Monday 1 November 2021 (afternoon)
Candidate session number
2 hours
Write your session number in the boxes above.
Do not open this examination paper until instructed to do so.
You are not permitted access to any calculator for this paper.
Section A: answer all questions. Answers must be written within the answer boxes provided.
Section B: answer all questions in the answer booklet provided. Fill in your session number
on the front of the answer booklet, and attach it to this examination paper and your
cover sheet using the tag provided.
Unless otherwise stated in the question, all numerical answers should be given exactly or
A clean copy of the
this paper.
The maximum mark for this examination paper is
is required for
.
8821 – 7101
© International Baccalaureate Organization 2021
14 pages
16EP01
–2–
Please
write on this page.
Answers written on this page
will not be marked.
16EP02
8821 – 7101
–3–
8821 – 7101
Full marks are not necessarily awarded for a correct answer with no working. Answers must be
supported by working and/or explanations. Where an answer is incorrect, some marks may be given
for a correct method, provided this is shown by written working. You are therefore advised to show all
working.
Answer
questions. Answers must be written within the answer boxes provided. Working may be
continued below the lines, if necessary.
[Maximum mark: 4]
Given that
dy
dx
cos x
4
and y
2 when x
3
, find y in terms of x .
4
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16EP03
–4–
8821 – 7101
[Maximum mark: 9]
The function f is defined by f ( x ) =
(a)
(b)
2x + 4
, where x
3− x
, x
3.
Write down the equation of
(i)
the vertical asymptote of the graph of f ;
(ii)
the horizontal asymptote of the graph of f .
[2]
Find the coordinates where the graph of f crosses
(i)
the x-axis;
(ii)
the y-axis.
[2]
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16EP04
–5–
(c)
8821 – 7101
Sketch the graph of f on the axes below.
[1]
y
15
10
5
15
10
5
0
5
10
15
x
5
10
15
The function g is defined by g ( x ) =
(d)
Given that g x
g
1
ax + 4
, where x
3− x
, x
3 and a
.
x , determine the value of a .
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16EP05
[4]
–6–
8821 – 7101
[Maximum mark: 5]
Solve the equation log 3 x =
1
+ log 3 ( 4 x3 ) , where x
2 log 2 3
0.
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16EP06
–7–
8821 – 7101
[Maximum mark: 5]
Box 1 contains 5 red balls and 2 white balls.
Box 2 contains 4 red balls and 3 white balls.
(a)
A box is chosen at random and a ball is drawn. Find the probability that the ball is red.
[3]
Let A be the event that “box 1 is chosen” and let R be the event that “a red ball is drawn”.
(b)
Determine whether events A and R are independent.
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16EP07
[2]
–8–
8821 – 7101
[Maximum mark: 7]
The function f is defined for all x
the graph of f at x 4 .
. The line with equation y
6x
1 is the tangent to
(a)
Write down the value of f 4 .
[1]
(b)
Find f 4 .
[1]
The function g is defined for all x
where g x
x2
f g x .
3x and h x
(c)
Find h 4 .
(d)
Hence find the equation of the tangent to the graph of h at x
[2]
4.
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16EP08
[3]
–9–
8821 – 7101
[Maximum mark: 7]
6
2 x2 5x 3
, x
x 1
(a)
Show that 2 x 3
(b)
Hence or otherwise, solve the equation 2 sin 2θ − 3 −
x 1
, x
1.
6
= 0 for 0
sin 2θ − 1
[2]
,θ≠
π
.
4
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16EP09
[5]
– 10 –
8821 – 7101
[Maximum mark: 7]
The equation 3px2
2px
1
p has two real, distinct roots.
(a)
Find the possible values for p .
(b)
Consider the case when p
the form x
a
13
6
[5]
4 . The roots of the equation can be expressed in
, where a
. Find the value of a .
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16EP10
[2]
– 11 –
8821 – 7101
[Maximum mark: 7]
dy ln 2 x
x2
dx
Give your answer in the form y f x .
Solve the differential equation
2y
, x 0 , given that y
x
4 at x
1
.
2
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16EP11
– 12 –
8821 – 7101
[Maximum mark: 7]
Consider the expression
1
1 ax
1 x where a
, a
0.
The binomial expansion of this expression, in ascending powers of x , as far as the term in x2
is 4bx bx2 , where b
.
(a)
Find the value of a and the value of b .
[6]
(b)
State the restriction which must be placed on x for this expansion to be valid.
[1]
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16EP12
– 13 –
Do
8821 – 7101
write solutions on this page.
Answer
questions in the answer booklet provided. Please start each question on a new page.
[Maximum mark: 16]
A particle P moves along the x-axis. The velocity of P is v m s 1 at time t seconds,
where v t
4 4t 3t2 for 0 t 3 . When t 0 , P is at the origin O .
(a)
(b)
(c)
(i)
Find the value of t when P reaches its maximum velocity.
(ii)
Show that the distance of P from O at this time is
88
metres.
27
[7]
Sketch a graph of v against t , clearly showing any points of intersection
with the axes.
[4]
Find the total distance travelled by P .
[5]
[Maximum mark: 14]
(a)
(b)
(c)
dn 2 x
x e ) =  x 2 + 2nx + n ( n − 1)  e x for n
.
n (
dx
Hence or otherwise, determine the Maclaurin series of f x
x2ex in ascending powers
4
of x , up to and including the term in x .
Prove by mathematical induction that
 ( x 2e x − x 2 )3 
.
Hence or otherwise, determine the value of lim 
x →0 

x9


16EP13
[7]
[3]
[4]
– 14 –
Do
8821 – 7101
write solutions on this page.
[Maximum mark: 22]
Consider the equation z 1 3
where Im 2
0 and Im 3
(a)
(i)
Verify that ω1 = 1 + e
(ii)
Find
2
and
The roots 1 , 2 and
Argand diagram.
3,
3
i, z
0.
i
π
6
. The roots of this equation are
1,
2
and
3,
is a root of this equation.
expressing these in the form a
ei , where a
and
0.
[6]
are represented by the points A, B and C respectively on an
(b)
Plot the points A, B and C on an Argand diagram.
[4]
(c)
Find AC.
[3]
Consider the equation z
(d)
1
3
iz3 , z
.
By using de Moivre’s theorem, show that α =
1
i
1− e
(e)
Determine the value of Re
.
π
6
is a root of this equation.
[3]
[6]
16EP14
Please
write on this page.
Answers written on this page
will not be marked.
16EP15
Please
write on this page.
Answers written on this page
will not be marked.
16EP16
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Markscheme
November 2021
Mathematics: analysis and approaches
Higher level
Paper 1
pages
2
N21/5/MATHX/HP1/ENG/TZ0/XX/M
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Instructions to Examiners
Abbreviations
M
Marks awarded for attempting to use a correct Method.
A
Marks awarded for an Answer or for Accuracy; often dependent on preceding M marks.
R
Marks awarded for clear Reasoning.
AG Answer given in the question and so no marks are awarded.
FT
Follow through. The practice of awarding marks, despite candidate errors in previous parts,
for their correct methods/answers using incorrect results.
Using the markscheme
1
General
Award marks using the annotations as noted in the markscheme eg M1, A2.
2
Method and Answer/Accuracy marks
Do not automatically award full marks for a correct answer; all working must be checked,
and marks awarded according to the markscheme.
It is generally not possible to award M0 followed by A1, as A mark(s) depend on the
preceding M mark(s), if any.
Where M and A marks are noted on the same line, e.g. M1A1, this usually means M1 for
an attempt to use an appropriate method (e.g. substitution into a formula) and A1 for
using the correct values.
Where there are two or more A marks on the same line, they may be awarded
independently; so if the first value is incorrect, but the next two are correct, award
A0A1A1.
Where the markscheme specifies A3, M2 etc., do not split the marks, unless there is a
note.
The response
does not need to restate the AG line, unless a
Note makes this explicit in the markscheme.
Once a correct answer to a question or part question is seen, ignore further working even
if this working is incorrect and/or suggests a misunderstanding of the question. This will
encourage a uniform approach to marking, with less examiner discretion. Although some
candidates may be advantaged for that specific question item, it is likely that these
candidates will lose marks elsewhere too.
An exception to the previous rule is when an incorrect answer from further working is used
in a subsequent part. For example, when a correct exact value is followed by an incorrect
decimal approximation in the first part and this approximation is then used in the second
part. In this situation, award FT marks as appropriate but do not award the final A1 in the
first part.
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Examples:
Correct
answer
seen
Further
working seen
5.65685...
1.
8 2
2.
3
35
72
(incorrect
decimal value)
(incorrect
decimal value)
Any FT issues?
Action
No.
Last
part
in
question.
Yes.
Value is used in
subsequent parts.
Award A1 for the final mark
(condone the incorrect further
working)
Award A0 for the final mark
(and full FT is available in
subsequent parts)
Implied marks
Implied marks appear in brackets e.g. (M1),and can only be awarded if correct work is
seen or implied by subsequent working/answer.
4
Follow through marks (only applied after an error is made)
Follow through (FT) marks are awarded where an incorrect answer from one part of a
question is used correctly in subsequent part(s) (e.g. incorrect value from part (a) used in
part (d) or incorrect value from part (c)(i) used in part (c)(ii)). Usually, to award FT marks,
there must be working present and not just a final answer based on an incorrect answer
to a previous part. However, if all the marks awarded in a subsequent part are for the answer
or are implied, then FT marks should be awarded for their correct answer, even when
working is not present.
For example: following an incorrect answer to part (a) that is used in subsequent parts,
where the markscheme for the subsequent part is (M1)A1, it is possible to award full marks
for their correct answer, without working being seen. For longer questions where all but
the answer marks are implied this rule applies but may be overwritten by a Note in the
Markscheme.
Within a question part, once an error is made, no further A marks can be awarded for
work which uses the error, but M marks may be awarded if appropriate.
If the question becomes much simpler because of an error then use discretion to award
fewer FT marks, by reflecting on what each mark is for and how that maps to the simplified
version.
If the error leads to an inappropriate value (e.g. probability greater than 1, sin
1.5 , noninteger value where integer required), do not award the mark(s) for the final answer(s).
may be using an incorrect value.
question, it is not appropriate to award any FT marks in the subsequent parts. This
N21/5/MATHX/HP1/ENG/TZ0/XX/M
subsequent parts use their incorrect answer rather than the given value.
Exceptions to these FT rules will be explicitly noted on the markscheme.
If a candidate makes an error in one part but gets the correct answer(s) to subsequent
5
Mis-read
If a candidate incorrectly copies values or information from the question, this is a mis-read
(MR). A candidate should be penalized only once for a particular misread. Use the MR
stamp to indicate that this has been a misread and do not award the first mark, even if this
is an M mark, but award all others as appropriate.
If the question becomes much simpler because of the MR, then use discretion to award
fewer marks.
If the MR leads to an inappropriate value (e.g. probability greater than 1, sin
1.5 , noninteger value where integer required), do not award the mark(s) for the final answer(s).
not constitute a misread, it is an error.
If a candidate
accuracy than given in the question, this is NOT a misread and full marks may be scored
in the subsequent part.
MR can only be applied when work is seen. For calculator questions with no working and
incorrect answers, examiners should not infer that values were read incorrectly.
6
Alternative methods
Candidates will sometimes use methods other than those in the markscheme. Unless the
question specifies a method, other correct methods should be marked in line with the
methods are not permitted unless covered by a note in the mark scheme.
Alternative methods for complete questions are indicated by METHOD
METHOD 2, etc.
Alternative solutions for parts of questions are indicated by EITHER . . . OR.
1,
N21/5/MATHX/HP1/ENG/TZ0/XX/M
7
Alternative forms
Unless the question specifies otherwise, accept equivalent forms.
As this is an international examination, accept all alternative forms of notation for
example 1.9 and 1,9 or 1000 and 1,000 and 1.000 .
Do not accept final answers written using calculator notation. However, M marks and
intermediate A marks can be scored, when presented using calculator notation, provided
the evidence clearly reflects the demand of the mark.
In the markscheme, equivalent numerical and algebraic forms will generally be written
in brackets immediately following the answer.
In the markscheme, some equivalent answers will generally appear in brackets. Not all
equivalent notations/answers/methods will be presented in the markscheme and
examiners are asked to apply appropriate discretion to judge if the candidate work is
equivalent.
8
Format and accuracy of answers
If the level of accuracy is specified in the question, a mark will be linked to giving the answer
to the required accuracy. If the level of accuracy is not stated in the question, the general
rule applies to final answers: unless otherwise stated in the question all numerical answers
must be given exactly or correct to three significant figures.
Where values are used in subsequent parts, the markscheme will generally use the exact
value, however candidates may also use the correct answer to 3 sf in subsequent parts.
The markscheme will often explicitly include the subsequent values that come from the use
of 3 sf values
Simplification of final answers: Candidates are advised to give final answers using good
mathematical form. In general, for an A mark to be awarded, arithmetic should be completed, and
any values that lead to integers should be simplified; for example,
25
5
should be written as .
2
4
An exception to this is simplifying fractions, where lowest form is not required (although the
numerator and the denominator must be integers); for example,
written as
10
may be left in this form or
4
5
10
. However,
should be written as 2, as it simplifies to an integer.
2
5
Algebraic expressions should be simplified by completing any operations such as addition and
multiplication, e.g. 4e 2 x e3 x should be simplified to 4e5 x , and 4e 2 x e3 x e 4 x e x should be
simplified to 3e5 x . Unless specified in the question, expressions do not need to be factorized, nor
do factorized expressions need to be expanded, so x ( x 1) and x 2 x are both acceptable.
Please note: intermediate A marks do NOT need to be simplified.
N21/5/MATHX/HP1/ENG/TZ0/XX/M
9
Calculators
No calculator is allowed. The use of any calculator on this paper is malpractice and will
result in no grade awarded. If you see work that suggests a candidate has used any
calculator, please follow the procedures for malpractice.
10. Presentation of candidate work
Crossed out work: If a candidate has drawn a line through work on their examination
script, or in some other way crossed out their work, do not award any marks for that work
unless an explicit note from the candidate indicates that they would like the work to be
marked.
More than one solution: Where a candidate offers two or more different answers to the
same question, an examiner should only mark the first response unless the candidate
indicates otherwise. If the layout of the responses makes it difficult to judge, examiners
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Section A
1.
METHOD 1
recognition that y
y sin x
cos x
4
dx
c
4
(A1)
substitute both x and y values into their integrated expression including c
2 sin
2
(M1)
(M1)
c
c 1
y sin x
4
1
A1
[4 marks]
METHOD 2
y
x
dy
2
cos x
4
3
4
y 2 sin x
y sin x
4
4
1
dx
sin
(M1)(A1)
2
A1
A1
[4 marks]
N21/5/MATHX/HP1/ENG/TZ0/XX/M
2.
(a)
(i)
x 3
(ii)
y
A1
2
A1
[2 marks]
(b)
(i)
(ii)
2,0 (accept x
0,
4
(accept y
3
2)
4
and f 0
3
A1
4
)
3
A1
[2 marks]
continued
1
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 2 continued.
(c)
A1
Note: Award A1 for completely correct shape: two branches in correct
quadrants with asymptotic behaviour.
[1 mark]
c
1
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 2 continued.
(d)
METHOD 1
g x
ax 4
3 x
y
attempt to find x in terms of y
(M1)
OR exchange x and y and attempt to find y in terms of x
3 y xy
ax 4
A1
ax xy 3 y 4
x a
3y 4
3y 4
y a
x
g
y
1
3x 4
x a
x
A1
Note: Condone use of y
g x
ax 4
3 x
a
1
g
x
3x 4
x a
3
A1
[4 marks]
c
1
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 2 continued.
METHOD 2
ax 4
3 x
g x
attempt to find an expression for g g x
ax 4
4
3 x
ax 4
3
3 x
and equate to x
(M1)
a
gg x
a ax 4
9 3x
x
4 3 x
ax 4
x
a ax 4 4 3 x
5 3 a x
x
a ax 4
x 5
4 3 x
A1
3 a x
A1
equating coefficients of x 2 (or similar)
a
3
A1
[4 marks]
Total [9 marks]
1
3.
N21/5/MATHX/HP1/ENG/TZ0/XX/M
attempt to use change the base
log 3 x
(M1)
log 3 2
log 3 4 x3
2
attempt to use the power rule
log 3 x
log 3 2 log 3 4 x3
attempt to use product or quotient rule for logs, ln a ln b
log3 x
(M1)
ln ab
(M1)
log3 4 2 x3
Note: The M marks are for attempting to use the relevant log rule and may be
applied in any order and at any time during the attempt seen.
4 2 x3
x
x 32 x 6
x5
x
1
32
1
2
(A1)
A1
[5 marks]
1
4.
(a)
N21/5/MATHX/HP1/ENG/TZ0/XX/M
valid approach to find P R
(M1)
tree diagram (must include probabilty of picking box) with correct required probabilities
OR P R
B1
P R
B2 OR P R | B1 P B1
P R | B2 P B2
5 1
7 2
4 1
7 2
(A1)
P R
9
14
A1
[3 marks]
(b)
events A and R are not independent, since
9 1
14 2
5
5
OR
7
14
5
9
OR
9
14
OR an explanation e.g. different number of red balls in each box
1
2
A2
Note: Both conclusion and reasoning are required. Do not split the A2.
[2 marks]
Total [5 marks]
1
5.
(a)
f (4) 6
N21/5/MATHX/HP1/ENG/TZ0/XX/M
A1
[1 mark]
(b)
f (4) 6 4 1 23
A1
[1 mark]
(c)
h(4)
f g (4)
h(4)
f (42 3 4)
(M1)
f (4)
h(4) 23
A1
[2 marks]
(d)
attempt to use chain rule to find h
f g x
h (4)
g x
OR
x 2 3x
(M1)
f x 2 3x
2 4 3 f 42 3 4
A1
30
y 23 30 x 4
OR y
30 x 97
A1
[3 marks]
Total [7 marks]
1
6.
(a)
N21/5/MATHX/HP1/ENG/TZ0/XX/M
METHOD 1
attempt to write all LHS terms over a common denominator of x 1
2x 3
2x x 1
6
x 1
3 x 1
x 1
6
OR
2 x 2 2 x 3x 3 6
2 x2 5x 3
OR
x 1
x 1
2x 3 x 1
x 1
(M1)
6
x 1
6
A1
x 1
2 x2 5x 3
x 1
AG
[2 marks]
METHOD 2
attempt to use algebraic division on RHS
correctly obtains quotient of 2 x 3 and remainder
2x 3
6
x 1
(M1)
6
A1
as required.
AG
[2 marks]
c
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 6 continued.
(b)
consider the equation
2sin 2 2
2sin 2 2 5sin 2
sin 2 1
5sin 2
3
0
(M1)
3 0
EITHER
attempt to factorise in the form 2sin 2
a sin 2
b
(M1)
Note: Accept any variable in place of sin 2 .
2sin 2
1 sin 2
3
0
OR
attempt to substitute into quadratic formula
sin 2
5
(M1)
49
4
THEN
1
or sin 2
2
sin 2
3
Note: Award A1 for sin 2
one of
7
6
OR
11
6
(A1)
1
only.
2
(accept 210 or 330)
7
,
(must be in radians)
12 12
(A1)
A1
Note: Award A0 if additional answers given.
[5 marks]
Total [7 marks]
7.
(a)
attempt to use discriminant b 2
2p
2
4 3p 1 p
16 p 2 12 p
p 4p 3
4ac
N21/5/MATHX/HP1/ENG/TZ0/XX/M
0
M1
0
0
(A1)
0
attempt to find critical values
p 0, p
recognition that discriminant
0
p
1
0 or p
3
4
3
4
M1
(M1)
A1
Note:
[5 marks]
(b)
p 4
12 x2 8x 3 0
valid attempt to use x
x
x
a
8
b
b 2 4ac
(or equivalent)
2a
M1
208
24
2
13
6
2
A1
[2 marks]
Total [7 marks]
N21/5/MATHX/HP1/ENG/TZ0/XX/M
8.
dy
dx
2y
x
ln 2 x
x2
(M1)
attempt to find integrating factor
e
x2
2
dx
x
e2ln x
dy
dx
2 xy
d 2
x y
dx
x2 y
x2
(A1)
ln 2 x
ln 2 x
ln 2 x dx
attempt to use integration by parts
x2 y
x ln 2 x x
ln 2 x
x
y
(M1)
1
x
substituting x
c
(M1)
A1
c
x2
1
,y
2
4 into an integrated equation involving c
M1
4 0 2 4c
c
y
3
2
ln 2 x
x
1
x
3
2x2
A1
[7 marks]
2
9.
(a)
attempt to expand binomial with negative fractional power
1
1 ax
1 ax
1 x
1 x
1
1 ax
1
2
1 x
1
2
1
3a 2 x 2
8
1
ax
2
x
2
x2
...
8
1 a
x
2
1 a
2
x:
4b ; x 2 :
1
,b
3
x
1
1
6
A1
A1
(M1)
3a 2 1
b
8
attempt to solve simultaneously
a
(M1)
3a 2 1 2
x ...
8
attempt to equate coefficients of x or x 2
(b)
N21/5/MATHX/HP1/ENG/TZ0/XX/M
(M1)
A1
[6 marks]
A1
[1 mark]
Total [7 marks]
2
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Section B
10.
(a)
(i)
valid approach to find turning point ( v
4 6t 0 OR
t
(ii)
4
2
b
, average of roots)
2a
(M1)
2
2
3
2
OR
3
0,
2
(s)
3
A1
attempt to integrate v
v dt
(M1)
4 4t 3t 2 dt
Note: Award A1 for 4t
4t 2t 2 t 3
c
A1A1
2t 2 , A1 for t 3 .
attempt to substitute their t into their solution for the integral
distance
8 8
3 9
2
4
3
2
2
3
2
2
3
(M1)
3
8
(or equivalent)
27
A1
88
(m)
27
AG
[7 marks]
c
2
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 10 continued.
(b)
valid approach to solve 4 4t 3t 2
2 t 2 3t
OR
4
0 (may be seen in part (a))
(M1)
16 48
6
correct x- intercept on the graph at t
2
A1
Note: The following two A marks may only be awarded if the shape is a
concave down parabola. These two marks are independent of each
other and the (M1).
correct domain from 0 to 3 starting at 0, 4
A1
Note: The 3 must be clearly indicated.
vertex in approximately correct place for t
2
and v 4
3
A1
[4 marks]
c
2
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 10 continued.
3
(c)
4 4t 3t 2 dt
recognising to integrate between 0 and 2, or 2 and 3 OR
(M1)
0
2
4 4t 3t 2 dt
0
8
A1
3
4 4t 3t 2 dt
2
5
A1
valid approach to sum the two areas (seen anywhere)
2
3
v dt
0
2
v dt OR
2
(M1)
3
v dt
0
total distance travelled
v dt
2
13 (m)
A1
[5 marks]
Total [16 marks]
2
11.
(a)
N21/5/MATHX/HP1/ENG/TZ0/XX/M
For n 1
LHS:
d 2 x
xe
dx
RHS: x
2
x 2e x 2 xe x
ex x2 2 x
2 1 x 1 1 1 ex
A1
e x x2 2 x
A1
so true for n 1
now assume true for n
k ; i.e.
dk 2 x
xe
dx k
x 2 2kx k k 1 e x
M1
n k
Note: Do not award M1
marks can still be awarded.
attempt to differentiate the RHS
dk 1 2 x
xe
dx k 1
d
dx
2 x 2k e x
x2
M1
x 2 2kx k k 1 e x
x 2 2kx k k 1 e x
A1
2 k 1 x k k 1 ex
so true for n
A1
k implies true for n k 1
therefore n 1 true and n
k true
n k 1 true
therefore, true for all n
R1
Note: Award R1 only if three of the previous four marks have been
awarded
[7 marks]
c
2
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 11 continued.
(b)
METHOD 1
dn 2 x
xe
dx n
attempt to use
Note: For x
f 0
0,
0, f 0
use of f x
f x
x2
dn 2 x
xe
dx n
0, f
f 0
x3
x 2 2nx n n 1 e x
n n 1 may be seen.
|x 0
0
xf 0
(M1)
2, f
x2
f
2!
0
6, f
0
4
x3
f
3!
0
12
0
x4
f
4!
4
0
(M1)
1 4
x
2
A1
[3 marks]
METHOD 2
x2
Maclaurin series of e x
x2 1 x
f x
(M1)
x2
2!
x2
(A1)
x3
1 4
x
2
A1
[3 marks]
c
2
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 11 continued.
(c)
METHOD 1
2 x
attempt to substitute x e
x 2e x
x
x
2 3
2
x
x
1 4
x
2
x9
3
x9
2
x
3
x 2e x x 2
1 4
x into
2
x9
3
M1
3
x
2
(A1)
EITHER
x
3
1 4
x
2
x9
3
x9
A1
higher order terms
x9
OR
x
3
1 4
x
2
x3
3
1
1
x
2
...
A1
3
...
THEN
1 ( higher order terms)
So lim
x
0
x 2e x
x9
x2
3
1
A1
[4 marks]
c
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 11 continued.
METHOD 2
x 2e x
lim
x
x2
x9
0
ex 1
lim
x 0
x
3
x 2e x x 2
lim
x 0
x3
lime x
x
1
M1
3
(A1)
attempt to use L'Hôpital's rule
ex 0
lim
x 0
1
3
M1
3
3
0
A1
[4 marks]
Total [14 marks]
N21/5/MATHX/HP1/ENG/TZ0/XX/M
12.
(a)
(i)
1 e
e
i
i
3
1
6
3
6
A1
i
e2
A1
cos
isin
2
2
i
AG
Note: Candidates who solve the equation correctly can be awarded the
above two marks. The working for part (i) may be seen in part (ii).
(ii)
z 1
3
z 1 e
k 1
k
2
e
i
i
2
2 k
(M1)
k
6
6
(M1)
i
2
1 e
3
1 e
5
6
i
A1
9
6
A1
[6 marks]
c
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 12 continued.
(b)
EITHER
attempt to express e
i
6
,e
i
5
6
,e
i
9
6
in Cartesian form and translate 1 unit in the
positive direction of the real axis
(M1)
OR
attempt to express w1 , w2 and w3 in Cartesian form
(M1)
THEN
Note: To award A marks, it is not necessary to see A,B or C, the wi , or
the solid lines
A1A1A1
[4 marks]
c
3
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 12 continued.
(c)
valid attempt to find
1
1
3
3
2
1
i
2
valid attempt to find
3
4
AC
1
3 (or
1 i
3
2
M1
1)
3
3
2
3
isin
isin
i OR cos
6
6
2
2
3
i
2
M1
9
4
3
A1
[3 marks]
c
3
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 12 continued.
(d)
METHOD 1
3
z 1
iz
3
z 1
z
z 1
z
3
3
i
M1
i
e2
1
A1
i
e6
A1
Note: This step to change from z to
may occur at any point in MS.
i
1
e6
i
e6 1
i
1 e6
1
1
AG
i
1 e6
c
3
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 12 continued.
METHOD 2
z 1
1
1
z
1
1
z
3
iz
3
3
z 1
z
3
i
M1
i
e2
A1
i
e6
Note: This step to change from z to
A1
may occur at any point in MS.
1
i
1 e6
1
AG
i
1 e6
c
3
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 12 continued.
METHOD 3
3
1
LHS= ( z 1)3
1 e
e
i
i
1
6
3
6
i
1 e6
i
i
i
3
5 3 3
3 3
i
2
2
2
1 e6
Note: Award M1
M1A1
5
2
(may be seen in
modulus- argument form.)
3
RHS= iz 3
i
1
i
1 e6
i
i
A1
3
1 e6
( z 1)3 iz 3
AG
c
3
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 12 continued.
METHOD 4
( z 1)3 iz 3
z 3 3 z 2 3 z 1 iz 3
(1 i) z 3 3z 2 3z 1 0
(M1)
3
(1 i)
2
1
1 e
i
3
1
1 e
6
(1 i) 3 1 e
i
i
6
(1 i) 3 1 e 6
i
1
3
1 e
6
3 1 e
i
i
1
6
2
1 e
6
i
3 1 2e 6
i
e3
i
3
6
(A1)
i
1 3e 6
i
3e 3
i
e2
A1
0
AG
Note: If the candidate does not interpret their conclusion, award
(M1)(A1)A0 as appropriate.
[3 marks]
c
3
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 12 continued.
(e)
METHOD 1
1
1
i
1 e6
1
2
2
3 i
cos
6
M1
isin
6
A1
attempt to use conjugate to rationalise
M1
4 2 3 2i
2
3
2
A1
1
4 2 3 2i
8 4 3
1
2
A1
1
i
4 2 3
Re
1
2
A1
Note: Their final imaginary part does not have to be correct in order for the
final three A marks to be awarded
[6 marks]
c
3
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 12 continued.
METHOD 2
1
1
i
1 e6
1
M1
cos +isin
6
6
attempt to use conjugate to rationalise
1 cos
1
1 cos
isin
6
1 cos
isin
6
1 cos
6
6
6
isin
isin
M1
6
A1
6
6
A1
2
1 cos
sin
6
1 cos
1 2cos
1 cos
6
cos 2
6
isin
6
2 2cos
isin
1
2
6
isin
6
6
sin 2
6
6
A1
6
6
2 2cos
Re
2
6
1
2
A1
Note: Their final imaginary part does not have to be correct in order for the
final three A marks to be awarded
[6 marks]
c
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 12 continued.
METHOD 3
attempt to multiply through by
e
e
1
e
i
i
i
12
M1
12
12
i
e 12 e
1 e6
i
i
A1
12
attempting to re-write in r-cis form
cos
cos
cos
12
12
isin
isin
2isin
cos
12
12
12
isin
A1
12
12
A1
12
1 1
cot
2 2i 12
Re
isin
12
M1
2
2
icot
12
1
2
A1
[6 marks]
c
N21/5/MATHX/HP1/ENG/TZ0/XX/M
Question 12 continued.
METHOD 4
attempt to multiply through by
1 e
1 e
1
1 e
i
1 e6
1 e
i
6
i
i
i
6
M1
6
6
A1
i
e6 1
attempting to re-write in r-cis form
1 cos
6
isin
M1
6
A1
attempt to re-write in Cartesian form
M1
2 2cos
3
2
1
2
Re( )
1
i
2
3
6
2
3
2
2
3
i
1
2
2
3
1
2
A1
Note: Their final imaginary part does not have to be correct in order for the
final A mark to be awarded
[6 marks]
Total [22 marks]
Mathematics: analysis and approaches
Higher level
Paper 2
Tuesday 2 November 2021 (morning)
Candidate session number
2 hours
Instructions to candidates
Write your session number in the boxes above.
Do not open this examination paper until instructed to do so.
A graphic display calculator is required for this paper.
Section A: answer all questions. Answers must be written within the answer boxes provided.
Section B: answer all questions in the answer booklet provided. Fill in your session number
on the front of the answer booklet, and attach it to this examination paper and your
cover sheet using the tag provided.
Unless otherwise stated in the question, all numerical answers should be given exactly or
A clean copy of the mathematics: analysis and approaches formula booklet is required for
this paper.
The maximum mark for this examination paper is [110 marks].
8821 – 7102
© International Baccalaureate Organization 2021
14 pages
16EP01
–2–
8821 – 7102
Full marks are not necessarily awarded for a correct answer with no working. Answers must be
supported by working and/or explanations. Solutions found from a graphic display calculator should be
supported by suitable working. For example, if graphs are used to find a solution, you should sketch
these as part of your answer. Where an answer is incorrect, some marks may be given for a correct
method, provided this is shown by written working. You are therefore advised to show all working.
Section A
Answer all questions. Answers must be written within the answer boxes provided. Working may be
continued below the lines, if necessary.
1.
[Maximum mark: 7]
In Lucy’s music academy, eight students took their piano diploma examination and achieved
scores out of 150. For her records, Lucy decided to record the average number of hours per
week each student reported practising in the weeks prior to their examination. These results
are summarized in the table below.
Average weekly practice
time h
28
13
45
33
17
29
39
36
Diploma score D
115
82
120
116
79
101
110
121
(a)
Find Pearson’s product-moment correlation coefficient, r , for these data.
[2]
(b)
The relationship between the variables can be modelled by the regression
equation D ah b . Write down the value of a and the value of b .
[1]
One of these eight students was disappointed with her result and wished she had
practised more. Based on the given data, determine how her score could have been
expected to alter had she practised an extra five hours per week.
[2]
Lucy asserts that the number of hours a student practises has a direct effect on their
final diploma result. Comment on the validity of Lucy’s assertion.
[1]
(c)
(d)
Lucy suspected that each student had not been practising as much as they reported. In order
to compensate for this, Lucy deducted a fixed number of hours per week from each of the
students’ recorded hours.
(e)
State how, if at all, the value of r would be affected.
(This question continues on the following page)
16EP02
[1]
–3–
8821 – 7102
(Question 1 continued)
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16EP03
–4–
Please do not write on this page.
Answers written on this page
will not be marked.
16EP04
8821 – 7102
–5–
2.
8821 – 7102
[Maximum mark: 5]
Consider a triangle ABC , where AC
12 , CB
ˆ
25º .
7 and BAC
Find the smallest possible perimeter of triangle ABC .
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16EP05
–6–
3.
8821 – 7102
[Maximum mark: 7]
A factory manufactures lamps. It is known that the probability that a lamp is found to be
defective is 0.05 . A random sample of 30 lamps is tested.
(a)
Find the probability that there is at least one defective lamp in the sample.
[3]
(b)
Given that there is at least one defective lamp in the sample, find the probability that
there are at most two defective lamps.
[4]
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16EP06
–7–
4.
8821 – 7102
[Maximum mark: 6]
The following diagram shows a semicircle with centre O and radius r . Points P , Q and R
ˆ
lie on the circumference of the circle, such that PQ 2r and ROQ
, where 0
.
R
P
O
Q
(a)
Given that the areas of the two shaded regions are equal, show that
(b)
Hence determine the value of
2 sin .
[5]
.
[1]
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16EP07
–8–
5.
8821 – 7102
[Maximum mark: 9]
r
2 7
The sum of the first n terms of a geometric sequence is given by S n = ∑   .
r =1 3  8 
(a) Find the first term of the sequence, u1 .
n
(b)
Find S .
(c)
Find the least value of n such that S
[2]
[3]
Sn
0.001 .
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16EP08
[4]
–9–
6.
8821 – 7102
[Maximum mark: 8]
(a)
Prove the identity p
The equation 2x2
5x
1
q
3
3pq p
q
p3
0 has two real roots,
q3 .
and
[2]
.
Consider the equation x2
mx
n
0 , where m , n
Without solving 2x2
5x
1
0 , determine the values of m and n .
(b)
and which has roots
1
3
and
1
3
.
[6]
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16EP09
– 10 –
7.
8821 – 7102
[Maximum mark: 6]
A continuous random variable X has a probability density function given by
arccos x 0 ≤ x ≤ 1
f ( x) = 
otherwise
 0
The median of this distribution is m .
(a)
Determine the value of m .
(b)
Given that P X
m
a
[2]
0.3 , determine the value of a .
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16EP10
[4]
– 11 –
8.
8821 – 7102
[Maximum mark: 8]
Consider the curve C given by y
(a)
(b)
x
xy ln xy where x
0, y
0.
dy  dy

+  x + y  (1 + ln ( xy ) ) = 1 .
dx  dx

Hence find the equation of the tangent to C at the point where x
[3]
Show that
1.
[5]
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16EP11
– 12 –
8821 – 7102
Do not write solutions on this page.
Section B
Answer all questions in the answer booklet provided. Please start each question on a new page.
9.
[Maximum mark: 15]
The height of water, in metres, in Dungeness harbour is modelled by the
d , where t is the number of hours after midnight,
function H t
a sin b t c
and a , b , c and d are constants, where a 0 , b 0 and c 0 .
The following graph shows the height of the water for 13 hours, starting at midnight.
H t
10
5
t
0
5
10
The first high tide occurs at 04:30 and the next high tide occurs 12 hours later. Throughout
the day, the height of the water fluctuates between 2.2 m and 6.8 m .
All heights are given correct to one decimal place.
π
.
6
(a)
Show that b =
(b)
Find the value of a .
[2]
(c)
Find the value of d .
[2]
(d)
Find the smallest possible value of c .
[3]
(e)
Find the height of the water at 12:00.
[2]
(f)
Determine the number of hours, over a 24-hour period, for which the tide is higher
than 5 metres.
[3]
[1]
(This question continues on the following page)
16EP12
– 13 –
8821 – 7102
Do not write solutions on this page.
(Question 9 continued)
A fisherman notes that the water height at nearby Folkestone harbour follows the same
sinusoidal pattern as that of Dungeness harbour, with the exception that high tides (and low
tides) occur 50 minutes earlier than at Dungeness.
(g)
10.
Find a suitable equation that may be used to model the tidal height of water at
Folkestone harbour.
[2]
[Maximum mark: 18]
x 2 − x − 12
, x∈
Consider the function f ( x ) =
2 x − 15
(a)
,x≠
15
.
2
Find the coordinates where the graph of f crosses the
(i)
x-axis;
(ii)
y-axis.
[3]
(b)
Write down the equation of the vertical asymptote of the graph of f .
(c)
The oblique asymptote of the graph of f can be written as y
(d)
(e)
ax
[1]
b where a , b
.
Find the value of a and the value of b .
[4]
Sketch the graph of f for 30 x
each axis and any asymptotes.
[3]
(i)
Express
30 , clearly indicating the points of intersection with
1
in partial fractions.
f ( x)
3
(ii)
Hence find the exact value of
1
∫ f ( x ) dx , expressing your answer as a
0
single logarithm.
[7]
Turn over
16EP13
– 14 –
8821 – 7102
Do not write solutions on this page.
11.
[Maximum mark: 21]
Three points A 3 , 0 , 0 , B 0 , 2 , 0 and C 1 , 1 , 7 lie on the plane
(a)
(i)
Find the vector AB and the vector AC .
(ii)
Hence find the equation of 1 , expressing your answer in the
form ax by cz d , where a , b , c , d
.
Plane
(b)
1.
2
has equation 3x
y
2z
[7]
2.
The line L is the intersection of
1
and
2.
Verify that the vector equation of L can be
1
 0
 
 
written as r =  −2  + λ  1  .
 −1
 0
 
 
(c)
The plane
(i)
(ii)
(d)
3
is given by 2x
[2]
2z
3 . The line L and the plane
3
intersect at the point P .
3
.
4
Hence find the coordinates of P .
Show that at the point P , λ =
[3]
The point B 0 , 2 , 0 lies on L .
(i)
Find the reflection of the point B in the plane
(ii)
Hence find the vector equation of the line formed when L is reflected in the
plane 3 .
16EP14
3.
[9]
Please do not write on this page.
Answers written on this page
will not be marked.
16EP15
Please do not write on this page.
Answers written on this page
will not be marked.
16EP16
N21/5/MATHX/HP2/ENG/TZ0/XX/M
Markscheme
November 2021
Mathematics: analysis and approaches
Higher level
Paper 2
3 pages
2
N21/5/MATHX/HP2/ENG/TZ0/XX/M
N21/5/MATHX/HP2/ENG/TZ0/XX/M
Instructions to Examiners
Abbreviations
M
Marks awarded for attempting to use a correct Method.
A
Marks awarded for an Answer or for Accuracy; often dependent on preceding M marks.
R
Marks awarded for clear Reasoning.
AG Answer given in the question and so no marks are awarded.
FT
Follow through. The practice of awarding marks, despite candidate errors in previous parts,
for their correct methods/answers using incorrect results.
Using the markscheme
1
General
Award marks using the annotations as noted in the markscheme eg M1, A2.
2
Method and Answer/Accuracy marks
Do not automatically award full marks for a correct answer; all working must be checked,
and marks awarded according to the markscheme.
It is generally not possible to award M0 followed by A1, as A mark(s) depend on the
preceding M mark(s), if any.
Where M and A marks are noted on the same line, e.g. M1A1, this usually means M1 for
an attempt to use an appropriate method (e.g. substitution into a formula) and A1 for
using the correct values.
Where there are two or more A marks on the same line, they may be awarded
independently; so if the first value is incorrect, but the next two are correct, award
A0A1A1.
Where the markscheme specifies A3, M2 etc., do not split the marks, unless there is a
note.
The response
does not need to restate the AG line, unless a
Note makes this explicit in the markscheme.
Once a correct answer to a question or part question is seen, ignore further working even
if this working is incorrect and/or suggests a misunderstanding of the question. This will
encourage a uniform approach to marking, with less examiner discretion. Although some
candidates may be advantaged for that specific question item, it is likely that these
candidates will lose marks elsewhere too.
An exception to the previous rule is when an incorrect answer from further working is used
in a subsequent part. For example, when a correct exact value is followed by an incorrect
decimal approximation in the first part and this approximation is then used in the second
part. In this situation, award FT marks as appropriate but do not award the final A1 in the
first part.
N21/5/MATHX/HP2/ENG/TZ0/XX/M
Examples:
Correct
answer
seen
1.
8 2
2.
3
35
72
Further
working seen
5.65685...
(incorrect
decimal value)
(incorrect
decimal value)
Any FT issues?
Action
No.
Last
part
in
question.
Yes.
Value is used in
subsequent parts.
Award A1 for the final mark
(condone the incorrect further
working)
Award A0 for the final mark
(and full FT is available in
subsequent parts)
Implied marks
Implied marks appear in brackets e.g. (M1),and can only be awarded if correct work is
seen or implied by subsequent working/answer.
4
Follow through marks (only applied after an error is made)
Follow through (FT) marks are awarded where an incorrect answer from one part of a
question is used correctly in subsequent part(s) (e.g. incorrect value from part (a) used in
part (d) or incorrect value from part (c)(i) used in part (c)(ii)). Usually, to award FT marks,
there must be working present and not just a final answer based on an incorrect answer
to a previous part. However, if all the marks awarded in a subsequent part are for the answer
or are implied, then FT marks should be awarded for their correct answer, even when
working is not present.
For example: following an incorrect answer to part (a) that is used in subsequent parts,
where the markscheme for the subsequent part is (M1)A1, it is possible to award full marks
for their correct answer, without working being seen. For longer questions where all but
the answer marks are implied this rule applies but may be overwritten by a Note in the
Markscheme.
Within a question part, once an error is made, no further A marks can be awarded for
work which uses the error, but M marks may be awarded if appropriate.
If the question becomes much simpler because of an error then use discretion to award
fewer FT marks, by reflecting on what each mark is for and how that maps to the simplified
version.
If the error leads to an inappropriate value (e.g. probability greater than 1, sin
1.5 , noninteger value where integer required), do not award the mark(s) for the final answer(s).
The markscheme
may be using an incorrect value.
question, it is not appropriate to award any FT marks in the subsequent parts. This
N21/5/MATHX/HP2/ENG/TZ0/XX/M
subsequent parts use their incorrect answer rather than the given value.
Exceptions to these FT rules will be explicitly noted on the markscheme.
If a candidate makes an error in one part but gets the correct answer(s) to subsequent
5
Mis-read
If a candidate incorrectly copies values or information from the question, this is a mis-read
(MR). A candidate should be penalized only once for a particular misread. Use the MR
stamp to indicate that this has been a misread and do not award the first mark, even if this
is an M mark, but award all others as appropriate.
If the question becomes much simpler because of the MR, then use discretion to award
fewer marks.
If the MR leads to an inappropriate value (e.g. probability greater than 1, sin
1.5 , noninteger value where integer required), do not award the mark(s) for the final answer(s).
not constitute a misread, it is an error.
If a candidate
accuracy than given in the question, this is NOT a misread and full marks may be scored
in the subsequent part.
MR can only be applied when work is seen. For calculator questions with no working and
incorrect answers, examiners should not infer that values were read incorrectly.
6
Alternative methods
Candidates will sometimes use methods other than those in the markscheme. Unless the
question specifies a method, other correct methods should be marked in line with the
methods are not permitted unless covered by a note in the mark scheme.
Alternative methods for complete questions are indicated by METHOD
METHOD 2, etc.
Alternative solutions for parts of questions are indicated by EITHER . . . OR.
1,
N21/5/MATHX/HP2/ENG/TZ0/XX/M
7
Alternative forms
Unless the question specifies otherwise, accept equivalent forms.
As this is an international examination, accept all alternative forms of notation for
example 1.9 and 1,9 or 1000 and 1,000 and 1.000 .
Do not accept final answers written using calculator notation. However, M marks and
intermediate A marks can be scored, when presented using calculator notation, provided
the evidence clearly reflects the demand of the mark.
In the markscheme, equivalent numerical and algebraic forms will generally be written
in brackets immediately following the answer.
In the markscheme, some equivalent answers will generally appear in brackets. Not all
equivalent notations/answers/methods will be presented in the markscheme and
examiners are asked to apply appropriate discretion to judge if the candidate work is
equivalent.
8
Format and accuracy of answers
If the level of accuracy is specified in the question, a mark will be linked to giving the answer
to the required accuracy. If the level of accuracy is not stated in the question, the general
rule applies to final answers: unless otherwise stated in the question all numerical answers
must be given exactly or correct to three significant figures.
Where values are used in subsequent parts, the markscheme will generally use the exact
value, however candidates may also use the correct answer to 3 sf in subsequent parts.
from the use
of 3 sf values
Simplification of final answers: Candidates are advised to give final answers using good
mathematical form. In general, for an A mark to be awarded, arithmetic should be completed, and
any values that lead to integers should be simplified; for example,
25
5
should be written as .
2
4
An exception to this is simplifying fractions, where lowest form is not required (although the
numerator and the denominator must be integers); for example,
written as
10
may be left in this form or
4
5
10
. However,
should be written as 2, as it simplifies to an integer.
2
5
Algebraic expressions should be simplified by completing any operations such as addition and
multiplication, e.g. 4e 2 x e3 x should be simplified to 4e5 x , and 4e 2 x e3 x e 4 x e x should be
simplified to 3e5 x . Unless specified in the question, expressions do not need to be factorized, nor
do factorized expressions need to be expanded, so x ( x 1) and x 2 x are both acceptable.
Please note: intermediate A marks do NOT need to be simplified.
N21/5/MATHX/HP2/ENG/TZ0/XX/M
9
Calculators
A GDC is required for this paper, but If you see work that suggests a candidate has used
any calculator not approved for IB DP examinations (eg CAS enabled devices), please follow
the procedures for malpractice.
10. Presentation of candidate work
Crossed out work: If a candidate has drawn a line through work on their examination
script, or in some other way crossed out their work, do not award any marks for that work
unless an explicit note from the candidate indicates that they would like the work to be
marked.
More than one solution: Where a candidate offers two or more different answers to the
same question, an examiner should only mark the first response unless the candidate
indicates otherwise. If the layout of the responses makes it difficult to judge, examiners
N21/5/MATHX/HP2/ENG/TZ0/XX/M
Section A
1.
(a)
use of GDC to give
r
0.883529...
r
0.884
(M1)
A1
Note: Award the (M1) for any correct value of r, a, b or r 2
0.780624...
seen in part (a) or part (b).
[2 marks]
(b)
a 1.36609... , b 64.5171...
a 1.37 , b 64.5
A1
[1 mark]
(c)
attempt to find their difference
5 1.36609... OR 1.36609... h 5
(M1)
64.5171...
1.36609...h 64.5171...
6.83045...
6.83 (6.85 from 1.37)
the student could have expected her score to increase by 7 marks.
A1
Note: Accept an increase of 6, 6.83 or 6.85.
[2 marks]
(d)
Lucy is incorrect in suggesting there is a causal relationship.
This might be true, but the data can only indicate a correlation.
R1
Note:
[1 mark]
(e)
no effect
A1
[1 mark]
Total [7 marks]
N21/5/MATHX/HP2/ENG/TZ0/XX/M
2.
EITHER
attempt to use cosine rule
122
AB2
(M1)
2 12 cos 25º AB 7 2 OR AB2 21.7513...AB 95 0
at least one correct value for AB
(A1)
(A1)
AB 6.05068... OR AB 15.7007...
using their smaller value for AB to find minimum perimeter
(M1)
12 7 6.05068...
OR
attempt to use sine rule
(M1)
sin B
12
(A1)
sin 25º
OR sin B 0.724488... OR B 133.573...º OR B 46.4263...º
7
at least one correct value for C
C
21.4263...º OR C 108.573...º
(A1)
using their acute value for C to find minimum perimeter
12 7
122 72 2 12 7cos 21.4263... OR 12 7
(M1)
7sin 21.4263...º
sin 25º
THEN
25.0506
minimum perimeter = 25.1 .
A1
Total [5 marks]
1
3.
(a)
N21/5/MATHX/HP2/ENG/TZ0/XX/M
recognize that the variable has a Binomial distribution
X
(M1)
B 30,0.05
attempt to find P X
1 P X
1
(M1)
0 OR 1 0.9530 OR 1 0.214638... OR 0.785361...
Note: The two M marks are independent of each other.
P X 1
0.785
A1
[3 marks]
(b)
recognition of conditional probability
P X
(M1)
2 | X 1 OR P at most 2 defective | at least 1 defective
Note: Recognition must be shown in context either in words or symbols but
not just P A | B .
P 1 X 2
P X 1
OR
P X
1 P X
P X 1
2
0.597540...
0.812178... 0.214638...
0.338903... 0.258636...
OR
OR
0.785361...
0.785361...
0.785361...
(A1)
(A1)
0.760847...
P X
2| X 1
0.761
A1
[4 marks]
Total [7 marks]
1
4.
(a)
N21/5/MATHX/HP2/ENG/TZ0/XX/M
attempt to find the area of either shaded region in terms of r and
(M1)
Note: Do not award M1 if they have only copied from the booklet and not
applied to the shaded area.
Area of segment =
1 2
r
2
1 2
r sin
2
1
2
Area of triangle = r 2 sin
correct equation in terms of
sin
sin
sin
sin
A1
A1
only
(A1)
A1
2sin
AG
Note: Award a maximum of M1A1A0A0A0 if a candidate uses degrees
(i.e.,
1 2
r sin 180
2
), even if later work is correct.
Note: If a candidate directly states that the area of the triangle is
1 2
r sin , award a maximum of M1A1A0A1A1.
2
[5 marks]
(b)
1.89549...
1.90
A1
Note: Award A0 if there is more than one solution. Award A0 for an answer
in degrees.
[1 mark]
Total [6 marks]
1
5.
(a)
u1
2 7
3 8
S1
14
24
N21/5/MATHX/HP2/ENG/TZ0/XX/M
7
12
(M1)
0.583333
A1
[2 marks]
(b)
7
8
r
0.875
(A1)
u1
substituting their values for u1 and r into S
14
3
(M1)
1 r
4.66666
A1
[3 marks]
(c)
attempt to substitute their values into the inequality or formula for S n
14
3
n
r
2 7
1 3 8
r
0.001 OR Sn
7
1
12
1
7
8
(M1)
n
7
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
S63 0.001036... OR S
OR S
S63 0.001 0.0000363683... OR S
least value is n
64
S64
(A1)
0.000906...
S64 0.001
0.0000931777...
A1
[4 marks]
Total [9 marks]
1
6.
(a)
N21/5/MATHX/HP2/ENG/TZ0/XX/M
METHOD 1
p q
3
p3
3 pq p q
attempts to expand p
q
q3
3
M1
p3 3 p 2 q 3 pq 2 q3
p q
3
p 3 3 p 2 q 3 pq 2
3 pq p q
q 3 3 pq p q
p3 3 p 2q 3 pq 2 q3 3 p 2q 3 pq 2
p
3
q
A1
3
AG
Note: Condone the use of equals signs throughout.
METHOD 2
p q
3
p3
3 pq p q
attempts to factorise p
p q
p3
2
p q
p2q
pq 2
q
q3
3
3 pq
3 pq p q
p q
p2
M1
pq q 2
p 2 q pq 2 q3
p3 q3
A1
AG
Note: Condone the use of equals signs throughout.
METHOD 3
p3
q3
p q
3
3 pq p q
attempts to factorise p
p2
p q
p q
p q
q3
M1
pq q 2
p q
3
3
2
3 pq
3 pq p q
A1
AG
Note: Condone the use of the equals sign throughout.
[2 marks]
1
N21/5/MATHX/HP2/ENG/TZ0/XX/M
(b)
Note: Award a maximum of A1M0A0A1M0A0 for m
5
,
by using
17
,
4
95 and n 8 found
0.219..., 2.28... .
Condone, as appropriate, solutions that state but clearly do not use the
values of
and .
Special case: Award a maximum of A1M1A0A1M0A0 for m
95 and
n 8 obtained by solving simultaneously for and from product of
roots and sum of roots equations.
product of roots of x 2
5
1
x
2
2
0
1
(seen anywhere)
2
1
1
considers
3
3
A1
by stating
1
n
3
M1
Note: Award M1 for attempting to substitute their value of
1
1
into
3
.
1
3
3
1
2
n 8
A1
sum of roots of x 2
5
1
x
2
2
0
5
(seen anywhere)
2
considers
1
1
3
3
A1
3
by stating
3
3
2
Note: Award M1 for attempting to substitute their values of
expression. Award M0 for use of
5
2
3
3
2
5
2
1
8
m
95
3
3
3
3
and
m
M1
into their
only.
125 30 95
A1
x 2 95 x 8 0
[6 marks]
Total [8 marks]
1
N21/5/MATHX/HP2/ENG/TZ0/XX/M
m
7.
(a)
arccos x dx
recognises that
0.5
(M1)
0
1 m2
m arccos m
0
1
0.5
m 0.360034...
m 0.360
(b)
A1
[2 marks]
METHOD 1
attempts to find at least one endpoint (limit) both in terms of m (or their m ) and a
P m a
X
m a
(M1)
0.3
0.360034... a
arccos x dx
0.3
(A1)
0.360034... a
m a
arccos x dx
Note: Award (A1) for
0.3 .
m a
x arccos x
1 x2
0.360034... a
0.360034... a
attempts to solve their equation for a
(M1)
Note: The above (M1) is dependent on the first (M1).
a 0.124861...
a 0.125
A1
METHOD 2
a
arccos x 0.360034... dx
(M1)(A1)
0.3
a
Note: Only award (M1) if at least one limit has been translated
correctly.
a
arccos x m dx
Note: Award (M1)(A1) for
0.3 .
a
attempts to solve their equation for a
a 0.124861...
a 0.125
(M1)
A1
1
N21/5/MATHX/HP2/ENG/TZ0/XX/M
METHOD 3
EITHER
a
arccos x 0.360034... dx
(M1)(A1)
0.3
a
Note: Only award (M1) if at least one limit has been translated
correctly.
a
arccos x m dx
Note: Award (M1)(A1) for
0.3 .
a
OR
2 0.360034... a
arccos x 0.360034... dx
(M1)(A1)
0.3
2 0.360034... a
Note: Only award (M1) if at least one limit has been translated
correctly.
2m a
arccos x m dx
Note: Award (M1)(A1) for
0.3 .
2m a
THEN
attempts to solve their equation for a
(M1)
Note: The above (M1) is dependent on the first (M1).
a 0.124861...
a 0.125
A1
[4 marks]
Total [6 marks]
N21/5/MATHX/HP2/ENG/TZ0/XX/M
8.
(a)
METHOD 1
attempts to differentiate implicitly including at least one application of the product rule
u
xy , v ln xy ,
dy
1
dx
xy dy
x
xy dx
du
dx
x
dy
dx
y,
y
x
dy
dx
y ln xy
dv
dx
x
dy
dx
y
dy
1
dx
x
dy
1
dx
dy
dx
x
x
xy dy
x
xy dx
dy
dx
dy
dx
dy
dx
y
x
y
x
dy
dx
1
xy
A1
Note: Award (M1)A1 for implicitly differentiating y
dy
1
dx
dy
ln xy
dx
y ln xy
x 1 y ln xy
and obtaining
.
y ln xy
y 1 ln xy
y 1 ln xy
(M1)
A1
1
AG
METHOD 2
y x xy ln x
xy ln y
attempts to differentiate implicitly including at least one application of the product rule
dy
1
dx
xy
x
x
dy
dx
xy dy
y dx
y ln x
x
dy
dx
y ln y
(M1)
A1
or equivalent to the above, for example
dy
1
dx
x ln x
dy
dx
1 ln x y
dy
dy
1 x
ln x ln y 1
dx
dx
y ln y x ln y
y ln x ln y 1
dy
dx
dy
dx
A1
or equivalent to the above, for example
dy
dy
1 x
ln xy 1 y ln xy
dx
dx
dy
dy
x
y 1 ln xy
1
dx
dx
1
AG
N21/5/MATHX/HP2/ENG/TZ0/XX/M
METHOD 3
attempt to differentiate implicitly including at least one application of the product rule
u
x ln xy , v
dy
1
dx
x
y,
dy
ln xy
dx
du
dx
ln xy
y ln xy
x
dy
dx
xy dy
x
xy dx
dy
dy
1 x
ln xy 1 y ln xy
dx
dx
dy
dy
x
y 1 ln xy
1
dx
dx
x dv
,
xy dx
y
M1
dy
dx
y
A1
1
A1
AG
METHOD 4
lets w
dy
dw dw
1
ln w
dx
dx dx
dw
dy
x
y
dx
dx
dy
dy
dy
1 x
y x
dx
dx
dx
dy
dx
dy
where y
dx
dw
1
1 ln w
dx
xy and attempts to find
x
dy
dx
y 1 ln xy
x w ln w
M1
A1
A1
y ln xy
1
1
x
dy
dx
y 1 ln xy
AG
[3 marks]
N21/5/MATHX/HP2/ENG/TZ0/XX/M
(b)
METHOD 1
substitutes x
1 into y
y 1 y ln y
x xy ln xy
(M1)
y 1
A1
substitutes x 1 and their non-zero value of y into
2
dy
dx
0
dy
dx
dy
dx
x
dy
dx
y 1 ln xy
0
1
(M1)
A1
equation of the tangent is y
1
A1
METHOD 2
substitutes x 1 into
dy
dy
dx
dx
EITHER
dy
dx
x
y 1 ln y
dy
dx
0
y 1 ln xy
1
(M1)
1
dy
1 y
into
dx
y
correctly substitutes ln y
dy
1
1
dx
y
dy
dx
dy
dx
y 1 ln y
A1
1
A1
0 y 1
OR
correctly substitutes y
dy
2 ln y
dx
THEN
substitutes x
y 1 y ln y
0
dy
dx
1 into y
y ln y 1 into
0 y 1
x xy ln xy
dy
dx
dy
dx
y 1 ln y
1
A1
A1
(M1)
y 1
equation of the tangent is y
1
A1
[5 marks]
Total [8 marks]
2
N21/5/MATHX/HP2/ENG/TZ0/XX/M
Section B
9.
(a)
12
b
2
b
OR b
2
12
A1
AG
6
[1 mark]
(b)
a
6.8 2.2
OR a
2
max min
2
2.3 (m)
(M1)
A1
[2 marks]
(c)
d
6.8 2.2
OR d
2
4.5 (m)
max min
2
(M1)
A1
[2 marks]
continued
2
N21/5/MATHX/HP2/ENG/TZ0/XX/M
Question 9 continued.
(d)
METHOD 1
4.5 and H
substituting t
6.8 2.3sin
4.5 c
6
6.8 for example into their equation for H
(A1)
4.5
attempt to solve their equation
(M1)
c 1.5
A1
METHOD 2
using horizontal translation of
12
4
(M1)
4.5 c 3
(A1)
c 1.5
A1
METHOD 3
H t
2.3
cos
t c
6
6
attempts to solve their H 4.5
2.3
6
cos
6
4.5 c
(A1)
0 for c
(M1)
0
c 1.5
A1
[3 marks]
(e)
attempt to find H when t
H
2.87365...
H
2.87 m
12 or t
0 , graphically or algebraically
(M1)
A1
[2 marks]
2
N21/5/MATHX/HP2/ENG/TZ0/XX/M
Question 9 continued.
(f)
attempt to solve 5
times are t
2.3sin
6
t 1.5
1.91852... and t
total time is 2
4.5
(M1)
7.08147... , t 13.9185..., t 19.0814...
(A1)
7.081... 1.919...
10.3258...
10.3 (hours)
A1
Note: Accept 10.
[3 marks]
(g)
METHOD 1
11
and H
3
substitutes t
6.8 2.3sin
H t
c
6 3
2.3sin
6
t
3
6.8 into their equation for H and attempts to solve for c
4.5
c
(M1)
3
4.5
A1
METHOD 2
uses their horizontal translation
11
c 3
3
H t
c
2.3sin
12
4
3
(M1)
2
3
6
t
3
4.5
A1
[2 marks]
Total [15 marks]
2
10.
(a)
N21/5/MATHX/HP2/ENG/TZ0/XX/M
(i)
Note: In part (a), penalise once only, if correct values are given instead of
correct coordinates.
attempts to solve x 2
x 12
0
3, 0 and 4, 0
0,
(ii)
(M1)
A1
4
5
A1
[3 marks]
(b)
15
2
x
A1
Note: Award A0 for x
15
.
2
Award A1 in part (b), if x
15
is seen on their graph in part (d).
2
[1 mark]
(c)
METHOD 1
ax b 2 x 15
x 2 x 12
attempts to expand ax b 2 x 15
2ax 2 15ax 2bx 15b x 2
1
a
2
equates coefficients of x
15
1
2b
2
13
b
4
y
x 13
2 4
(M1)
x 12
A1
(M1)
A1
2
N21/5/MATHX/HP2/ENG/TZ0/XX/M
METHOD 2
attempts division on
x 2 x 12
2 x 15
x 13
...
2 4
1
a
2
13
b
4
x 13
y
2 4
M1
M1
A1
A1
METHOD 3
1
2
a
A1
x 2 x 12
2 x 15
x
c
b
2
2 x 15
2 x 15 x
x 2 x 12
2 x 15 b c
2
equates coefficients of x :
15
1
2b
2
13
b
4
y
x 13
2 4
M1
(M1)
A1
2
N21/5/MATHX/HP2/ENG/TZ0/XX/M
METHOD 4
x 2 x 12
attempts division on
2 x 15
13x
12
2
x x 12 x
2
2 x 15
2 2 x 15
1
a
2
13x
12
13
2
...
2 x 15
4
13
b
4
x 13
y
2 4
M1
A1
M1
A1
[4 marks]
2
N21/5/MATHX/HP2/ENG/TZ0/XX/M
(d)
two branches with approximately correct shape ( for 30 x 30 )
A1
their vertical and oblique asymptotes in approximately correct positions with both branches
showing correct asymptotic behaviour to these asymptotes
A1
their axes intercepts in approximately the correct positions
A1
Note: Points of intersection with the axes and the equations of asymptotes
are not required to be labelled.
[3 marks]
N21/5/MATHX/HP2/ENG/TZ0/XX/M
(e)
(i)
attempts to split into partial fractions:
2 x 15
x 3 x 4
2 x 15
A
B
x 3
x 4
A x 4
B x 3
A 3
B
1
3
1
x 3 x 4
3
(ii)
0
3
1
x 3
x 4
(M1)
A1
A1
dx
attempts to integrate and obtains two terms involving
3ln x 3 ln x 4
3
0
(M1)
A1
3ln 6 ln1 3ln 3 ln 4
3ln 2 ln 4
ln 8 ln 4
A1
ln 32
A1
5ln 2
Note: The final A1 is dependent on the previous two A marks.
[7 marks]
Total [18 marks]
N21/5/MATHX/HP2/ENG/TZ0/XX/M
11.
(a)
(i)
attempts to find either AB or AC
3
2 and AC
0
AB
(ii)
(M1)
2
1
7
A1
METHOD 1
attempts to find AB
AC
(M1)
14
21
7
AB AC
A1
EITHER
equation of plane is of the form 14 x 21y 7 z
d
2x 3y z
d
substitutes a valid point e.g 3, 0, 0 to obtain a value of d
d
42 d
r
r
2
3
1
M1
6
OR
attempts to use r n
14
21
7
(A1)
3
0
0
3
0
0
a n
14
21
7
2
3
1
14
21
7
r
r
(M1)
2
3
1
42
A1
6
THEN
14 x 21y 7 z
42 2 x 3 y z 6
A1
METHOD 2
equation of plane is of the form
x
y
z
3
0
0
3
s 2
0
2
t 1
7
attempts to form equations for x, y, z in terms of their parameters
x 3 3s 2t , y 2s t , z
7t
A1
(M1)
A1
eliminates at least one of their parameters
2x 3y 6 z
for example, 2 x 3 y 6 7t
2x 3y z
6
(M1)
A1
[7 marks]
N21/5/MATHX/HP2/ENG/TZ0/XX/M
(b)
METHOD 1
0
2
0
substitutes r
2
1:
1
1 into their
1
3 2
6 and
1
and
3
2:
2
(given)
2
2
(M1)
2
Note: Award (M1)A0 for correct verification using a specific value of
so the vector equation of L can be written as r
0
2
0
1
1
1
A1
.
AG
METHOD 2
EITHER
attempts to find
2
3
1
3
1
2
M1
7
7
7
OR
2
3
1
1
1
1
2 3 1
0 and
3
1
2
1
1
1
3 1 2
0
M1
2
A1
THEN
substitutes 0, 2, 0 into
2 0
3 2
1
and
2
3 0
2
2 0
so the vector equation of L can be written as r
0
2
0
1:
0
6 and
METHOD 3
attempts to solve 2 x 3 y
for example, x
,y
z
2
2:
6 and 3x
,z
y 2z
2
1
1
1
AG
(M1)
A1
3
Note: Award A1 for substituting x
N21/5/MATHX/HP2/ENG/TZ0/XX/M
0 (or y
2 or z
and solving simultaneously. For example, solving
y 2z
2 to obtain y
2 and z
0 ) into
3y z
1
and
2
6 and
0.
so the vector equation of L can be written as r
0
2
0
1
1
1
AG
[2 marks]
(c)
(i)
substitutes the equation of L into the equation of
2
(ii)
2
3
4
3
4
(M1)
3
3
A1
AG
P has coordinates
3 5 3
,
,
4 4 4
A1
[3 marks]
(d)
(i)
normal to
3
is n
2
0
2
(A1)
Note: May be seen or used anywhere.
considers the line normal to
0
2
0
r
3
passing through B 0, 2,0
2
0
2
(M1)
A1
EITHER
finding the point on the normal line that intersects
attempts to solve simultaneously with plane 2 x 2 z
4
4
3
(M1)
3
3
8
point is
3
A1
3
3
, 2,
4
4
3
N21/5/MATHX/HP2/ENG/TZ0/XX/M
OR
2
2
2
5
4
3
4
3
4
2
3
8
4
2
0
2
3
4
3
2
0
(M1)
0
A1
OR
attempts to find the equation of the plane parallel to
solve simultaneously with L
2
2
3
containing B
x z 3 and
(M1)
3
3
4
A1
THEN
so, another point on the reflected line is given by
0
2
0
r
(A1)
3
3
, 2,
2
2
B
(ii)
2
3
0
4
2
A1
EITHER
attempts to find the direction vector of the reflected line using their P and B (M1)
3
4
PB
3
4
3
4
OR
attempts to find their direction vector of the reflected line using a vector approach (M1)
PB
1
3
1
4
1
PB BB
1
3
0
2
1
THEN
3
2
r
3
4
2
3
2
3
4
3
4
(or equivalent)
A1
3
Note: Award A0
r
x
y
z
N21/5/MATHX/HP2/ENG/TZ0/XX/M
Award A0 for L
[9 marks]
Total [21 marks]
Mathematics: analysis and approaches
Higher level
Paper 3
Tuesday 9 November 2021 (morning)
1 hour
Instructions to candidates
Do not open this examination paper until instructed to do so.
A graphic display calculator is required for this paper.
Answer all the questions in the answer booklet provided.
Unless otherwise stated in the question, all numerical answers should be given exactly or
A clean copy of the mathematics: analysis and approaches formula booklet is required for
this paper.
The maximum mark for this examination paper is [55 marks].
3 pages
8821 – 7103
© International Baccalaureate Organization 2021
–2–
8821 – 7103
Answer all questions in the answer booklet provided. Please start each question on a new page. Full
marks are not necessarily awarded for a correct answer with no working. Answers must be supported
by working and/or explanations. Solutions found from a graphic display calculator should be supported
by suitable working. For example, if graphs are used to find a solution, you should sketch these as part
of your answer. Where an answer is incorrect, some marks may be given for a correct method, provided
this is shown by written working. You are therefore advised to show all working.
1.
[Maximum mark: 25]
In this question you will explore some of the properties of special functions f and g
and their relationship with the trigonometric functions, sine and cosine.
Functions f and g are defined as f ( z ) =
Consider t and u , such that t , u
(a)
Verify that u
(b)
Show that
f t
(c)
Using eiu
cos u
e z + e− z
e z − e− z
and g ( z ) =
, where z
2
2
.
f t satisfies the differential equation
2
g t
2
.
d 2u
dt 2
u.
[2]
f 2t .
[3]
i sin u , find expressions, in terms of sin u and cos u , for
(i)
f iu ;
[3]
(ii)
g iu .
[2]
(d)
Hence find, and simplify, an expression for
(e)
Show that
f t
2
g t
2
f iu
2
f iu
2
g iu
2
.
g i u 2.
[2]
[4]
The functions cos x and sin x are known as circular functions as the general
defines points on the unit circle with equation x2 y2 1 .
point cos , sin
The functions f x and g x are known as hyperbolic functions, as the general
,g
defines points on a curve known as a hyperbola with
point f
equation x2 y2 1 . This hyperbola has two asymptotes.
(f)
Sketch the graph of x2 y2 1 , stating the coordinates of any axis intercepts and the
equation of each asymptote.
The hyperbola with equation x2
by xy k , k
.
(g)
y2
Find the possible values of k .
[4]
1 can be rotated to coincide with the curve defined
[5]
–3–
2.
8821 – 7103
[Maximum mark: 30]
In this question you will be exploring the strategies required to solve a system of
linear differential equations.
Consider the system of linear differential equations of the form:
dx
dt
where x , y , t
(i)
y,
dy
By solving the differential equation
dt
a constant.
dx
dt
x
y , show that y
Aet where A is
[3]
Aet .
(ii)
Show that
(iii)
Solve the differential equation in part (a)(ii) to find x as a function of t .
[1]
(iii)
(iv)
Hence show that x
By differentiating
Now consider the case where a
(i)
[4]
1.
dy
d2 y
dy
x y with respect to t , show that 2 2 .
dt
dt
dt
dy
, show that Y Be2t where B is a constant.
By substituting Y
dt
Hence find y as a function of t .
(i)
(ii)
(c)
ax
0.
Now consider the case where a
(b)
dy
dt
and a is a parameter.
First consider the case where a
(a)
x y and
B 2t
e C , where C is a constant.
2
4.
d2 y
dy
2
3y
Show that
2
dt
dt
0.
[3]
[3]
[2]
[3]
[3]
From previous cases, we might conjecture that a solution to this differential equation
is y Fe t,
and F is a constant.
(ii)
Find the two values for
d2 y
dy
that satisfy
2
3y
2
dt
dt
Let the two values found in part (c)(ii) be
(iii)
Verify that y = Fe 1t + Ge
where G is a constant.
2t
1
and
0.
[4]
2.
is a solution to the differential equation in (c)(i),
[4]
N21/5/MATHX/HP3/ENG/TZ0/XX/M
Markscheme
November 2021
Mathematics: analysis and approaches
Higher level
Paper 3
1 pages
2
N21/5/MATHX/HP3/ENG/TZ0/XX/M
N21/5/MATHX/HP3/ENG/TZ0/XX/M
Instructions to Examiners
Abbreviations
M
Marks awarded for attempting to use a correct Method.
A
Marks awarded for an Answer or for Accuracy; often dependent on preceding M marks.
R
Marks awarded for clear Reasoning.
AG Answer given in the question and so no marks are awarded.
FT
Follow through. The practice of awarding marks, despite candidate errors in previous parts,
for their correct methods/answers using incorrect results.
Using the markscheme
1
General
Award marks using the annotations as noted in the markscheme eg M1, A2.
2
Method and Answer/Accuracy marks
Do not automatically award full marks for a correct answer; all working must be checked,
and marks awarded according to the markscheme.
It is generally not possible to award M0 followed by A1, as A mark(s) depend on the
preceding M mark(s), if any.
Where M and A marks are noted on the same line, e.g. M1A1, this usually means M1 for
an attempt to use an appropriate method (e.g. substitution into a formula) and A1 for
using the correct values.
Where there are two or more A marks on the same line, they may be awarded
independently; so if the first value is incorrect, but the next two are correct, award
A0A1A1.
Where the markscheme specifies A3, M2 etc., do not split the marks, unless there is a
note.
The response
does not need to restate the AG line, unless a
Note makes this explicit in the markscheme.
Once a correct answer to a question or part question is seen, ignore further working even
if this working is incorrect and/or suggests a misunderstanding of the question. This will
encourage a uniform approach to marking, with less examiner discretion. Although some
candidates may be advantaged for that specific question item, it is likely that these
candidates will lose marks elsewhere too.
An exception to the previous rule is when an incorrect answer from further working is used
in a subsequent part. For example, when a correct exact value is followed by an incorrect
decimal approximation in the first part and this approximation is then used in the second
part. In this situation, award FT marks as appropriate but do not award the final A1 in the
first part.
N21/5/MATHX/HP3/ENG/TZ0/XX/M
Examples:
Correct
answer
seen
Further
working seen
5.65685...
1.
8 2
2.
3
35
72
(incorrect
decimal value)
(incorrect
decimal value)
Any FT issues?
Action
No.
Last
part
in
question.
Yes.
Value is used in
subsequent parts.
Award A1 for the final mark
(condone the incorrect further
working)
Award A0 for the final mark
(and full FT is available in
subsequent parts)
Implied marks
Implied marks appear in brackets e.g. (M1),and can only be awarded if correct work is
seen or implied by subsequent working/answer.
4
Follow through marks (only applied after an error is made)
Follow through (FT) marks are awarded where an incorrect answer from one part of a
question is used correctly in subsequent part(s) (e.g. incorrect value from part (a) used in
part (d) or incorrect value from part (c)(i) used in part (c)(ii)). Usually, to award FT marks,
there must be working present and not just a final answer based on an incorrect answer
to a previous part. However, if all the marks awarded in a subsequent part are for the answer
or are implied, then FT marks should be awarded for their correct answer, even when
working is not present.
For example: following an incorrect answer to part (a) that is used in subsequent parts,
where the markscheme for the subsequent part is (M1)A1, it is possible to award full marks
for their correct answer, without working being seen. For longer questions where all but
the answer marks are implied this rule applies but may be overwritten by a Note in the
Markscheme.
Within a question part, once an error is made, no further A marks can be awarded for
work which uses the error, but M marks may be awarded if appropriate.
If the question becomes much simpler because of an error then use discretion to award
fewer FT marks, by reflecting on what each mark is for and how that maps to the simplified
version.
If the error leads to an inappropriate value (e.g. probability greater than 1, sin
1.5 , noninteger value where integer required), do not award the mark(s) for the final answer(s).
may be using an incorrect value.
question, it is not appropriate to award any FT marks in the subsequent parts. This
N21/5/MATHX/HP3/ENG/TZ0/XX/M
subsequent parts use their incorrect answer rather than the given value.
Exceptions to these FT rules will be explicitly noted on the markscheme.
If a candidate makes an error in one part but gets the correct answer(s) to subsequent
5
Mis-read
If a candidate incorrectly copies values or information from the question, this is a mis-read
(MR). A candidate should be penalized only once for a particular misread. Use the MR
stamp to indicate that this has been a misread and do not award the first mark, even if this
is an M mark, but award all others as appropriate.
If the question becomes much simpler because of the MR, then use discretion to award
fewer marks.
If the MR leads to an inappropriate value (e.g. probability greater than 1, sin
1.5 , noninteger value where integer required), do not award the mark(s) for the final answer(s).
not constitute a misread, it is an error.
If a candidate
accuracy than given in the question, this is NOT a misread and full marks may be scored
in the subsequent part.
MR can only be applied when work is seen. For calculator questions with no working and
incorrect answers, examiners should not infer that values were read incorrectly.
6
Alternative methods
Candidates will sometimes use methods other than those in the markscheme. Unless the
question specifies a method, other correct methods should be marked in line with the
methods are not permitted unless covered by a note in the mark scheme.
Alternative methods for complete questions are indicated by METHOD
METHOD 2, etc.
Alternative solutions for parts of questions are indicated by EITHER . . . OR.
1,
N21/5/MATHX/HP3/ENG/TZ0/XX/M
7
Alternative forms
Unless the question specifies otherwise, accept equivalent forms.
As this is an international examination, accept all alternative forms of notation for
example 1.9 and 1,9 or 1000 and 1,000 and 1.000 .
Do not accept final answers written using calculator notation. However, M marks and
intermediate A marks can be scored, when presented using calculator notation, provided
the evidence clearly reflects the demand of the mark.
In the markscheme, equivalent numerical and algebraic forms will generally be written
in brackets immediately following the answer.
In the markscheme, some equivalent answers will generally appear in brackets. Not all
equivalent notations/answers/methods will be presented in the markscheme and
examiners are asked to apply appropriate discretion to judge if the candidate work is
equivalent.
8
Format and accuracy of answers
If the level of accuracy is specified in the question, a mark will be linked to giving the answer
to the required accuracy. If the level of accuracy is not stated in the question, the general
rule applies to final answers: unless otherwise stated in the question all numerical answers
must be given exactly or correct to three significant figures.
Where values are used in subsequent parts, the markscheme will generally use the exact
value, however candidates may also use the correct answer to 3 sf in subsequent parts.
from the use
of 3 sf values
Simplification of final answers: Candidates are advised to give final answers using good
mathematical form. In general, for an A mark to be awarded, arithmetic should be completed, and
any values that lead to integers should be simplified; for example,
25
5
should be written as .
2
4
An exception to this is simplifying fractions, where lowest form is not required (although the
numerator and the denominator must be integers); for example,
written as
10
may be left in this form or
4
5
10
. However,
should be written as 2, as it simplifies to an integer.
2
5
Algebraic expressions should be simplified by completing any operations such as addition and
multiplication, e.g. 4e 2 x e3 x should be simplified to 4e5 x , and 4e 2 x e3 x e 4 x e x should be
simplified to 3e5 x . Unless specified in the question, expressions do not need to be factorized, nor
do factorized expressions need to be expanded, so x ( x 1) and x 2 x are both acceptable.
Please note: intermediate A marks do NOT need to be simplified.
N21/5/MATHX/HP3/ENG/TZ0/XX/M
9
Calculators
A GDC is required for this paper, but If you see work that suggests a candidate has used
any calculator not approved for IB DP examinations (eg CAS enabled devices), please follow
the procedures for malpractice.
10. Presentation of candidate work
Crossed out work: If a candidate has drawn a line through work on their examination
script, or in some other way crossed out their work, do not award any marks for that work
unless an explicit note from the candidate indicates that they would like the work to be
marked.
More than one solution: Where a candidate offers two or more different answers to the
same question, an examiner should only mark the first response unless the candidate
indicates otherwise. If the layout of the responses makes it difficult to judge, examiners
N21/5/MATHX/HP3/ENG/TZ0/XX/M
1.
(a)
(b)
et e t
2
t
e et
f (t )
2
f (t )
A1
f (t )
METHOD 1
2
f (t )
g (t )
A1
AG
[2 marks]
2
M1
substituting f and g
(et e t )2 (et e t )2
4
t 2
(e ) 2 (e t )2 (et ) 2 2 (e t ) 2
4
t 2
t 2
(e ) (e )
e 2t e 2t
2
2
f (2t )
(M1)
A1
AG
METHOD 2
f (2t )
e2t
e
2t
2
(e ) (e t )2
2
t
(e e t )2 (et e t )2
4
2
2
f (t )
g (t )
t 2
M1
M1A1
AG
[3 marks]
Note: Accept combinations of METHODS 1 & 2 that meet at equivalent expressions.
N21/5/MATHX/HP3/ENG/TZ0/XX/M
(c)
substituting ei u cos u isin u into the expression for f
obtaining e i u cos u isin u
cos u isin u cos u isin u
f (i u )
2
(i)
(M1)
(A1)
Note: The M1 can be awarded for the use of sine and cosine being odd and even
respectively.
2cos u
2
cosu
(ii)
A1
[3 marks]
cos u isin u cos u isin u
2
g (i u )
substituting and attempt to simplify
2isin u
2
isinu
(d)
METHOD 1
2
f (i u )
g (i u )
A1
[2 marks]
2
substituting expressions found in part (c)
2
(M1)
2
cos u sin u ( cos 2u)
METHOD 2
e2i u e 2i u
f (2iu )
2
cos 2u isin 2u cos 2u isin 2u
2
cos2u
(M1)
A1
M1
A1
Note: Accept equivalent final answers that have been simplified removing all
2
imaginary parts eg 2cos u 1 etc
[2 marks]
1
(e)
f (t )
2
g (t )
(e2t e
2t
(et
2
2) (e2t
4
N21/5/MATHX/HP3/ENG/TZ0/XX/M
e t )2 (et e t ) 2
4
2t
e
2)
M1
A1
4
1
4
A1
Note: Award A1 for a value of 1 obtained from either LHS or RHS of given
expression.
f (i u )
1
2
g (i u )
2
hence f (t )
cos 2 u sin 2 u
2
g (t )
2
f (i u )
Note: Award full marks for showing that
M1
2
g (i u )
f ( z)
2
2
g ( z)
AG
2
1, z
.
[4 marks]
(f)
A1A1A1A1
Note: Award A1 for correct curves in the upper quadrants, A1 for correct curves in
the lower quadrants , A1 for correct x-intercepts of ( 1, 0) and (1, 0) (condone
x.
x 1 and 1), A1 for y x and y
[4 marks]
1
(g)
N21/5/MATHX/HP3/ENG/TZ0/XX/M
(M1)
attempt to rotate by 45 in either direction
Note: Evidence of an attempt to relate to a sketch of xy
this (M1).
attempting to rotate a particular point, eg (1, 0)
(1, 0) rotates to
hence k
1
2
1
,
2
1
2
(or similar)
k would be sufficient for
(M1)
(A1)
A1A1
[5 marks]
Total [25 marks]
1
2.
(a)
(i)
METHOD 1
dy
y
dt
dy
dt
y
ln y t c OR ln y
N21/5/MATHX/HP3/ENG/TZ0/XX/M
(M1)
t c
A1A1
Note: Award A1 for ln y and A1 for t and c.
Aet
y
AG
METHOD 2
rearranging to
ye
t
y
0 AND multiplying by integrating factor e
t
M1
A
A1A1
Aet
y
(ii)
dy
dt
AG
[3 marks]
Aet into differential equation in x
substituting y
M1
dx
x Aet
dt
dx
x
Aet
dt
AG
[1 mark]
(iii)
integrating factor (IF) is e
t
xe
x
(M1)
(A1)
t
e
e
1dt
dx
dt
t
xe
At
t
A
(A1)
A1
D
At D e
t
Note: The first constant must be A, and the second can be any constant for the final
A1 to be awarded. Accept a change of constant applied at the end.
[4 marks]
1
(b)
(i)
d2 y
dt 2
dx
dt
N21/5/MATHX/HP3/ENG/TZ0/XX/M
dy
dt
A1
EITHER
x y
dy
dt
dy
dt
(M1)
dy
dt
A1
OR
x y
2
x y
(M1)
x y
A1
THEN
dy
2
dt
AG
[3 marks]
(ii)
dY
2Y
dt
dY
2dt
Y
ln Y 2t c OR ln Y
Y
(iii)
Be2t
A1
M1
2t c
A1
AG
[3 marks]
dy
dt
Be2t
y
Be2t dt
M1
y
B 2t
e C
2
A1
Note: The first constant must be B, and the second can be any constant for the final
A1 to be awarded. Accept a change of constant applied at the end.
[2 marks]
1
N21/5/MATHX/HP3/ENG/TZ0/XX/M
(iv) METHOD 1
substituting
B e2t
x
x
dy
dy
B e 2t and their (iii) into
dt
dt
B 2t
e C
2
M1(M1)
x y
A1
B 2t
e C
2
AG
Note: Follow through from incorrect part (iii) cannot be awarded if it does not lead to
the AG.
METHOD 2
dx
B 2t
x
e C
dt
2
dx
B 2t
x
e C
dt
2
d (x e t )
B t
e Ce t
dt
2
B
xe t
et C e t dt
2
B
xe t
et C e t D
2
B 2t
x
e C Det
2
dy
B 2t
x y B e 2t
e C Det
dt
2
B 2t
x
e C
2
M1
A1
B 2t
e C
2
D 0
M1
AG
[3 marks]
1
(c)
(i)
dy
dt
d2 y
dt 2
4x
4
N21/5/MATHX/HP3/ENG/TZ0/XX/M
y
dx
dt
dy
seen anywhere
dt
METHOD 1
dy
d2 y
4( x y )
2
dt
dt
attempt to eliminate x
1
dy
4
y
y
4
dt
dy
2
3y
dt
d2 y
dy
2
3y 0
2
dt
dt
M1
M1
dy
dt
A1
AG
METHOD 2
M1
rewriting LHS in terms of x and y
2
d y
dy
2
3y
2
dt
dt
0
(ii)
8x 5 y
2
4x y
d2 y
F 2e t
2
dt
2 t
F e
2 F e t 3Fe t 0
2
2
3 0 (since e t 0 )
1 and
2 are 3 and -1 (either order)
dy
dt
F e t,
3y
A1
AG
[3 marks]
(A1)
(M1)
A1
A1
[4 marks]
1
(iii)
N21/5/MATHX/HP3/ENG/TZ0/XX/M
METHOD 1
y Fe3t Ge
t
dy
d2 y
3Fe3t Ge t , 2 9 Fe3t Ge t
dt
dt
2
d y
dy
2
3 y 9 Fe3t Ge t 2 3Fe3t Ge
2
dt
dt
3t
9 Fe
Ge t 6 Fe3t 2Ge t 3Fe3t 3Ge t
0
(A1)(A1)
t
3 Fe3t Ge
t
M1
A1
AG
METHOD 2
y Fe 1t Ge 2t
dy
d2 y
F 1e 1t G 2e 2t , 2 F 12e 1t G 22e 2t
dt
dt
2
d y
dy
2
3 y F 12e 1t G 22e 2t 2 F 1e 1t G 2e
2
dt
dt
2
1t
Fe
2 3 Ge 2t 2 2 3
0
(A1)(A1)
2t
3 Fe 1t Ge
2t
M1
A1
AG
[4 marks]
Total [30 marks]