ANSWERS
Analysis & Approaches HL Exam Practice Workbook
Answers to Practice Questions
1 Number and algebra
1
Just plug the numbers into your calculator. The answer is 4
2
3
3
We can write this expression as
4 10
10
10 30 4
26 10
2.6 10 10
2.6 10
6
8
10
10
10
0.75
7.5
7.5
3
10
10
4
10
10
10
10
4 The first term is 20. The common difference is 3. Therefore
𝑢
20
3
25
20 3 24
52
1
Tip: If this had been a calculator question, you could have just used 𝑢
your calculator sequence function.
5
1602
𝑢
1581
6
21
17 𝑛
93
𝑛
94
𝑛
𝑢
𝑢
17 𝑛
20, 𝑢
𝑢
3 in
1
1
1
10
𝑢
3𝑑
34
𝑢
9𝑑
Subtracting gives;
24
4
6𝑑
𝑑
Substituting into the first equation:
10
3
𝑢
𝑢
4
2
Therefore
𝑢
2
74
19
4
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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7
𝑢
𝑆
8
9
13, 𝑑
30
2
2
𝑆
4
3 so
13
3
130
30
1
915
1340
The easiest way to deal with sigma notation is to write out the first few terms by substituting
in 𝑟 1, 𝑟 2, 𝑟 3, etc.:
𝑆
16
21
26 …
So the first term is 16 and the common difference is 5.
10 Your GDC should have a sum function, which can be used for this. The answer will be
26350, but you should still write down the first term and common difference found in
question 9 as part of your working.
11 This is an arithmetic sequence with first term 500 and common difference 100. The question
is asking for 𝑆 .
𝑆
28
2
2
500
100
27
51800 m
Tip: The hardest part of this question is realising that it is looking for the sum of the sequence,
rather than just how far Ahmed ran on the 28th day.
12 2.4% of $300 is $7.20. This is the common difference. After one year, there is $307.20 in
the account, so this is the ‘first term’.
𝑢
307.20
9
7.20
$372
Tip: The hardest part of this question is being careful with what ‘after 10 years’ means – it is
very easy to be out by one year.
13 a The differences in velocity are 1.1, 0.8 and 0.8. Their average is 0.9. When 𝑡
are looking for the sixth term of the sequence which would be
𝑢
b





0
0.9
5
4.5 m s
There are many criticisms which could be made about this model – for example:
There is too little data for it to be reliable.
There is no theoretical reason given for it being an arithmetic sequence.
The ball will eventually hit the ground.
The model predicts that the ball’s velocity grows without limit.
There seems to be a pattern with smaller differences later on.
14 The first term is 32 and the common ratio is
𝑢
0.5 we
32
1
2
.
1
16
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Analysis & Approaches HL Exam Practice Workbook
Tip: If this had been a calculator question, you could have solved the simultaneous equations
using your GDC.
𝑢
1
Analysis & Approaches HL Exam Practice Workbook
15 The first term is 1 and the common ratio is 2 so
2
If 𝑢
4096 then
4096
2
There are four ways you should be able to solve this:
 On a non-calculator paper you might be expected to figure out that 4096 2
 You can take logs of both sides to get ln 4096
𝑛 1 ln 2 and solve for n.
and intersect it with 𝑦 4096
 You can graph 𝑦 2
 You can create a table showing the sequence and determine which row 4096 is in.
Whichever way, the answer is 13.
16 𝑢
𝑢 𝑟
16
𝑢
𝑢 𝑟
256
Dividing the two equations:
256
16
𝑢 𝑟
𝑢 𝑟
𝑟
16
𝑟
2
𝑢
4 for both possible values of 𝑟.
17 The first term is 162, the common ratio is
1
3
162 1
𝑆
1
3
1
6560
27
243
Tip: You can always use either sum formula, but generally if r is between 0 and 1 the second
formula avoids negative numbers.
18 The easiest way to deal with sigma notation is to write out the first few terms by substituting
in 𝑟 1, 𝑟 2, 𝑟 3, etc.:
𝑆
10
50
250 …
So the first term is 10 and the common ratio is 5.
19 Your GDC should have a sum function, which can be used for this. The answer will be
24414060, but you should still write down the first term and common ratio found in
question 18 as part of your working.
20 a This is a geometric sequence with first term 50,000 and common ratio 1.2. ‘After 12
days’ corresponds to the 13th term of the sequence so:
𝑢
50,000
1.2
445805
b The model suggests that the number of bacteria can grow without limit.
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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𝐹𝑉
2000
4
1
100
£2981.67
12
However, the general expectation is that you would use the TVM package on your
calculator for this type of question. Make sure you can get the same answer using your
package as different calculators have slightly different syntaxes.
22 Using the TVM package, 𝑖
5.10%
23 Unless stated otherwise, you should assume that the compound interest is paid annually.
Using 𝐹𝑉 200 and 𝑃𝑉 100, the TVM package suggests that 33.35 years are required.
There 33 years would be insufficient so 34 complete years are required.
24 12% annual depreciation is modelled as compound interest with an interest rate of –12%.
Using the TVM package the value is $24000 to three significant figures.
25 When adjusting for inflation, the ‘real’ interest rate is 3.2
the TVM package we find a final value of $2081
2
26 This expression is 2
2
27 2𝑥
𝑥
2.4
0.8%. Using this value in
16
8𝑥
28 10
This is equivalent to 𝑥
log
29 The given statement is equivalent to
2𝑥
6
ln 5
So
2𝑥
ln 5
6
𝑥
1
ln 5
2
3
Tip: The answer could be written in several different ways – for example, ln √5 𝑒 . Generally
speaking any correct and reasonably simplified answer would be acceptable.
30 Using appropriate calculator functions:
ln 10
log
𝑒
2.30
0.434
2.74
31 LHS
𝑚 1 𝑚 1
𝑚
𝑚 𝑚 1
2𝑚
𝑚
1
RHS
32 a 𝑥
𝑥 1
𝑥
𝑎𝑥
𝑎
𝑏
𝑏
𝑏
1
𝑎
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21 You could use the formula:
1
𝑎
Comparing constant terms:
0
𝑏
Tip: If you are not familiar with comparing coefficients, you can also substitute in 𝑥
𝑥 1 to set up some simultaneous equations to get the same results.
8
33 8
2
√8
0 and
4
34 If 𝑥 log 32 this is equivalent to 4
32. There are many ways to proceed, but we could
write everything in terms of powers of 2:
2
2
2
2
Therefore, 2𝑥
35 𝑥
5 so 𝑥
2.5
log 2 5
log 2 log 5
1 𝑦
36 log 12
37 ln 5
ln 4 3
𝑥 1 ln 5 ln 4
𝑥 ln 5 ln 5 ln 4
𝑥 ln 5 2𝑥 ln 3 ln 4
𝑥 ln 5 2 ln 3
ln 4
𝑥
ln 3
2𝑥 ln 3
ln 5
ln 5
ln 4 ln 5
ln 5 2 ln 3
ln 20
ln 5 ln 9
ln 20
5
ln
9
Tip: In the calculator paper, if an exact form is not required then this type of equation is best
solved using an equation solver or a graphical method.
38 The first term is 2, the common ratio is , so
𝑆
2
1
3
1
3
39 The common ratio is 2𝑥. This will converge if |2𝑥|<1 which is when |𝑥|
𝑥
also write this as
40 2
𝑥
2
𝐶
2
16
16
. You could
.
𝑥 𝐶
2
𝑥
𝐶
4 8 𝑥 6 4 𝑥
32𝑥 24𝑥
8𝑥
𝑥
2
4
𝑥
2
𝑥
𝑥
𝑥
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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Analysis & Approaches HL Exam Practice Workbook
b Comparing coefficients of x:
2
0.01, then
16 32 0.01
16 0.32
16.32
0.01
24
0.0001 …
42 You could find the full expansion, but that would waste time (especially if the brackets were
raised to a larger exponent). The general term is
1
𝑥
𝐶 2𝑥
𝐶2
𝐶2
𝑥 𝑥
1
𝑥
1
For this to be a constant, we need 4𝑟
The term is then 𝐶 2
12
0, so 𝑟
3
32
1
!
43 𝐶
1
1
6
1
6
7
! !
2
2
7
2
7
6
8
3
3
8
3
8
4 5 6 7 8
1 2 3 4 5
56
Tip: You could also find this by looking at the appropriate number in Pascal’s triangle.
44 There are 7 letters so there are 7!
5040 possible arrangements.
45 The order in which people are picked does not change the committee, so this is
𝐶
210
46 The order here matters. For example, 134 is different from 413. There are 𝑃
60 ways.
47 There are two possibilities – either it ends with a 2 or a 4. For each of these there are
𝑃
24 ways to select the first three digits, making 48 ways in total.
48
1
2𝑥
1
𝑥
1
𝑥
2
2𝑥
!
⋯
⋯
49
𝑥
1
1
2
2
1
1
1
2
1 𝑥 𝑥
2 4 8
𝑥
2
1
2
2!
𝑥
2
⋯
⋯
50 The expansion is valid if
1 so |𝑥|
2
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Analysis & Approaches HL Exam Practice Workbook
41 Use 𝑥
𝑥
4≡𝐴 𝑥
1
2∶6
3𝐴
If 𝑥
𝐴
𝐵 𝑥
Analysis & Approaches HL Exam Practice Workbook
≡
51
2
2
If 𝑥
1
3
3𝐵
𝐵
1
Therefore
𝑥
𝑥 4
2 𝑥
1
1
2𝑖 2
52 𝑤𝑧
≡
2
𝑥
1
2
𝑖
𝑥
2
1
𝑖
4𝑖
2 𝑖
4
3𝑖
Therefore
2𝑧
𝑤𝑧
4
2𝑖
4
3𝑖
8
𝑖
𝑖
53
So, the real part is
54 Let 𝑧
𝑎
𝑖𝑏
𝑎
𝑖𝑏
2 𝑎
3𝑎
𝑖𝑏
4
𝑖𝑏
4
6𝑖
6𝑖
Comparing real and imaginary parts:
4
3
3𝑎
4 so 𝑎
𝑏
6 so 𝑏
So, 𝑧
55 |𝑧|
6
6𝑖
1
tan arg 𝑧
arg 𝑧
√3
√4
2
√3
1
π 2π
or
3
3
Considering the position of 𝑧 on the complex plane, arg 𝑧
(
would also be an acceptable answer)
56 |𝑧|
√2
π
arg 𝑧
4
So, 𝑧
2
√8
√8 cis
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𝑧
4 cis
Analysis & Approaches HL Exam Practice Workbook
57 Write 𝑧
in cartesian form.
4𝑖 sin
4 cos
4√3
2
4𝑖
2
= 2√3
2𝑖
58 First, write each part of the sum in cartesian form.
𝑒
cos
π
3
𝑖 sin
𝑒
cos
π
6
𝑖 sin
Re 𝑒
2𝑒
59 2 cis
60
2
π
3
1
2
π
6
1
2
√3
2
√3
2
1
𝑖
𝑖
1
2
√3
3 cis
𝑖 𝑒
√3
2
𝑖
6 cis
2
𝑖
𝑖
1
2
√3
6 cis π
6
√
61 Since the cubic has real coefficients, the roots occur in conjugate pairs, so 𝑥 1 𝑖 is also
a solution. This means that 𝑥 1 𝑖 and 𝑥 1 𝑖 are both factors, so together they
form the factor:
𝑥
1
𝑖 𝑥
1
𝑖
𝑥
1
4𝑥
Dividing into the polynomial, 𝑥
So, the final root is 𝑥
62 𝑧
𝑖
𝑥
6𝑥
2𝑥
4
2
𝑥
2𝑥
2 𝑥
2
2
2 cis 4
16 cis π
16
63 From DeMoivre’s theorem:
cos 3𝑥
𝑖 sin 3𝑥
cos 𝑥
𝑖 sin 𝑥
Using the binomial expansion:
cos 𝑥
3𝑖 cos 𝑥 sin 𝑥
3 cos 𝑥 sin 𝑥
𝑖 sin 𝑥
Comparing real parts:
cos 3𝑥
cos 𝑥
3 cos 𝑥 sin 𝑥
Using the identity sin 𝑥
cos 3𝑥
cos 𝑥
4 cos 𝑥
1
3 cos 𝑥 1
cos 𝑥:
cos 𝑥
3 cos 𝑥
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8
8𝑖
8 cis
𝑧
8𝑖
5π
9π
π
or 8 cis
or 8 cis
2
2
2
8 cis
8 cis
Analysis & Approaches HL Exam Practice Workbook
64 First write 8𝑖 in polar form, finding three different forms since we are effectively cube
rooting each side:
π
2
π
2
1
,
3
/
or 8 cis
8 cis
5π
2
5π
2
/
or 8 cis
1
,
3
8 cis
9π
2
𝑖, 𝑧
√3
𝑖, 𝑧
9π
2
/
1
3
5π
3π
π
, 2 cis
2 cis , 2 cis
6
2
6
Tip: You could write this as 𝑧
65 When 𝑛
1:
LHS
2
1
RHS
1
1
√3
1
1
So, the statement is true when 𝑛
1
Assume that the statement is true when 𝑛
2𝑟
2𝑖.
1
𝑘
𝑘
Then:
2𝑟
1
2𝑟
𝑘
2𝑘
2
𝑘
2𝑘
1
𝑘
1
2 𝑘
1
1
1
1
So, if 𝑛 𝑘 is true then 𝑛
for all positive integer 𝑛.
𝑘
1 is also true. Since 𝑛
1 is true, the statement is true
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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1:
2
7
5
So, the statement is true when 𝑛
1
Assume that the statement is true when 𝑛
2
7
Analysis & Approaches HL Exam Practice Workbook
66 When 𝑛
𝑘
5𝐴
where A is an integer.
Then:
2
7
7
7
5𝐴
2
35𝐴
7
2
35𝐴
5
2
5 7𝐴
7
2
2
2
2
2
2
2
which is divisible by 5.
So, if 𝑛 𝑘 is true then 𝑛
for all positive integer 𝑛.
67 When 𝑛
LHS
𝑘
1 is also true. Since 𝑛
1 is true, the statement is true
1:
cos 𝜃
𝑖 sin 𝜃
RHS
So, the statement is true when 𝑛
1
Assume that the statement is true when 𝑛
cos 𝜃
𝑖 sin 𝜃
cos 𝑘𝜃
𝑘
𝑖 sin 𝑘𝜃
Then:
cos 𝜃
𝑖 sin 𝜃
cos 𝑘𝜃
cos 𝜃
𝑖 sin 𝑘𝜃 cos 𝜃
cos 𝑘𝜃 cos 𝜃
𝑖 sin 𝜃
cos 𝜃
𝑖 sin 𝜃
𝑖 sin 𝜃
sin 𝑘𝜃 sin 𝜃
𝑖 cos 𝑘𝜃 sin 𝜃
sin 𝑘𝜃 cos 𝜃
Using the compound angle theorem:
cos 𝑘
1 𝜃
𝑖 sin 𝑘
So, if 𝑛 𝑘 is true then 𝑛
for all positive integer 𝑛.
68 Suppose that
𝑘
1 is also true. Since 𝑛
1 is true, the statement is true
√5 where 𝑝 and 𝑞 have no common factors above 1.
5
Then
𝑝
1 𝜃
5𝑞
So, 𝑝 is divisible by 5, therefore 𝑝 is divisible by 5 so let 𝑝
5 so 𝑞
5𝑟
25𝑟 therefore 𝑞 is divisible by 5. This contradicts the fact that p and q
have no common factor above 1, so √5 cannot be written as a fraction, so is irrational.
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70 Using the GDC, 𝑥
1, 𝑦
2, 𝑧
Analysis & Approaches HL Exam Practice Workbook
69 When 𝑛 41 all three terms are divisible by 41 so the sum is divisible by 41, therefore
this is not prime.
3
71 Adding the first two equations:
2𝑥
4𝑦
6𝑧
2
Dividing by 2:
𝑥
2𝑦
3𝑧
1
This contradicts the third equation, so the system is inconsistent.
72 Eliminating 𝑧, 1
2𝑥
1
𝑦
7
2:
4
3:
4𝑥
2𝑦
2𝑥
𝑦
14
7
5
Since equations [4] and [5] are identical, there are infinite solutions.
Setting 𝑥
𝑦
2𝜆
𝜆 in equation [4]:
7
Substituting into 1:
𝜆
𝑧
2𝜆
3𝜆
7
𝑧
2
9
So, the general solution is 𝑥
𝜆, 𝑦
2𝜆
7, 𝑧
3𝜆
9
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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1 Rearrange into the form 𝑦
3𝑥
4𝑦
5
4𝑦
0
3𝑥
3
𝑥
4
𝑦
So, 𝑚
2 Use 𝑦
𝑦
𝑦
𝑚𝑥
5
5
4
𝑚 𝑥
𝑦
𝑥 :
3 𝑥 2
3𝑥 6
3𝑥 2
3 Find the gradient using 𝑚
1
9
5
3
Use 𝑦
𝑦
𝑦
1
2𝑦
2𝑦
2
7
𝑥
4
𝑦
2𝑥
:
1
2
𝑚 𝑥
𝑥 :
1
𝑥 9
2
𝑥 9
0
4 Gradient of parallel line is 𝑚
𝑦
𝑐:
,𝑐
4
4
𝑦
𝑚
2 𝑥
2
1
2
5 Gradient of perpendicular line is 𝑚
𝑦
3
𝑦
4𝑥
4 𝑥
7 2𝑥
𝑥
2
1
1
2
4
2
11
6 Substitute 𝑥
f
Analysis & Approaches HL Exam Practice Workbook
2 Functions
3
8
2 into the function:
2
4
0
8 Graph the function using the GDC:
f 1
2, so range is f 𝑥
2
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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12
4
3𝑥
3𝑥
𝑥
8 , solve f 𝑥
8:
8
12
4
10 Reflect the graph in the line 𝑦
𝑥:
11 Put in the vertical and horizontal asymptotes and the 𝑥-intercepts (zeros of the function).
Since f 𝑥
2, it must tend to ∞ as it approaches the vertical asymptote from either side.
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Analysis & Approaches HL Exam Practice Workbook
9 To find f
The 𝑦-intercept is 0, 2 . Now sketch the graph from the plot on the GDC:
13 a Graph the function and use ‘min’ and ‘max’ to find the coordinates of the vertices, moving
the cursor as necessary:
3.12, 20.3 ,
Coordinates of vertices:
1, 0 , 1.12, 20.3
b From the graph you can see there is a line of symmetry through the maximum point:
Line of symmetry: 𝑥
1
14 Graph the function and look for values where there appear to be asymptotes:
Vertical asymptotes occur at values of 𝑥 where the 𝑦-values appears as ‘error’:
Vertical asymptotes: 𝑥
3 and 𝑥
2
The 𝑦-value approaches 1 as the 𝑥 value gets big and positive or big and negative:
Horizontal asymptote: 𝑦
1
15 Graph the function and use ‘root’, moving the cursor from one to the other:
Zeros: 𝑥
0.311, 1.92
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Analysis & Approaches HL Exam Practice Workbook
12 Graph the function using the GDC:
Analysis & Approaches HL Exam Practice Workbook
16 Graph the function and use ‘isct’, moving the cursor from one intersection point to the
other:
0.467, 0.599 and 1.83, 7.50
Points of intersection:
17 a Substitute g 𝑥 into f 𝑥 :
1
3𝑥 4
1
3𝑥 6
f g 𝑥
2
b Substitute f 𝑥 into g 𝑥 :
g f 𝑥
3
𝑥 2
3
4
4
18 Domain of f is 𝑥
𝑥
3
𝑥
19 Let 𝑦
2
5
f 𝑥 and rearrange to make 𝑥 the subject:
𝑦
𝑥𝑦
𝑥
𝑥 1
2 so domain of fg is
2𝑦
𝑥𝑦
𝑦
𝑥
𝑥
𝑥
𝑥
1
1
1
1
1
2
1
2𝑦
2𝑦
2𝑦
𝑦
So,
f
20
𝑥
1 2𝑥
1 𝑥
Graph the function and use ‘min’ to find the coordinates of the minimum point:
The turning point has 𝑥-coordinate 𝑥
1.
1, so largest possible domain of given form is 𝑥
21 Graph A is the only negative quadratic so that is equation b.
Graph B has a negative 𝑦-intercept so that is equation c.
Graph C has a positive 𝑦-intercept so that is equation a.
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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For 𝑥-intercepts solve 3𝑥
5𝑥 4
0
3 𝑥
3 𝑥 1 𝑥 4
0
𝑥 1 or 4
23 a 𝑥
𝑥
4𝑥
2
7
3
𝑥
2
15𝑥
4
12
Analysis & Approaches HL Exam Practice Workbook
22 Negative quadratic with 𝑦-intercept 0, 12
0:
7
b Positive quadratic with 𝑦-intercept 0, 7
Vertex at ℎ, 𝑘
7𝑥 15
3 𝑥 5
24 2𝑥
2𝑥
2, 3
0
0
So
2𝑥
3
0 so 𝑥
5
0 so 𝑥
or
𝑥
5
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5𝑥
3
3
13
4
5
2
𝑥
𝑥
Analysis & Approaches HL Exam Practice Workbook
25 a 𝑥
b Use the completed square form from part a and solve for 𝑥:
5𝑥
3
13
5
4
2
5
𝑥
2
5
𝑥
2
𝑥
𝑥
𝑥
0
0
13
4
√13
2
5 √13
2
26 Use the quadratic formula with 𝑎
4
𝑥
4 3
4
2 3
3, 𝑏
4, 𝑐
2:
2
4
√40
6
4 2√10
6
2 √10
3
27 Solve the equation 𝑥
𝑥
4 𝑥
3
𝑥
0
4 or
𝑥
12
0:
3
Sketch the graph:
So, 𝑥
28 Δ
3 or 𝑥
4
4 4 3
5
25 48
23 0
So, no real roots
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4
4 3𝑘 12𝑘
4 36𝑘
𝑘
1
3
0
0
1
9
𝑘
30 Vertical asymptote: 𝑥
Horizontal asymptote: 𝑦
31 𝑦
0 (where 𝑎
3𝑘, 𝑏
4, 𝑐
12𝑘)
Analysis & Approaches HL Exam Practice Workbook
29 Two distinct real roots, so Δ
1
3
0
0
has
𝑥-intercept – , 0
𝑦-intercept 0,
Vertical asymptote 𝑥
Horizontal asymptote 𝑦
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Analysis & Approaches HL Exam Practice Workbook
32
33
34 𝑦
0.5 is decreasing since 0.5
𝑦 0.5 and 𝑦
line 𝑦 𝑥
35 2.8
e
e
log
.
1.
𝑥 are inverse functions, so one is a reflection of the other in the
.
.
So, 𝑘
1.03
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𝑥 ln 𝑥
𝑥 ln 𝑥
So
𝑥
4𝑥
4
Analysis & Approaches HL Exam Practice Workbook
36 Rearrange so RHS is 0 and then factorise:
0
0
0
or
ln 𝑥
𝑥
37 Let 𝑦
4
e
√𝑥
7𝑦
2 𝑦
𝑦
𝑦
10
5
𝑦
0
0
2 or 5
So,
√𝑥
𝑥
2 or 5
4 or 25
38 Combine the log terms using log 𝑥
quadratic equation:
log 𝑥 𝑥 2
𝑥
2𝑥
𝑥
2𝑥 8
𝑥 4 𝑥 2
𝑥
log 𝑦
log 𝑥𝑦 and then undo the log leaving a
3
2
0
0
4 or 2
However, checking both possible solutions in the original equation, you can see that
𝑥
4 is not valid, as you cannot have a log of a negative number.
So, 𝑥
2.
39 Graph the function and use ‘root’, moving the cursor from one to the other:
𝑥
1.27 or 1.25
40 This could be solved as above by rearranging to the form f 𝑥
𝑥 1:
intersection of the curves 𝑦 2 sin 𝑥 and 𝑦 𝑥
𝑥
0 or by finding the
1.67, 0.353 or 1.31
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f 𝑥
4 is a vertical translation by 4:
42 A translation 3 units to the right is given by 𝑦
𝑦
𝑥
𝑥
𝑥
3
6𝑥
8𝑥
44 𝑦
45 𝑦
3 :
2 𝑥 3
5
9 2𝑥 6 5
20
43 A vertical stretch by scale factor 2 is given by 𝑦
𝑦
f 𝑥
Analysis & Approaches HL Exam Practice Workbook
41 𝑦
2f 𝑥 :
𝑥 2
2 3𝑥
6𝑥
2𝑥 4
f 2𝑥 is a horizontal stretch with scale factor :
f 𝑥 is a reflection in the 𝑥-axis:
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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𝑦
𝑥
𝑥
47 𝑦
4f 𝑥
3
3𝑥
𝑥
4𝑥
4
1
𝑥
f
𝑥 :
1
1 is a vertical stretch with scale factor 4 followed by a vertical translation by 1.
So, 3, 2 ⟶ 3, 2
4
1
3, 7
Note that the order matters for two vertical transformations.
48 𝑦
3f
𝑥 is vertical stretch with scale factor 3 and a horizontal stretch with scale factor 2
(in either order):
49 Graph A is a polynomial with even degree since for large positive and large negative
values of 𝑥 the behaviour is the same.
Graphs B and C are both polynomials with odd degree since in each case the behaviour for
large positive values of 𝑥 is opposite to that for large negative values of 𝑥.
Since graph B is large and positive when 𝑥 is large and negative, the highest power must
have a negative coefficient.
So, graph A has equation b
graph B has equation c
graph C has equation a
50 The term with the highest power of 𝑥 is 2𝑥 so the graph has negative cubic shape.
There are roots at 𝑥 1 and 𝑥
touches the 𝑥-axis at 𝑥 1.
The 𝑦-intercept is 𝑦
2 0
3 but since the factor 𝑥
1
0
3
1 is squared, the graph only
6.
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Analysis & Approaches HL Exam Practice Workbook
46 A reflection in the 𝑦-axis is given by 𝑦
𝑦
𝑎 𝑥
4 𝑥
When 𝑥
So, 𝑦
12:
12
1
3 𝑥
4 𝑥
1
52 The remainder is given by f
𝑟
3
0:
4 is a factor then f
5
3
40
𝑎
:
4
2
12
53 If 3𝑥
𝑎
1
1
0, 𝑦
𝑎 0 4 0
4𝑎 12
𝑎
3
1 corresponding to the roots 𝑥 4 and 𝑥
1 is zero, the factor 𝑥 1 is cubed:
42
𝑎
0
0
40
54 a Use the factor theorem:
f
2
3
22
2
Therefore, 𝑥
2
20
2
24
0
2 is a factor of f 𝑥 .
b f 𝑥 is the product of 𝑥 2 and a quadratic factor of the form
3𝑥
𝑎𝑥 12: 3𝑥
22𝑥
20𝑥 24
𝑥 2 3𝑥
𝑎𝑥
12
Equating coefficients of 𝑥 :
22 𝑎
𝑎 16
So, f 𝑥
6
𝑥
2 3𝑥
16𝑥
12
Factorize the quadratic and solve:
𝑥
𝑥
2 3𝑥
16𝑥
2 3𝑥 2 𝑥
2
2, , 6
3
𝑥
55 Use the results sum
Sum
–
12
0
6
0
and product
:
–
Product
80
1
6
40
3
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Analysis & Approaches HL Exam Practice Workbook
51 The graph has factors 𝑥 4 and 𝑥
respectively. Since the gradient at 𝑥
𝛼
Analysis & Approaches HL Exam Practice Workbook
56 First find the sum and product of the roots 𝛼 and 𝛽:
𝛽
3
4
𝛼𝛽
Then use these to find the sum and product of the roots
and :
Finally relate these to the coefficients of the new quadratic 𝑎𝑥
𝑏𝑥
𝑐
0:
⇒𝑏
⇒𝑐
Let 𝑎
3, so that 𝑏
4 and 𝑐
So, the equation is 3𝑥
4𝑥
57 The sum of the roots will be –
3
2
𝑎
3 2
3
𝑎
2
i
2
16
0
:
i
The product of the roots will be
3 2 i 2
3
𝑐
5
2
2
𝑐 15
16
:
i
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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has

𝑥-intercept – , 0

𝑦-intercept 0,

Horizontal asymptote 𝑦

Vertical asymptotes for any real roots of 𝑐𝑥
𝑥
𝑥
𝑥
Analysis & Approaches HL Exam Practice Workbook
58 𝑦
3𝑥
4
4 𝑥
0
𝑑𝑥
𝑒
𝑐
0
0
0
1
0
4, 1
So, vertical asymptotes at 𝑥
59 𝑦
4 and 𝑥
1
has
 𝑥-intercepts for any real roots of 𝑎𝑥
𝑏𝑥
 𝑦-intercept 0,
 Vertical asymptote 𝑥
 An oblique asymptote of the form 𝑦
𝑥
𝑥
𝑥
𝑥
12
3 𝑥
𝑝𝑥
𝑞
0
4
0
3, 4
So, 𝑥-intercepts 3, 0 and
4, 0
Divide the rational function out to find the
oblique asymptote:
𝑥
𝑥
12
𝑥
𝑥
1 𝑥
2
10
2
So, the oblique asymptote is 𝑦
𝑥
2
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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25
f
𝑥 and see whether it is the same as
f 𝑥 , f 𝑥 or neither:
𝑥
cos 𝑥
𝑥
f 𝑥
So, f 𝑥 is an odd function.
61 Graph A is not symmetric in the origin and not symmetric in the 𝑦-axis, so it is neither odd
nor even
Graph B is symmetric in the 𝑦-axis so it is an even function
Graph C is symmetric in the origin so it is an odd function
62 Find the 𝑥-values of the turning points:
3𝑥
6𝑥 9 0
𝑥
2𝑥 3 0
𝑥 3 𝑥 1
0
𝑥
3, 1
Sketch the graph:
The function is one-to-one between the two turning points so this is the largest interval of
the required form:
So, 3
𝑥
𝑥 :
63 Find f
𝑦
𝑥𝑦
𝑥𝑦
𝑥
f
f 𝑥
1
3𝑦 3𝑥
3𝑥 3𝑦
3𝑦 1
𝑦 3
1
1
𝑥
f
𝑥 and so f is self-inverse.
64 Graph A isn’t symmetric in the line 𝑦
𝑥 so it isn’t self-inverse
Graph B isn’t symmetric in the line 𝑦
𝑥 so it isn’t self-inverse
Graph C is symmetric in the line 𝑦
𝑥 so it is self-inverse
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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Analysis & Approaches HL Exam Practice Workbook
60 Find f
𝑥 𝑥 5 𝑥 2
𝑥 0, 5, 2
3𝑥
10𝑥
0 and solve the equation 𝑥
3𝑥
10𝑥
0:
Analysis & Approaches HL Exam Practice Workbook
65 Rearrange so that 𝑥
0
Sketch the graph:
So, 𝑥
66 Graph 𝑦
5 or 0
𝑥
and 𝑦
2
e
using the GDC and find the points of intersection:
Read off the required region (noting the asymptote at 𝑥
So, 0.924 𝑥 1 or 𝑥 1.56
67 Sketch the graph of 𝑦
axis:
𝑥
𝑥
1):
6 and reflect the part below the 𝑥-axis to be above the 𝑥-
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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69 Sketch the graphs of 𝑦
𝑥
|𝑥
𝑥
6 for 𝑥
1| and 𝑦
0 and reflect that in the 𝑦-axis:
4
Analysis & Approaches HL Exam Practice Workbook
68 Sketch the graph of 𝑦
3𝑥 on the same axes:
There is just one intersection point 𝑃 (solution to the equation), which is on the original
part of 𝑦 |𝑥 1|:
𝑥 1 4 3𝑥
4𝑥 3
3
𝑥
4
70 Sketch the graphs of 𝑦
|𝑥
2| and 𝑦
|2𝑥
3| on the same axes:
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
© Paul Fannon, Vesna Kadelburg and Stephen Ward 2021
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At 𝑄
𝑥 2
𝑥 2 2𝑥 3
3𝑥
1
1
𝑥
3
At 𝑃
𝑥 2 2𝑥
𝑥
5
So,
71 𝑦
5
𝑥
3
2𝑥
2
Analysis & Approaches HL Exam Practice Workbook
The graphs intersect at two points, 𝑃 and 𝑄, both of which are on the reflected part of
𝑦 |𝑥 2|:
3
2𝑥
3
𝑥
𝑓 𝑥 has
 an 𝑥-intercept at 𝑎, 0 so 𝑦
has a vertical asymptote at 𝑥
 a 𝑦-intercept at 0, 𝑏 so 𝑦
 a vertical asymptote at 𝑥
 a horizontal asymptote at 𝑦
has a 𝑦-intercept at 0,
has an 𝑥-intercept at 𝑐, 0
𝑐 so 𝑦
has a horizontal asymptote at 𝑦
𝑑 so 𝑦
 a minimum point at 𝑝, 𝑞 so 𝑦
 𝑦 → ∞ as 𝑥 → ∞ so 𝑦
𝑎
has a maximum point at 𝑝,
has a horizontal asymptote at 𝑦
0
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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29
f
factor
73 𝑦
𝑥
3 is a horizontal translation by 3 followed by a horizontal stretch with scale
2:
𝑓 𝑥 has
 an 𝑥-intercept at 𝑎, 0 so 𝑦
𝑓 𝑥
has a minimum point at 𝑎, 0
 a 𝑦-intercept at 0, 𝑏 so 𝑦
𝑓 𝑥
has a 𝑦-intercept at 0, 𝑏
 a vertical asymptote at 𝑥
𝑐 so 𝑦
 a horizontal asymptote at 𝑦
All negative values on 𝑦
𝑓 𝑥
𝑑 so 𝑦
𝑓 𝑥
also has a vertical asymptote at 𝑥
has a horizontal asymptote at 𝑦
𝑓 𝑥 become positive on 𝑦
𝑦
𝑓 𝑥
𝑐
𝑑
:
𝑑
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Analysis & Approaches HL Exam Practice Workbook
72 𝑦
𝑑
1
7
√25
10.5
2
3
2
49
𝑀
4
1
Analysis & Approaches HL Exam Practice Workbook
3 Geometry and trigonometry
5
36
,
,
2, 5, 0.5
3 Radius = 8 cm
π𝑟
Volume
4
π
3
8
2140 cm
4π𝑟
Surface area
4π
8
804 cm
π𝑟 ℎ
4 Volume
1
π
3
6
15
565 cm
Slope length, 𝑙, is given by:
𝑙
𝑟
ℎ
15
6
3√29
π𝑟𝑙
Surface area
π
π𝑟
6
3√29
π
6
418 cm
5 Volume
𝑥 ℎ
5
9
75.0 cm
𝑙
2.5
√349
2
Surface area
9
5
4
5
√
118 cm
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
© Paul Fannon, Vesna Kadelburg and Stephen Ward 2021
31
π𝑟 ℎ
π𝑟
π
30
5
π
Analysis & Approaches HL Exam Practice Workbook
6 Volume
5
2620 m
7 Draw the lines and label the angle required 𝜃:
Since the gradient of 𝒚
tan 𝛼
𝛼
4
1
tan
4
𝟒𝒙
76.0°
Since the gradient of 𝑦
𝛼
3
1
tan
3
So, 𝜃
180
76.0
tan 𝛽
𝟑 is 𝟒,
5
3𝑥 is 3,
71.6°
71.6
32.4°
8 Draw in the diagonal 𝐴𝐺. Angle needed is 𝐺𝐴𝐶
𝜃
First work in triangle 𝐴𝐵𝐶:
𝐴𝐶
8
√80
4
Then in triangle 𝐴𝐶𝐺:
tan 𝜃
𝜃
5
√80
tan
5
√80
29.2°
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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Analysis & Approaches HL Exam Practice Workbook
9 Draw in the two diagonals – they will intersect at the midpoint of each, 𝑀.
From triangle 𝐴𝐺𝐶:
𝐴𝐺
5
√80
√105
By symmetry, 𝐸𝐶
And 𝐴𝑀
√105
√
𝐶𝑀
Using the cosine rule in triangle 𝐴𝑀𝐶:
cos 𝑀
𝑀
𝐴𝑀
𝐶𝑀
𝐴𝐶
2 𝐴𝑀 𝐶𝑀
26.25 26.25 80
52.5
121.6°
So, acute angle between 𝐴𝐺 and 𝐸𝐶 is 180
10 sin 𝜃
𝜃
121.6
58.4°
.
.
sin
21.6°
1.8
4.9
11 By the sine rule,
3.8
sin 80
𝐴𝐶
𝐴𝐶
sin 55
3.8
sin 55
sin 80
3.16 cm
12 By the cosine rule,
cos 𝐶
𝐶
13 𝐴
𝑎
𝑏
𝑐
2𝑎𝑏
10
12
9
2 10 12
10
2
cos
6
15
12
9
10 12
47.2°
sin 42
30.1
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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Analysis & Approaches HL Exam Practice Workbook
14 Draw a diagram:
ℎ
6.5
6.5 tan 68
16.1 m
tan 68
ℎ
15 Start by drawing the situation described:
𝑥
30° by alternate angles
𝑦
180
So, 𝑃𝐴𝐵
110
30
70°
70
100°
By the cosine rule,
𝑑
150
𝑑
80
150
80
182 km
16 a 55°
55
150
2
80
150
80
cos 100
cos 100
radians
b 1.2 radians
17 𝑠
2
1.2
68.8°
𝑟𝜃
6 0.7
4.2 cm
18 𝐴
𝑟 𝜃
1
10
2
90 cm
1.8
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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Analysis & Approaches HL Exam Practice Workbook
19
Using the unit circle:
a sin 𝜃
π
sin 𝜃
b cos 𝜃
20 tan 2π
sin 𝜃
0.4
0.4
𝜃
sin 𝜃
cos 𝜃
sin 𝜃
cos 𝜃
tan 𝜃
since sin and cos are 2π periodic
Note that certain relationships, such as sin 𝜃
sin 𝜃 and cos 𝜃
cos 𝜃, are used
so often that you should just know them. You don’t want to have to go back to the unit
circle each time to derive them.
21 Relate
cos
4π
3
to one of the ‘standard’ angles: cos
π
π
3
π
using the unit circle or remembering cos 𝜃
cos
3
1
2
cos
π
cos 𝜃
22 By the sine rule,
sin 𝜃
14
sin 38
11
𝜃
𝜃
sin
sin 38
11
51.6° or 𝜃
14
180
51.6
128.4°
Check that each value of 𝜃 is possible by making sure that the angle sum in each case is less
than 180°:
51.6
38
128.4
38
89.6
166.4
180
180
So both are possible:
𝜃
51.6° or 128.4°
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
© Paul Fannon, Vesna Kadelburg and Stephen Ward 2021
35
cos 𝜃
1
sin 𝜃
9
16
1
7
16
√7
4
cos 𝜃
But cos 𝜃
So,
sin 𝜃 ≡ 1 to relate the value of cos 𝜃 to the value of sin 𝜃:
Analysis & Approaches HL Exam Practice Workbook
23 Use the identity cos 𝜃
0 for
π
√
cos 𝜃
24 cos 𝜃
𝜃
sin 𝜃
≡ cos 𝜃
2 cos 𝜃 sin 𝜃 sin 𝜃
≡ 2 sin 𝜃 cos 𝜃 cos 𝜃
sin 2𝜃 1
sin 𝜃
25 Relate 15° to a ‘standard’ angle:
cos 2
15
So, use cos 2𝜃
cos 30
√3
2
cos 30
2 cos 𝜃
2 cos 15
1:
1
cos 30 1
2
√3
1
2
2
√3 2
4
cos 15
So, taking the positive square root (since cos 15°
0):
√3 2
4
cos 15°
26
Period
2π
Amplitude
1
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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36
tan is a transformation of the form 𝑦
f
𝑥 where f 𝑥
tan 𝑥
So, it is a horizontal stretch of the tan curve with scale factor 2.
28 𝑦
3sin 𝜃
5 is a transformation of the form 𝑦
3f 𝜃
5 where f 𝜃
sin 𝜃
So, vertically there is a stretch with scale factor 3 followed by a translation by 5 of the sin
curve.
Horizontally, max points of sin 𝜃 occur at 𝜃
,
,
…
So,
π
3
π
𝑥
3
2 𝑥
𝑥
π 5π 9π
, , .
2 2 2
π 5π 9π
, , …
4 4 4
7π 19π
,
12 12
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37
Analysis & Approaches HL Exam Practice Workbook
27 𝑦
0.5
𝑎
0.16
Analysis & Approaches HL Exam Practice Workbook
29 The amplitude is half the difference between the minimum and maximum heights:
0.17
2
The period is twice the time from the maximum height to the minimum height:
2
Period
0.3
0.6
But period
So,
2π
𝑏
10π
3
0.6
𝑏
𝑐 is halfway between the maximum and minimum heights:
0.5
𝑐
0.16
0.33
2
So,
ℎ
0.17 cos
30 Graph 𝑦
𝑥
7 cos 2𝑥
√
and tan
√3
3
tan
The graph of 𝑦
𝜃
0.33
and 𝑦
4 and find the 𝑥-values of the intersection points:
𝜃
tan 𝜃,
0.795, 2.97
31 Since tan
𝜃
10π
𝑡
3
𝜋
π
6
tan 𝜃 for π
So, 𝜃
𝜃
π shows there is one other solution:
,
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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38
𝑥
Then, 0
75
𝑥
360 ⇒ 75
√
sin 𝐴
So,
𝐴
√3
2
sin
The graph of 𝑦
𝐴
𝐴
180 60
60 360
So, 𝐴
for 75°
𝑥
75
𝐴
360
Analysis & Approaches HL Exam Practice Workbook
32 Let 𝐴
75.
435°
60°
sin 𝐴 for 75°
𝐴
435° shows there are two solutions (𝐴 and 𝐴 ):
120°
420°
120°, 420°
𝑥
45°, 345°
33 Use the sine double angle formula so that everything is a function of 𝜃:
sin 2𝜃
2 sin 𝜃 cos 𝜃
2 sin 𝜃 cos 𝜃 sin 𝜃
sin 𝜃 2 cos 𝜃 1
sin 𝜃
sin 𝜃
0
0
So,
sin 𝜃
𝜃
0
0, π, 2π
or
2 cos 𝜃
1
cos 𝜃
𝜃
𝜃
0
1
2
π 5π
,
3 3
5π
π
0, , π, , 2π
3
3
Note that while it is tempting to cancel sin 𝜃 on the second line of the working, this would
lose the solutions resulting from sin 𝜃 0.
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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39
2 sin 𝑥
2 1 cos 𝑥
2 2 cos 𝑥
2 cos 𝑥
cos 𝑥 so that only cos 𝑥 is involved:
3 cos 𝑥
3 cos 𝑥
3 cos 𝑥
3 cos 𝑥
3
3
3
1
Analysis & Approaches HL Exam Practice Workbook
34 Use sin 𝑥 ≡ 1
0
0
0
0
This is a quadratic in cos 𝑥:
2 cos 𝑥
1 cos 𝑥
1
0
So
𝑥
1
2
120°, 120°
cos 𝑥
𝑥
1
180°, 180°
cos 𝑥
or
𝑥
180°, 120°, 120°, 180°
35 Use the definition sec 𝜃
:
sec
1
1
√2
√2
36
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
© Paul Fannon, Vesna Kadelburg and Stephen Ward 2021
40
cot 𝜃 ≡ cosec 𝜃 to relate the value of tan 𝜃 to the value of sin 𝜃:
1
tan 𝜃
cot 𝜃
cosec 𝜃
1
2
3
√13
2
cosec 𝜃
38 𝑦
So,
𝜃
2𝜋
√
√
arcsin
Since
√13
0 for
sin 𝜃
sin 𝑦
2
1
cosec 𝜃
But, sin 𝜃
So,
3
2
3
2
1
9
1
4
13
4
sin 𝜃
Analysis & Approaches HL Exam Practice Workbook
37 Use the identity 1
√3
2
𝑦
,𝑦
√
arcsin
39
Domain: 𝑥 ∈ ℝ
Range:
𝑦
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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41
𝐵 ≡ sin 𝐴 cos 𝐵
cos 𝐴 sin 𝐵 with the standard
sin 105° sin 60° 45°
sin 60° cos 45° cos 60° sin 45°
√3 √2 1 √2
2 2
2
2
√6 √2
4
41 tan 2𝜃
2 tan 𝜃
4
3 1 tan 𝜃
4 tan 𝜃 4 6 tan 𝜃
2 tan 𝜃 3 tan 𝜃 2 0
2 tan 𝜃 1 tan 𝜃 2
0
1
or 2
tan 𝜃
2
42 cos π 𝑥
cos π cos 𝑥 sin π sin 𝑥
1 cos 𝑥 0 sin 𝑥
cos 𝑥
43 The vector goes 5 units to the left and 2 units up:
5
2
𝐯
4
2
1
44
4𝐢
2𝐣
𝐤
1
0
6
46 Trace a path from P to S via Q and R, noting that going backwards along an arrow means
the vector needs to be negative:
45 𝐢
6𝐤
𝑃𝑆⃗ 𝑃𝑄⃗ 𝑄𝑅⃗
𝐚 𝐛 𝐜
𝑅𝑆⃗
47 Since M is the midpoint of YZ, 𝑌𝑀⃗
𝑌𝑍⃗ .
So, start by finding an expression for 𝑌𝑍⃗ :
𝑌𝑍⃗
𝑌𝑋⃗ 𝑋𝑍⃗
𝐚 𝐛
𝑋𝑀⃗
𝑋𝑌⃗
𝑌𝑍⃗
1
𝐚 𝐛
𝐚
2
1
1
𝐚
𝐛
2
2
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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Analysis & Approaches HL Exam Practice Workbook
40 Use the compound angle identity sin 𝐴
angles 𝐴 60° and 𝐵 45°:
𝑝
𝑞
𝑡
2𝑡
12
4𝑡
1
2
3
From 3 : 𝑡
3
So, 𝑝
6
3, 𝑞
49 |𝐯|
3
√35
5
1
2
6
3
50 Unit vector
1
7
𝑡𝐚 for some scalar 𝑡:
Analysis & Approaches HL Exam Practice Workbook
48 If two vectors are parallel, then 𝐛
𝑝
1
𝑞
𝑡 2
12
4
𝑡
2𝑡
4𝑡
2
6
3
51 𝐴𝐵⃗
𝐛 𝐚
4
5
3
1
1
2
9
2
1
52 𝐴𝐵 |𝐛 𝐚|
| 5𝐢 3𝐣 2𝐤
|4𝐢 𝐣 4𝐤|
1
4
4
√33
𝐢
4𝐣
2𝐤 |
53 For 𝐴𝐵𝐶𝐷 to be a parallelogram, need 𝐴𝐵⃗
𝐴𝐵⃗
𝐛
𝐷𝐶⃗ :
𝐚
5𝐢 4𝐣 𝐤
2𝐢 4𝐣 3𝐤
3𝐢
2𝐤
𝐷𝐶⃗
𝐜 𝐝
6𝐢 3𝐣 3𝐤
2𝐢 4𝐣 2𝐤
𝐴𝐵⃗
54 𝐚 ∙ 𝐛
4𝐢
𝐣
𝐤
𝐷𝐶⃗ so 𝐴𝐵𝐶𝐷 is not a parallelogram
𝑎 𝑏
4 2
𝑎 𝑏
3 5
𝑎 𝑏
1
3
4
55 𝐚 ∙ 𝐛 |𝐚||𝐛| cos 𝜃
3 8 cos 45°
24
√
12√2
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43
|𝐚||𝐛|
Analysis & Approaches HL Exam Practice Workbook
56 cos 𝜃
4
√14√69
𝜃
cos
√
97.39°
√
Required angle is acute so, 180
57 2𝐚
3𝐛 ∙ 2𝐚
3𝐛
4𝐚 ∙ 𝐚
4𝐚 ∙ 𝐚
4|𝐚|
4 25
64
97.39
82.6°
2𝐚 ∙ 2𝐚 2𝐚 ∙ 3𝐛
3𝐛 ∙ 2𝐚 3𝐛 ∙ 3𝐛
6𝐚 ∙ 𝐛 6𝐛 ∙ 𝐚 9𝐛 ∙ 𝐛
9𝐛 ∙ 𝐛
since 𝐚 ∙ 𝐛 𝐛 ∙ 𝐚
9|𝐛|
since 𝐚 ∙ 𝐚 |𝐚| and 𝐛 ∙ 𝐛 |𝐛|
9 4
58 For two vectors to be perpendicular, their scalar product must be zero:
2
𝑡
1
3 ∙ 4𝑡 1
𝑡
5
2 𝑡 3 4𝑡 1
1 6𝑡 0
1
𝑡
6
59 Use 𝐫
𝐫
𝐚
0
5𝑡
0
5
4 and 𝐝
7
𝜆𝐝 with 𝐚
5
4
7
1
2 :
0
1
2
0
𝜆
60 A direction vector for the line is given by the vector 𝐴𝐵⃗
𝐝
Use 𝐫
𝐫
3
5
4
2
6
1
𝐚
𝐛
𝐚:
5
11
3
𝜆𝐝:
2
6
1
𝜆
5
11
3
𝑥
61 Write 𝐫 as 𝑦 and form a separate equation for each component:
𝑧
𝑥
3
1
3 𝜆
𝑦
𝜆 4
0
4𝜆
𝑧
5
2
5 2𝜆
So, 𝑥
3
𝜆, 𝑦
4𝜆, 𝑧
5
2𝜆
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44
𝜆
𝑦
2
𝜆
𝜆
𝑥
𝑦
𝑧
4
3𝜆
2 𝜆
6 5𝜆
Write as the components of the vector 𝐫 and split into a part without 𝜆 and a part with 𝜆 as a
factor:
4
3𝜆
𝜆
6 5𝜆
4
3
So, 𝐫
𝜆 1
2
6
5
63 The angle between the lines is the angle between their direction vectors:
𝐫
2
cos 𝜃
10
√6√35
𝜃
cos
√ √
46.4°
64 a Speed is magnitude of the velocity vector:
Speed
1
3
2
√14 m s
b The position vector at time 𝑡 is given by 𝐫
𝐫
5𝐢
25𝐢
𝐤
10𝐣
10 3𝐢
21𝐤
𝐣
𝐫
𝑡𝐯:
2𝐤
65 First check whether one direction vector is a multiple of the other:
4
2
2
10
5
6
3
So, the lines have the same direction.
Then check whether the given point on 𝑙 also lies on 𝑙 (or vice versa):
4
0
6
𝜇
10
8
7
6
5
4
0 6 4𝜇
⇒ 𝜇 1.5
8
7 10𝜇 ⇒ 𝜇 1.5
5 4 6𝜇 ⇒ 𝜇 1.5
Since 𝑙 and 𝑙 have the same direction and share a common point, they are coincident.
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Analysis & Approaches HL Exam Practice Workbook
62 Set each expression equal to 𝜆 and rearrange to express 𝑥, 𝑦 and 𝑧 in terms of 𝜆:
5
2
3
1
1
2
𝜆
5
2
3
2
13
11
𝜆
2 2𝜇
𝜆 13 3𝜇
2𝜆 11 4𝜇
Analysis & Approaches HL Exam Practice Workbook
66 Set the position vectors to be equal and try to solve for 𝜆 and 𝜇:
2
3
4
𝜇
1
2
3
Solve any pair simultaneously, say 1 and 2 :
𝜆 2𝜇
𝜆 3𝜇
7
11
From GDC: 𝜆
1, 𝜇
4
Check in 3 :
3
2 1
11
4
5
4
5
So, the lines are skew.
67 Set the position vectors to be equal and solve for 𝜆 and 𝜇:
1
7
5
1
7
5
𝜆
3
1
4
3𝜆
𝜆
4𝜆
2
3
6
2
3
6
𝜇
2𝜇
𝜇
𝜇
1
2
1
1
2
3
Solve any pair simultaneously, say 1 and 2 :
3𝜆
𝜆
𝜇
2𝜇
3
4
From GDC: 𝜆
2, 𝜇
3
Check in 3 :
5
4
6
2
3
3
3
So, the lines intersect.
Substitute 𝜆
𝐫
1
7
5
2 into 𝑙 (or 𝜇
2
3
1
4
3 into 𝑙 ) to find the point of intersection:
5
9
3
So, point of intersection is 5, 9, 3
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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1
4
3
𝑏
𝑏
𝑏
4
2
3 5
1 2
5
2
2
𝐛
5𝐚
𝑏 𝑎
𝑏 𝑎
𝑏 𝑎
:
2 3
2 1
5 4
14
17
18
69 |𝐚 𝐛| |𝐚||𝐛| sin 𝜃
4 5 sin 30°
1
20
2
10
70 2𝐚
𝑎 𝑏
𝑎 𝑏
𝑎 𝑏
Analysis & Approaches HL Exam Practice Workbook
𝑎
68 Use the result that 𝑎
𝑎
3𝐛
2𝐚 5𝐚 2𝐚 3𝐛 𝐛 5𝐚 𝐛 3𝐛
10𝐚 𝐚 6𝐚 𝐛 5𝐛 𝐚 3𝐛 𝐛
0 6𝐚 𝐛 5𝐛 𝐚 0
since 𝐚 𝐚 𝐛 𝐛 0
6𝐚 𝐛 5𝐚 𝐛
since 𝐚 𝐛
𝐛 𝐚
𝐚 𝐛
71 Any two vectors (starting at the same vertex) are suitable for defining the triangle, for
example 𝑃𝑄⃗ and 𝑃𝑅⃗ :
𝑃𝑄⃗
𝑃𝑅⃗
𝑃𝑄⃗
4
2
5
1
0
3
1
3
2
1
3
2
5
5
7
𝑃𝑅⃗
4
11
5
𝑃𝑄⃗
Area
1
4
2
9√2
2
72 Use 𝐫
𝐫
𝐚
1
3
6
2
3
5
2
3
5
𝑃𝑅⃗
11
5
𝜇𝐝 with 𝐚
𝜆𝐝
𝜆
5
5
7
3
0
2
𝜇
1
3 ,𝐝
6
3
0 and 𝐝
2
4
5 :
1
4
5
1
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
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47
4
6
0
4
6
0
3
1
4
𝐝
𝐝
2
2
1
Use 𝐫
𝐚
𝐴𝐶⃗ :
1
7
4
6
8
1
𝜇𝐝 :
𝜆𝐝
4
6
0
𝐫
𝐴𝐵⃗ and 𝐝
Analysis & Approaches HL Exam Practice Workbook
73 Two vectors parallel to the plane are 𝐝
1
7
4
𝜆
6
8
1
𝜇
Note that the position vector of any of the three points can be used as 𝐚 in the equation of
the plane.
74 Use 𝐫 ⋅ 𝐧
𝐫⋅
3
1 and 𝐚
2
𝐚 ⋅ 𝐧 with 𝐧
3
1
2
5
8 ⋅
4
3
1
2
5
8 :
4
15
3
15
1
2
75 Finding the cross product of the two vectors parallel to the plane will give a vector
perpendicular to the plane, i.e. a normal 𝐧:
So, 𝐫 ⋅
𝐧
2
2
1
Use 𝐫 ⋅ 𝐧
𝐫⋅
9
2 :
5
𝐚 ⋅ 𝐧 with 𝐚
7
11
8
So, 𝐫 ⋅
7
11
8
3
1
4
7
11
8
9
2 ⋅
5
7
11
8
1
1
𝑥
𝑦
76 Use the scalar product form but replace 𝐫 with
:
𝑧
𝑥
4
3
4
𝑦 ⋅ 1
0 ⋅ 1
𝑧
2
5
2
4𝑥
4𝑥
𝑦
𝑦
2𝑧
2𝑧
12
2
0
10
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𝑥
𝑦
𝑧
1
3
2
Analysis & Approaches HL Exam Practice Workbook
77 Convert the equation of the line to parametric form:
4𝜆
𝜆
6𝜆
Substitute into the equation of the plane and solve for 𝜆:
2 1 4𝜆
4
12 2𝜆 16
𝜆
2
3
𝜆
2
6𝜆
16
Use this value of 𝜆 in the equation of the line to find the point of intersection:
1
3
2
4
1
6
2
7
5
10
7, 5, 10
Point of intersection is
78 Eliminate any one of the variables between the two equations:
𝑥 2𝑦
4𝑥 𝑦
1
3𝑧
2𝑧
5
11
1
2
2 2 :
9𝑥
𝑧
Let 𝑧
9𝑥
𝜆:
𝜆
𝑥
27
3
27
𝜆
9
Substitute into 1 and rearrange to find an expression for 𝑦 in terms of 𝜆:
3
18𝑦
𝑦
𝜆
9
2𝑦
3𝜆
5
18 28𝜆
14
𝜆
1
9
3
𝑥
𝑦
𝑧
1
3
1
0
𝜆
𝜆
1/9
𝜆 14/9
1
3
1
𝜆
1
14
0
9
79 Solve the system of equation using the GDC:
So, the line of intersection is 𝐫
𝑥
3, 𝑦
2, 𝑧
1
So, the point of intersection is 3, 2, 1
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𝑥 3𝑦 2𝑧
7
4𝑥 𝑦 𝑧
5
6𝑥 5𝑦 3𝑧 1
4
1
5
3 2
1
2
3
2 2
3 :
9𝑥 𝑦
17
18𝑥 2𝑦
14
6𝑥 5𝑦 3𝑧 1
4
5
3
Now eliminate 𝑦 from 4 :
4
0
5 :
10
So, the system is inconsistent and therefore the planes don’t intersect.
The normals to the three planes are:
4
6
1
3 ,𝐧
1 ,𝐧
5
2
1
3
Since none of these normals are parallel to each other (none is a multiple of another), no two
planes are parallel.
𝐧
Therefore, the planes form a triangular prism.
Note that the only other possibilities for non-intersecting planes are that either one plane
intersects two parallel planes or the three planes are all parallel.
81 First find a normal to the plane:
3𝑥 4𝑦 2𝑧 10
𝑥
3
𝑦 ∙
10
4
𝑧
2
3
𝐫∙
10
4
2
3
So, 𝐧
4
2
The angle between a line with direction vector 𝐝 and a plane with normal 𝐧 is 𝜃
|𝐝∙𝐧|
where cos 𝜙 |𝐝||𝐧|:
90
𝜙,
∙
cos 𝜙
√
21
𝜙
𝜃
√
√30√29
44.6°
90
44.6
45.4°
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Analysis & Approaches HL Exam Practice Workbook
80 Attempt to solve the simultaneous equations. Start by eliminating a variable from any two
equations, say 𝑧 from the first two:
2𝑥
𝑥
3𝑦
5𝑧
2𝑦
4𝑧
4⇒𝐧
9⇒𝐧
Analysis & Approaches HL Exam Practice Workbook
82 First find a normal to each plane:
2
3
5
1
2
4
The angle between two planes is the angle between their normals, so use cos 𝜃
𝐧 ∙𝐧
:
|𝐧 ||𝐧 |
∙
cos 𝜃
√
24
𝜃
√
√38√21
148.2°
So, the acute angle between the planes is 180
148.2
31.8°
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51
1 a Discrete – it can only take certain values.
b Continuous (although its measurement might be discrete since it will be to a particular
accuracy)
c Discrete
2 a There are many possibilities. It could be all of his patients, all of his ill patients, all
people in his area, or all people in the world.
b This is not a random sample of the population since there are some people (those who do
not attend the clinic) who cannot possibly be included.
3 Yes – people who do less exercise might be less likely to choose to participate.
4 Yes – there appears to be consistency when the observation is repeated (within a reasonable
statistical noise).
5 Item D is not a possible human height. It might have been a participant not taking the test
seriously or it might have been a misread (e.g. giving height in feet and inches). You could
either return to participant D and ask them to check their response, or if this were not
possible you would discard the data item.
6 The IQR is 4. Anything above 11 1.5 4 17 is an outlier. Just because a data item is
an outlier does not mean that it should be excluded. It should be investigated carefully to
ensure that it is still a valid member of the population of interest.
7 Convenience sampling.
8 The proportion from Italy is
so it should contain
20
. The stratified sample must be in the same proportion,
8 students from Italy.
9 The proportion is
10 This is all of the 30 to 40 group and half of the 20 to 30 group, which is 7
9
11 a The total frequency is 160. Reading off half of this frequency (80) on the frequency axis
is about 42 on the x-axis, which is the median.
b The lower quartile corresponds to a frequency of 40, which is an x-value of
approximately 30.
The upper quartile corresponds to a frequency of 120 which is an x-value of
approximately 60. Therefore, the interquartile range is 60 30 30
c The 90th percentile corresponds to a frequency of 0.9
of about 72 which is the 90th percentile.
160
144. This has an x-value
12 Putting the data into the GDC, the following summary statistics can be found:
Min: 12, lower quartile: 14, median: 16, upper quartile: 18.5, max: 20
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4 Statistics and probability
b B is more likely to be normally distributed as it has a symmetric distribution.
14 The mode is 14 as it is the only one which occurs twice. From the GD, the median is 18 and
the mean is 18.5
Tip: You should also be able calculate the median without a calculator.
7
15 The mean is
So 23
𝑥
35, therefore 𝑥
12
16 a The midpoints are 15, 25, 40, 55.
𝑛
10
12
15
13
50
𝑥̅
35.3
b We are not using the original data.
17 The modal class is 15
𝑥
20
18 a From the GDC, 𝑄
18, 𝑄
7 so IQR
11
b From the GDC, the standard deviation is 5.45
29.7
c The variance is 5.45
19 The new mean is 12
2
4
28
The new standard deviation is 10
2
20.
20 Using the GDC, the lower quartile is 16 and the upper quartile is 28.5
21 a From the GDC: 𝑟
0.910
b There is strong positive correlation between 𝑥 and 𝑦
22 a Something like:
b Approximately 7.5
23 From GDC, 𝑦
0.916𝑥
4.89
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13 a Both have the similar spread (same IQR (4) and range excluding outliers (10) but
population A is higher on average (median of 16 versus 10).
13.1
(ii)
23.21
(iii)
5.58
b Only (i). Part (ii) is extrapolation and part (iii) is using a y-on-x line inappropriately.
25 a This is the expected number of text messages sent by a pupil who does not spend any
time on social media in a day.
b For every additional hour spent on social media, the model predicts that the pupil will
send 1.4 additional texts.
26 a Split the data into the first four points and the next five points and do a regression for
each part separately.
𝐿
4.19𝐴 0.259
0.830𝐴 25.4
𝐴
𝐴
6
6
b Using the first part of the piecewise graph:
𝐿
4.19
3
0.259
12.3
So expect a length of 12.3cm
0.67
27
28 There are six possible outcomes of which three (2, 3 and 5) are prime, so the probability is
0.5
29 P 𝐴
30 30
1
P 𝐴
0.05
1.5
0.4
Tip: Remember that expected values should not be rounded to make them actually achievable.
31 We can illustrate this in a Venn diagram:
There are 14
4
10 students who study only French
There are 18
4
14 students who study only Spanish
Therefore, there are 10
14
4
28 students who study either French or Spanish. This
leaves 2 students who do not study either, so the probability is
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24 a (i)
Analysis & Approaches HL Exam Practice Workbook
32 a There will be 7 socks left, of which 5 are black so it is
b This is best illustrated using a tree diagram:
There are two branches relevant to the question.
White then black:
3
8
5
7
15
56
Black then white:
5
8
3
7
15
56
The total probability is
33 a This can be best illustrated using a sample space diagram:
2nd Roll
1st Roll
1
2
3
4
1
2
3
4
5
2
3
4
5
6
3
4
5
6
7
4
5
6
7
8
There are 16 places in the sample space diagram: 6 of them have a score above 5
(shaded in the diagram) therefore the probability is
b In the sample space diagram, there are 6 scores above 5. Two of them are 7 so
P score
34 a 𝑥
7|score
100
40
5
30
20
10
b There are 60 out of 100 students who prefer Maths, so
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P 𝐴
P 𝐵
P 𝐴∩𝐵
36 P 𝐴 ∪ 𝐵
P 𝐴
P 𝐵
0
0.5
0.7
0.3
0.9
Analysis & Approaches HL Exam Practice Workbook
35 P 𝐴 ∪ 𝐵
0.6
37 There are 70 people who prefer soccer. Out of these 40 prefer maths. So
P Maths|Soccer
38 P 𝐴 ∩ 𝐵
P 𝐴 P 𝐵
0.24
39 𝑊 can take three possible values: 0, 1 or 2
3
6
2
7
P 𝑊
0
P 𝐵𝐵
4
7
P 𝑊
1
P 𝐵𝑊
P 𝑊𝐵
P 𝑊
2
P 𝑊𝑊
4
7
1
7
2
6
3
7
3
6
3
7
4
6
4
7
So
𝒘
𝐏 𝑾
0
1
2
2
7
4
7
1
7
0
1
2
𝑘
2𝑘
3𝑘
𝒘
40 Create a table:
𝒙
𝐏 𝑿
𝒙
The total probability is 𝑘
41 𝐸 𝑋
0.5
0.5
1
2𝑘
0.4
3𝑘
2.5
6𝑘 which must equal 1 so 𝑘
0.1
0.9
42 a The probability of winning a prize is 0.095
0.005
of winning $2000 is 0.01, so the conditional probability is
b 𝐸 𝑋
P 𝑋
43 𝐸 𝑋
0
0.9
10
0.095
10.95
P 𝑋
2000
1
0
0.6
0.3
If the game is fair then 𝐸 𝑋
0.1𝑘
0.6
0.1𝑘
0.6
𝑘
0.1𝑘
2000
0.005
0.1. Out of this, the probability
.
.
0.05
10.95
0.005
0.1𝑘
0.6
0 so:
0
6
44 The outcome of each trial is not independent of the previous trial.
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a P 𝑋
2
0.3456
b To use the calculator, you need to write the given question into a cumulative probability:
P 𝑋
3
1
P 𝑋
46 a If 𝑋 is the number of heads then 𝐸 𝑋
b Var 𝑋
𝑛𝑝 1
𝑝
10
0.6
2
1
0.683
𝑛𝑝
10
0.6
0.4
0.317
6
2.4 so the standard deviation is √2.4
1.55
47 We know that about 68% of the data occurs within one standard deviation of the mean, but
this takes us to negative values of time which is not possible.
48 a It is a symmetric, bell-shaped curve.
b The line of symmetry is approximately at 50, so this is a good estimate of the mean.
49 P 11 𝑋 15 can be found on the calculator – either directly or as P 𝑋
11 . It equals 0.625
15
P 𝑋
50 Some calculators can deal with the given information directly, but some require you to first
convert the information into a cumulative probability: P 𝑋 𝑘
0.3. Using the inverse
normal function on the calculator gives 𝑘 92.1
51 Enter the 𝑥-data in the 𝑦 column and the 𝑦-data in the 𝑥 column. Remember that the output
is in the form 𝑥 𝑎𝑦 𝑏:
Regression line is:
𝑥
0.617𝑦
13.3
52 a Substitute each value of 𝑦 into the equation:
(i)
𝑥
1.82
20
11.5
24.9
(ii)
𝑥
1.82
35
11.5
52.2
b The prediction when 𝑦 20 can be considered as reliable since 20 is within the range
of known 𝑦-values and the correlation coefficient is close to 1 suggesting a good linear
relationship.
The prediction when 𝑦 35 cannot be considered as reliable since the relationship needs to
be extrapolated significantly beyond the range of the given data.
53 Use the standard formula with 𝐴 and 𝐵 swapped around. Note that 𝐴 ∩ 𝐵 is the
same as 𝐵 ∩ 𝐴:
P 𝐵|𝐴
P 𝐵∩𝐴
P 𝐴
0.4
0.6
2
3
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Analysis & Approaches HL Exam Practice Workbook
45 Your calculator should have two functions – one which finds P 𝑋 𝑥 , which we will use
in part a and one which finds P 𝑋 𝑥 which we will use in part b.
P 𝐴∪𝐵
0.7
P 𝐴∩𝐵
P 𝐴
P 𝐴
P 𝐴∩𝐵
0.2 0.8 P 𝐴 ∩ 𝐵
0.3
P 𝐴 P 𝐵
0.2
P 𝐴∩𝐵
P 𝐴 P 𝐵
0.8
P 𝐴 P 𝐵 so first find P 𝐴 ∩ 𝐵 :
0.16
So 𝐴 and 𝐵 are not independent
55 The 𝑧-value is the number of standard deviations from the mean:
𝑧
𝑥
𝜇
𝜎
17 10
4.8
1.46
17 is 1.46 standard deviations from the mean.
56 Since the mean and variance are unknown, write each probability statement in terms of the
standard normal 𝑍~𝑁 0, 1 and use the inverse normal to find 𝑧 and 𝑧 :
P 𝑋
P 𝑍
12
𝑧
𝑧
0.3
0.3
0.52440
P 𝑋
P 𝑍
34
𝑧
𝑧
0.2
0.2
0.84162
to form two equations in 𝜇 and 𝜎:
Use 𝑧
12
0.52440
𝜇
0.52440𝜎
0.84162
𝜇
12
34
0.84162𝜎
𝜇
𝜎
1
𝜇
𝜎
34
2
Solve 1 and 2 simultaneously on the GDC:
𝜇
20.4, 𝜎
16.1
57 a Putting all the information into the Bayes’ Theorem formula,
0.4:
noting that 𝑃 𝐵
𝑃 𝐵|𝐴
.
.
.
.
.
.
b Replacing 𝐴 by 𝐴′ in the formula, we now also need 𝑃 𝐴 |𝐵
𝑃 𝐵|𝐴
.
.
.
.
.
0.6 and 𝑃 𝐴 |𝐵
0.2.
.
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Analysis & Approaches HL Exam Practice Workbook
𝐴 and 𝐵 are independent if P 𝐴 ∩ 𝐵
54
P late and Aline
P late)
0.2
0.2
0.05
So P Aline | late
59 First we need 𝑘
E 𝑋
Var 𝑋
0
1
1
7
60 We need f 𝑥
0.01
0.35 0.16 0.45
0.01
0.133
0.075
2𝑘
1
7
0
0.05
4𝑘
2
7
2
2
7
1
0 and
4
10
7
4
7
f 𝑥 d𝑥
0.02
0.075
. Then
1 so 𝑘
4
7
Analysis & Approaches HL Exam Practice Workbook
58 Using a tree diagram may be helpful.
26
49
10
7
1.
The former can be seen from the graph:
𝑘 1
cos 𝑘𝑥
2 𝑘
𝑘
sin 𝑘𝑥 d𝑥
2
𝑘
cos π
2𝑘
cos 0
1
2
1
1
1
61 We first need to find 𝑘:
2𝑘
𝑥
3
𝑘 √𝑥 d𝑥
2𝑘
3
3
2
1 ⇒ 𝑘
Then
𝑃 𝑋
1
4
3
√𝑥 d𝑥
2
1
4
𝑌
3
4
1
1
8
7
8
g 𝑦 d𝑦 into two parts:
62 You need to split
P
𝑥
4π
4 π
sin π𝑦 dy
2
2𝑦 dy
0.726 (3 s.f.)
63 The largest value of the pdf in the given interval is either at a local maximum point or at one
of the endpoints. Sketching the graphs shows that the mode of 𝑋 is , but the mode of 𝑌 is 0.
64 You can solve the following equation for 𝑚 using GDC:
1
ln 𝑥
d𝑥
2
ln 64 2
The median is 3.15
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59
𝑘
2
𝑘 d𝑥
𝑘𝑥 d𝑥
𝑘
1 ⇒ 𝑘
Analysis & Approaches HL Exam Practice Workbook
65 We first need to find the value of 𝑘:
2
3
Now check whether the median is less than 1:
1
3
2
𝑥 d𝑥
3
1
2
So, the median is between 1 and 2, and satisfies:
2
1 1 1
d𝑥
3
2 3 6
2
𝑚
3
2
𝑥
3
⇔
∴ 𝑚
5
4
66 a E 𝑋
2
E 𝑋
3
E 𝑋
67 𝐸 𝑋
4
9
4
b Var 𝑋
16
0.467
d𝑥
0
0.467
d𝑥
𝑥
2
𝑥
d𝑥
6
E 𝑋
0.799
0.467
E 𝑋
E 𝑋
√0.639
0
𝑥
cos 𝑥 d𝑥
2
68 E 𝑋
0.639 so SD 𝑋
E 𝑋
cos 𝑥 d𝑥
E 𝑋
Var 𝑋
1
6
1
Var 𝑋
26
3
8
3
SD 𝑋
14
9
1.25 (3 s.f.)
𝑥
d𝑥
12
26
3
14
9
69 a For a fair game, the cost should equal the expected value of the winnings.
.
E 𝑋
d𝑥
8
The charge should be 8 Yen.
b To make a profit, a throw should be longer than 8#m.
P 𝑋
.
8
d𝑥
0.569
Around 57% or players make a profit.
70 E 80
Var 𝑋
3𝑋
80
3
3E 𝑋
24
44
216
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Analysis & Approaches HL Exam Practice Workbook
5 Calculus
1
sin 𝟑𝒙
𝟎. 𝟐𝒙
0.25
0.2588
0.2617
0.2618
x
10
5
1
0.1
The limit is 0.26
2 Look at the values on the graph close to 𝑥
2.
The limit is 0.5
12
3 The derivative is
4 ‘rate’ means
dA
d𝑡
5
7.
; ‘decreases’ means that the rate of change is negative.
𝑘𝐴
5 The y value is f 𝑥 and the gradient is f′ 𝑥 . So when 𝑦
6 Use Δ𝑥
4, Δ𝑦
𝑥
𝑥
𝑦
𝑦
Δ𝑥
1.
2 and gradient
Δ𝑦
Gradient
of PQ
5
2.236
1
0.236
0.236
4.1
2.025
0.1
0.025
0.248
4.01
2.002
0.01
0.002
0.250
4.001
2.000
0.001
0.000
0.250
The gradient is
4, f 𝑥
0.25
7 f′ 𝑥 is where the graph is decreasing, which is between the two turning points.
1.29
𝑥
1.29
8 The gradient starts positive but decreasing, then changes to negative, then back to positive
and then to negative again.
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61
8𝑥
10
𝑥
11 a f 𝑥
3
12𝑥
b f 𝑥
3𝑥 so f 𝑥
1
𝑥
𝑥
12 f 𝑥
8𝑥
2𝑥
f 2
16
2
4
12
13
5
𝑥
, so f 𝑥
𝑥
c f 𝑥
[or 6𝑥
, so f 𝑥
18𝑥
]
[or
𝑥
]
16.5
5𝑥
2
1
2
10, 𝑥
1
𝑥
√2
2𝑥
14
8, 𝑦
Tangent: 𝑦
3
15
2𝑥
Normal: 𝑦
5
2
𝑥
7
3
13
8 𝑥
4
𝑥
2
3
5
6
𝑦
16
13
2
2
𝑦
39
7
2𝑥, so the tangent at 𝑎, 𝑎
16
𝑦
𝑎
When 𝑥
9
𝑎
24𝑥
𝑎
3
2𝑎 𝑥
0, 𝑦
3 is:
𝑎
12:
2𝑎
3
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Analysis & Approaches HL Exam Practice Workbook
9 The gradient starts off negative, so f 𝑥 is decreasing. It then increases, and then decreases
again.
Analysis & Approaches HL Exam Practice Workbook
17 Use GDC to find the gradient and to draw the tangent.
0.021
a
b 𝑦
0.021𝑥
0.13
and intersect it with 𝑦
18 Use GDC to sketch the graph of
2. The coordinates are
0.5, 0.098 .
3𝑥
19 3𝑥
𝑥
20
𝑐
𝑥
d𝑥
21 Integrate: 𝑦
Use 𝑦
So 𝑦
4𝑥
3, 𝑥
2𝑥
2𝑥
24 f 𝑥
f 9
𝑐
2 2
9
5 sin 𝑥
√
1
3
3
9
1
3𝑥
2
0
1
6 sin 5𝑥
6𝑥
𝑢 and 𝑢
3𝑥
1:
3𝑥
√3𝑥
5 cos 5𝑥
1
2𝑢 and 𝑢
sin 5𝑥 , and remembering that sin 5𝑥
30 sin 5𝑥 cos 5𝑥
12𝑥e
26 4e
27 Using the quotient rule with 𝑢
d𝑦
d𝑥
𝑐⇒𝑐
31.5
b Using the chain rule with 𝑦
differentiates to 5 cos 5𝑥 :
d𝑦
d𝑥
𝑐
9
25 a Using the chain rule with 𝑦
d𝑦
d𝑥
2𝑥
2𝑥
2 2
1 d𝑥
2𝑥
3 cos 𝑥
𝑥
2 d𝑥
2: 3
22 Use GDC:
23
𝑥
4𝑥
1
4 ln 𝑥
𝑥
16𝑥
4
4 ln 𝑥
16𝑥
ln 𝑥 and 𝑣
1
4𝑥:
ln 𝑥
4𝑥
28 6𝑥
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
© Paul Fannon, Vesna Kadelburg and Stephen Ward 2021
63
𝑥
0 there.
30 Concave-down means that the graph curves downwards, which is at the points B, D and E.
31 Concave up means f
3𝑥
32
33 f 𝑥
𝑥
𝑦
f
34 𝑆
𝑥
sin 𝑥
3π
4
√2
2
4
𝑥
2
√
0
√2
2
√2
2
0
64 so 𝑥
4.
0, so this is a local minimum.
60𝑥
48 cm
4
120𝑥
0 when 𝑥
60𝑥 𝑥
2
0 or 2.
Check for change in sign of
𝒙
1
𝐝𝟐 𝒚
𝐝𝒙𝟐
0
:
1
15 cos 5𝑡 , 𝑎
When 𝑡
2, 𝑎
(or find
at 𝑡
3
0
The only point of inflexion is at 𝑥
36 𝑣
0∴𝑥
𝑥
Min surface area 𝑆
So
3π
4
4
√2
0 when 𝑥
2
30𝑥
√
cos 𝑥 , f
√2
2
𝑥
24 ⇔ 𝑥
sin 𝑥 , f
2𝑥
35
0. So f
0 ⇔ 3𝑥
cos 𝑥
f
𝑥
0
2.
75 sin 5𝑡
40.8 m s
2 using GDC)
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
© Paul Fannon, Vesna Kadelburg and Stephen Ward 2021
64
Analysis & Approaches HL Exam Practice Workbook
29 The curve is concave-down in the middle section so f
(or find
.
0.6e
Analysis & Approaches HL Exam Practice Workbook
37 Velocity: 𝑣
0.270
using GDC)
So speed = 0.270 m s–1
38 4
d𝑡
√
5.18 m
39 13.8 m
ln |𝑥|
𝑥
40 2
e
41
42 a
6𝑥
ln|𝑥|
9e
𝑐
sin 𝑥
4
cos 2𝑥
43
𝑐
ln 𝑥
d𝑥
b
𝑐
3
𝑐
1
44 1.79 (3 s.f.)
45
sin 𝑥
1 d𝑥
π
2
1
0
cos 𝑥
0
1
𝑥
π
2
1.
So area
,e
46 e
3e
3; 𝑥
1 d𝑥
2
3e
Area = 2
1 d𝑥
ln 3
3e
ln 3. The coordinates are (ln 3, 0)
3e
3
𝑥
e
ln 3
3e
3
3
0
ln 3
3
ln 3
3e
𝑥
1
ln 3
3
3e
2
2 ln 3
3e
3
3
ln 3
3e
1
ln 3
47 a
Using GDC, points of intersection are
4.82, 0.180 , 2.69, 7.69
b Subtract the bottom curve from the top curve and integrate:
.
𝐴
𝑥 5 2e . d𝑥 14.6
.
48 Put 𝑥
3 into both expressions; equating those gives 𝑘
6
49 In question, 2x2 should be 2x3.
Evaluate g 2 and g′ 2 for both expressions. They give the same answer, so the function is
differentiable at 𝑥 2.
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
© Paul Fannon, Vesna Kadelburg and Stephen Ward 2021
65
0 gives
Analysis & Approaches HL Exam Practice Workbook
50 a Setting 𝑥
which tends to infinity.
; when 𝑥 tends to infinity, this tends to .
b
and then take lim :
51 Simplify
3 𝑥
d𝑦
d𝑥
→
4
ℎ
ℎ
lim 6𝑥
3𝑥
6𝑥
3ℎ
→
6𝑥ℎ
4
3ℎ
ℎ
3ℎ
6𝑥
52 Differentiate three times:
f 𝑥
2 cos 2𝑥 , f
𝑥
So f
8 cos
4 sin 2𝑥 , f
𝑥
8 cos 2𝑥
4
53 a Differentiate top and bottom:
When 𝑥 → 0 this tends to
.
b Differentiate top and bottom:
When 𝑥 → 2, top tends to ∞ and bottom tends to sec
So, the expression diverges to infinity.
54 The Maclaurin series of cos 3𝑥 is
cos 3𝑥
1
3𝑥
2
1
1
3𝑥
24
⋯
1
9𝑥
2
27𝑥
8
So, the expression becomes:
1
9𝑥
2
27𝑥
8
2𝑥
Which tends to
⋯
1
9
4
27𝑥
16
⋯
when 𝑥 → 0
55 It is necessary to use the rules twice:
e .
lim
→ 2𝑥
0.2e .
lim
→
4𝑥
0.04e
lim
→
4
lim 0.01e
.
.
→
The exponential diverges, so the limit is not finite.
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
© Paul Fannon, Vesna Kadelburg and Stephen Ward 2021
66
2𝑥
9𝑦
d𝑦
d𝑥
d𝑦
d𝑥
Analysis & Approaches HL Exam Practice Workbook
56 a Using implicit differentiation:
0
2𝑥
9𝑦
b Using implicit differentiation and the product rule:
sin 𝑦
cos 𝑥
𝑥 cos 𝑦
𝑦
d𝑦
𝑥 cos 𝑦
d𝑥
At 0,
sin 𝑥
cos 𝑥
d𝑦
d𝑥
0
𝑦 sin 𝑥
sin 𝑦
𝑦 sin 𝑥
𝑥 cos 𝑦
sin 𝑦
cos 𝑥
,
1
3
57 Area of circle decreases at rate of 3 ⇒
2𝜋𝑟
And
Using the chain rule:
d𝑟 d𝐴
d𝐴 d𝑡
1
3
2𝜋𝑟
d𝑟
d𝑡
So, rate of decrease is when 𝑟
58 a For stationary points,
3𝑥
12 is
cm s−1
0:
6𝑥 0
𝑥 0 or 2
Stationary points are 0, 0 and 2, 4
In question part b, interval changed from 3
b
The largest value cannot be attained at 𝑥
check at 𝑥 4:
𝑦
4
3
4
Comparing to 𝑦
4 as
𝑥
3 to 4
4
3
4
𝑥
4
0 so only need to
16
0 and 𝑦
4 at the two stationary points, the maximum value is 16
59 a Using the product rule:
arctan 2𝑥
𝑥
arctan 2𝑥
b Using the chain rule:
1
sec 𝑥 ln 2
60
cosec 𝑥 cot 𝑥
cosec 𝑥 d𝑥
tan 𝑥
ln 2
sec 𝑥 tan 𝑥
cosec 𝑥
cot 𝑥
𝑐
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
© Paul Fannon, Vesna Kadelburg and Stephen Ward 2021
67
2
1
2𝑥
1
arcsin 2𝑥
1
62 a 𝐴 𝑥
1
d𝑥
1
𝑐
𝐵 𝑥
2
Substituting in 𝑥
d𝑥
b
1
arcsin 2𝑥
2
2
Analysis & Approaches HL Exam Practice Workbook
61 Complete the square inside the square root:
2𝑥
5
2 and 𝑥
3ln|𝑥
1 gives 𝐴
2|
ln|𝑥
1|
3, 𝐵
1
𝑐
Substituting in the limits and sing rules of logs:
3ln 7
4
63 𝑥
d𝑥
3ln 1
ln 10
4 sec 𝑢
1
ln 4
ln
4 tan 𝑢
2 sec 𝑢 tan 𝑢 d𝑢
So, the integral becomes:
1
d𝑢
2
1
sec
2
𝑢
3 √𝑢 d𝑢
64
𝑥
2
𝑐
𝑢
3 𝑢
144
5
65 a Taking 𝑢
𝑥,
b Taking 𝑢
𝑥,
e d𝑦
68
π sin 𝑥 d𝑥
69
π
70
| |
2 e
d𝑦
𝑐
𝑥
𝑐
𝑥 sin 2𝑥 d𝑥
sin 2𝑥 to get
1
𝑥 cos 2𝑥
2
67
𝑦
e
cos 2𝑥 to get 𝑥 sin 2𝑥
Then take 𝑢
1
𝑥 sin 2𝑥
2
gives 𝑥e
𝑥 gives 𝑥 ln 5𝑥
ln 5𝑥 ,
𝑥 ,
66 Take 𝑢
e
1
sin 2𝑥
4
𝑐
e
4.93 (from GDC)
π
40π
𝑘𝑣
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
© Paul Fannon, Vesna Kadelburg and Stephen Ward 2021
68
x
y
0
2.0000
0.1
2.0909
0.2
2.1723
0.3
2.2419
0.4
2.2983
So, 𝑦
2.298
𝑥 𝑦
72
1
d𝑦
⇒
𝑥 d𝑥
1
𝑦
1
𝑥
2
⇒ arctan 𝑦
𝑦
Analysis & Approaches HL Exam Practice Workbook
71
tan
1
𝑥
2
𝑐
𝑥
1 d𝑥
73
ln 𝑦
1
𝑥
2
2
Using 𝑦
𝑦
2
𝑦
3e
74 a 𝑣
𝑐
1
1, 𝑥
1: 𝑐
𝑐
ln 3
e
2
𝑣
𝑥
√𝑣, which simplifies to the required equation.
b Separating variables:
⇒ 2√𝑣
ln 𝑥
𝑐
Initial conditions: 𝑥
1, 𝑦
0, 𝑣
√
0, so 𝑐
0; gives 𝑦
ln 𝑥
75 Integrating factor:
𝐼
e
e
𝑥
∴𝑦
1 𝑦
𝑥
4𝑥 𝑥
2𝑥
𝑥
𝑥
1
1 d𝑥
𝑥
2𝑥
𝑐
𝑐
1
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
© Paul Fannon, Vesna Kadelburg and Stephen Ward 2021
69
ln 1
𝑥 .
Analysis & Approaches HL Exam Practice Workbook
76 Let f 𝑥
Then:
𝑓 0
𝑓
1
0, 𝑓 0
1
1
0
1
1
0
2
𝑓 0
1
2
0
6
𝑓
1
𝑥
77 cos 3𝑥
1
1
78 e 1
!
9𝑥
2
27𝑥
8
𝑥
1
!
𝑥
1
𝑥
𝑥
3𝑥
2
1
79 1
6
0
So, f 𝑥
1
0
𝑥
1
𝑥
𝑥
1
arcsin 𝑥
√1
arcsin 0
d𝑥
𝑥
1
𝑥
6
𝑥
3
𝑥
40
𝐶
0, so
arcsin 𝑥
𝑥
𝑥
6
3𝑥
40
80 a Write the general Maclaurin series for 𝑦 and substitute into the differential equation:
∑
𝑘
1 𝑎
𝑥
𝑥
2𝑦
When 𝑥
0, 𝑦
𝐴, so 𝑎
𝐴.
𝑘
0: 𝑎
2𝑎
𝑘
1: 2𝑎
1
𝑘
2: 𝑘
1 𝑎
𝑥
∑
1
4𝐴
2𝐴
2𝑎 ⇒ 𝑎
2𝑎
⇒𝑎
b Looking at the first few terms show that 𝑎
solution is
𝑦
𝐴
2𝐴𝑥
2𝑎 𝑥
∑
!
1
𝑎
!
1
4𝐴 for all 𝑘
2. Hence the
4𝐴 𝑥
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
© Paul Fannon, Vesna Kadelburg and Stephen Ward 2021
70
1
cos 𝑦
0
sin 𝑦
b Hence 𝑦
Analysis & Approaches HL Exam Practice Workbook
sin
81 a
cos 𝑦
0
𝑥
Note: The next term is
𝑥 .
Exam Practice Workbook for Mathematics for the IB Diploma: analysis and approaches HL
© Paul Fannon, Vesna Kadelburg and Stephen Ward 2021
71