Assessment [53 marks]
1.
[Maximum mark: 6]
Consider any three consecutive integers, n − 1, n and n + 1.
(a)
22M.1.SL.TZ2.3
Prove that the sum of these three integers is always divisible by
3.
[2]
Markscheme
(n − 1)+n +(n + 1)
(A1)
A1
= 3n
which is always divisible by 3
AG
[2 marks]
(b)
Prove that the sum of the squares of these three integers is
never divisible by 3.
[4]
Markscheme
(n − 1)
2
2
+ n
+ (n + 1)
2
2
(= n
2
− 2n + 1 + n
2
+ n
+ 2n + 1)
A1
attempts to expand either (n − 1) or (n + 1)
(M1)
n + 1)
2
2
(do not accept n
2
− 1
2
2
= 3n
+ 2
A1
demonstrating recognition that 2 is not divisible by 3 or
correct expression divided by 3
2
3n
is divisible by 3 and so 3n
2
R1
+ 2
2
3
is never divisible by 3
seen after
or
OR the first term is divisible by 3, the second is not
OR
2
3(n
+
2
3
)
OR
2
3n +2
3
2
= n
+
2
3
hence the sum of the squares is never divisible by 3
[4 marks]
AG
2.
[Maximum mark: 5]
Consider an arithmetic sequence where u = S = 8. Find the value
of the first term, u , and the value of the common difference, d.
8
21M.1.SL.TZ1.3
8
[5]
1
Markscheme
METHOD 1 (finding u first, from S )
1
8
(A1)
4(u1 + 8)= 8
A1
u1 = −6
u1 + 7d = 8
OR
4(2u1 + 7d)= 8
(may be seen with their value of u )
1
(A1)
attempt to substitute their u
d = 2
1
(M1)
A1
METHOD 2 (solving simultaneously)
(A1)
u1 + 7d = 8
4(u1 + 8)= 8
OR
4(2u1 + 7d)= 8
OR
u1 = −3d
attempt to solve linear or simultaneous equations
u1 = −6, d = 2
[5 marks]
A1A1
(A1)
(M1)
3.
[Maximum mark: 4]
21M.1.SL.TZ1.2
The diameter of a spherical planet is 6 × 10
(a)
4
km
.
Write down the radius of the planet.
[1]
Markscheme
3 × 10
4
OR
30000 (km)
(accept 3 ∙ 10 ) A1
4
[1 mark]
(b)
The volume of the planet can be expressed in the form
k
π(a × 10 ) km
3
where 1
≤ a < 10
and k ∈
Z
.
Find the value of a and the value of k.
[3]
Markscheme
4
3
4
3
π(3 × 10 )
=
4
3
OR
π × 27 × 10
12
4
3
π(30000)
3
(= π(36 × 10
(A1)
12
))
OR
=
4
3
π × 27000000000000
(A1)
= π(36 × 10
[3 marks]
13
3
) (km )
OR
a = 3. 6, k = 13
A1
4.
[Maximum mark: 5]
21M.1.SL.TZ2.4
7
In the expansion of (x + k) , where k ∈
R
, the coefficient of the term
in x is 63.
5
Find the possible values of k.
[5]
Markscheme
EITHER
attempt to use the binomial expansion of (x + k)
7
7
0
C x k
0
7
7
C k x
0
0
+
+
7
6
1
C x k
1
7
5
C k x
1
1
+
+
7
5
2
5
2
C x k
7
C k x
)
+ …
2
identifying the correct term
(or
+ …
2
(M1)
7
7
5
2
C x k
2
(or
7
2
C k x
5
5
)
(A1)
OR
attempt to use the general term
r = 2
(or r
= 5
)
7
r
7−r
Cr x k
(or
7
r
Cr k x
(A1)
THEN
7
C
2
= 21
5
2
21x k
(or
7
C
= 63x
k = ±√3
5
5
= 21
2
(21k
(seen anywhere)
2
= 63 , k
= 3)
(A1)
A1
A1
Note: If working shown, award M1A1A1A1A0 for k =
√3
.
7−r
)
(M1)
[5 marks]
5.
[Maximum mark: 4]
22M.2.AHL.TZ2.9
Consider the set of six-digit positive integers that can be formed from the digits
0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 .
Find the total number of six-digit positive integers that can be formed such that
(a)
the digits are distinct.
[2]
Markscheme
9 × 9 × 8 × 7 × 6 × 5 (= 9 ×
= 136080 (= 9 ×
9!
4!
)
9P5 )
(M1)
A1
Note: Award M1A0 for
10 × 9 × 8 × 7 × 6 × 5 (=
Note: Award M1A0 for
9P6
10P6
= 151200 =
10!
4!
.
)
= 60480
[2 marks]
(b)
the digits are distinct and are in increasing order.
Markscheme
METHOD 1
EITHER
every unordered subset of 6 digits from the set of 9 non-zero digits can
be arranged in exactly one way into a 6-digit number with the digits in
increasing order.
A1
[2]
OR
A1
9
C (×1)
6
THEN
= 84
A1
METHOD 2
EITHER
removes 3 digits from the set of 9 non-zero digits and these 6 remaining
digits can be arranged in exactly one way into a 6-digit number with the
digits in increasing order.
A1
OR
A1
9
C (×1)
3
THEN
= 84
[2 marks]
A1
6.
[Maximum mark: 9]
The sum of the first n terms of a geometric sequence is given by
n
Sn =
2
Σ
r=1
(a)
3
(
7
8
r
)
21N.2.SL.TZ0.6
.
Find the first term of the sequence, u .
[2]
1
Markscheme
u1 = S 1 =
=
14
24
(=
2
3
7
12
×
7
8
(M1)
= 0. 583333 …)
A1
[2 marks]
(b)
Find S .
[3]
∞
Markscheme
7
r =
8
(= 0. 875)
(A1)
substituting their values for u and r into S
1
=
14
3
(= 4. 66666 …)
∞
=
(M1)
u1
1−r
A1
[3 marks]
(c)
Find the least value of n such that S
∞
− S n < 0. 001
.
[4]
Markscheme
attempt to substitute their values into the inequality or formula for S
(M1)
n
7
14
3
n
− Σ
r=1
2
3
(
7
8
r
)
< 0. 001
OR
Sn =
n
7
( 1− (
12
)
)
8
7
( 1−
)
8
attempt to solve their inequality using a table, graph or logarithms
(must be exponential)
(M1)
Note: Award (M0) if the candidate attempts to solve S
∞
− un < 0. 001
correct critical value or at least one correct crossover value
63. 2675 …
OR
S ∞ − S 63 = 0. 001036 …
OR
S ∞ − S 64 = 0. 000906 …
OR
S ∞ − S 63 − 0. 001 = 0. 0000363683 …
S ∞ − S 64 − 0. 001 = 0. 0000931777 …
least value is n =
[4 marks]
64
A1
OR
(A1)
.
7.
[Maximum mark: 6]
22M.2.SL.TZ2.3
Gemma and Kaia started working for different companies on January 1st 2011.
Gemma’s starting annual salary was $45
2% on January 1st each year after 2011.
(a)
000
, and her annual salary increases
Find Gemma’s annual salary for the year 2021, to the nearest
dollar.
Markscheme
METHOD 1
using geometric sequence with r
= 1. 02
correct expression or listing terms correctly
45000 × 1. 02
10
OR
45000 × 1. 02
11−1
(M1)
(A1)
OR listing terms
Gemma’s salary is $54855 (must be to the nearest dollar)
A1
METHOD 2
N = 10
PV = ∓45000
I% = 2
P/Y = 1
C/Y = 1
F V = ±54854. 7489 …
(M1)(A1)
Gemma’s salary is $54855 (must be to the nearest dollar)
A1
[3]
[3 marks]
Kaia’s annual salary is based on a yearly performance review. Her salary for the
years 2011, 2013, 2014, 2018, and 2022 is shown in the following table.
(b)
Assuming Kaia’s annual salary can be approximately modelled
by the equation S = ax + b, show that Kaia had a higher
salary than Gemma in the year 2021, according to the model.
Markscheme
finds a =
1096. 89 …
b = −2. 16 × 10
6
)
and b =
−2160753. 8 …
(accept
(A1)(A1)
Note: Award (A1)(A1) for S
, or
= 1096. 89 … x + 33028. 49 …
, or
S = 1096. 89 … x + 43997. 4 …
S = 1096. 89 … x + 45094. 3 …
Kaia’s salary in 2021 is $56063. 21 (accept $56817. 09 from
b = −2. 16 × 10
6
)
A1
Kaia had a higher salary than Gemma in 2021
[3 marks]
AG
[3]
8.
[Maximum mark: 6]
18N.1.AHL.TZ0.H_6
n
Use mathematical induction to prove that ∑
r (r!) = (n + 1) ! − 1
,
r=1
for n ∈
Z
+
.
[6]
Markscheme
* This question is from an exam for a previous syllabus, and may contain
minor differences in marking or structure.
consider n = 1.
1 (1!) = 1
and 2! − 1
= 1
therefore true for n =
1
R1
Note: There must be evidence that n = 1 has been substituted into both
expressions, or an expression such LHS=RHS=1 is used. “therefore true for
n = 1 ” or an equivalent statement must be seen.
k
assume true for n = k, (so that ∑
r (r!) = (k + 1) ! − 1
r=1
Note: Assumption of truth must be present.
consider n =
k + 1
k+1
k
∑ r (r!) = ∑ r (r!) + (k + 1) (k + 1) !
r=1
r=1
= (k + 1) ! − 1 + (k + 1) (k + 1) !
= (k + 2) (k + 1) ! − 1
M1
Note: M1 is for factorising (k + 1) !
= (k + 2) ! − 1
A1
(M1)
)
M1
= ((k + 1) + 1) ! − 1
so if true for n = k, then also true for n =
then true for all n (∈ Z ) R1
k + 1
, and as true for n =
1
+
Note: Only award final R1 if all three method marks have been awarded.
Award R0 if the proof is developed from both LHS and RHS.
[6 marks]
9.
[Maximum mark: 8]
EXN.2.AHL.TZ0.7
Consider the identity
(a)
2+7x
≡
( 1+2x ) ( 1−x )
A
1+2x
+
B
1−x
, where A,
B ∈ Z
.
Find the value of A and the value of B.
[3]
Markscheme
* This sample question was produced by experienced DP mathematics
senior examiners to aid teachers in preparing for external assessment in the
new MAA course. There may be minor differences in formatting compared
to formal exam papers.
2 + 7x ≡ A(1 − x)+B(1 + 2x)
EITHER
substitutes x
= 1
and attempts to solve for B and substitutes x
attempts to solve for A
9 = 3B ⇒ B = 3 ;
1
2
and
(M1)
3A
2
= −
= −
3
2
⇒ A = −1
OR
forms A + B =
(M1)
2
and −A + 2B =
THEN
A = −1
and B =
3
A1A1
7
and attempts to solve for A and B
[3 marks]
(b)
Hence, expand
2+7x
( 1+2x ) ( 1−x )
in ascending powers of x, up to
[4]
and including the term in x .
2
Markscheme
uses the binomial expansion on either 3(1 − x)
3(1 − x)
(1 + 2x)
−1
= 3(1 + x + x
−1
=(1 − 2x +
2
+ …)
( −1 ) ( −2 )
2!
−1
or (1 + 2x)
−1
M1
A1
(2x)
2
+ …)(= 1 − 2x + 4x
2
+ …)
A1
3 + 3x + 3x
2
2
−(1 − 2x + 4x )
so the expansion is 2 + 5x − x (in ascending powers of x)
2
A1
[4 marks]
(c)
Give a reason why the series expansion found in part (b) is not
valid for x
=
3
4
.
[1]
Markscheme
(1 + 2x)
−1
(is convergent) requires |x|<
the expansion is not valid
[1 mark]
R1
1
2
and x
=
3
4
is outside this so
© International Baccalaureate Organization, 2023