1.

(a) u1 = 27
81 27

2 1 r M1
1
r= 3 A1

(b) v2 = 9
v4 = 1
2d = – 8  d = –4 (A1)
v1 = 13 (A1)
N
2 (2 × 13 – 4(N – 1)) > 0 (accept equality) M1
N
2 (30 – 4N) > 0
N(15 – 2N) > 0
N < 7.5 (M1)
N=7 A1
Note: 13 + 9 + 5 + 1 – 3 – 7 – 11 > 0  N = 7 or equivalent
receives full marks.
[7]

2
2. (a) let the first three terms of the geometric sequence be given by u1, u1 r, u1r
 u1 = a + 2d, u1r = a + 3d and u1r2 = a + 6d (M1)
a  6d a  3d

a  3d a  2d A1
2 2 2 2
a + 8ad + 12d = a + 6ad + 9d A1
2a + 3d = 0
3
a=–2d AG

IB Questionbank Mathematics Higher Level 3rd edition 1

 d 3d  9d 
, u1 r  ,  u1 r 2  
(b) u1 = 2 2  2  M1
r=3 A1
27d
th 3
geometric 4 term u1r = 2 A1
3
th

d  15d
arithmetic 16 term a + 15d = 2 M1
27d
= 2 A1
Note: Accept alternative methods.
[8]

3. u4 = u1 + 3d = 7, u9 = u1 + 8d = 22 A1A1
Note: 5d = 15 gains both above marks
u1 = –2, d = 3 A1
n
Sn = 2 (–4 + (n – 1)3) > 10000 M1
n = 83 A1
[5]

1
4. P (six in first throw) = 6 (A1)
25 1

P (six in third throw) = 36 6 (M1)(A1)
2
 25  1
  
P (six in fifth throw) =  36  6
2
1 25 1  25  1
       ...
P(A obtains first six) = 6 36 6  36  6 (M1)
25
recognizing that the common ratio is 36 (A1)
1
6
25
1
P(A obtains first six) = 36 (by summing the infinite GP) M1
6
= 11 A1
[7]

IB Questionbank Mathematics Higher Level 3rd edition 2

5. (100 + 101 + 102 + ... + 999) – (102 + 105 + ... + 999) (M1)
900 300
(100  999)  (102  999)
= 2 2 M1A1A1
= 329 400 A1 N5
Note: A variety of other acceptable methods may be seen including
300 600
(201 1995) or (100  998)
for example 2 2 .
[5]

6. (a) 18n – 10 (or equivalent) A1

 (18r  10)
(b) 1 (or equivalent) A1

(c) by use of GDC or algebraic summation or sum of an AP (M1)

15

 (18r  10)
1 = 2010 A1
[4]

7. METHOD 1
5(2a + 9d) = 60 (or 2a + 9d = 12) M1A1
10(2a + 19d) = 320 (or 2a + 19d = 32) A1
solve simultaneously to obtain M1
a = –3, d = 2 A1
th
the 15 term is – 3 + 14 × 2 = 25 A1
Note: FT the final A1 on the values found in the penultimate line.

IB Questionbank Mathematics Higher Level 3rd edition 3

 METHOD 2
with an AP the mean of an even number of consecutive terms equals
the mean of the middle terms (M1)
a10  a11
2 = 16 (or a10 + a11 = 32) A1
a5  a6
2 = 6 (or a5 + a6 = 12) A1
a10 – a5 + a11 – a6 = 20 M1
5d + 5d = 20
d = 2 and a = –3 (or a5 = 5 or a10 = 15) A1
th
the 15 term is –3 + 14 × 2 = 25 (or 5 + 10 × 2 = 25 or 15 + 5 × 2 = 25) A1
Note: FT the final A1 on the values found in the penultimate line.
[6]

8. METHOD 1
(a) un = Sn – Sn–1 (M1)
n n n 1 n 1
7 a 7 a

= 7 n
7 n 1 A1

(b) EITHER
a
u1 = 1 – 7 A1
2
 a
a
 1   2
u2 = 1 – 7  7 M1
a a
1  
= 7 7 A1
a
common ratio = 7 A1

OR
n n 1
a a
  1  
un = 1 –  7  7 M1
n 1
a  a
  1  
= 7  7
n 1
7a a
 
= 7 7 A1
7a a
u1 = 7 , common ratio = 7 A1A1

IB Questionbank Mathematics Higher Level 3rd edition 4

 (c) (i) 0 < a < 7 (accept a < 7) A1
(ii) 1 A1

METHOD 2
n 1
 7  a  a 
  
un = br =  7  7 
n–1
(a) A1A1

(b) for a GP with first term b and common ratio r

b(1  r n )  b   b  n
  r
Sn = 1  r 1 r  1 r  M1
n
7n  an a
n
 1  
as Sn = 7 7
comparing both expressions M1
b a
1  r = 1 and r = 7
a 7a
 
b=1 7 7
7a a
u1 = b = 7 , common ratio = r = 7 A1A1
Note: Award method marks if the expressions for b and r are deduced
in part (a).

(c) (i) 0 < a < 7 (accept a < 7) A1

(ii) 1 A1
[8]

9. (a) (i) a, 2a, 3a, ..., na are n consecutive terms of an AP

with first term a and common difference a
n(n  1)
a
a  2a  3a  ...  na 2

so their mean is n n M1A1
a ( n  1)
= 2 AG N0

IB Questionbank Mathematics Higher Level 3rd edition 5

 4(n  1)
(ii) 4 + 2 × 4 + 3 × 4 + ... + 4n > 2 + 100 M1
4n(n  1)
2 > 2(n + 1) + 100 A1
2
2n + 2n > 2n + 102
attempt to solve (M1)
2
n > 51
so the minimum value of n that satisfies the condition is 8 A1 N0
Note: Award M1A1(M1)A1 for use of equations if there is a
clear conversion to an inequality.

x1  ...  x m  y1  ...  y n
(b) (i) M= mn M1
0  m  1 n
= mn A1
n
= mn AG N0

EITHER
2 2
 n   n 
0    m  1   n
 mn  mn
S= mn M1A1
attempt to simplify
m2n  n2m
( m  n) 2 mn(m  n)

mn ( m  n) 3
S=
mn
(m  n 2 )
= A1
mn
= mn AG N0

IB Questionbank Mathematics Higher Level 3rd edition 6

 OR
m n


i 1
x i2  y
i 1
2
i
M 2
Var(x) = mn M1A1
attempt to simplify M1
n n2

m  n ( m  n) 2
Var(x) =
n  n 
1  
= m  n  m  n 
n m

= mn mn
mn
2
= ( m  n) A1
mn
S 
mn AG N0

n mn
 
(ii) M=S m  n m n A1
attempt to solve M1
 n  mn
 n = m, as n > 0 A1
so, then the set has 2n numbers, x1, ..., xn, y1, ..., yn
from which the first n are all 0 and the last n are all 1 (M1)
x n  y1 1

hence the value of the median is 2 2 A1 N0
[17]

IB Questionbank Mathematics Higher Level 3rd edition 7

10. (a)

A3
Note: Award A1 for each correct shape,
A1 for correct relative position.

–x
(b) e sin (4x) = 0 (M1)
sin (4x) = 0 A1
4x = 0, π, 2π, 3π, 4π, 5π A1
π 2 π 3π 4 π 5 π
, , , ,
x = 0, 4 4 4 4 4 AG

–x –x
(c) e = e sin (4x) or reference to graph
sin 4x = 1 M1
π 5π 9 π
, ,
4x = 4 2 2 A1
π 5π 9 π
, ,
x= 8 8 8 A1 N3

–x
(d) (i) y = e sin 4x
dy
dx = –e–x sin 4x + 4e–x cos 4x M1A1
–x
y=e
dy
dx = –e–x A1
verifying equality of gradients at one point R1
verifying at the other two R1

IB Questionbank Mathematics Higher Level 3rd edition 8

 dy
(ii) since dx ≠ 0 at these points they cannot be local maxima R1

–x –x
(e) (i) maximum when y′ = 4e cos 4x – e sin 4x = 0 M1
arctan(4) arctan(4)  π arctan(4)  2 π
, , , ...
x= 4 4 4
maxima occur at
arctan(4) arctan(4)  2π arctan(4)  4π
, ,
x= 4 4 4 A1
1
 (arctan( 4))
so y1 = e 4 sin(arctan (4)) (= 0.696) A1
1
 (arctan( 4 )  2 π )
y2 = e 4 sin(arctan (4) + 2π) A1
  (arctan(4)  2 π )
1

 e 4 sin(arctan(4))  0.145
 
 
1
 (arctan( 4 )  4 π )
y3 = e 4 sin(arctan (4) + 4π) A1
  (arctan(4)  4 π )
1

 e 4 sin(arctan(4))  0.0301
 
  N3

y3 y
and 2
(ii) for finding and comparing y 2 y1 M1
π

r= e 2 A1
Note: Exact values must be used to gain the M1 and the A1.
[22]

6
 81  (2a  5d )
11. (a) S6 = 81 2 M1A1
 27 = 2a + 5d
11
 231  (2a  10d )
S11 = 231 2 M1A1
 21 = a + 5d
solving simultaneously, a = 6, d = 3 A1A1

IB Questionbank Mathematics Higher Level 3rd edition 9

 (b) a + ar = 1 A1
2 3
a + ar + ar + ar = 5 A1
 (a + ar) + ar2(1 + r) = 5
1
 1  ar 2   5
a
2
obtaining r – 4 = 0 M1
 r = ±2
r = 2 (since all terms are positive) A1
1
a= 3 A1

th
(c) AP r term is 3r + 3 A1
1 r 1
th
2
GP r term is 3 A1
1
r–1 r–1
3(r + 1) × 3 2 = (r + 1)2 M1AG

 (r  1)2 r 1

= n2 , n 
n +
(d) prove: Pn : r 1
show true for n = 1, i.e.
0
LHS = 2 × 2 = 2 = RHS A1
assume true for n = k, i.e. M1
k

 (r  1)2
r 1
r 1
 k2k , k 
+

consider n = k + 1
k 1

 (r  1)2 r 1
 k 2 k  (k  1)2 k
r 1 M1A1
k
= 2 (k + k + 2)
k
= 2(k + 1)2 A1
k+1
= (k + 1)2 A1
hence true for n = k + 1
Pk+1 is true whenever Pk is true, and P1 is true, therefore Pn is true R1
for n 
+

[21]

IB Questionbank Mathematics Higher Level 3rd edition 10

12. METHOD 1
If the areas are in arithmetic sequence, then so are the angles. (M1)
n
 a  l   12  θ  2θ   18θ
 Sn = 2 2 M1A1
 18q = 2p (A1)
π
θ
9 (accept 20°) A1
METHOD 2
a12 = 2a1 (M1)
12
 a1  2a1   πr 2
2 M1A1
πr2
3a1 
6
3 2 πr 2
r θ
2 6 (A1)
2π π
θ 
18 9 (accept 20°) A1
METHOD 3
Let smallest angle = a, common difference = d
a + 11d = 2a (M1)
a = 11d A1
12
Sn   2a  11d   2π
2 M1
6(2a + a) = 2p (A1)
18a = 2p
π
a= 9 (accept 20°) A1
[5]

x
13. (a) 0<2 <1 (M1)
x<0 A1 N2

IB Questionbank Mathematics Higher Level 3rd edition 11

 35
 40
(b) 1 r M1
 40  40  r = 35
 40  r =  5 (A1)
1
x
r=2 = 8 A1
1
 x = log2 8 (= 3) A1
x
Note: The substitution r = 2 may be seen at any stage in the solution.
[6]

1

14. (a) r= 3 (A1)
27
S 
1
1
3 M1
81
S    20.25
4 A1 N1

IB Questionbank Mathematics Higher Level 3rd edition 12

 (b) Attempting to show that the result is true for n = 1 M1
a 1  r 
a
LHS = a and RHS = 1  r A1
Hence the result is true for n = 1
Assume it is true for n = k

a  ar  ar 2  ...  ar k 1 

a 1 r k 
1 r M1
Consider n = k + 1:

a  ar  ar 2  ...  ar k 1  ar k 

a 1 r k 
 ar k
1 r M1


 
a 1  r k  ar k 1  r 
1 r

a  ar k  ar k  ar k  1
= 1 r A1
Note: Award A1 for an equivalent correct intermediate step.
a  ar k  1

1 r


a 1  r k 1 
= 1 r A1
Note: Illogical attempted proofs that use the result to be proved
would gain M1A0A0 for the last three above marks.
The result is true for n = k  it is true for n = k + 1 and as it is
true for n =1, the result is proved by mathematical induction. R1 N0
Note: To obtain the final R1 mark a reasonable attempt must
have been made to prove the k + 1 step.
[10]

n
15. 81 = 2 (1.5 + 7.5) M1
 n = 18 A1
1.5 + 17d = 7.5 M1
6
d 
17 A1 N0
[4]

IB Questionbank Mathematics Higher Level 3rd edition 13

 n–1
16. 2 × 1.05 > 500 M1
log 250
n – 1 > log 1.05 M1
n – 1 > 113.1675... A1
n = 115 (A1)
u115 = 521 A1 N5
Note: Accept graphical solution with appropriate sketch.
[5]

17. (a) Let p = 2,  8 + 4 – 10 – 2 = 0 M1

Since this fits p = 2 is a solution. R1

3 2 2
(b) p + p – 5p – 2 = (p – 2)(p + ap + b)
3 2 2
= p + ap + bp – 2p – 2ap – 2b M1A1
3 2
= p + p (a – 2) + p(b – 2a) – 2b
Equate constants  – 2 = – 2b
b=1 A1
Equate coefficients of p  a – 2 = 1
2

a=3 A1

2
(c) p + 3p + 1 = 0 M1
3 9 4 3 5

p= 2 2 A1A1

(d) (i) Arithmetic sequence: 1, 1 + p, 1 + 2p, 1 + 3p A1

2 3
Geometric sequence: 1, p, p , p A1

2 3
(ii) (1 + 2p) + (1 + 3p) = p + p M1A1
 p3 + p2 – 5p – 2 = 0 A1
3 5
Therefore, from part (i), p = 2, p = 2 R1

(iii) The sum to infinity of a geometric series exists if │p│ < 1. R1

3 5
Hence, p = 2 is the only such number. A1

IB Questionbank Mathematics Higher Level 3rd edition 14

 (iv) The sum of the first 20 terms of the arithmetic series can
be found by applying the sum formula
S20 = 10(2a + 19d) = 10(2 + 19 p) M1A1

  5  3 
10 2  19    265  95 5
 2 
So, S20 =    A1A1A1
[22]

1 2iq
e
2
iq
18. (a) z= e (M1)
1 iq
e
z= 2 A1 N2

1

(b) │z│ 2 A2
│z│< 1 AG

a
(c) Using S∞ = 1  r (M1)
iq
e
1
1  e iq
S∞ = 2 A1 N2

e iq cisq

1 1
1  e iq 1  cisq
(d) (i) S∞ = 2 2 (M1)
cosq  i sinq
1
1  (cosq  i sinq )
2 (A1)
1 2iq 1 3iq
iθ
e  e  ...
Also S∞ = e + 2 4
1 1
cis2q  cis3q  ...
= cis θ + 2 4 (M1)
 1 1   1 1 
 cosq  cos 2q  cos 3q  ...  i sinq  sin 2q  sin 3q  ...
S∞ =  2 4   2 4  A1

IB Questionbank Mathematics Higher Level 3rd edition 15

 (ii) Taking real parts,
 
 
1 1 cos q  i sinq
cosq  cos 2q  cos 3q  ...  Re 
2 4  1 
 1  (cosq  i sinq ) 
 2  A1
 1 1 
 1  cosq  i sinq 
(cosq  i sinq )
Re 2 2 

 1 1   1 1 
 1  cosq  i sinq  1  cosq  i sinq  
=  2 2   2 2  M1
1 1
cosq  cos 2 q  sin 2 q
2 2
2
 1  1
1  cosq   sin q
2

=  2  4 A1
 1
 cosq  
 2
1
1  cosq  (sin 2 q  cos 2 q )
= 4 A1
(2 cosq  1)  2 4(2 cosq  1)

= (4  4 cosq  1)  4 2(5  4 cosq ) A1
4 cosq  2
= 5  4 cosq A1AG N0
[25]

n
19. (a) 5000(1.063) A1 N1

5
(b) Value = $ 5000(1.063) (= $ 6786.3511…)
= $ 6790 to 3 s.f. (accept $ 6786, or $ 6786.35) A1 N1

n n
(c) (i) 5000(1.063) >10 000 (or (1 .063) > 2) A1 N1

(ii) Attempting to solve the above inequality n log (1.063) > log 2 (M1)
n > 11.345... (A1)
12 years A1 N3
Note: Candidates are likely to use TABLE or LIST on a GDC to
find n. A good way of communicating this is suggested below.
x
Let y = 1.063 (M1)
When x = 11, y = 1.9582, when x = 12, y = 2.0816 (A1)
x = 12 i.e. 12 years A1 N3
[6]

IB Questionbank Mathematics Higher Level 3rd edition 16