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