question paper. For examiner

reference the scale has been added here and the numerical

answers are minima at x = –3 and 2, maximum at x = 0 and

points of inflexion at x = –1.79 and 1.12. [3]

22.

(a) Area = ( – 2x) sin x. (M1)(A1) (C2)

(b) Maximum Area = 1.12 units2 (A1) (C1)

[3]

23. By the remainder theorem,

f (–1) = 6 – 11 – 22 – a + 6 (M1)

= –20 (M1)

a = –1 (A2) (C4) [4]

–4 –3 –2 –1

–10

–20

10

1 2 3

min minmax

inf

inf

1.0

0.8

0.6

0.4

0.2

0.0x

9

24. Attempting to find f (2) = 8 + 12 + 2a + b (M1)

= 2a + b + 20 A1

Attempting to find f (–1) = –1 + 3 – a + b. (M1)

= 2 – a + b A1

Equating 2a + 20 = 2 – a A1

a = –6. A1 (N2) [6]

25. 9 log5 x = 25 logx 5

9 log5 x = M1A1

(log5 x)2 = M1

log5 x = A1

x = or x = (accept p = 5, q = 3) A1A1 N0 [6]

26. (M2)

4 = (A1)

64 = 100 – x2 (M1)

x2 = 36

x = ±6 (A1)(A1) (C6)

Note: Award (C1) if only x = 6 is given without working. [6]

27. = 0

(k – 4)(k + 1) + 6 = 0 (M1)

k2 – 3k + 2 = 0 (M1)

(k – 2)(k – 1) = 0

k = 2 or k = 1 (A1) (C3) [3]

x5log

25

9

25

3

5

3

5

5 3

5

5

3 22

1

10016 x

3 2100 x

12

34

k

k

10

28. A= 2(2k + 4) + (–5) + k(4 –3k) (M1)(A1)

A= 0 (M1)

3k2 – 8k – 3 = 0 (A1)

k = 3, – (A1)(A1) (C6)

Note: Award (A2) for k = 3 with no working. [6]

29. AA–1

XB = ΑC (M1)(A1)

IXBB–1

= ACB–1

(M1)(A1)

X = ACB–1

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

30. Area sector OAB (M1)(A1)

Area of OAB = (M1)(A1)

Shaded area = area of sector OAB – area of OAB (M1)

= 20.6 (cm2) (A1) (C6)

[6]

31. sin C = (M1)(A1)

= 56.4° or 123.6° (A1)(A1)

= 93.6° or 26.4° (A1)(A1) (C6)

Note: Award (C1) for one correct answer with no working. [6]

3

1

8

755

4

3

2

1 2

4

225

4

3sin)5)(5(

2

1

3

5.05sin

a

Ac

C

B

11

32.

(a) Using the cosine rule (a2 = b

2 + c

2 – 2bc cos A) (M1)

Substituting correctly

BC2 = 65

2 + 104

2 – 2(65)(104) cos 60° A1

= 4225 + 10816 – 6760 = 8281

BC = 91 m A1 N2 3

(b) Finding the area using = bc sin A (M1)

Substituting correctly, area = (65)(104) sin 60° A1

= 1690 (Accept p = 1690) A1 N2 3

(c) (i) Smaller area A1 = (65)(x) sin 30° (M1)A1

= AG N0

Larger area A2 = (104)(x) sin 30° M1

= 26x A1 N1

A

A

A

B

C

D

65 m

104 m

30°

30°

x

2

1

21

21

3

21

465x

21

12

(ii) Using A1 + A2 = A (M1)

Substituting + 26x = 1690 A1

Simplifying = 1690 A1

Solving

x = 40 (Accept q = 40) A1 N1 8

(d) using sin rule in ΔADB and ΔACD (M1)

Substituting correctly A1

and A1

Since + = 180° R1

It follows that sin = sin R1

A1

AG N0 6

[20]

33. (M1)(M1)

(A1)

(A1)

(A1)

≡ tanθ (AG) [5]

465x 3

4169x 3

169

316904x

3

BDAsin

30sin65BD

BDAsin

6530sin

BD

CDAsin

30sin104DC

CDAsin

10430sin

DC

BDA CDA

BDA CDA

10465

DCBD

104DC

65BD

85

DCBD

)2sin2(cos1(2cos

))sin(cos1(2cos2sin2

)4cos1(2cos

)2cos1(4sin22

22

2sin2cos1

)sincos1(2sin222

22

2sin2

)sin2(2sin22

2

2sin

sin2 2

cossin2

sin2 2

cos

sin

13

34. (a) f ( ) = R cos cos + R sin sin (M1)

R cos = 4, R sin = 3 (M1)

R = 5, = arctan = 0.644 (A1)(A1)

f ( ) = 5 cos ( – 0.644) (C4)

(b) f ( ) is maximum when = (M1)

ie = 0.644 radians (A1) (C2) [6]

35. 2 sin x = tan x

2 sin x cos x – sin x = 0

sin x(2 cos x – 1) = 0 (M1)

sin x = 0, cos x =

x = 0, x = ± or ±1.05 (3 s. f.) (A1)(A1) (C3)

OR

x = 0, x = ± (or ±1.05 (3 s. f.)) (G1)(G1)(G1) (C3)

Note: Award (G2) for x = 0, ± 60°. [3]

36. (a) = sec2 x – 8 cos x (A1) (C1)

(b) (M1)

= 0

cos x = (A1) (C2)

[3]

37. f (x) = x2 ln x

f (x) = 2x ln x + x2 (M1)(M1)

= 2x ln x + x (A1) (C3)

f : x 2x ln x + x [3]

38. y = e3x

sin (πx)

4

3

2

1

3

π

3

π

x

y

d

d

x

x

x

y2

3

cos

cos81

d

d

x

y

d

d

2

1

x

1

14

(a) = 3e3x

sin (πx) + πe3x

cos (πx) (M1)(A1)(A1) (C3)

(b) 0 = e3x

(3 sin (πx) + π cos (πx))

tan (πx) = – (M1)

πx = –0.80845 + (M1)

x = 0.7426… (0.743 to 3 sf) (A1) (C3) [6]

39. By implicit differentiation,

(2x2 – 3y

2 = 2) 4x – 6y = 0 (M1)

(A1)

When x = 5, 50 – 3y2 = 2 (M1)

y2 = 16

y = ±4

Therefore (A1)(C2)(C2)

Note: This can be done explicitly. [4]

40.

= (M1)

= + C (M1)(A1) (C3)

Note: Do not penalize for the absence of +C. [3]

41.

(M1)

k – (A1)

2k2 – 3k – 2 = 0

(2k + 1)(k – 2) = 0 (M1)

k = 2 since k > 1 (A1) (C4) [4]

x

y

d

d

3

π

xd

d

x

y

d

d

y

x

x

y

6

4

d

d

6

5

d

d

x

y

tt

tt

t

t d2

1d

2

11

3

5

3

1

3

53

1

tt

t d2

3

4

3

1

3

1

3

4

2

3

4

3

tt

k

xx1 2 2

3d

11

2

31

1

k

xx

2

31

k

15

42. y = ex – x

2 + C (A1)(A1)(A1)

3 = e0 – 0 + C (M1)

C = 2 (A1)

y = ex – x

2 + 2 (A1) (C6)

[6]

43. (a) Using the chain rule f (x) = (M1)

= 10 cos A1 2

(b) f (x) =

= + c A1

Substituting to find c, f = – + c = 1 M1

c = 1 + cos 2 = 1 + = (A1)

f (x) = – cos +

A1 N2 4 [6]

52

5cos2

x

25 x

xxf d)(

2

π5cos

5

2x

2π

2

π

2

π5cos

52

52

52

57

52

25 x

57

16

44. (a) v(t) = t sin = 0 when t = 0, t = 3 or t = 6 (C1)(C1)(C1) 3

(b) (i) The required distance, d = (M1)

(ii) d = 2.865 + 8.594 (G2)

= 11.459

= 11.5 m. (A1)

OR

(i) The required distance, d = . (M1)

(ii) d = 11.5 m. (G2)(A1)

OR

(i) The required distance, d = (M1)

(ii) d = (C1)

=

[from (i)] (C1)

= (sin – cos – sin 2 + 2 cos 2 + sin – cos )

= m (11.5 m). (A1) 4

Note: Award (A1) for which is obtained by

integrating v from 0 to 6. [7]

45. (a) s = 50t = 10t2 + 1000

(M1)

= 50 – 20t A1 (N2) 2

(b) Displacement is max when v = 0, M1

ie when t = . A1

Substituting t = , s = 50 × – 10 × + 1000 (M1)

s = 1062.5 m A1 (N2) 4 [6]

t

3

π

3

0

6

3d

3

πsind

3

πsin tttttt

6

0d

3

πsin ttt

3

0

6

3d

3

πsind

3

πsin tttttt

3

0

6

3

22

2d

3

πsin

3

πd

3

πsin

3

π

π

9tttttt

6

3

3

0

2 3

πcos

3

π

3

πsin

3

πcos

3

π

3

πsin

π

9tttttt

2π

9

π

36

)73.5(18

tsv

dd

25

25

25

2

25

17

46. a(t) = – t + 2

v(t) = – t2 + 2t + c (M1)

v = 0 when t = 0, and so c = 0

Thus, v(t) = – t2 + 2t = – t(t – 80). (A1)

Since v(t) 0 for 0 t 80, the distance travelled = (M1)

=

= 602

= 1800 m. (A1) (C4) [4]

47. = 8 (cm3s–1), V = r3

=> = 4r2

(M1)(A1)

= × => = ÷ (M1)

When r = 2, = 8 ÷ (4 × 22) (M1)(A1)

= (cm s–1

) (do not accept 0.159) (A1) (C6)

[6]

48. tan θ =

sec2 θ (M1)

when θ = , x2 = 3 and sec

2 θ = 4 (A1)(A1)

(M1)

km s–1

= –240 km h–1

(A1)

The airplane is moving towards him at 240 km h–1

(A1) (C6)

Note: Award (C5) if the answer is given as –240 km h–1

. [6]

20

1

40

1

40

1

40

1

60

0d)( ttv

60

0

23

120

1

tt

2

11

t

V

d

d

3

4

r

V

d

d

t

V

d

d

r

V

d

d

t

r

d

d

t

r

d

d

t

V

d

d

r

V

d

d

t

r

d

d

π2

1

x

3

t

x

xt d

d3

d

d2

3

π

t

x

t

x

d

d

3

sec

d

d 22

60

1

3

)4(3

d

d

t

x

15

1

d

d

t

x

t

x

d

d

18

49. x sin (x2) = 0 when x

2 = 0 (+k, k ), ie x = 0 (A1)

The required area = dx (M1)

= 1 (G1) (C3)

OR

Area = dx

= – (M1)

= – (–1 – 1)

= 1 (A1) (C3) [3]

50. Solving sinx = x2 – 2x + 1.5

gives x = 0.6617 or 1.7010 (using a graphic display calculator) (A1)

Then A = dx (M1)

= 0.271 units2 (using a graphic display calculator) (A1)

[3]

51. Let the volume of the solid of revolution be V.

V = dx (M1)

= (a2x

2 + 4ax + 4 – x

4 – 4x

2 –4)dx (M1)

= (M1)

= units3 (A1)

= (a2 + 5) (C4)

Note: The last line is not required [4]

)π( k

π

0

2 )(sin xx

π

0

2 )sin(xx

π

0

2 )cos(2

1x

2

1

7010.1

6617.0

2 ))5.12((sin xxx

a

xax0

222 )2()2(

a

0

a

xxaxxa0

35232

3

4

5

12

3

1

35

3

2

15

2aa

15

π2 3a

19

52. (a) Given = –kv

ln v = –kt + C (M1)

v = Ae–kt

(A = eC)

At t = 0, v = v0 A = v0

v = v0e–kt

(A1) (C2)

(b) Put v =

then = v0e–kt

(M1)

= e–kt

ln = –kt

t = (A1) (C2)

Note: Accept equivalent forms, eg t =

[4]

53. If A g is present at any time, then = kA where k is a constant.

Then,

ln A = kt + c

A = ekt +c

= c1ekt

When t = 0, c1 = 50 48 = 50e10k

. (A1)

= k or k = –0.00408(2) (A1)

For half life, 25 = 50ekt

ln 0.5 = kt

t = = 169.8.

Therefore, half-life = 170 years (3 sf) (A1) (C3) [3]

54. xy = 1 + y2 (M1)

ln (1 + y2) = ln x + ln c (M1)

1 + y2 = kx

2 (k = c

2)

y = 0 when x = 2, and so 1 = 4k

Thus, 1 + y2 = x

2 or x

2 – 4y

2 = 4. (A1) (C3)

t

v

d

d

tkv

vd

d

2

0v

2

0v

2

1

2

1

k

2ln

k

2

1ln

t

A

d

d

tkA

Ad

d

10

96.0ln

96.0ln

5.0ln10

x

y

d

d

x

xy

y

yd

1d

1 2

2

1

4

1