L E AV I N G C E R T M AT H S ( H L )
PAST EXAM PAPER
SOLUTIONS
2 0 21 - 2 012
POCKETTUTOR.IE
2022
First published in 2022 by PocketTutor.ie
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© Pocket Tutor 2022
1
Exam Top Tips
1. Timing
• Time for each question = Marks for the question ÷ 2.
• I.e. for a 30 mark question, spend max 15 mins on it.
2. Attempt marks
• If you write anything even vaguely relevant, you have a good chance of
picking up some ‘partial credits’ (aka attempt marks)
• Every question is marked on a scale, the further you progress towards
the right answer the more marks you can get.
• It is very possible to have very few correct answers and still get a H2.
3. Choice
• There will be a choice on the 2022 paper, as there was on 2021. You
no longer have to answer every question on your paper.
• If you have the time, you may decide to answer all questions, however be careful as this
will put you under pressure. Figure out your strategy early in the exam.
4. Topics – crossover is possible. I.e. Trigonometry appeared in Paper 1 in 2017
Paper 1
Paper 2
• Algebra
• Probability
• Functions
• Statistics
• Geometry (Incl.
• Differentiation
constructions/theorems)
• Integration
• The Line
• Logs & Indices
• The Circle
• Series & Sequences
• Trigonometry
• Financial Maths
• Length, Area, Volume
• Complex Numbers
• Transformation & Enlargements
• Proofs by Induction
5. Teach what you study
• Learning by teaching others is extremely effective. This strategy works because it requires
you to retrieve the information, consider the ideas and express them in a logical way. If you
cannot do this, then you need to review the topic again before starting your imaginary
lesson. This works for all subjects, and can be used as a study tool for any exam.
• After studying a topic, test yourself by pretending to explain the topics from the beginning.
Answer questions like ‘What part of the course are we looking at? What are the essentials to
know? What are the hardest parts of this section? Then proceed to tackle a question and
explain what you are doing for each step.
© Pocket Tutor 2022
2
Table of Contents
2021 PAPER 1
10
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
QUESTION 9
QUESTION 10
10
15
17
20
22
25
27
30
35
39
2020 PAPER 1
45
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
QUESTION 9
45
48
51
55
58
60
62
66
69
2019 PAPER 1
73
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
QUESTION 9
73
75
77
79
81
82
85
87
90
2018 PAPER 1
94
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
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94
96
98
99
101
3
QUESTION 6
QUESTION 7
QUESTION 8
QUESTION 9
103
105
109
111
2017 PAPER 1
114
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
QUESTION 9
114
116
118
120
121
123
125
128
133
2016 PAPER 1
135
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
QUESTION 9
135
137
139
141
143
145
147
150
154
2015 PAPER 1
158
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
QUESTION 9
158
159
160
162
163
165
166
169
172
2014 PAPER 1
175
QUESTION 1
QUESTION 2
QUESTION 3
175
177
179
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4
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
QUESTION 9
181
183
185
188
191
194
2013 PAPER 1
197
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
QUESTION 9
197
199
201
202
203
204
205
207
210
2012 PAPER 1
213
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
QUESTION 9
213
215
216
218
220
221
223
226
231
PAPER 2
235
2021 PAPER 2
236
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
QUESTION 9
QUESTION 10
236
238
242
245
247
249
251
255
259
264
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5
2020 PAPER 2
267
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
QUESTION 9
267
269
272
274
276
278
281
285
290
2019 PAPER 2
292
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
QUESTION 9
292
293
295
297
299
301
303
307
309
2018 PAPER 2
312
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
QUESTION 9
312
313
315
316
318
321
322
325
328
2017 PAPER 2
331
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
331
332
334
337
339
341
342
344
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6
QUESTION 9
347
2016 PAPER 2
350
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
QUESTION 9
350
352
354
355
357
358
359
362
365
2015 PAPER 2
367
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
QUESTION 9
367
368
369
372
374
375
377
380
383
2014 PAPER 2
386
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 7
QUESTION 8
QUESTION 9
386
387
388
389
391
394
397
399
2013 PAPER 2
404
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
404
405
407
409
411
413
414
© Pocket Tutor 2022
7
QUESTION 8
QUESTION 9
417
419
2012 PAPER 2
423
QUESTION 1
QUESTION 2
QUESTION 3
QUESTION 4
QUESTION 5
QUESTION 6
QUESTION 7
QUESTION 8
423
424
427
429
430
432
433
435
© Pocket Tutor 2022
8
Paper 1
Usual Paper 1 topics:
Algebra
Integration
Financial Maths
Functions
Logs & Indices Series
Complex Numbers
Differentiation
& Sequences
Proofs by Induction
© Pocket Tutor 2022
9
2021 Paper 1
Question 1
a)
𝒌𝒌 = −𝟏𝟏
4 − 2𝑖𝑖
= 0 + 𝑘𝑘𝑘𝑘
2 + 4𝑖𝑖
To solve for 𝑘𝑘 we start by getting rid of the fraction.
4 − 2𝑖𝑖 = 2𝑘𝑘𝑘𝑘 + 4𝑘𝑘𝑖𝑖 2
Multiplying out the brackets.
4 − 2𝑖𝑖 = (0 + 𝑘𝑘𝑘𝑘)(2 + 4𝑖𝑖)
4 − 2𝑖𝑖 = 2𝑘𝑘𝑘𝑘 + 4𝑘𝑘(−1)
4 − 2𝑖𝑖 = 2𝑘𝑘𝑘𝑘 − 4𝑘𝑘
4 = −4𝑘𝑘
4
= 𝑘𝑘
−4
𝒌𝒌 = −𝟏𝟏
© Pocket Tutor 2022
So, multiplying across by 2 + 4𝑖𝑖.
Remember 𝑖𝑖 2 = −1
Now we can let the real numbers equal each other.
Therefore, we can solve for 𝑘𝑘 by dividing across by
−4.
10
b)
𝟐𝟐 + 𝟑𝟑𝟑𝟑 𝐨𝐨𝐨𝐨 − 𝟐𝟐 − 𝟑𝟑𝟑𝟑
√−5 + 12𝑖𝑖 = 𝑎𝑎 + 𝑏𝑏𝑏𝑏
To rewrite this in the form 𝑎𝑎 + 𝑏𝑏𝑏𝑏 we can let the
square root equal 𝑎𝑎 + 𝑏𝑏𝑏𝑏 and then square both
sides.
−5 + 12𝑖𝑖 = (𝑎𝑎 + 𝑏𝑏𝑏𝑏)2
−5 + 12𝑖𝑖 = 𝑎𝑎2 + 2𝑎𝑎𝑎𝑎𝑎𝑎 − 𝑏𝑏 2
Squaring both sides.
𝑎𝑎2 − 𝑏𝑏 2 = −5
Letting the real numbers equal each other.
equation 1
2𝑎𝑎𝑎𝑎 = 12
Letting the imaginary numbers equal each other.
12
𝑏𝑏 =
2𝑎𝑎
𝑏𝑏 =
6
𝑎𝑎
Dividing across by 2𝑎𝑎 to get 𝑏𝑏 in terms of 𝑎𝑎.
6 2
𝑎𝑎2 − � � = −5
𝑎𝑎
𝑎𝑎2 −
36
= −5
𝑎𝑎2
(𝑎𝑎2 )𝑎𝑎2
−
(𝑎𝑎2 )
36
= −5(𝑎𝑎2 )
𝑎𝑎2
© Pocket Tutor 2022
Now subbing in this expression for 𝑏𝑏 in equation 1.
6
6
Squaring the fraction ( × =
𝑎𝑎
𝑎𝑎
36
𝑎𝑎2
).
Multiplying across by 𝑎𝑎2 to get rid of the fraction.
11
𝑎𝑎4 − 36 = −5𝑎𝑎2
Now that we have multiplied out the brackets,
we are left with an equation which resembles
a quadratic equation.
𝑎𝑎4 + 5𝑎𝑎2 − 36 = 0
(𝑎𝑎2 + 9)(𝑎𝑎2 − 4) = 0
𝑎𝑎2 + 9 = 0,
𝑎𝑎2 − 4 = 0
𝑎𝑎2 = −9,
𝑏𝑏 =
𝑏𝑏 =
6
𝑎𝑎
6
2
𝑏𝑏 = 3
𝑎𝑎2 = 4
𝑎𝑎 = ±2
or
or
b=
6
−2
b = −3
𝑎𝑎 + 𝑏𝑏𝑏𝑏
→ 𝟐𝟐 + 𝟑𝟑𝟑𝟑 𝐨𝐨𝐨𝐨 − 𝟐𝟐 − 𝟑𝟑𝟑𝟑
© Pocket Tutor 2022
We can factorise it in the same way as a
quadratic by starting each bracket with 𝑎𝑎2 .
Letting each bracket equal 0.
A square number cannot equal a negative, so
we ignore 𝑎𝑎2 = −9.
The square root of 4 is plus or minus 2, so
these are the two values of 𝑎𝑎.
Now subbing each of these values into the
expression for 𝑎𝑎 we found earlier.
Rewriting the two values of 𝑎𝑎 with the
corresponding values of 𝑏𝑏 in the form 𝑎𝑎 + 𝑏𝑏𝑏𝑏.
12
c)
𝟏𝟏 + √𝟑𝟑𝒊𝒊, −𝟐𝟐, 𝟏𝟏 − √𝟑𝟑𝒊𝒊
𝑧𝑧 3 = −8 + 0𝑖𝑖
1
𝑧𝑧 = (−8 + 0𝑖𝑖)3
→ −8 + 0𝑖𝑖
Modulus:
𝑟𝑟 = �(−8)2 + (0)2 = 8
Argument:
tan 𝜃𝜃 =
𝑏𝑏
𝑎𝑎
tan 𝜃𝜃 =
0
−8
tan 𝜃𝜃 = 0
𝜃𝜃 = tan
𝜃𝜃 = 0
−1
To start we cube root both sides. A cube root can be
written as being to the power of
1
3
To find the cube root of a complex number we use De
Moivre’s theorem.
To use De Moivre’s theorem we need to find the
modulus and the argument of the complex number.
We say complex numbers are in the form 𝑎𝑎 + 𝑏𝑏𝑏𝑏.
0
−8 + 0𝑖𝑖 is in the second quadrant
→ 𝜃𝜃 = 𝜋𝜋 + 0 = 𝜋𝜋
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𝑏𝑏
The modulus = √𝑎𝑎2 + 𝑏𝑏 2 . The argument: 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = .
𝑎𝑎
Finding the argument by finding the tan inverse of both
sides.
If the point is in the second quadrant we add pie to the
reference angle.
13
𝑟𝑟(cos(𝜃𝜃 + 2𝑛𝑛𝑛𝑛) + 𝑖𝑖 sin(𝜃𝜃 + 2𝑛𝑛𝑛𝑛))𝑛𝑛
8(cos(𝜋𝜋 + 2𝑛𝑛𝑛𝑛) + 𝑖𝑖 sin(𝜋𝜋 +
1
2𝑛𝑛𝑛𝑛))3
1
1
1
83 �cos � (𝜋𝜋 + 2𝑛𝑛𝑛𝑛)� + 𝑖𝑖 sin � (𝜋𝜋 + 2𝑛𝑛𝑛𝑛)��
3
3
2 �cos �
𝜋𝜋 + 2𝑛𝑛𝑛𝑛
𝜋𝜋 + 2𝑛𝑛𝑛𝑛
� + 𝑖𝑖 sin(
)�
3
3
Now we sub the values we found for 𝑟𝑟, the
modulus, and 𝜃𝜃, the argument, into the
1
expression and subbing in for 𝑛𝑛.
3
Using De Moivre’s theorem we put the
1
modulus to the power of and we multiply
1
the argument by .
3
3
𝑛𝑛 = 0 → 2 �cos �
𝜋𝜋 + 2(0)𝜋𝜋
𝜋𝜋 + 2(0)𝜋𝜋
� + 𝑖𝑖 sin �
��
3
3
Now we sub in 0 for 𝑛𝑛 and find the first
solution.
𝑛𝑛 = 1 → 2 �cos �
𝜋𝜋 + 2(1)𝜋𝜋
𝜋𝜋 + 2(1)𝜋𝜋
� + 𝑖𝑖 sin �
��
3
3
Doing the same for 𝑛𝑛 = 1 and 𝑛𝑛 = 2.
= 𝟏𝟏 + √𝟑𝟑𝒊𝒊
= −𝟐𝟐
𝑛𝑛 = 2
2 �cos �
𝜋𝜋 + 2(2)𝜋𝜋
𝜋𝜋 + 2(2)𝜋𝜋
� + 𝑖𝑖 sin �
��
3
3
= 𝟏𝟏 − √𝟑𝟑
© Pocket Tutor 2022
14
Question 2
a)
𝒑𝒑 = 𝟖𝟖,
𝒑𝒑 = −𝟐𝟐
|𝑥𝑥 + 𝑝𝑝| = 5
|(−3) + 𝑝𝑝| = 5
(−3 + 𝑝𝑝)2 = (5)2
9 − 3𝑝𝑝 − 3𝑝𝑝 + 𝑝𝑝2 = 25
𝑝𝑝2 − 6𝑝𝑝 + 9 − 25 = 0
𝑝𝑝2 − 6𝑝𝑝 − 16 = 0
(𝑝𝑝 − 8)(𝑝𝑝 + 2) = 0
𝑝𝑝 − 8 = 0,
𝒑𝒑 = 𝟖𝟖,
𝑝𝑝 + 2 = 0
𝒑𝒑 = −𝟐𝟐
© Pocket Tutor 2022
We have been told that 𝑥𝑥 = −3 is a solution of this
equation, so to find 𝑝𝑝, we sub in −3 for 𝑥𝑥 and solve
the equation.
We can square both sides to get rid of the modulus
bars.
Squaring out the bracket.
Rearranging the left hand side and taking 25 from
both sides. This gives us a quadratic which we can
solve for 𝑝𝑝.
Factorising and letting each bracket equal 0.
Solving for the two values of 𝑝𝑝.
15
b)
𝒙𝒙 = −𝟒𝟒, 𝟕𝟕, 𝟐𝟐
𝑓𝑓(𝑥𝑥) = 𝑥𝑥 3 + 𝑞𝑞𝑥𝑥 2 − 22𝑥𝑥 + 56
𝑥𝑥 + 4 is a factor
→ 𝑥𝑥 = −4
𝑓𝑓(−4) = 0
𝑓𝑓(−4) = (−4)3 + 𝑞𝑞(−4)2 − 22(−4) + 56 = 0
−64 + 16𝑞𝑞 + 88 + 56 = 0
16𝑞𝑞 + 80 = 0
If 𝑥𝑥 + 4 is a factor, then we know that
𝑥𝑥 = −4 is a root.
So, we can sub in −4 for 𝑥𝑥 in the function
and it will equal 0.
Therefore, by subbing in −4 for 𝑥𝑥 and
letting the function equal 0 we can solve
for 𝑞𝑞.
Dividing across by 16 shows that 𝑞𝑞 = −5.
16𝑞𝑞 = −80
𝑞𝑞 = −5
𝑥𝑥 2 − 9𝑥𝑥 + 14
To find the other two factors, we divide
the function by the factor we were given.
𝑥𝑥 3 + 4𝑥𝑥 2
First subbing in −5 for 𝑞𝑞.
𝑥𝑥 + 4 |𝑥𝑥 3 − 5𝑥𝑥 2 − 22𝑥𝑥 + 56
−9𝑥𝑥 2 − 22𝑥𝑥
2
We then use algebraic long division.
−9𝑥𝑥 − 36𝑥𝑥
14𝑥𝑥 + 56
14𝑥𝑥 + 56
0
𝑥𝑥 2 − 9𝑥𝑥 + 14
(𝑥𝑥 − 7)(𝑥𝑥 − 2)
→ 𝑥𝑥 = 7, 𝑥𝑥 = 2
Roots:
Taking the resulting quadratic and
factorising it gives us the other two
factors.
If 𝑥𝑥 − 7 is a factor, then 𝑥𝑥 = 7 is a root.
Similarly, if 𝑥𝑥 − 2 is a factor then 𝑥𝑥 = 2 is
a root.
−𝟒𝟒, 𝟕𝟕, 𝟐𝟐
© Pocket Tutor 2022
16
Question 3
a)
𝟖𝟖√𝟔𝟔 𝒄𝒄𝒎𝒎𝟑𝟑
Volume = length × height × width
𝑉𝑉 = 𝑦𝑦 × 𝑧𝑧 × 𝑥𝑥
8√6 = 𝑦𝑦 × 𝑧𝑧
2√2 = 𝑥𝑥 × 𝑧𝑧
4√3 = 𝑥𝑥 × 𝑦𝑦
2√2
= 𝑧𝑧
𝑥𝑥
4√3
= 𝑦𝑦
𝑥𝑥
8√6 =
8√6 =
𝑥𝑥 = 1
4√3 2√2
×
𝑥𝑥
𝑥𝑥
8√6
𝑥𝑥 2
𝑦𝑦 = 4√3
𝑧𝑧 = 2√2
𝑉𝑉 = 𝑦𝑦 × 𝑧𝑧 × 𝑥𝑥
𝑉𝑉 = 4√3 × 2√2 × 1
We can find each of these values using the areas of each
of the sides given.
We can see from the diagram that the area of the large
side on the right is equal to 8√6. We also know that area
is gotten by multiplying the two sides, 𝑦𝑦 and 𝑧𝑧.
We can do this for the two other sides as well.
Now rearranging two of these equations to get them in
terms of 𝑥𝑥.
Subbing these expressions in for 𝑦𝑦 and 𝑧𝑧 respectively in
the first equation.
Multiplying the two fractions on the right together.
Solving for 𝑥𝑥.
Subbing this value of 𝑥𝑥 into the expressions we found
for 𝑧𝑧 and 𝑦𝑦.
Now multiplying 𝑥𝑥, 𝑧𝑧 and 𝑦𝑦 together to find the
volume.
𝑉𝑉 = 𝟖𝟖√𝟔𝟔 𝒄𝒄𝒎𝒎𝟑𝟑
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17
b) i)
𝒙𝒙 =
𝟕𝟕
, 𝒙𝒙 = −𝟓𝟓
𝟑𝟑
3𝑥𝑥 2 + 8𝑥𝑥 − 35 = 0
(3𝑥𝑥 − 7)(𝑥𝑥 + 5) = 0
3𝑥𝑥 − 7 = 0, 𝑥𝑥 + 5 = 0
3𝑥𝑥 = 7,
𝟕𝟕
𝒙𝒙 =
𝟑𝟑
𝒙𝒙 = −𝟓𝟓
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We can find the roots by factorising the quadratic,
letting each factor equal 0 and solving for 𝑥𝑥.
Getting 𝑥𝑥 on one side and the constant on the other.
Dividing across by 3.
18
ii)
𝒎𝒎 = 𝒍𝒍𝒍𝒍𝒍𝒍𝟑𝟑 𝟕𝟕 − 𝟏𝟏
32𝑚𝑚+1 = 35 − 8(3𝑚𝑚 )
To answer this question, it is helpful to have the laws
of indices open on page 21 of the Maths Tables Book.
32𝑚𝑚 . 3 = 35 − 8(3𝑚𝑚 )
First, we can rewrite 32𝑚𝑚+1 as 3 × 32𝑚𝑚 .
3. (3𝑚𝑚 )2 = 35 − 8(3𝑚𝑚 )
Now we can rewrite 32𝑚𝑚 as (3𝑚𝑚 )2
𝑙𝑙𝑙𝑙𝑙𝑙 𝑥𝑥 = 3𝑚𝑚
Now, we can use substitution to solve the equation.
→ 3. (𝑥𝑥)2 = 35 − 8(𝑥𝑥)
So, we sub in 𝑥𝑥 everywhere there’s a 3𝑚𝑚 .
3𝑥𝑥 2 + 8𝑥𝑥 − 35 = 0
We know the solutions for 𝑥𝑥, as we found in them in
the last part.
3𝑥𝑥 2 = 35 − 8𝑥𝑥
7
𝑥𝑥 = −5, 𝑥𝑥 =
3
3𝑚𝑚 = −5,
3𝑚𝑚 =
7
𝑚𝑚 = log 3
3
𝑚𝑚 = log 3 7 − log 3 3
𝒎𝒎 = 𝒍𝒍𝒍𝒍𝒍𝒍𝟑𝟑 𝟕𝟕 − 𝟏𝟏
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Now subbing 3𝑚𝑚 back in for 𝑥𝑥.
7
3
As 3𝑚𝑚 is positive it is always greater than 0, so we
disregard 3𝑚𝑚 = −5.
Now using the laws of logs on page 21 of the Maths
Tables Book.
𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥
𝑎𝑎 = 3, 𝑥𝑥 = 𝑚𝑚, 𝑦𝑦 =
𝑥𝑥
7
3
Also, log 𝑎𝑎 = log 𝑎𝑎 𝑥𝑥 − log 𝑎𝑎 𝑦𝑦.
𝑦𝑦
Finally, log 3 3 = 1
19
Question 4
a)
23𝑛𝑛−1 + 3 is divisible by 7 for all 𝑛𝑛 ∈ 𝑁𝑁
1. Showing it is true for 𝒏𝒏 = 𝟏𝟏
→ 23(1)−1 + 3
Subbing in 1 for 𝑛𝑛
2. Assuming it is true for 𝒏𝒏 = 𝒌𝒌
→ 23(𝑘𝑘)−1 + 3 = 7𝑀𝑀
Subbing in 𝑘𝑘 for 𝑛𝑛 and letting it equal any
number divisible by 7 (7𝑀𝑀).
3. Proving it is true for 𝒏𝒏 = 𝒌𝒌 + 𝟏𝟏
Subbing in (𝑘𝑘 + 1) for 𝑛𝑛
23𝑘𝑘+2 + 3
Letting it equal 𝑛𝑛 = 𝑘𝑘 times 23 , as this is
equivalent.
22 + 3 = 7
23𝑘𝑘−1 = 7𝑀𝑀 − 3
23(𝑘𝑘+1)−1 + 3
23𝑘𝑘+2 + 3 = 23 �23𝑘𝑘−1 � + 3
3𝑘𝑘+2
2
3𝑘𝑘+2
2
= 8(7𝑀𝑀 − 3) + 3
= 56𝑀𝑀 − 24 + 3
23𝑘𝑘+2 = 56𝑀𝑀 − 21
7 is divisible by 7.
Now, subbing in our assumption for 𝑛𝑛 = 𝑘𝑘
Multiplying out the brackets.
This is divisible by 7.
𝑃𝑃(𝐾𝐾) is divisible by 7.
True for 𝑛𝑛 = 1 and, if true for 𝑛𝑛 = 𝑘𝑘, then true for 𝑛𝑛 = 𝑘𝑘 + 1. Therefore true for all 𝑛𝑛 ≥ 1.
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20
b) i)
𝑻𝑻𝒏𝒏 = 𝒑𝒑 + 𝟕𝟕𝟕𝟕 − 𝟕𝟕
𝑝𝑝, 𝑝𝑝 + 7, 𝑝𝑝 + 14, 𝑝𝑝 + 21
𝑇𝑇𝑛𝑛 = 𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑
𝑎𝑎 = 𝑝𝑝
𝑑𝑑 = (𝑝𝑝 + 7) − 𝑝𝑝
𝑑𝑑 = 7
𝑇𝑇𝑛𝑛 = 𝑝𝑝 + (𝑛𝑛 − 1)7
𝑻𝑻𝒏𝒏 = 𝒑𝒑 + 𝟕𝟕𝟕𝟕 − 𝟕𝟕
To find an expression for the 𝑛𝑛𝑡𝑡ℎ term of an arithmetic
sequence we use the expression on page 22 of the Maths
Tables Book.
𝑎𝑎 = the first term, 𝑑𝑑 = the common difference which we can
get by taking the first term from the second.
Plugging in the values for 𝑎𝑎 and 𝑑𝑑 and then multiplying out
the bracket.
ii)
𝒑𝒑 = 𝟓𝟓
𝑇𝑇𝑛𝑛 = 𝑝𝑝 + 7𝑛𝑛 − 7
To find this we let the expression we found in the last
part equal 2021.
𝑝𝑝 + 7𝑛𝑛 = 2028
Adding 7 to both sides.
𝑝𝑝 + 7𝑛𝑛 − 7 = 2021
7𝑛𝑛 = 2028 − 𝑝𝑝
Taking 𝑝𝑝 away from both sides.
So, 2028 − 𝑝𝑝 is the nearest multiple of 7 that is less than2028. This is 2023. So:
2028 − 𝑝𝑝 = 2023
𝒑𝒑 = 𝟓𝟓
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21
Question 5
a) i)
𝒂𝒂 = 𝟔𝟔, 𝒃𝒃 = 𝟏𝟏, 𝒄𝒄 = −𝟏𝟏𝟏𝟏
𝑓𝑓(𝑥𝑥) = 2𝑥𝑥 3 + 6𝑥𝑥 2 − 12𝑥𝑥 + 3
Differentiating by rule to find the derivative.
𝑓𝑓 ′ (𝑥𝑥) = 6𝑥𝑥 2 + 12𝑥𝑥 − 12
To write the derivative in the form given in the
question we first factorise out the six.
6(𝑥𝑥 2 + 2𝑥𝑥 + 1 − 1 − 2)
Now we complete the square by adding and
subtracting half the coefficient of the 𝑥𝑥, squared.
So, we half the 2 and add and subtract 12 .
𝑓𝑓 ′ (𝑥𝑥) = (3)2𝑥𝑥 2 + (2)6𝑥𝑥 1 − 12
2
6(𝑥𝑥 + 2𝑥𝑥 − 2)
6((𝑥𝑥 + 1)(𝑥𝑥 + 1) − 3)
6(𝑥𝑥 + 1)2 − 18
𝒂𝒂 = 𝟔𝟔, 𝒃𝒃 = 𝟏𝟏, 𝒄𝒄 = −𝟏𝟏𝟏𝟏
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Now factorising the first 3 terms.
Multiplying the −3 by the 6 to remove the bracket.
22
ii)
𝒙𝒙 < −𝟒𝟒, 𝒙𝒙 > 𝟐𝟐
𝑔𝑔(𝑥𝑥) = 36𝑥𝑥 + 5
𝑔𝑔′ (𝑥𝑥) = 36
𝑓𝑓 ′ (𝑥𝑥) > 𝑔𝑔′(𝑥𝑥)
6𝑥𝑥 2 + 12𝑥𝑥 − 12 > 36
2
6𝑥𝑥 + 12𝑥𝑥 − 48 > 0
𝑥𝑥 2 + 2𝑥𝑥 − 8 > 0
To find the values of 𝑥𝑥 for which 𝑓𝑓′(𝑥𝑥) > 𝑔𝑔′(𝑥𝑥) we first
differentiate 𝑔𝑔(𝑥𝑥).
Now taking the expression we found for 𝑓𝑓′(𝑥𝑥) in the last
part and putting it as greater than 𝑔𝑔′ (𝑥𝑥).
Taking 36 from both sides.
Dividing across by 6.
(𝑥𝑥 + 4)(𝑥𝑥 − 2) > 0
Factorising.
Let = 0
Letting the expression equal 0 and solving for the two values
of 𝑥𝑥.
𝑥𝑥 + 4 = 0,
𝑥𝑥 = −4,
𝑥𝑥 − 2 = 0
𝑥𝑥 = 2
𝒙𝒙 < −𝟒𝟒, 𝒙𝒙 > 𝟐𝟐
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Now sketching the shape of the quadratic to see where it is
above 0. It is above 0 when 𝑥𝑥 is less than −4 and when 𝑥𝑥 is
greater than 2.
23
b)
𝒌𝒌 = 𝟎𝟎. 𝟔𝟔𝟔𝟔
Equation of tangent:
ℎ(𝑥𝑥) = 2 sin 2𝑥𝑥
To find (0, 𝑘𝑘) we need the equation of the
tangent.
ℎ′ (𝑥𝑥) = 2(cos 2𝑥𝑥) × 2
To find the equation of a line we need the
slope and a point on the line.
𝜋𝜋
𝜋𝜋
ℎ′ � � = 4 cos �2 � ��
6
6
Now subbing in the 𝑥𝑥 coordinate at which the
tangent touches the curve to find its slope.
𝜋𝜋
𝜋𝜋
ℎ � � = 2 sin 2 � � = √3
6
6
Finding a point on the line by subbing the 𝑥𝑥
coordinate at which the tangent touches the
curve into the equation of the curve.
ℎ′ (𝑥𝑥) = 4 cos 2𝑥𝑥
=2
𝜋𝜋
� , √3� , 𝑚𝑚 = 2
6
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
𝜋𝜋
𝑦𝑦 − √3 = 2 �𝑥𝑥 − �
6
𝑥𝑥 = 0
𝜋𝜋
→ 𝑦𝑦 − √3 = 2 �0 − �
6
𝜋𝜋
𝑦𝑦 = 2 �− � + √3
6
𝑦𝑦 = 0.68
We can find the slope of the tangent by
differentiating the equation of the curve.
Now we have coordinates on the line and its
slope.
Subbing these into the equation of a line from
page 18 of the Maths Tables Book.
Now subbing in 0 for 𝑥𝑥 in this line to find the
corresponding coordinate 𝑘𝑘.
Adding √3 to both sides.
Subbing the right hand side into the calculator.
𝒌𝒌 = 𝟎𝟎. 𝟔𝟔𝟔𝟔
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24
Question 6
a)
Roots:
𝑥𝑥 = −1, 𝑥𝑥 = 3
Factors:
(𝑥𝑥 + 1), (𝑥𝑥 − 3)
ℎ′ (𝑥𝑥) = 𝑎𝑎(𝑥𝑥 + 1)(𝑥𝑥 − 3)
We can form the equation of a quadratic from
its roots.
If 𝑥𝑥 = −1 is a root, then 𝑥𝑥 + 1 is a factor of
the equation.
So, we multiply the factors together.
2
𝑎𝑎(𝑥𝑥 − 3𝑥𝑥 + 𝑥𝑥 − 3)
𝑎𝑎(𝑥𝑥 2 − 2𝑥𝑥 − 3)
𝑎𝑎((0)2 − (0) − 3) = 6
We have to multiply this by 𝑎𝑎, to make sure
that the equation gives us a graph with the
same maximum point as the one in the
question.
𝑎𝑎(−3) = 6
𝑎𝑎 = −2
We can see that the 𝑦𝑦 −intercept is 6 so when
we let 𝑥𝑥 equal 0, the equation must equal 6.
−2(𝑥𝑥 2 − 2𝑥𝑥 − 3)
Now subbing in the value, we just found for 𝑎𝑎
and multiplying out the bracket.
ℎ′ (𝑥𝑥) = −2𝑥𝑥 2 + 4𝑥𝑥 + 6
b)
8
ℎ′ (𝑥𝑥) = −2𝑥𝑥 2 + 4𝑥𝑥 + 6
ℎ′′ (𝑥𝑥) = −4𝑥𝑥 + 4
−4𝑥𝑥 + 4 = 0
4 = 4𝑥𝑥
𝑥𝑥 = 1
ℎ′ (1) = −2(1)2 + 4(1) + 6
= 𝟖𝟖
To find the maximum positive value of a
slope, we differentiate the equation for the
slope of a tangent.
So, differentiating ℎ′ (𝑥𝑥).
Now we let this derivative equal 0 and solve
for the value of 𝑥𝑥, which will give a maximum
value.
Finally, we plug this into ℎ′ (𝑥𝑥), the equation
for the slope of a tangent, to find the
maximum slope.
c)
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25
𝟐𝟐
𝒉𝒉(𝒙𝒙) = − 𝒙𝒙𝟑𝟑 + 𝟐𝟐𝒙𝒙𝟐𝟐 + 𝟔𝟔𝟔𝟔 − 𝟐𝟐
𝟑𝟑
ℎ′ (𝑥𝑥) = −2𝑥𝑥 2 + 4𝑥𝑥 + 6
To find the equation of ℎ(𝑥𝑥) we need to integrate ℎ′ (𝑥𝑥).
4
2
− 𝑥𝑥 3 + 𝑥𝑥 2 + 6𝑥𝑥 + 𝑐𝑐 = ℎ(𝑥𝑥)
2
3
Integrating by adding 1 to the power of each 𝑥𝑥 and dividing
by the new power. The constant (6) becomes 6𝑥𝑥.
Remember when integrating we always add ′𝑐𝑐′.
� −2𝑥𝑥 2 + 4𝑥𝑥 + 6 𝑑𝑑𝑑𝑑 = ℎ(𝑥𝑥)
2
− 𝑥𝑥 3 + 2𝑥𝑥 2 + 6𝑥𝑥 + 𝑐𝑐 = ℎ(𝑥𝑥)
3
2
− (0)3 + 2(0)2 + 𝑐𝑐 = −2
3
𝑐𝑐 = −2
𝟐𝟐
→ 𝒉𝒉(𝒙𝒙) = − 𝒙𝒙𝟑𝟑 + 𝟐𝟐𝒙𝒙𝟐𝟐 + 𝟔𝟔𝟔𝟔 − 𝟐𝟐
𝟑𝟑
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Now, to find the value of ′𝑐𝑐′, we sub in the given point
(0, −2).
Subbing in 0 for each 𝑥𝑥 and letting the equation equal −2,
then solving for 𝑐𝑐.
Subbing in this value for 𝑐𝑐.
26
Question 7
a) i)
Swing
Length of Arc
(cm)
𝟏𝟏
45
𝟐𝟐
40.5
𝟑𝟑
729
20
𝟒𝟒
6561
200
𝟓𝟓
29.5245
Multiply each length by 0.9 to find 90% of it, i.e. the next
length.
ii)
𝑇𝑇𝑛𝑛 = 45(0.9)𝑛𝑛−1
𝑇𝑇25 = 45(0.9)
𝑇𝑇25 = 𝟑𝟑. 𝟔𝟔𝒄𝒄𝒄𝒄
25−1
To find the arc length on the 25th swing, we sub in
25 for 𝑛𝑛 in the expression given and plug it into the
calculator.
iii)
𝟒𝟒𝟒𝟒𝟒𝟒 𝒄𝒄𝒄𝒄
𝑆𝑆𝑛𝑛 =
𝑎𝑎(1 − 𝑟𝑟 𝑛𝑛 )
1 − 𝑟𝑟
𝑎𝑎 = 45, 𝑟𝑟 = 0.9, 𝑛𝑛 = 40
𝑆𝑆40 =
45(1 − (0.9)40 )
1 − 0.9
𝑆𝑆40 = 𝟒𝟒𝟒𝟒𝟒𝟒 𝒄𝒄𝒄𝒄
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To find the total distance travelled by the tip of the pendulum
we need to find the sum of the distances travelled on each
swing. We can do this using the equation for the sum of a
geometric series on page 22 of the Maths Tables Book.
𝑎𝑎 = the first term, 𝑟𝑟 = the common ratio, 𝑛𝑛 = the number of
swings.
Subbing in each of these values and plugging the expression
into the calculator gives us our answer.
27
iv)
𝒑𝒑 = 𝟑𝟑𝟑𝟑
𝑇𝑇𝑛𝑛 = 45(0.9)𝑛𝑛−1
To find 𝑝𝑝, we take the equation for the length of the
arc length from part ii) and let it equal 2𝑐𝑐𝑐𝑐.
2 = 45(0.9)𝑝𝑝−1
We then sub in 𝑝𝑝 for 𝑛𝑛 and solve for 𝑝𝑝.
𝑛𝑛 = 𝑝𝑝, 𝑇𝑇𝑛𝑛 = 2
2
= (0.9)𝑝𝑝−1
45
log 0.9
2
= 𝑝𝑝 − 1
45
29.55 = 𝑝𝑝 − 1
30.55 = 𝑝𝑝
Dividing across by 45.
We can use logs to solve for 𝑝𝑝.
Taking the rule: 𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 from page 21 of the
Maths Tables Book. 𝑎𝑎 = 0.9, 𝑦𝑦 =
2
45
Plugging the log into the calculator.
, 𝑥𝑥 = (𝑝𝑝 − 1).
Adding 1 to both sides.
The 31st swing is the first one which will be shorter
than 2cm.
𝒑𝒑 = 𝟑𝟑𝟑𝟑
b) i)
𝜃𝜃
�
𝑙𝑙 = 2𝜋𝜋𝜋𝜋 �
360
𝑙𝑙 = 0.45, 𝑟𝑟 = 1
0.45 = 2𝜋𝜋(1) �
𝜃𝜃
0.45
=
360
2𝜋𝜋
𝜃𝜃
�
360
0.45
× 360 = 𝜃𝜃
2𝜋𝜋
𝜃𝜃 = 25.78°
𝜃𝜃 = 26°
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1𝑚𝑚
We can find the angle using the
formula for the length of an arc on
page 9 of the Maths Tables Book.
The radius (length of the
pendulum) is 1𝑚𝑚 and we convert
the arc length of 45𝑐𝑐𝑐𝑐 to 0.45𝑚𝑚.
We then sub these into the
equation and solve for 𝜃𝜃.
Dividing across by 2𝜋𝜋 then
multiplying across by 360.
Rounding to the nearest whole
number.
28
ii)
𝑆𝑆𝑛𝑛 =
𝑎𝑎
1 − 𝑟𝑟
𝑎𝑎 = 26, 𝑟𝑟 = 0.9
26
= 𝟐𝟐𝟐𝟐𝟐𝟐°
1 − 0.9
The total accumulated angle that the pendulum swings through can
be found using the formula for the sum to infinity of a geometric
series from page 22 of the Maths Tables Book.
Subbing in the angle we found in the last part for 𝑎𝑎, the first term,
and 0.9 as the common ratio and plugging into the calculator.
iii)
𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
𝑆𝑆𝑛𝑛 =
𝑎𝑎
1 − 𝑟𝑟
𝑎𝑎 = 45, 𝑟𝑟 = 0.9
45
= 450𝑐𝑐𝑐𝑐
1 − 0.9
450
= 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
2
We can find half the total distance by finding the full distance
and then dividing it by 2.
So, using the formula for the sum to infinity of a geometric
series from page 22 of the Maths Tables Book.
Subbing in 45 for 𝑎𝑎, the length of the first arc, and 0.9 for 𝑟𝑟, the
common ratio.
Dividing the result by 2 gives us our answer.
𝑶𝑶𝑶𝑶
Half the accumulated angle:
260
= 130°
2
𝜃𝜃
�
𝑙𝑙 = 2𝜋𝜋𝜋𝜋 �
360
𝜃𝜃 = 130, 𝑟𝑟 = 100𝑐𝑐𝑐𝑐
𝑙𝑙 = 2𝜋𝜋(100) �
𝑙𝑙 = 𝟐𝟐𝟐𝟐𝟐𝟐𝒄𝒄𝒄𝒄
130
�
360
© Pocket Tutor 2022
We first find half the accumulated angle by dividing our
answer from the previous part by 2.
Now we can use the formula for the length of an arc
from page 9 of the Maths Tables Book, to find the
distance travelled.
Converting the radius to centimetres as we want our
answer in centimetres.
Subbing in the radius and the angle and plugging into
the calculator.
29
Question 8
a) i)
ℎ(𝑥𝑥) = 0.001𝑥𝑥 3 − 0.12𝑥𝑥 2 + 𝑝𝑝𝑝𝑝 + 5
To find the value of 𝑝𝑝, we sub in 10 for 𝑥𝑥
and let the expression equal 30.
ℎ(10) = 30
ℎ(10) = 0.001(10)3 − 0.12(10)2 + 𝑝𝑝(10) + 5 = 30
Multiplying out the brackets.
10𝑝𝑝 = 30 + 6
Adding 6 to both sides.
Adding and subtracting the constants.
1 − 12 + 10𝑝𝑝 + 5 = 30
10𝑝𝑝 = 36
Dividing across by 10 gives us the answer.
𝑝𝑝 = 3.6
ii)
ℎ(𝑥𝑥) = 0.001𝑥𝑥 3 − 0.12𝑥𝑥 2 + 3.6𝑥𝑥 + 5
𝑥𝑥
ℎ(𝑥𝑥)
0
5
10
30
20
37
30
32
40
21
50
10
60
5
70
12
75
21.875
To fill out the table we use our calculator. Press MODE, then 3. Table, then input the equation
ℎ(𝑥𝑥) = 0.001𝑥𝑥 3 − 0.12𝑥𝑥 2 + 3.6𝑥𝑥 + 5. Press enter, then start 0, end 75, step 5. Use this to fill
in the values in the table above.
Then plot the values in the table as the 𝑥𝑥 and 𝑦𝑦 coordinates on the graph and join them with a
curve as shown below.
© Pocket Tutor 2022
30
© Pocket Tutor 2022
31
b) i)
ℎ(𝑥𝑥) = 0.001𝑥𝑥 3 − 0.12𝑥𝑥 2 + 3.6𝑥𝑥 + 5
ℎ′ (𝑥𝑥) = 3(0.001𝑥𝑥 2 ) − 2(0.12𝑥𝑥) + 3.6
𝒉𝒉′ (𝒙𝒙) = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝒙𝒙𝟐𝟐 − 𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐 + 𝟑𝟑. 𝟔𝟔
To find the derivative we multiply each term by the
power of the 𝑥𝑥 and then decrease the power by 1.
Constants disappear.
ii)
ℎ′ (𝑥𝑥) = 0.003𝑥𝑥 2 − 0.24𝑥𝑥 + 3.6
0.003𝑥𝑥 2 − 0.24𝑥𝑥 + 3.6 = 0
𝑥𝑥 =
𝑥𝑥 =
−𝑏𝑏 ± √𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎
2𝑎𝑎
−(−0.24) ± �(−0.24)2 − 4(0.003)(3.6)
2(0.003)
𝑥𝑥 = 60, 𝑥𝑥 = 20
ℎ′ (𝑥𝑥) = 0.003𝑥𝑥 2 − 0.24𝑥𝑥 + 3.6
ℎ′′ (𝑥𝑥) = 0.006𝑥𝑥 − 0.24
0.006(60) − 0.24 = 0.12
0.12 > 0 ∴ 𝑀𝑀𝑀𝑀𝑀𝑀
0.006(20) − 0.24 = −0.12
−0.12 < 0 ∴ 𝑀𝑀𝑀𝑀𝑀𝑀
© Pocket Tutor 2022
To find when a graph reaches its maximum height, we
let the derivative equal 0 and solve for 𝑥𝑥.
Taking the derivative from the last part and letting it
equal 0.
We can use the −𝑏𝑏 formula from page 20 of the Maths
Tables Book to solve this quadratic.
Remember quadratics are written in the form 𝑎𝑎𝑥𝑥 2 +
𝑏𝑏𝑏𝑏 + 𝑐𝑐 = 0
Plugging the fraction into the calculator, once with a
plus in front of the square root and then with a minus,
to find the two values of 𝑥𝑥.
To find out which is the maximum and which is the
minimum value we get the second derivative of the
equation.
Differentiating ℎ′(𝑥𝑥) to get the second derivative.
Now subbing in 60 for 𝑥𝑥. This gives us a positive value,
so it is a minimum.
Subbing in 20 for 𝑥𝑥 gives us a negative value so it is the
maximum.
32
iii)
𝟐𝟐𝟐𝟐𝟐𝟐
ℎ′′ (𝑥𝑥) = 0.006𝑥𝑥 − 0.24
0.006𝑥𝑥 − 0.24 = 0
0.006𝑥𝑥 = 0.24
𝑥𝑥 =
0.24
0.006
𝑥𝑥 = 40
ℎ(𝑥𝑥) = 0.001𝑥𝑥 3 − 0.12𝑥𝑥 2 + 3.6𝑥𝑥 + 5
ℎ(40) = 0.001(40)3 − 0.12(40)2 + 3.6(40) + 5
ℎ(40) = 𝟐𝟐𝟐𝟐𝟐𝟐
© Pocket Tutor 2022
To find an inflection point, we let the second
derivative equal 0 and solve for 𝑥𝑥.
So, letting the second derivative, we found
in the last part, equal 0.
Adding 0.24 to both sides and then dividing
across by 0.006.
Now that we have the value of 𝑥𝑥 we sub it
into the original equation to find the height
at this value.
Subbing in 40 for 𝑥𝑥 and plugging into the
calculator gives us the answer.
33
c)
𝟐𝟐𝟐𝟐. 𝟒𝟒𝟒𝟒𝟒𝟒
𝑏𝑏
1
� 0.001𝑥𝑥 3 − 0.12𝑥𝑥 2 + 3.6𝑥𝑥 + 5 𝑑𝑑𝑑𝑑
𝑏𝑏 − 𝑎𝑎 𝑎𝑎
To find the average height we
integrate the equation between
the given limits as shown.
75
1
� 0.001𝑥𝑥 3 − 0.12𝑥𝑥 2 + 3.6𝑥𝑥 + 5 𝑑𝑑𝑑𝑑
75 − 0 0
Integrating by increasing the
power of each 𝑥𝑥 by 1 and dividing
by the new power.
75
1 0.001 4 0.12 3 3.6 2
�
𝑥𝑥 −
𝑥𝑥 +
𝑥𝑥 + 5𝑥𝑥�
75 4
3
2
0
0.12
3.6
1 0.001
(75)4 −
(75)3 +
(75)2 + 5(75)�
��
4
3
2
75
0.12
3.6
0.001
(0)3 +
(0)2 + 5(0)��
(0)4 −
−�
3
2
4
1
[1535.15625 − 0]
75
= 𝟐𝟐𝟐𝟐. 𝟒𝟒𝟕𝟕𝒎𝒎
© Pocket Tutor 2022
Now subbing in the max and min
limits and taking the minimum
away from the maximum.
Plugging each bracket into the
calculator.
Multiplying in by the fraction.
34
Question 9
a) i)
𝑇𝑇(𝑡𝑡) = 𝐴𝐴𝑒𝑒 −0.081𝑡𝑡 + 20
𝑇𝑇(0) = 95
𝐴𝐴𝑒𝑒 −0.081(0) + 20 = 95
𝐴𝐴𝑒𝑒 −0.081(0) = 95 − 20
𝐴𝐴(1) = 75
𝐴𝐴 = 75
We know that the temperature of the
coffee at the start (0 minutes) is 95°.
So, we can sub in 0 for 𝑡𝑡 and let the
equation equal 95.
This allows us to solve for 𝐴𝐴.
Taking 20 from both sides.
Anything to the power of 0 equals 1.
ii)
This is the lowest temperature the coffee will cool to/The room temperature.
iii)
𝟒𝟒𝟒𝟒°
𝑇𝑇(𝑡𝑡) = 75𝑒𝑒 −0.081(𝑡𝑡) + 20
𝑇𝑇(10) = 75𝑒𝑒 −0.081(10) + 20
𝑇𝑇(10) = 53°
95 − 53 = 𝟒𝟒𝟒𝟒°
© Pocket Tutor 2022
To find the decrease in temperature we find the temperature
after 10 minutes and take this away from the original
temperature.
We find the temperature after 10 minutes by subbing in 10
for 𝑡𝑡.
Plugging the expression into the calculator.
Subtracting the result from the original temperature gives us
the difference.
35
b)
𝟏𝟏𝟏𝟏𝟏𝟏 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬
𝑇𝑇(𝑡𝑡) = 75𝑒𝑒 −0.081(𝑡𝑡) + 20
82 = 75𝑒𝑒 −0.081(𝑡𝑡) + 20
82 − 20 = 75𝑒𝑒 −0.081(𝑡𝑡)
62 = 75𝑒𝑒 −0.081(𝑡𝑡)
62
= 𝑒𝑒 −0.081(𝑡𝑡)
75
log 𝑒𝑒
62
= −0.081(𝑡𝑡)
75
−0.1904 = −0.081𝑡𝑡
−0.1904
= 𝑡𝑡
−0.081
To find the time at which the coffee is at this
temperature we let the equation for the temperature
equal 82.
We then solve for 𝑡𝑡.
Taking 20 away from both sides.
Dividing across by 75.
We can use logs to solve for 𝑡𝑡.
Taking the rule: 𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥 from page 21 of
the Maths Tables Book. 𝑎𝑎 = 𝑒𝑒, 𝑦𝑦 =
Plugging the log into the calculator.
62
75
, 𝑥𝑥 = −0.081𝑡𝑡.
2.35 mins = t
Dividing across by −0.081.
2.35 × 60 = 𝟏𝟏𝟏𝟏𝟏𝟏 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬
Multiplying the result by 60 to get our answer in seconds.
© Pocket Tutor 2022
36
c)
𝟕𝟕𝟕𝟕°
𝑇𝑇(𝑡𝑡) = 75𝑒𝑒 −0.081(𝑡𝑡) + 20
𝑇𝑇 ′ (𝑡𝑡) = (−0.081)75𝑒𝑒 −0.081(𝑡𝑡)
(−0.081)75𝑒𝑒 −0.081(𝑡𝑡)
𝑒𝑒 −0.081𝑡𝑡 =
𝑒𝑒 −0.081𝑡𝑡 =
= −4.05
−4.05
(−0.081)75
2
3
−0.081𝑡𝑡 = log 𝑒𝑒
2
3
−0.081𝑡𝑡 = −0.405
𝑡𝑡 = 5
𝑇𝑇(𝑡𝑡) = 75𝑒𝑒 −0.081(𝑡𝑡) + 20
𝑇𝑇(5) = 75𝑒𝑒 −0.081(5) + 20
= 𝟕𝟕𝟕𝟕°
© Pocket Tutor 2022
To find the temperature of the coffee when 𝑇𝑇 ′ (𝑡𝑡) =
−4.05, we need to find the derivative and let it
equal −4.05. This allows us to solve for 𝑡𝑡, the time
at which this is true.
We can then plug this time into the original
equation to find the temperature.
So, differentiating 𝑇𝑇(𝑡𝑡) to find the derivative and
letting the derivative equal −4.05.
Dividing across by (−0.081)75.
We can use logs to solve for 𝑡𝑡.
Taking the rule: 𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥 from page 21
2
of the Maths Tables Book. 𝑎𝑎 = 𝑒𝑒, 𝑦𝑦 = , 𝑥𝑥 =
−0.081𝑡𝑡.
3
Plugging the log into the calculator and then
dividing across by −0.081.
Now subbing the value we found for 𝑡𝑡 into the
original equation to find the temperature.
37
d)
−
𝟒𝟒
𝒄𝒄𝒄𝒄/𝒔𝒔𝒔𝒔𝒔𝒔
𝟏𝟏𝟏𝟏
Find = Given × Need
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
=
×
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
We use the phrase Find = Given × Need, with what we
need to find to answer the question on the left, equal to
what the question has Given us times what is Needed to
fulfil the equation.
𝑉𝑉 = 𝑥𝑥 3
To find
𝑑𝑑𝑑𝑑
= 3𝑥𝑥 2
𝑑𝑑𝑑𝑑
cube.
1
𝑑𝑑𝑑𝑑
= 2
𝑑𝑑𝑑𝑑 3𝑥𝑥
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
, we can take the equation for the volume of a
We then differentiate this to find
To find
3 1
1
1
→ 𝑥𝑥 = � =
Volume =
64 4
64
1
𝑑𝑑𝑑𝑑
=− ×
20
𝑑𝑑𝑑𝑑
We can use related rates of change to find the rate of
change of the side length with respect to time.
1
1 2
3� �
4
𝟒𝟒
𝑑𝑑𝑑𝑑
=−
𝒄𝒄𝒄𝒄/𝒔𝒔𝒔𝒔𝒔𝒔
𝟏𝟏𝟏𝟏
𝑑𝑑𝑑𝑑
© Pocket Tutor 2022
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
.
, we put the result under 1.
We have been told the volume is
1
64
. So, we find the
length of the side, 𝑥𝑥, by cube rooting this.
Now, going back to our word equation and plugging in
what we were given and what we need, subbing in the
value of 𝑥𝑥. The rate
cube is decreasing.
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
is −
1
20
as the volume of the sugar
Multiplying out gives us our answer.
38
Question 10
a) i)
𝒕𝒕 = 𝟏𝟏𝟏𝟏 days
𝑉𝑉(𝑡𝑡) = 60 + 41𝑡𝑡 − 3𝑡𝑡 2
0 = 60 + 41𝑡𝑡 − 3𝑡𝑡
2
3𝑡𝑡 2 − 41𝑡𝑡 − 60 = 0
(3𝑡𝑡 + 4)(𝑡𝑡 − 15) = 0
3𝑡𝑡 + 4 = 0,
3𝑡𝑡 = −4,
𝑡𝑡 = −
4
3
𝑡𝑡 − 15 = 0
𝒕𝒕 = 𝟏𝟏𝟏𝟏 days
To find the value of 𝑡𝑡 when the tank is empty, we let
the equation for the volume equal 0 and solve for 𝑡𝑡.
Rewriting the equation so we have a positive 𝑡𝑡 2 .
Factorising the quadratic. (If you struggle finding the
factors you can use the ‘−𝑏𝑏′ formula on page 20 of
the Maths Tables Book.
Letting each bracket equal 0 and solving for the value
of 𝑡𝑡.
We are told in the question that 𝑡𝑡 ≥ 0, so we only
take the positive value of 𝑡𝑡.
ii)
𝟏𝟏𝟏𝟏 litres/day
𝑉𝑉(𝑡𝑡) = 60 + 41𝑡𝑡 − 3𝑡𝑡 2
𝑉𝑉 ′ (𝑡𝑡) = 41 − (2)3𝑡𝑡
𝑉𝑉 ′ (𝑡𝑡) = 41 − 6𝑡𝑡
𝑉𝑉 ′ (5) = 41 − 6(5)
𝑉𝑉 ′ (5) = 𝟏𝟏𝟏𝟏 litres/day
© Pocket Tutor 2022
To find the rate of change of volume at the time
given, we differentiate the equation for the volume
in the tank.
Differentiating by rule, the constant (60)
disappears, and we multiply by the power of the 𝑡𝑡
and reduce it by 1.
Now subbing 5 in for 𝑡𝑡 in the derivative gives us the
rate of change of volume when 𝑡𝑡 = 5.
39
iii)
𝟒𝟒𝟒𝟒
𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝
𝟔𝟔
𝑀𝑀𝑀𝑀𝑀𝑀 → 𝑉𝑉 ′ (𝑡𝑡) = 0
To find when the volume is at a maximum, we
let the derivative of the equation equal 0 and
solve for 𝑡𝑡.
𝑉𝑉 ′ (𝑡𝑡) = 41 − 6𝑡𝑡
Taking the derivative we found in the last part
and letting it equal 0.
41 − 6𝑡𝑡 = 0
41 = 6𝑡𝑡
Adding 6𝑡𝑡 to both sides.
𝟒𝟒𝟒𝟒
𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝
𝑡𝑡 =
𝟔𝟔
Dividing across by 6.
iv)
𝑉𝑉(𝑡𝑡) = 60 + 41𝑡𝑡 − 3𝑡𝑡 2
𝑉𝑉 �
41
41
41
� = 60 + 41 � � − 3 � �
6
6
6
𝑉𝑉 �
41
� = 𝟐𝟐𝟐𝟐𝟐𝟐 𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥𝐥
6
© Pocket Tutor 2022
2
To find the maximum volume in the tank, we sub in the
value we found for the time at which the volume is at its
maximum.
So, subbing the value we found for 𝑡𝑡 in the last part, into
the original equation.
Plugging into the calculator.
40
b) i)
Show that 𝐼𝐼(𝑡𝑡) > 0, for 1 ≤ 𝑡𝑡 ≤ 10
𝐼𝐼(𝑡𝑡) = 1.5 + sin
sin 𝐴𝐴 ≥ −1
𝜋𝜋𝜋𝜋
5
So 𝐼𝐼(𝑡𝑡) ≥ 0.5
Radius increases every year, as 𝐼𝐼(𝑡𝑡) > 0.
ii)
𝐼𝐼(6) < 𝐼𝐼(5)
𝐼𝐼(𝑡𝑡) = 1.5 + sin
𝜋𝜋𝜋𝜋
5
To show that 𝐼𝐼(6) is less than 𝐼𝐼(5), we simply sub in 6
for 𝑡𝑡 in the equation 𝐼𝐼(𝑡𝑡) and then we sub in 5 for 𝑡𝑡
and compare the results.
𝐼𝐼(6) = 1.5 + sin
𝜋𝜋(6)
5
Subbing in 6 for 𝑡𝑡.
𝐼𝐼(5) = 1.5 + sin
𝜋𝜋(5)
5
𝐼𝐼(6) = 0.91
𝐼𝐼(5) = 1.5
0.91 < 1.5
∴ 𝐼𝐼(6) < 𝐼𝐼(5)
Plugging into the calculator.
Subbing in 5 for 𝑡𝑡.
Plugging into the calculator.
We can see that the value for 𝐼𝐼(6) is less than the
value for 𝐼𝐼(5).
This means that the radius increases more in the 5th year than it increases in the 6th year.
© Pocket Tutor 2022
41
iii)
𝒓𝒓(𝟐𝟐) = 𝟏𝟏𝟏𝟏 + 𝐬𝐬𝐬𝐬𝐬𝐬
𝝅𝝅
𝟐𝟐𝟐𝟐
+ 𝐬𝐬𝐬𝐬𝐬𝐬
𝟓𝟓
𝟓𝟓
𝑟𝑟(𝑡𝑡 + 1) = 𝑟𝑟(𝑡𝑡) + 𝐼𝐼(𝑡𝑡 + 1)
𝑟𝑟(2) = 𝑟𝑟(1 + 1)
To find 𝑟𝑟(2), we use the expression given:
𝑟𝑟(𝑡𝑡 + 1) = 𝑟𝑟(𝑡𝑡) + 𝐼𝐼(𝑡𝑡 + 1)
We can rewrite 𝑟𝑟(2) as 𝑟𝑟(1 + 1).
→ 𝑟𝑟(1 + 1) = 𝑟𝑟(1) + 𝐼𝐼(1 + 1)
So, we do this and sub in 1 for 𝑡𝑡, in the equation above.
𝑟𝑟(1) = 𝑟𝑟(0 + 1)
Now we need to find the value of 𝑟𝑟(1) and the value of
𝐼𝐼(1 + 1).
→ 𝑟𝑟(0 + 1) = 𝑟𝑟(0) + 𝐼𝐼(0 + 1)
𝑟𝑟(1) = (10) + �1.5 + sin
𝑟𝑟(1) = 11.5 + sin
𝜋𝜋
5
𝐼𝐼(1 + 1) = 1.5 + sin
𝐼𝐼(1 + 1) = 1.5 + sin
𝜋𝜋(1 + 1)
5
2𝜋𝜋
5
2𝜋𝜋
𝜋𝜋
+ �1.5 + sin �
5
5
𝝅𝝅
𝟐𝟐𝟐𝟐
+ 𝐬𝐬𝐬𝐬𝐬𝐬
𝟓𝟓
𝟓𝟓
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We are told in the question that 𝑟𝑟(0) = 10.
We calculate 𝐼𝐼(0 + 1) by subbing (0 + 1) in for 𝑡𝑡 in the
𝜋𝜋𝜋𝜋
expression 𝐼𝐼(𝑡𝑡) = 1.5 + sin .
5
𝑟𝑟(1 + 1) = 11.5 + sin
𝒓𝒓(𝟐𝟐) = 𝟏𝟏𝟏𝟏 + 𝐬𝐬𝐬𝐬𝐬𝐬
𝜋𝜋(1)
�
5
First, we find the value of 𝑟𝑟(1) by rewriting it as 𝑟𝑟(0 + 1)
and subbing in 0 for 𝑡𝑡 in the original equation.
Now we find the value of 𝐼𝐼(1 + 1) by subbing (1 + 1) in for
𝜋𝜋𝜋𝜋
𝑡𝑡 in the expression 𝐼𝐼(𝑡𝑡) = 1.5 + sin .
5
Finally, we go back to our expression, 𝑟𝑟(1 + 1) = 𝑟𝑟(1) +
𝐼𝐼(1 + 1) and we sub in the values we found for 𝑟𝑟(1) and
𝐼𝐼(1 + 1).
42
iv)
𝒌𝒌 = 𝟔𝟔. 𝟐𝟐𝟐𝟐
𝑉𝑉2 = 𝑘𝑘𝑉𝑉1
Taking the formula of a cylinder from page 10 of
the Maths Tables Book and writing an expression
out the volume of each cylindrical tree trunk.
𝑅𝑅2 = 𝑘𝑘𝑟𝑟 2
We can cancel out the height and 𝜋𝜋 on both
sides.
𝜋𝜋𝑅𝑅 2 ℎ = 𝑘𝑘𝑘𝑘𝑟𝑟 2 ℎ
𝑟𝑟 = 10 at beginning of first year.
We know that the radius of the smaller tree trunk
is 10 at the beginning of the first year.
𝑅𝑅2 = 𝑘𝑘(10)2
So, we can say that the radius of the larger tree
squared is equal to 𝑘𝑘 times 102 .
𝑟𝑟(𝑡𝑡 + 𝑖𝑖) = 𝑟𝑟(𝑡𝑡) + 𝐼𝐼(𝑡𝑡 + 1)
From part iii).
𝑅𝑅2 = 100𝑘𝑘
𝑟𝑟(9 + 1) = 𝑟𝑟(9) + 𝐼𝐼(9 + 1)
𝑟𝑟(10) = 𝑟𝑟(9) + 𝐼𝐼(10)
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Subbing in for 𝑟𝑟(10).
43
𝑟𝑟(8 + 1) = 𝑟𝑟(8) + 𝐼𝐼(8 + 1)
𝑟𝑟(9) = 𝑟𝑟(8) + 𝐼𝐼(9)
If we keep going back to find the value of
𝑟𝑟(9) and then 𝑟𝑟(8) and so on, we see a
pattern emerging.
𝑟𝑟(8) = 𝑟𝑟(7) + 𝐼𝐼(7 + 1)
𝑟𝑟(8) = 𝑟𝑟(7) + 𝐼𝐼(8)
𝑟𝑟(10) = 𝐼𝐼(1) + 𝐼𝐼(2) + 𝐼𝐼(3) + ⋯ + 𝐼𝐼(10)
𝑟𝑟(10) = 10 + 2.09 + 2.45 + 2.45 + 2.09 + 1.5
+0.91 + 0.55 + 0.55 + 0.91 + 1.5
From this pattern we can calculate the value
of 𝑟𝑟.
𝜋𝜋𝜋𝜋
Subbing 1,2,3…, 10 into 𝐼𝐼(𝑡𝑡) = 1.5 + sin .
5
Adding up the results gives us the length of
the radius.
= 25𝑐𝑐𝑐𝑐
𝑟𝑟 2 = 100𝑘𝑘
(25)2 = 100𝑘𝑘
625 = 100𝑘𝑘
Subbing this radius into the expression we
found earlier.
Squaring the bracket.
Dividing across by 100.
𝒌𝒌 = 𝟔𝟔. 𝟐𝟐𝟐𝟐
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44
2020 Paper 1
Question 1
a) i) 𝟔𝟔 = 𝒑𝒑
𝑓𝑓(𝑥𝑥) = 𝑥𝑥 2 + 5𝑥𝑥 + 𝑝𝑝
𝑥𝑥 + 3
→ 𝑥𝑥 = −3
𝑓𝑓(−3) = (−3)2 + 5(−3) + 𝑝𝑝 = 0
9 − 15 + 𝑝𝑝 = 0
−6 + 𝑝𝑝 = 0
𝟔𝟔 = 𝒑𝒑
ii) 𝒑𝒑 = 𝟒𝟒
root = 𝑦𝑦
second root = y + 3
factor = 𝑥𝑥 − 𝑦𝑦
second factor = 𝑥𝑥 − 𝑦𝑦 − 3
𝑥𝑥 2 + 5𝑥𝑥 + 𝑝𝑝 = (𝑥𝑥 − 𝑦𝑦)(𝑥𝑥 − 𝑦𝑦 − 3)
𝑥𝑥 2 + 5𝑥𝑥 + 𝑝𝑝 = 𝑥𝑥 2 − 2𝑥𝑥𝑥𝑥 − 3𝑥𝑥 + 𝑦𝑦 2 + 3𝑦𝑦
−2𝑥𝑥𝑥𝑥 − 3𝑥𝑥 = 5𝑥𝑥
−2𝑥𝑥𝑥𝑥 = 8𝑥𝑥
If 𝑥𝑥 + 3 is a factor of the function, then 𝑥𝑥 =
−3 is a root.
Therefore, when we plug −3 in for 𝑥𝑥 the
function equals 0.
So, subbing in −3 for 𝑥𝑥 and letting the
equation equal 0.
Multiplying out and solving for 𝑝𝑝.
We can write two roots which differ by 3 as 𝑦𝑦
and 𝑦𝑦 + 3.
If 𝑦𝑦 is a root, then 𝑥𝑥 − 𝑦𝑦 is a factor.
Similarly, if 𝑦𝑦 + 3 is a root then 𝑥𝑥 − 𝑦𝑦 − 3 is a
factor.
We can multiply these two factors together and
let them equal the function given.
Now matching up the 𝑥𝑥’s,
Adding 3𝑥𝑥 to both sides.
𝑦𝑦 = −4
Dividing across by −2𝑥𝑥 gives us the value of 𝑦𝑦.
𝑝𝑝 = 𝑦𝑦 2 + 3𝑦𝑦
Now matching up the constants.
𝑝𝑝 = (−4)2 + 3(−4)
𝒑𝒑 = 𝟒𝟒
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Subbing in the value we found for 𝑦𝑦 and
multiplying out gives us the value of 𝑝𝑝.
45
iii)
𝒑𝒑 = 𝟕𝟕, 𝒑𝒑 = 𝟖𝟖
𝑓𝑓(𝑥𝑥) = 𝑥𝑥 2 + 5𝑥𝑥 + 𝑝𝑝
𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎 < 0
𝑎𝑎 = 1, 𝑏𝑏 = 5, 𝑐𝑐 = 𝑝𝑝
(5)2 − 4(1)(𝑝𝑝) < 0
25 − 4𝑝𝑝 < 0
25 < 4𝑝𝑝
6.25 < 𝑝𝑝
𝒑𝒑 = 𝟕𝟕, 𝒑𝒑 = 𝟖𝟖
© Pocket Tutor 2022
If the graph of a function does not cross the 𝑥𝑥 − axis we say
that its roots are not real. If the roots of a quadratic are not
real, then the discriminate, the ′𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎′ is less than 0.
Labelling the parts of the equation as quadratics are written
in the form 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐.
Subbing our values in.
Adding 4𝑝𝑝 to both sides.
Dividing across by 4.
So, 𝑝𝑝 is greater than 6.25. However, at the start of the
question it tells us that 𝑝𝑝 is an Integer and that it is not
greater than 8. So, we are left with 𝑝𝑝 = 7, 𝑝𝑝 = 8.
46
b)
−𝟑𝟑 ≤ 𝒙𝒙 ≤ −𝟐𝟐
|2𝑥𝑥 + 5| − 1 ≤ 0
|2𝑥𝑥 + 5| ≤ 1
(2𝑥𝑥 + 5)2 ≤ (1)2
(2𝑥𝑥 + 5)(2𝑥𝑥 + 5) ≤ 1
Adding 1 to both sides.
To get rid of the modulus
bars we square both sides.
Squaring out the brackets.
4𝑥𝑥 2 + 20𝑥𝑥 + 25 ≤ 1
Taking 1 from both sides.
4𝑥𝑥 2 + 20𝑥𝑥 + 24 ≤ 0
Factorising the quadratic.
𝑥𝑥 2 + 5 + 6 ≤ 0
(𝑥𝑥 + 2)(𝑥𝑥 + 3) ≤ 0
Let = 0
(𝑥𝑥 + 2)(𝑥𝑥 + 3) = 0
𝑥𝑥 = −2,
𝑥𝑥 = −3
−𝟑𝟑 ≤ 𝒙𝒙 ≤ −𝟐𝟐
© Pocket Tutor 2022
Letting the quadratic equal
0 and solving for 𝑥𝑥.
By sketching the graph, we
can see that the |2𝑥𝑥 + 5| −
1 is less than (below) 0
when between −3 and −2.
47
Question 2
a)
𝑖𝑖𝑧𝑧1 = −4 + 3𝑖𝑖
𝑖𝑖(𝑖𝑖𝑧𝑧1 ) = 𝑖𝑖(−4 + 3𝑖𝑖)
−𝑧𝑧1 = −4𝑖𝑖 − 3
𝒛𝒛𝟏𝟏 = 𝟑𝟑 + 𝟒𝟒𝟒𝟒
3𝑧𝑧1 − 𝑧𝑧2 = 11 + 17𝑖𝑖
3(3 + 4𝑖𝑖) − 𝑧𝑧2 = 11 + 17𝑖𝑖
9 + 12𝑖𝑖 − 𝑧𝑧2 = 11 + 17𝑖𝑖
9 + 12𝑖𝑖 − 11 − 17𝑖𝑖 = 𝑧𝑧2
−𝟐𝟐 − 𝟓𝟓𝟓𝟓 = 𝒛𝒛𝟐𝟐
© Pocket Tutor 2022
To solve for 𝑧𝑧1 we multiply across by 𝑖𝑖 to get 𝑧𝑧1 by
itself as 𝑖𝑖 2 = −1.
Multiplying across by −1 and writing in the form of
𝑎𝑎 + 𝑏𝑏𝑏𝑏.
Now subbing the value we found for 𝑧𝑧1 into the
second equation.
Multiplying out.
Adding 𝑧𝑧2 to both sides and taking 11 + 17𝑖𝑖 from
both sides.
Subtracting real from real and imaginary from
imaginary numbers.
48
b) i)
𝒓𝒓 = 𝟏𝟏 − 𝒊𝒊
3 + 2𝑖𝑖, 5 − 𝑖𝑖
𝑟𝑟 =
𝑟𝑟 =
𝑇𝑇2
𝑇𝑇1
5 − 𝑖𝑖
3 + 2𝑖𝑖
5 − 𝑖𝑖 3 − 2𝑖𝑖
×
3 + 2𝑖𝑖 3 − 2𝑖𝑖
𝑟𝑟 =
15 − 3𝑖𝑖 − 10𝑖𝑖 + 2𝑖𝑖 2
9 + 6𝑖𝑖 − 6𝑖𝑖 − 4𝑖𝑖 2
𝑟𝑟 =
15 − 13𝑖𝑖 − 2
9+4
𝑟𝑟 =
13 − 13𝑖𝑖
13
𝒓𝒓 = 𝟏𝟏 − 𝒊𝒊
© Pocket Tutor 2022
The common ratio, 𝑟𝑟, is found by dividing the
second term by the first term.
To divide complex numbers, we multiply the top
and the bottom by the conjugate of the bottom.
To find the conjugate we change the sign before
the imaginary part of the complex number.
Multiplying the top by the top and the bottom by
the bottom.
Remember 𝑖𝑖 2 = −1.
Tidying up.
Dividing by the 13.
49
ii)
𝑻𝑻𝟗𝟗 = 𝟒𝟒𝟒𝟒 + 𝟑𝟑𝟑𝟑𝟑𝟑
𝑇𝑇9 = 𝑎𝑎𝑟𝑟 8
𝑎𝑎 = (3 + 2𝑖𝑖)
𝑇𝑇𝑛𝑛 = 𝑎𝑎𝑟𝑟 𝑛𝑛−1 , so 𝑇𝑇9 = 𝑎𝑎𝑟𝑟 9−1 (page 22 of the
Maths Tables Book).
𝑇𝑇9 = (3 + 2𝑖𝑖)(1 − 𝑖𝑖)8
Subbing them in.
𝑟𝑟 = (1 − 𝑖𝑖)
1 − 𝑖𝑖 → Polar form
𝑟𝑟 = �(1)2 + (−1)2
𝑟𝑟 = √2
𝜃𝜃 = tan−1
1 𝜋𝜋
=
1 4
(3 + 2𝑖𝑖)𝑟𝑟(cos 𝜃𝜃 + 𝑖𝑖 sin 𝜃𝜃)𝑛𝑛
7𝜋𝜋
7𝜋𝜋 8
(3 + 2𝑖𝑖) �√2 �cos
+ 𝑖𝑖 sin � �
4
4
8
To use De Moivre’s theorem we need to
write (1 − 𝑖𝑖) in polar form.
First, we get the modulus using the formula:
𝑟𝑟 = √𝑎𝑎2 + 𝑏𝑏 2
Next, we get the argument using the
π 7𝜋𝜋
Fourth quadrant → 2π − =
4
4
(3 + 2𝑖𝑖) �√2 �cos
𝑎𝑎 = the first term. 𝑟𝑟 = the common
difference we found in b) i).
7𝜋𝜋
7𝜋𝜋
(8) + 𝑖𝑖 sin
(8)��
4
4
(3 + 2𝑖𝑖)�16(1 + 0)�
𝑻𝑻𝟗𝟗 = 𝟒𝟒𝟒𝟒 + 𝟑𝟑𝟑𝟑𝟑𝟑
© Pocket Tutor 2022
𝑏𝑏
equation 𝜃𝜃 = tan−1 . As (1 − 𝑖𝑖) is in the
𝑎𝑎
fourth quadrant we ignore the sign and then
take the angle we get away from 2𝜋𝜋.
Now we sub the modulus and the argument
into 𝑟𝑟(cos 𝜃𝜃 + 𝑖𝑖 sin 𝜃𝜃)𝑛𝑛 .
Now using de Moivre’s theorem, multiplying
the argument by the power and putting the
modulus to the power.
Plugging into the calculator.
Multiplying out.
50
Question 3
a)
𝒇𝒇�𝒈𝒈(𝒙𝒙)� = 𝒙𝒙, 𝒈𝒈�𝒇𝒇(𝒙𝒙)� = 𝒙𝒙
𝑓𝑓(𝑥𝑥) = 6𝑥𝑥 − 5
𝑥𝑥 + 5
𝑔𝑔(𝑥𝑥) =
6
𝑓𝑓�𝑔𝑔(𝑥𝑥)� = 𝑓𝑓(
𝑥𝑥 + 5
)
6
𝑥𝑥 + 5
𝑥𝑥 + 5
� = 6�
�−5
𝑓𝑓 �
6
6
𝑓𝑓(𝑔𝑔(𝑥𝑥)) = 𝑥𝑥 + 5 − 5
𝒇𝒇�𝒈𝒈(𝒙𝒙)� = 𝒙𝒙
𝑔𝑔�𝑓𝑓(𝑥𝑥)� = 𝑔𝑔(6𝑥𝑥 − 5)
𝑔𝑔(6𝑥𝑥 − 5) =
𝑔𝑔�𝑓𝑓(𝑥𝑥)� =
(6𝑥𝑥 − 5) + 5
6
6𝑥𝑥
6
𝒈𝒈�𝒇𝒇(𝒙𝒙)� = 𝒙𝒙
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To find 𝑓𝑓�𝑔𝑔(𝑥𝑥)�, we sub the function 𝑔𝑔(𝑥𝑥) into the
function 𝑓𝑓(𝑥𝑥).
So, subbing
6𝑥𝑥 − 5.
𝑥𝑥+5
6
in for 𝑥𝑥 in the expression 𝑓𝑓(𝑥𝑥) =
Multiplying out the bracket and subtracting 5.
Similarly, to find 𝑔𝑔�𝑓𝑓(𝑥𝑥)� we sub the function 𝑓𝑓(𝑥𝑥)
into the function 𝑔𝑔(𝑥𝑥).
So, subbing (6𝑥𝑥 − 5) in for 𝑥𝑥 in
𝑥𝑥+5
6
.
Dividing by the 6.
51
b) i)
𝒂𝒂 = 𝟏𝟏, 𝒃𝒃 = 𝟐𝟐
𝑦𝑦 = 5𝑥𝑥 2
log 5 𝑦𝑦 = log 5 5𝑥𝑥 2
log 5 𝑦𝑦 = log 5 5 + log 5 𝑥𝑥 2
log 5 𝑦𝑦 = 1 + log 5 𝑥𝑥 2
log 5 𝑦𝑦 = 1 + 2 log 5 𝑥𝑥
𝒂𝒂 = 𝟏𝟏, 𝒃𝒃 = 𝟐𝟐
© Pocket Tutor 2022
The laws of logarithms on page 21 of the Maths tables book
will help us to answer this question.
Putting both sides to base log base 5 as they have in the
question.
log 𝑎𝑎 (𝑥𝑥𝑥𝑥) = log 𝑎𝑎 𝑥𝑥 + log 𝑎𝑎 𝑦𝑦
log 5 5 = 1
log 𝑎𝑎 𝑥𝑥 𝑞𝑞 = 𝑞𝑞𝑞𝑞𝑞𝑞𝑔𝑔𝑎𝑎 𝑥𝑥, so, we can take the power of 2 down and
put it in front.
In the question it asks us to write it in the form log 5 𝑦𝑦 = 𝑎𝑎 +
𝑏𝑏 log 5 𝑥𝑥, so 𝑎𝑎 = 1, 𝑏𝑏 = 2.
52
ii)
𝒚𝒚 =
𝟏𝟏
, 𝒚𝒚 = 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑
𝟓𝟓
126𝑥𝑥
− 1�
log 5 𝑦𝑦 = 2 + log 5 �
25
log 5 𝑦𝑦 = 1 + 2 log 5 𝑥𝑥
1 + 2 log 5 𝑥𝑥 = 2 + log 5 �
1 + log 5 𝑥𝑥 2 = 2 + log 5 �
log 5 𝑥𝑥 2 − log 5 �
log 5 �
�
126𝑥𝑥
− 1�
25
126𝑥𝑥
− 1� = 2 − 1
25
𝑥𝑥 2
�=1
126𝑥𝑥
−1
25
2
126𝑥𝑥
− 1�
25
1
𝑥𝑥
� =5
126𝑥𝑥
−1
25
𝑥𝑥 2
=5
126𝑥𝑥
−1
25
© Pocket Tutor 2022
We know from the previous part that, log 5 𝑦𝑦 = 1 +
2 log 5 𝑥𝑥. So, we can let 1 + 2 log 5 𝑥𝑥 equal 2 + log 5 �
1�.
126𝑥𝑥
25
−
According to page 21 of the Maths tables book:
log 𝑎𝑎 𝑥𝑥 𝑞𝑞 = 𝑞𝑞𝑞𝑞𝑞𝑞𝑔𝑔𝑎𝑎 𝑥𝑥, so we can write the two as a power of
the 𝑥𝑥.
Now putting the logs on one side and the numbers on the
other by subtracting.
According to the laws of logs listed on page 21 of the Maths
𝑥𝑥
Tables Book, log 𝑎𝑎 𝑥𝑥 − log 𝑎𝑎 𝑦𝑦 = log 𝑎𝑎 . So, rewriting the
line accordingly.
𝑦𝑦
Although this looks complicated, if we follow the basic rule
of logs: 𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥, we can rewrite the bracket
to the power of 1 and let it equal the base of the log.
53
𝑥𝑥 2 = 5 �
𝑥𝑥 2 =
126𝑥𝑥
− 1�
25
126𝑥𝑥
−5
5
Multiplying across by the bottom of the fraction.
5𝑥𝑥 2 = 126𝑥𝑥 − 25
Multiplying across by 5.
5𝑥𝑥 2 − 126𝑥𝑥 + 25 = 0
Taking 126𝑥𝑥 from both sides and adding 25 to both sides.
(5𝑥𝑥 − 1)(𝑥𝑥 − 25) = 0
Factorising the quadratic and solving for 𝑥𝑥.
𝑦𝑦 = 5𝑥𝑥 2
Now we can find the two values of 𝑦𝑦, by subbing these 𝑥𝑥
values into the equation given for 𝑦𝑦 at the start of 𝑏𝑏 𝑖𝑖).
𝑥𝑥 =
1
, 𝑥𝑥 = 25
5
1 2 𝟏𝟏
𝑦𝑦 = 5 � � =
𝟓𝟓
5
𝑦𝑦 = 5(25)2 = 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑
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54
Question 4
a)
𝑓𝑓(𝑥𝑥) = 𝑥𝑥 3 + 𝑘𝑘𝑥𝑥 2 + 15𝑥𝑥 + 8
𝑓𝑓 ′ (𝑥𝑥) = 3𝑥𝑥 2 + 2𝑘𝑘𝑘𝑘 + 15
𝑓𝑓′(3) = 3(3)2 + 2𝑘𝑘(3) + 15 = −12
27 + 6𝑘𝑘 + 15 = −12
6𝑘𝑘 + 42 = −12
6𝑘𝑘 = −54
𝑘𝑘 = −9
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Firstly, we need to differentiate the function
to find 𝑓𝑓′(𝑥𝑥).
Differentiating by rule.
Now to find 𝑘𝑘 we sub in 3 for 𝑥𝑥 and let the
equation equal −12.
Multiplying out.
Taking 42 from both sides.
Dividing across by 6.
55
b)
𝟖𝟖𝟖𝟖 + 𝒚𝒚 − 𝟐𝟐𝟐𝟐 = 𝟎𝟎
Turning points → 𝑓𝑓 ′ (𝑥𝑥) = 0
We can find the equation of a line once we have two
points on the line. We can find two points on the line
𝑔𝑔(𝑥𝑥) by finding the turning points of the curve.
3𝑥𝑥 2 − 18𝑥𝑥 + 15 = 0
Taking the derivative we found in the previous part
and subbing in the value for 𝑘𝑘 (−9).
𝑓𝑓 ′ (𝑥𝑥) = 3𝑥𝑥 2 − 18𝑥𝑥 + 15
𝑥𝑥 2 − 6𝑥𝑥 + 5 = 0
(𝑥𝑥 − 5)(𝑥𝑥 − 1) = 0
𝑥𝑥 = 5, 𝑥𝑥 = 1
𝑓𝑓(𝑥𝑥) = 𝑥𝑥 3 − 9𝑥𝑥 2 + 15𝑥𝑥 + 8
(5)3 − 9(5)2 + 15(5) + 8 = −17
(5, −17)
(1)3 − 9(1)2 + 15(1) + 8 = 15
(1,15)
15 − (−17)
= −8
𝑚𝑚 =
1−5
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
𝑦𝑦 − (−17) = −8(𝑥𝑥 − 5)
To find the turning points of a curve we let the
derivative equal 0 and solve for 𝑥𝑥.
Letting the derivative equal 0 and dividing across by 3
to simplify.
Factorising and solving for 𝑥𝑥.
Now to find the 𝑦𝑦 coordinate of each of these turning
points we sub the 𝑥𝑥 values we found into the equation
of the curve.
Taking the equation of the curve and subbing in the
value of 𝑘𝑘.
Subbing in 5 for 𝑥𝑥 to get the corresponding 𝑦𝑦
coordinate.
Now doing the same with 1.
Now that we have two points on the line, we need to
𝑦𝑦 −𝑦𝑦
find its slope using the equation; 𝑚𝑚 = 2 1 .
𝑥𝑥2 −𝑥𝑥1
Now subbing its slope and one of the points into the
equation of a line from page 18 of the Maths Tables
Book.
𝑦𝑦 + 17 = −8𝑥𝑥 + 40
𝒈𝒈(𝒙𝒙) = 𝟖𝟖𝟖𝟖 + 𝒚𝒚 − 𝟐𝟐𝟐𝟐 = 𝟎𝟎
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56
c)
Point of inflection 𝑓𝑓 ′′ (𝑥𝑥) = 0
𝑓𝑓 ′ (𝑥𝑥) = 3𝑥𝑥 2 − 18𝑥𝑥 + 15
To show that 𝑔𝑔(𝑥𝑥) contains the point of
inflection we first, need to find the point of
inflection.
6𝑥𝑥 − 18 = 0
To find the point of inflection we get the
second derivative of 𝑓𝑓(𝑥𝑥), let it equal 0 and
solve for 𝑥𝑥.
𝑥𝑥 = 3
This is the 𝑥𝑥 coordinate of the point of
inflection.
𝑓𝑓 ′′ (𝑥𝑥) = 6𝑥𝑥 − 18
6𝑥𝑥 = 18
𝑓𝑓(𝑥𝑥) = 𝑥𝑥 3 − 9𝑥𝑥 2 + 15𝑥𝑥 + 8
𝑓𝑓(3) = (3)3 − 9(3)2 + 15(3) + 8 = −1
(3, −1)
𝑔𝑔(𝑥𝑥) = 8𝑥𝑥 + 𝑦𝑦 − 23 = 0
8(3) + (−1) − 23 = 0
0=0
∴ 𝑔𝑔(𝑥𝑥) contains the point of inflection.
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To find the 𝑦𝑦 coordinate we sub this value
back into the original equation 𝑓𝑓(𝑥𝑥).
Now that we have the coordinates of the
point of inflection, we can verify that 𝑔𝑔(𝑥𝑥)
passes through it by subbing the 𝑥𝑥 and 𝑦𝑦
coordinates into the equation for 𝑔𝑔(𝑥𝑥) and
showing that it equals 0.
57
Question 5
a)
€𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎
𝐴𝐴 = 𝑃𝑃
𝑖𝑖(1 + 𝑖𝑖)𝑡𝑡
(1 + 𝑖𝑖)𝑡𝑡 − 1
𝑃𝑃 = 250,000,
𝑖𝑖 = 0.00287,
𝑡𝑡 = 25 × 12 = 300
𝐴𝐴 = 250000 �
(0.00287)(1 + 0.00287)300
�
(1 + 0.00287)300 − 1
Taking the amortisation formula from page
31 of the Maths Tables Book.
𝑃𝑃 = The size of the mortgage
𝑖𝑖 = the interest rate in decimal form
𝑡𝑡 = the time in months.
Subbing in our values and plugging it into the
calculator gives us the monthly repayment.
𝐴𝐴 = €𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏. 𝟎𝟎𝟎𝟎
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58
b)
€𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐. 𝟐𝟐𝟐𝟐
Series = Sum of the present values of the remaining 14 years
𝑡𝑡 = 12 × 14 = 168
Present value =
𝐹𝐹
(1 + 𝑖𝑖)𝑡𝑡
𝐹𝐹 = 1771, 𝑖𝑖 = 0.003
1771
1771
1771
1771
+
+ ⋯+
+
2
167
1.003 1.003
1.003
1.003168
𝑎𝑎(1 − 𝑟𝑟 𝑛𝑛 )
𝑆𝑆𝑛𝑛 =
1 − 𝑟𝑟
𝑎𝑎 =
𝑟𝑟 =
1771
1.003
1771
1
1771
÷
=
2
1.003 1.003
1.003
𝑛𝑛 = 168
𝑆𝑆168
1 168
1771
� �
�1 − �
1.003
1.003
=
1
1−
1.003
𝑆𝑆168 = €𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐. 𝟐𝟐𝟐𝟐
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The amount left to be repaid is equal to the
sum of the present value of each of the
remaining repayments.
The present value formula is on page 30 of
the Maths Tables Book.
The future value is the value of the
repayment and the interest is 0.3% in
decimal form.
Writing out the sum of the present values
from the next repayment to the final one in
168 months (12 × 14 years).
Now taking the equation for the sum of a
series from page 22 of the Maths Tables
Book. 𝑎𝑎 = the first term, 𝑟𝑟 = the second
term divided by the first term.
𝑛𝑛 = the number of months.
Subbing these values into the formula and
plugging it into the calculator gives us the
sum to be repaid.
59
Question 6
a)
𝟏𝟏𝟏𝟏𝟏𝟏 + 𝟐𝟐
To differentiate this from first
principles we first need to
multiply the brackets together.
(3𝑥𝑥 − 5)(2𝑥𝑥 + 4)
6𝑥𝑥 2 + 2𝑥𝑥 − 20
1. Sub in 𝑥𝑥 + ℎ for 𝑥𝑥 in the
equation.
𝑓𝑓(𝑥𝑥 + ℎ) = 6(𝑥𝑥 + ℎ)2 + 2(𝑥𝑥 + ℎ) − 20
6(𝑥𝑥 2 + 2𝑥𝑥ℎ + ℎ2 ) + 2𝑥𝑥 + 2ℎ − 20
2
Multiplying out the brackets.
2
6𝑥𝑥 + 12𝑥𝑥ℎ + 6ℎ + 2𝑥𝑥 + 2ℎ − 20
𝑓𝑓(𝑥𝑥 + ℎ) − 𝑓𝑓(𝑥𝑥) = 6𝑥𝑥 2 + 12𝑥𝑥ℎ + 6ℎ2 + 2𝑥𝑥 + 2ℎ − 20 − (6𝑥𝑥 2 + 2𝑥𝑥 − 20)
= 12𝑥𝑥ℎ + 6ℎ2 + 2ℎ
2. Take the original equation for
𝑓𝑓(𝑥𝑥) away from the previous
line.
𝑓𝑓(𝑥𝑥 + ℎ) − 𝑓𝑓(𝑥𝑥)
= 12𝑥𝑥 + 6ℎ + 2
ℎ
3. Divide the result by ℎ.
𝑓𝑓(𝑥𝑥 + ℎ) − 𝑓𝑓(𝑥𝑥)
= 12𝑥𝑥 + 2
ℎ→0
ℎ
4. Limit any remaining ℎ’s to 0.
lim
𝑓𝑓 ′ (𝑥𝑥) = 𝟏𝟏𝟏𝟏𝟏𝟏 + 𝟐𝟐
b) i)
ℎ(𝑥𝑥) =
1
ln(2𝑥𝑥 + 3) + 𝐶𝐶
2
1
1
�
�×2
2 2𝑥𝑥 + 3
𝒉𝒉
′ (𝒙𝒙)
𝟏𝟏
=
𝟐𝟐𝟐𝟐 + 𝟑𝟑
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Page 25 of the Maths Tables Book helps us to differentiate
1
log functions. �ln 𝑥𝑥 = �.
𝑥𝑥
We then need to multiply by the derivative of the bracket.
60
ii)
𝑨𝑨 = 𝟏𝟏𝟏𝟏
�
𝐴𝐴
0
1
𝑑𝑑𝑑𝑑 = ln 3
2𝑥𝑥 + 3
𝐴𝐴
1
� ln(2𝑥𝑥 + 3)� = ln 3
2
0
1
1
� ln(2(𝐴𝐴) + 3) − ln(2(0) + 3)� = ln 3
2
2
1
1
� ln(2𝐴𝐴 + 3) − ln 3� = ln 3
2
2
1 2𝐴𝐴 + 3
ln
= ln 3
3
2
1
2𝐴𝐴 + 3 2
� = ln 3
ln �
3
1
2
2𝐴𝐴 + 3
� =3
�
3
2𝐴𝐴 + 3
=9
3
2𝐴𝐴 + 3 = 27
We can find the area under a curve by integrating the
equation of the curve between certain limits. In this
case we know the area of the curve but want to find
one of the limits, so we can integrate the curve and
let it equal the given area.
As we differentiated the equation in the previous
part, to integrate it we just go back to the original
equation given, as integration is just
antidifferentiation, (we can leave out the plus C when
calculating area).
Now subbing in the upper and lower limits. Taking
the version with the lower limit away from the
version with the upper limit.
According to the laws of logs on page 21 of the Maths
𝑥𝑥
Tables Book, log 𝑎𝑎 𝑥𝑥 − log 𝑎𝑎 𝑦𝑦 = log 𝑎𝑎 .
𝑦𝑦
So, we can rewrite this line accordingly.
Page 21 also states that 𝑞𝑞 log 𝑎𝑎 𝑥𝑥 = log 𝑎𝑎 𝑥𝑥 𝑞𝑞 . So, we
1
can rewrite the as a power.
2
Cancelling out the logs
Squaring both sides.
Multiplying across by 3.
Taking 3 from both sides then dividing across by 2.
2𝐴𝐴 = 24
𝑨𝑨 = 𝟏𝟏𝟏𝟏
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61
Question 7
a) i)
Term
𝑻𝑻𝟏𝟏
Triangular
1
𝑻𝑻𝟐𝟐
3
𝑻𝑻𝟑𝟑
6
𝑻𝑻𝟒𝟒
10
𝑻𝑻𝟓𝟓
15
𝑻𝑻𝟔𝟔
21
𝑻𝑻𝟕𝟕
28
𝑻𝑻𝟖𝟖
36
Number
ii)
𝑇𝑇𝑛𝑛 =
𝑛𝑛(𝑛𝑛 + 1)
2
𝑛𝑛(𝑛𝑛 + 1)
1275 =
2
To find out if 1275 is a triangular number we
let it equal the equation for the nth triangular
number and solve for 𝑛𝑛.
1275(2) = 𝑛𝑛(𝑛𝑛 + 1)
Multiplying across by 2.
𝑛𝑛2 + 𝑛𝑛 − 2250 = 0
Taking 2550 away from both sides.
2550 = 𝑛𝑛2 + 𝑛𝑛
(𝑛𝑛 − 50)(𝑛𝑛 + 51) = 0
𝑛𝑛 = 50
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 is the 50 triangular number
th
Multiplying out the bracket.
Factorising the quadratic.
Solving for the positive value of 𝑛𝑛. As 𝑛𝑛 is a
whole number 1275 is a triangular number.
b) i)
(𝒏𝒏 + 𝟐𝟐)(𝒏𝒏 + 𝟏𝟏)
𝟐𝟐
𝑛𝑛(𝑛𝑛 + 1)
+ (𝑛𝑛 + 1)
2
𝑛𝑛(𝑛𝑛 + 1) 2(𝑛𝑛 + 1)
+
2
2
𝑛𝑛(𝑛𝑛 + 1) + 2(𝑛𝑛 + 1)
2
(𝒏𝒏 + 𝟐𝟐)(𝒏𝒏 + 𝟏𝟏)
𝟐𝟐
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To write it as a single fraction we multiply (𝑛𝑛 + 1) by 2
and put it over 2.
Putting the fractions together.
Factorising out the (𝑛𝑛 + 1)s.
62
ii)
𝑛𝑛(𝑛𝑛 + 1) (𝑛𝑛 + 2)(𝑛𝑛 + 1)
+
2
2
To prove that the sum will always be a square
number we add the expressions for 𝑇𝑇𝑛𝑛 and 𝑇𝑇𝑛𝑛+1 .
(𝑛𝑛 + (𝑛𝑛 + 2))(𝑛𝑛 + 1)
2
Factorising out the (𝑛𝑛 + 1)s.
𝑛𝑛(𝑛𝑛 + 1) + (𝑛𝑛 + 2)(𝑛𝑛 + 1)
2
(2𝑛𝑛 + 2)(𝑛𝑛 + 1)
2
2(𝑛𝑛 + 1)(𝑛𝑛 + 1)
2
(𝑛𝑛 + 1)(𝑛𝑛 + 1)
(𝑛𝑛 + 1)
2
Adding the 𝑛𝑛 and the (𝑛𝑛 + 2).
Factorising out the 2.
Dividing by the 2.
As the bracket is squared the sum will always be
a square number.
iii)
𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔
(𝑛𝑛 + 1)2 = 12544
𝑛𝑛 + 1 = √12544
𝑛𝑛 + 1 = 112
𝑛𝑛 = 111
𝑇𝑇𝑛𝑛 =
𝑛𝑛(𝑛𝑛 + 1)
2
𝑇𝑇111 =
111(111 + 1)
= 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔
2
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Taking the expression we just found for the sum of
two consecutive triangular numbers and letting
this equal the value given.
Square rooting both sides.
Taking 1 away from both sides.
This is the value of 𝑛𝑛.
Now subbing this into the equation for the nth
triangular number to find the triangular number.
63
c)
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝑁𝑁𝑘𝑘 = �
𝑘𝑘 = 3
𝑁𝑁3 = �
𝑘𝑘
�3 + 2√2� − �3 − 2√2�
4√2
3
�3 + 2√2� − �3 − 2√2�
= 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
4√2
𝑘𝑘
�
2
3 2
�
To find 𝑁𝑁3 we sub 3 in for 𝑘𝑘 and plug the
expression into the calculator.
d)
To prove:
12 + 22 + 32 + 42 + ⋯ + 𝑛𝑛2 =
𝑛𝑛(𝑛𝑛 + 1)(2𝑛𝑛 + 1)
6
Show that it is true for 𝑛𝑛 = 1:
1=
(1)(1 + 1)(2(1) + 1)
6
Subbing 1 in for 𝑛𝑛.
1=1
Assume that it is true for 𝑛𝑛 = 𝑘𝑘
1 + 4 + 9 + ⋯ + 𝑘𝑘 2 +=
𝑘𝑘(𝑘𝑘 + 1)(2𝑘𝑘 + 1)
6
Show that it is true for 𝑛𝑛 = 𝑘𝑘 + 1
(𝑘𝑘 + 1)�(𝑘𝑘 + 1) + 1�(2(𝑘𝑘 + 1) + 1)
1 + 4 + 9 + ⋯ + 𝑘𝑘 2 + (𝑘𝑘 + 1)2 =
6
1 + 4 + 9 + ⋯ + 𝑘𝑘 2 + (𝑘𝑘 + 1)2 =
(𝑘𝑘 + 1)(𝑘𝑘 + 2)(2𝑘𝑘 + 3)
6
(𝑘𝑘 + 1)(𝑘𝑘 + 2)(2𝑘𝑘 + 3)
𝑘𝑘(𝑘𝑘 + 1)(2𝑘𝑘 + 1)
+ (𝑘𝑘 + 1)2 =
6
6
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Subbing 𝑘𝑘 in for 𝑛𝑛.
Subbing in 𝑘𝑘 + 1 for 𝑛𝑛.
Subbing in our assumption for
𝑃𝑃(𝑘𝑘) on the left hand side.
64
𝑘𝑘(𝑘𝑘 + 1)(2𝑘𝑘 + 1) + 6(𝑘𝑘 + 1)2
6
Now focusing on the left hand side.
Expressing it as a single fraction by
multiplying (𝑘𝑘 + 1)2 by 6.
(𝑘𝑘 + 1)[2𝑘𝑘 2 + 𝑘𝑘 + 6𝑘𝑘 + 6]
6
Multiplying out the brackets on the right.
(𝑘𝑘 + 1)[(𝑘𝑘 + 2)(2𝑘𝑘 + 3)]
6
Factorising the quadratic.
(𝑘𝑘 + 1)[𝑘𝑘(2𝑘𝑘 + 1) + 6(𝑘𝑘 + 1)]
6
Factorising out (𝑘𝑘 + 1).
(𝑘𝑘 + 1)[2𝑘𝑘 2 + 7𝑘𝑘 + 6]
6
(𝑘𝑘 + 1)(𝑘𝑘 + 2)(2𝑘𝑘 + 3) (𝑘𝑘 + 1)(𝑘𝑘 + 2)(2𝑘𝑘 + 3)
=
6
6
Writing in the right hand side again.
The sides equal. Therefore, it is proven.
Thus, the proposition is true for 𝑛𝑛 = 𝑘𝑘 + 1 provided it is true for 𝑛𝑛 = 𝑘𝑘 but it is true for 𝑛𝑛 = 1 and
therefore it is true for all positive integers.
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65
Question 8
a) i)
𝒂𝒂 = 𝟓𝟓, 𝒃𝒃 = 𝟓𝟓
cos 𝜃𝜃 =
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 =
𝑎𝑎𝑎𝑎𝑎𝑎
ℎ𝑦𝑦𝑦𝑦
𝑥𝑥
5
5 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑥𝑥
𝑜𝑜𝑜𝑜𝑜𝑜
sin 𝜃𝜃 =
ℎ𝑦𝑦𝑦𝑦
sin 𝜃𝜃 =
𝑦𝑦
5
5 sin 𝜃𝜃 = 𝑦𝑦
(𝑥𝑥, 𝑦𝑦) = (5 cos 𝜃𝜃 , 5 sin 𝜃𝜃)
The 𝑥𝑥 coordinate is equal to the distance along the
𝑥𝑥 axis that the point is from the centre. We can
write this distance in terms of cos 𝜃𝜃 as it is equal to
the adjacent side in the triangle in the diagram
shown in the question.
Subbing in the length of the adjacent side (𝑥𝑥) and
the given length of the hypotenuse.
Similarly, we can write the 𝑦𝑦 coordinate in terms of
sin 𝜃𝜃 as the height (𝑦𝑦) is equal to the opposite side
of the triangle shown.
So, subbing in the opposite side (𝑦𝑦) and the length
given for the hypotenuse.
Writing the two coordinates together.
𝒂𝒂 = 𝟓𝟓, 𝒃𝒃 = 𝟓𝟓
ii)
Area of rectangle in first quadrant × 4 = area of full rectangle
Area of rectangle in first quadrant = 5 cos 𝜃𝜃 × 5 sin 𝜃𝜃
= 25 cos 𝜃𝜃 sin 𝜃𝜃
25 cos 𝜃𝜃 sin 𝜃𝜃 × 4
= 100 cos 𝜃𝜃 sin 𝜃𝜃
50 × 2 cos 𝜃𝜃 sin 𝜃𝜃
50 × sin 2𝜃𝜃
50 sin 2𝜃𝜃
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We can find the area of the rectangle in
the first quadrant by multiplying the
lengths of sides.
Multiplying the result by 4 as this is a
quarter of the full rectangle.
The question asks for it in the form
50 sin 2𝜃𝜃, so rewriting it as 50 ×
2 cos 𝜃𝜃 sin 𝜃𝜃 , which can be written as
sin 2𝜃𝜃 as shown on page 14 of the Maths
Tables Book.
66
iii)
Area = 50 sin 2𝜃𝜃
𝑑𝑑𝑑𝑑
= 50 cos 2 𝜃𝜃 × 2
𝑑𝑑𝑑𝑑
To find the value of 𝜃𝜃 which gives the maximum area we
differentiate the equation for the area.
100 cos 2𝜃𝜃 = 0
Now letting the derivative equal 0 to find the value of 𝜃𝜃
which gives a maximum area.
𝑑𝑑𝑑𝑑
= 100 cos 2𝜃𝜃
𝑑𝑑𝑑𝑑
cos 2𝜃𝜃 = 0
2𝜃𝜃 = cos −1 0
𝜋𝜋
2
𝜋𝜋
𝜃𝜃 =
4
2𝜃𝜃 =
𝜋𝜋
= 5√2
4
𝜋𝜋
10 sin = 5√2
4
10 cos
∴ square.
When differentiated sin 𝑎𝑎𝑎𝑎 → acos 𝑎𝑎𝑎𝑎. (Switching sin to cos
and multiplying by the derivative of the angle).
Dividing across by 100.
Finding the 𝑐𝑐𝑐𝑐𝑐𝑐 inverse of both sides.
Dividing across by 2.
Now plugging this value into the equation for the length and
the width (2 times the coordinates we found in the first
part).
Both give the same value ∴ length and width are equal ∴ it’s
a square.
iv)
50 sin 2𝜃𝜃
𝜋𝜋
50 sin 2 � � = 𝟓𝟓𝟓𝟓 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮
4
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Plugging the value of 𝜃𝜃 we found for the
maximum area into the equation for the area.
67
b)
𝟏𝟏𝟏𝟏\𝒔𝒔𝒔𝒔c
Find = Given × Need
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
=
×
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝑥𝑥
2
=
5 𝑥𝑥 + 𝑙𝑙
2=
5𝑥𝑥
𝑥𝑥 + 𝑙𝑙
2(𝑥𝑥 + 𝑙𝑙) = 5𝑥𝑥
2𝑥𝑥 + 2𝑙𝑙 = 5𝑥𝑥
2𝑙𝑙 = 3𝑥𝑥
𝑥𝑥 =
2
𝑙𝑙
3
𝑑𝑑𝑑𝑑 2
=
𝑑𝑑𝑑𝑑 3
2
𝑑𝑑𝑑𝑑
= 1.5 ×
3
𝑑𝑑𝑑𝑑
To find the rate of change of the length of the person’s
shadow (𝑥𝑥) with time (𝑡𝑡) we can multiply the rate of
change of the distance (𝑙𝑙) with time (𝑡𝑡), which is given, by
the rate of change of the length of the shadow (𝑥𝑥) with
distance (𝑙𝑙), which we need.
The triangle with base 𝑥𝑥 and the larger triangle with base
(𝑥𝑥 + 𝑙𝑙) are similar as they are right angled and share the
2
same top angle. Therefore, we can say that is equal to
𝑥𝑥
𝑥𝑥+𝑙𝑙
5
, i.e. the left side over left side and base over base.
Multiplying across by the bottom of each fraction and
rearranging to get 𝑥𝑥 on one side by itself.
Now we have 𝑥𝑥 in terms of 𝑙𝑙, so we can differentiate it to
find
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
, the rate of change of the length of the shadow with
distance.
Multiplying the result by the change in distance with
respect to time gives us our answer.
𝑑𝑑𝑑𝑑
= 𝟏𝟏𝟏𝟏\𝒔𝒔𝒔𝒔𝒔𝒔
𝑑𝑑𝑑𝑑
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68
Question 9
a) i)
𝑁𝑁(𝑡𝑡) = 450𝑒𝑒 0.065𝑡𝑡
𝑡𝑡 = 4.5
𝑁𝑁(4.5) = 450𝑒𝑒 0.065(4.5)
= 602.89
To find the number of bacteria in the colony
after 4.5 hours, we sub in 4.5 for 𝑡𝑡 and plug it
into the calculator.
= 𝟔𝟔𝟔𝟔𝟔𝟔
ii)
𝟖𝟖. 𝟕𝟕 𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡
790 = 450𝑒𝑒 0.065𝑡𝑡
790
= 𝑒𝑒 0.065𝑡𝑡
450
𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥
log 𝑒𝑒
790
= 0.065𝑡𝑡
450
To find the time it takes for there to be 790
bacteria, we let the equation given equal 790
and solve for 𝑡𝑡.
Dividing across by 450.
Using the law of logs from page 21 of the Maths
Tables Book.
0.56279 = 0.065𝑡𝑡
0.56279
= 𝑡𝑡
0.065
8.66 = 𝑡𝑡
Dividing across by 0.065 and then rounding to
one decimal place.
𝒕𝒕 = 𝟖𝟖. 𝟕𝟕 𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡
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69
b)
𝟕𝟕𝟕𝟕𝟕𝟕
𝑁𝑁(𝑡𝑡) = 450𝑒𝑒 0.065𝑡𝑡
12
1
� 450𝑒𝑒 0.065𝑡𝑡 𝑑𝑑𝑑𝑑
12 − 3 3
1 450 0.065𝑡𝑡 12
�
�
𝑒𝑒
9 0.065
3
450 0.065(3)
1 450 0.065(12)
�
�
𝑒𝑒
−
𝑒𝑒
0.065
9 0.065
1
(6688.8)
9
To find the average number of bacteria we
integrate the equation using the following
expression: where 𝑏𝑏 and 𝑎𝑎 are the limits; 12 and 3:
𝑏𝑏
1
� 𝑓𝑓(𝑥𝑥) 𝑑𝑑𝑑𝑑
𝑏𝑏 − 𝑎𝑎 𝑎𝑎
Integrating the expression following page 26 of the
1
Maths Tables Book, 𝑒𝑒 𝑎𝑎𝑎𝑎 → 𝑒𝑒 𝑎𝑎𝑎𝑎 .
𝑎𝑎
Now subbing in the upper and lower limits and
subtracting them.
Plugging into the calculator.
= 𝟕𝟕𝟕𝟕𝟕𝟕
c)
𝑁𝑁(𝑡𝑡) = 450𝑒𝑒 0.065𝑡𝑡
𝑑𝑑𝑑𝑑
= (0.065)450𝑒𝑒 0.065𝑡𝑡
𝑑𝑑𝑑𝑑
(0.065)450𝑒𝑒 0.065(12) = 63.8
To find the rate of change when 𝑡𝑡 = 12, we differentiate
the equation and then sub in 12 for 𝑡𝑡.
Differentiating according to page 25 of the Maths Tables
Book, 𝑒𝑒 𝑎𝑎𝑎𝑎 → 𝑎𝑎𝑒𝑒 𝑎𝑎𝑎𝑎 .
Subbing in 12 for 𝑡𝑡 and plugging into the calculator.
At hour 12 the population is growing at a rate of 63.8 bacteria per hour.
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70
d)
𝟏𝟏𝟏𝟏 = 𝒌𝒌
𝑑𝑑𝑑𝑑
= (0.065)450𝑒𝑒 0.065𝑡𝑡
𝑑𝑑𝑑𝑑
90 = (0.065)450𝑒𝑒 0.065𝑘𝑘
90
= 𝑒𝑒 0.065𝑘𝑘
(0.065)450
40
= 𝑒𝑒 0.065𝑘𝑘
13
log 𝑒𝑒
40
= 0.065𝑘𝑘
13
1.1239 = 0.065𝑘𝑘
17.29 = 𝑘𝑘
𝟏𝟏𝟏𝟏 = 𝒌𝒌
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Letting the derivative we found in the last part equal the
rate given and solving for 𝑘𝑘.
Dividing across by whatever is in front of the 𝑒𝑒 0.065𝑘𝑘
Using the law of logs from page 21 of the Maths Tables
Book: 𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥.
Dividing across by 0.065.
We round up to the nearest whole number as the question
asks when will the rate be greater than 90 and 𝑘𝑘 ∈ 𝑁𝑁.
71
e)
𝟕𝟕 hours
𝑁𝑁(𝑡𝑡) = 450𝑒𝑒 0.065𝑡𝑡
𝑃𝑃(𝑡𝑡) = 220𝑒𝑒 0.17𝑡𝑡
450𝑒𝑒 0.065𝑡𝑡 = 220𝑒𝑒 0.17𝑡𝑡
To find when the two colonies have the same number
of bacteria, we let the two equations equal each other
and solve for 𝑡𝑡.
450 0.065𝑡𝑡
𝑒𝑒
= 𝑒𝑒 0.17𝑡𝑡
220
Dividing across by 220.
450
= 𝑒𝑒 0.17𝑡𝑡−0.065𝑡𝑡
220
Using the rule of indices on page 21 of the Maths
𝑒𝑒 0.17𝑡𝑡
450
= 0.065𝑡𝑡
220 𝑒𝑒
450
= 𝑒𝑒 0.105𝑡𝑡
220
log 𝑒𝑒
450
= 0.105𝑡𝑡
220
0.7156 = 0.105𝑡𝑡
𝑡𝑡 = 6.82
𝑡𝑡 = 𝟕𝟕 hours
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Dividing across by 𝑒𝑒 0.065𝑡𝑡
Tables Book:
𝑎𝑎𝑝𝑝
𝑎𝑎𝑞𝑞
= 𝑎𝑎𝑝𝑝−𝑞𝑞 .
Using the law of logs from page 21 of the Maths
Tables Book: 𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥.
Dividing across by 0.105 and rounding to the nearest
hour.
72
2019 Paper 1
Question 1
(a)
𝒑𝒑 = 𝟐𝟐
(2𝑥𝑥 + 1)(𝑥𝑥 2 + 𝑝𝑝𝑝𝑝 + 4)
Multiplying out the brackets
2𝑥𝑥 3 + 2𝑝𝑝𝑥𝑥 2 + 𝑥𝑥 2 + 8𝑥𝑥 + 𝑝𝑝𝑝𝑝 + 4
Rearranging
2(1 + 2𝑝𝑝) = 8 + 𝑝𝑝
Letting the coefficient (the number in
front) of 𝑥𝑥 equal two times the
coefficient of 𝑥𝑥 2
3
2
2
2𝑥𝑥 + 2𝑝𝑝𝑥𝑥 + 8𝑥𝑥 + 𝑥𝑥 + 𝑝𝑝𝑝𝑝 + 4
2 + 4𝑝𝑝 = 8 + 𝑝𝑝
3𝑝𝑝 = 6
𝒑𝒑 = 𝟐𝟐
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73
𝒃𝒃) 𝒙𝒙 = −𝟑𝟑, 𝒙𝒙 =
𝟑𝟑
𝟒𝟒
2
2
3
+ =
2𝑥𝑥 + 1 5 3𝑥𝑥 − 1
2
2(2𝑥𝑥 + 1)
3 + (2𝑥𝑥 + 1) =
5
3𝑥𝑥 − 1
3(3𝑥𝑥 − 1) +
9𝑥𝑥 − 3 +
4𝑥𝑥 + 2
(3𝑥𝑥 − 1) = 2(2𝑥𝑥 + 1)
5
12𝑥𝑥 2 + 2𝑥𝑥 − 2
= 4𝑥𝑥 + 2
5
We want to get rid of the fractions, so
we are going to multiply across by the
bottom of each one.
Multiplying across by (2𝑥𝑥 + 1)
Multiplying across by (3𝑥𝑥 − 1)
5(9𝑥𝑥 − 3) + 12𝑥𝑥 2 + 2𝑥𝑥 − 2 = 5(4𝑥𝑥 + 2)
Multiplying across by 5
45𝑥𝑥 − 15 + 12𝑥𝑥 2 + 2𝑥𝑥 − 2 = 20𝑥𝑥 + 10
Rearranging
12𝑥𝑥 2 + 27𝑥𝑥 − 27 = 0
4𝑥𝑥 2 + 9 − 9 = 0
(𝑥𝑥 + 3)(4𝑥𝑥 − 3)
𝒙𝒙 = −𝟑𝟑, 𝒙𝒙 =
𝟑𝟑
𝟒𝟒
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Dividing across by 3 to make it easier
to solve
Solving the quadratic
74
Question 2
a) i)
ii)
𝑓𝑓(𝑥𝑥) = 3𝑥𝑥
(1.9)
𝑓𝑓(1.9) = 3
= 8.064
𝑔𝑔(𝑥𝑥) = 4𝑥𝑥 + 1
𝑔𝑔(1.9) = 4(1.9) + 1 = 8.6
∴ 𝑓𝑓(𝑥𝑥) < 𝑔𝑔(𝑥𝑥) for 𝑥𝑥 = 1.9
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Plugging in 1.9 for 𝑥𝑥 in both
equations.
We can see that 𝑓𝑓(𝑥𝑥) is less than
𝑔𝑔(𝑥𝑥) when 𝑥𝑥 = 1.9
75
b)
To prove:
3𝑛𝑛 ≥ 4𝑛𝑛 + 1, when 𝑥𝑥 ≥ 2
1. Prove true for 𝑛𝑛 = 2
𝑛𝑛 = 2
→ 32 ≥ 4(2) + 1
1. Plugging in 2 for 𝑛𝑛 to show that the
statement is true for this value.
9 ≥ 9`
2. Assume true for 𝑛𝑛 = 𝑘𝑘
𝑛𝑛 = 𝑘𝑘
2. Plugging in 𝑘𝑘 for 𝑛𝑛 and assuming that this
is true.
→ 3𝑘𝑘 ≥ 4(𝑘𝑘) + 1
3𝑘𝑘 ≥ 4𝑘𝑘 + 1
3. Prove true for 𝑛𝑛 = 𝑘𝑘 +1
𝑛𝑛 = 𝑘𝑘 + 1
→ 3𝑘𝑘+1 ≥ 4(𝑘𝑘 + 1) + 1
Plugging in 𝑘𝑘 + 1 for 𝑛𝑛
3𝑘𝑘 . 31 ≥ 4𝑘𝑘 + 4 + 1
3𝑘𝑘 . 31 ≥ 4𝑘𝑘 + 5
3𝑘𝑘 ≥ 4𝑘𝑘 + 1
3(4𝑘𝑘 + 1) ≥ 4𝑘𝑘 + 5
12𝑘𝑘 + 3 ≥ 4𝑘𝑘 + 5 for 𝑘𝑘 ≥ 2
Now using our assumption for 3𝑘𝑘
3𝑘𝑘+1 ≥ 3(4𝑘𝑘 + 1)
As 3𝑘𝑘 ≥ 4𝑘𝑘 + 1
We know that 12𝑘𝑘 + 3 is greater than 4𝑘𝑘 +
5 as long as 𝑘𝑘 ≥ 2 as is stated in the
question. Thus, it is proven.
True for 𝑛𝑛 = 𝑘𝑘 + 1 provided it is true for 𝑛𝑛 = 𝑘𝑘 but it is true for 𝑛𝑛 = 2
∴ True for all 𝑛𝑛 ≥ 2, 𝑛𝑛 ∈ 𝑁𝑁
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76
Question 3
a)
3𝑥𝑥𝑥𝑥 − 9𝑥𝑥 + 4𝑦𝑦 − 12
3𝑥𝑥(𝑦𝑦 − 3) + 4(𝑦𝑦 − 3)
(𝟑𝟑𝟑𝟑 + 𝟒𝟒)(𝒚𝒚 − 𝟑𝟑)
Factorising using factors by
grouping.
b) 𝒙𝒙 = 𝒆𝒆𝟑𝟑
𝑔𝑔(𝑥𝑥) = 3𝑥𝑥 ln 𝑥𝑥 − 9𝑥𝑥 + 4 ln 𝑥𝑥 − 12
3𝑥𝑥 ln 𝑥𝑥 − 9𝑥𝑥 + 4 ln 𝑥𝑥 − 12 = 0
3𝑥𝑥(ln 𝑥𝑥 − 3) + 4(ln 𝑥𝑥 − 3) = 0
(3𝑥𝑥 + 4)(ln 𝑥𝑥 − 3) = 0
3𝑥𝑥 + 4 = 0
3𝑥𝑥 = −4
𝑥𝑥 = −
4
3
ln 𝑥𝑥 − 3 = 0
ln 𝑥𝑥 = 3
𝒙𝒙 = 𝒆𝒆𝟑𝟑
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Letting the given equation equal 0
Factorising the same way as in part a) except with ln 𝑥𝑥 in place
of 𝑦𝑦.
Letting each bracket equal 0
ln 𝑥𝑥 is the same as log 𝑒𝑒 𝑥𝑥.
Using the identity 𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥 from pg 21 of The
Maths Tables Book, we can rewrite it as 𝑥𝑥 = 𝑒𝑒 3
4
𝑥𝑥 cannot equal − as logs are not defined for negative numbers
3
77
c) −𝟏𝟏. 𝟓𝟓𝟓𝟓
𝑔𝑔(𝑥𝑥) = 3𝑥𝑥 ln 𝑥𝑥 − 9𝑥𝑥 + 4 ln 𝑥𝑥 − 12
1
1
𝑔𝑔′ (𝑥𝑥) = 3𝑥𝑥. � � + 3. ln 𝑥𝑥 − 9 + 4.
𝑥𝑥
𝑥𝑥
3 + 3 ln 𝑥𝑥 − 9 +
𝑔𝑔′ (𝑥𝑥) = 3 ln 𝑥𝑥 +
4
𝑥𝑥
4
−6
𝑥𝑥
𝑔𝑔′ (𝑒𝑒) = 3 ln(𝑒𝑒) +
𝑔𝑔′ (𝑒𝑒) = −𝟏𝟏. 𝟓𝟓𝟓𝟓
4
−6
𝑒𝑒
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To find 𝑔𝑔′(𝑥𝑥) we need to differentiate
the expression.
To differentiate 3𝑥𝑥. ln 𝑥𝑥 we use the
product rule which can be found on
page 25 of The Maths Tables Book.
1
The derivative of ln 𝑥𝑥 is this is also
listed on page 25.
𝑥𝑥
Now we sub 𝑒𝑒 in for 𝑥𝑥 in the derivative
and plug it into the calculator
78
Question 4
a)
∫ (4𝑥𝑥 3 − 6𝑥𝑥 + 10)𝑑𝑑𝑑𝑑
To integrate we increase the
power of the 𝑥𝑥 by 1 and divide by
the new power.
4𝑥𝑥 4 6𝑥𝑥 2
−
+ 10𝑥𝑥 + 𝑐𝑐
4
2
Don’t forget to add c!
𝒙𝒙𝟒𝟒 − 𝟑𝟑𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟏𝟏 + 𝒄𝒄
b) i) 𝟐𝟐𝒙𝒙𝟑𝟑 − 𝟐𝟐𝟐𝟐𝒙𝒙𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟏𝟏𝟏𝟏
𝑓𝑓 ′ (𝑥𝑥) = 6𝑥𝑥 2 − 54𝑥𝑥 + 109
∫ (6𝑥𝑥 2 − 54𝑥𝑥 + 109)𝑑𝑑𝑑𝑑
6𝑥𝑥 3 54𝑥𝑥 2
−
+ 109𝑥𝑥 + 𝑐𝑐
3
2
2𝑥𝑥 3 − 27𝑥𝑥 2 + 109𝑥𝑥 + 𝑐𝑐
𝑓𝑓(2) = 0
→ 2(2)3 − 27(2)2 + 109(2) + 𝑐𝑐 = 0
16 − 108 + 218 + 𝑐𝑐 = 0
To find 𝑓𝑓(𝑥𝑥) we need to integrate 𝑓𝑓′(𝑥𝑥).
Now we have an expression for 𝑓𝑓(𝑥𝑥) but we
need to find what 𝑐𝑐 is.
We can see from the graph that 𝑓𝑓(𝑥𝑥) passes
through the point (2,0) so we can plug 2 in for
𝑥𝑥 and let the equation equal 0 in order to solve
for 𝑐𝑐
126 + 𝑐𝑐 = 0
𝑐𝑐 = −126
𝟑𝟑
𝟐𝟐
→ 𝑓𝑓(𝑥𝑥) = 𝟐𝟐𝒙𝒙 − 𝟐𝟐𝟐𝟐𝒙𝒙 + 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 − 𝟏𝟏𝟏𝟏𝟏𝟏
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Plugging our value for 𝑐𝑐 back into our
expression for 𝑓𝑓(𝑥𝑥)
79
ii) 𝑩𝑩(𝟒𝟒. 𝟓𝟓, 𝟎𝟎) 𝑪𝑪(𝟕𝟕, 𝟎𝟎)
𝑥𝑥 = 2 is a root
We know from the graph that 𝑥𝑥 = 2 is a root of
the equation. This means that 𝑥𝑥 − 2 is a factor of
the equation.
→ 𝑥𝑥 − 2 is a factor
2𝑥𝑥 2 − 23𝑥𝑥 + 63
𝑥𝑥 − 2 )2𝑥𝑥 3 − 27𝑥𝑥 2 + 109𝑥𝑥 − 126
So, we can divide the equation by 𝑥𝑥 − 2 to get a
quadratic which we can solve for the other two
roots.
2𝑥𝑥 3 − 4𝑥𝑥 2
−23𝑥𝑥 2 + 109𝑥𝑥
−23𝑥𝑥 2 + 46𝑥𝑥
63𝑥𝑥 − 126
63𝑥𝑥 − 126
2𝑥𝑥 2 − 23𝑥𝑥 + 63 = 0
(2𝑥𝑥 − 9)(𝑥𝑥 + 7)
𝑥𝑥 = 4.5 𝑥𝑥 = 7
𝑩𝑩(𝟒𝟒. 𝟓𝟓, 𝟎𝟎) 𝑪𝑪(𝟕𝟕, 𝟎𝟎)
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0
If you struggle to factorise the equation you can
use the minus b formula to solve for 𝑥𝑥
These are the two other 𝑥𝑥 coordinates where the
equation crosses the 𝑥𝑥 −axis.
Filling in the coordinates.
80
Question 5
a) 𝒑𝒑 = −𝟔𝟔, 𝒒𝒒 = 𝟏𝟏𝟏𝟏
If 3 + 2𝑖𝑖 is a root, then 3 − 2𝑖𝑖 is also a root
𝑧𝑧 2 − 𝑧𝑧(𝑠𝑠𝑠𝑠𝑠𝑠 𝑜𝑜𝑜𝑜 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟) + (𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑜𝑜𝑜𝑜 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟)
𝑧𝑧 2 − 𝑧𝑧((3 + 2𝑖𝑖) + (3 − 2𝑖𝑖) + ((3 + 2𝑖𝑖)(3 − 2𝑖𝑖)))
𝑧𝑧 2 − 𝑧𝑧(6) + (9 + 6𝑖𝑖 − 6𝑖𝑖 − 4𝑖𝑖 2 )
𝑧𝑧 2 − 6𝑧𝑧 + (9 − 4(−1))
𝑧𝑧 2 − 6𝑧𝑧 + 13
If a complex number is the root of an equation,
then its conjugate is also a root. We get the
conjugate by changing the sign in front of the 𝑖𝑖.
Subbing two roots into this expression for finding
a quadratic.
Remember 𝑖𝑖 2 = −1
Writing out 𝑝𝑝 and 𝑞𝑞 from our quadratic
𝒑𝒑 = −𝟔𝟔, 𝒒𝒒 = 𝟏𝟏𝟏𝟏
b) i)
𝟒𝟒 �𝐜𝐜𝐜𝐜𝐜𝐜
𝟓𝟓𝟓𝟓
𝟓𝟓𝟓𝟓
+ 𝒊𝒊 𝐬𝐬𝐬𝐬𝐬𝐬 �
𝟑𝟑
𝟑𝟑
𝑣𝑣 = 2 − 2√3
Modulus = √𝑎𝑎2 + 𝑏𝑏 2
2
→ �(2)2 + �−2√3� = 4
Argument = tan−1
2√3
𝜋𝜋
�=
3
2
→ tan−1 �
2𝜋𝜋 −
𝑏𝑏
𝑎𝑎
𝜋𝜋 5𝜋𝜋
=
3
3
𝑣𝑣 = 𝟒𝟒 �𝐜𝐜𝐜𝐜𝐜𝐜
𝟓𝟓𝟓𝟓
𝟓𝟓𝟓𝟓
+ 𝒊𝒊 𝐬𝐬𝐬𝐬𝐬𝐬 �
𝟑𝟑
𝟑𝟑
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To write a complex number in the form 𝑟𝑟(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖) we
need to find the modulus (𝑟𝑟) and the argument (𝜃𝜃)
Remember we say that 2 − 2√3𝑖𝑖 is in the form of 𝑎𝑎 + 𝑏𝑏𝑏𝑏
To find the argument we find the tan inverse of
𝑏𝑏
𝑎𝑎
As 2 − 2√3𝑖𝑖 is in the 4th quadrant, we take the angle away
from 2𝜋𝜋
Plugging our argument and modulus into the form
𝑟𝑟(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖)
Note: The argument could equally be found in degrees and
we would be left with: 𝑣𝑣 = 4(cos 300 ° + 𝑖𝑖 sin 300°)
81
ii) −√𝟑𝟑 + 𝒊𝒊, √𝟑𝟑 − 𝒊𝒊
𝑤𝑤 2 = 𝑣𝑣
If 𝑤𝑤 2 = 𝑣𝑣 then 𝑤𝑤 is equal to the square root of
1
𝑣𝑣, which can be written as 𝑣𝑣 2 .
𝑤𝑤 = √𝑣𝑣
1
𝑤𝑤 = 𝑣𝑣 2
1
𝑣𝑣 2
1
5𝜋𝜋
5𝜋𝜋 2
= 4 �cos
+ 𝑖𝑖 sin �
3
3
1
5𝜋𝜋 1
5𝜋𝜋 1
42 �cos � × � + 𝑖𝑖 sin � × ��
3 2
3 2
±2 �cos �
±2 �−
2 �−
5𝜋𝜋
5𝜋𝜋
� + 𝑖𝑖 sin � ��
6
6
√3 1
+ 𝑖𝑖�
2
2
√3 1
+ 𝑖𝑖�
2
2
−√𝟑𝟑 + 𝒊𝒊
Now we can use De Moivre’s theorem to solve
for 𝑣𝑣.
Remember the square root of 4 can be either
+2 or −2, as both numbers square equal 4.
Plugging cos
calculator.
or − 2 �−
√3 1
+ 𝑖𝑖�
2
2
5𝜋𝜋
6
and then sin
5𝜋𝜋
6
into the
Separating the plus and minus 2 to get our two
answers.
√𝟑𝟑 − 𝒊𝒊
Question 6
a) i) 𝟐𝟐√𝟐𝟐
𝑥𝑥 − √32 = √128 − 5𝑥𝑥
𝑥𝑥 − √16 × 2 = √64 × 2 − 5𝑥𝑥
𝑥𝑥 − √16. √2 = √64√2 − 5𝑥𝑥
𝑥𝑥 − 4√2 = 8√2 − 5𝑥𝑥
𝑥𝑥 + 5𝑥𝑥 = 8√2 + 4√2
6𝑥𝑥 = 12√2
𝑥𝑥 = 𝟐𝟐√𝟐𝟐
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To deal with the surds we try and rewrite them,
so they have the same base.
We can rewrite each of them as a perfect
square multiplied by 2. We can then separate
out the square roots. This leaves us with both
surds in the form 𝑎𝑎√2.
Now we rearrange to have 𝑥𝑥′s on one side and
surds on the other.
Dividing across by 6 gives us our answer.
82
ii)
𝐴𝐴 = ��32𝑘𝑘 2 , �50𝑘𝑘 2 , �128𝑘𝑘 2 , �98𝑘𝑘 2 �
Mean:
�32𝑘𝑘 2 = 4�2𝑘𝑘 2
�50𝑘𝑘 2 = 5�2𝑘𝑘 2
The first part of the question gives us a
hint that we should rewrite these surds,
so they have the same base.
�128𝑘𝑘 2 = 8�2𝑘𝑘 2
�98𝑘𝑘 2 = 7�2𝑘𝑘 2
4√2𝑘𝑘 2 + 5√2𝑘𝑘 2 + 8√2𝑘𝑘 2 + 7√2𝑘𝑘 2
4
Now finding the mean by adding the
terms and dividing by the number of
terms.
24√2𝑘𝑘 2
= 6�2𝑘𝑘 2
4
Median:
4�2𝑘𝑘 2 , 5�2𝑘𝑘 2 , 7�2𝑘𝑘 2 , 8�2𝑘𝑘 2
1
(𝑛𝑛 + 1)𝑡𝑡ℎ term
2
𝑛𝑛 = 4
1
(4 + 1)𝑡𝑡ℎ = 2.5
2
To find the median of data we get the
1
2
(𝑛𝑛 + 1)𝑡𝑡ℎ term, where 𝑛𝑛 is the number
of terms.
As this gives us 2.5, we get the mean of
the 2nd and 3rd terms.
2.5 → Mean of 2nd and 3rd terms
5√2𝑘𝑘 2 + 7√2𝑘𝑘 2 12√2𝑘𝑘 2
=
2
2
= 6�2𝑘𝑘 2
We can see that the median is the same
as the mean.
Mean = Median
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83
b)
We assume √2 is a rational number.
𝑎𝑎
If √2 is rational it can be written as , where 𝑎𝑎, 𝑏𝑏 ∈ 𝑍𝑍, 𝑏𝑏 ≠ 0 and 𝑎𝑎 and 𝑏𝑏 have no
common factor.
√2 =
2=
𝑏𝑏
(A rational number is one
which can be expressed as a
fraction)
𝑎𝑎
𝑏𝑏
𝑎𝑎2
𝑏𝑏 2
2𝑏𝑏 2 = 𝑎𝑎2
𝑎𝑎2 is even as it is equal to 2 × 𝑏𝑏 2 (i.e. it has a factor of two)
As 𝑎𝑎 is divisible by 2 it can be written as 𝑎𝑎 = 2𝑐𝑐
→ 𝑎𝑎2 = 4𝑐𝑐 2 = 2𝑏𝑏 2
4𝑐𝑐 2 = 2𝑏𝑏 2
2𝑐𝑐 2 = 𝑏𝑏 2
𝑏𝑏 is also even
∴ 𝑎𝑎 and 𝑏𝑏 have a common factor of 2 which contradicts our assumption that √2 is
rational
∴ √2 is an irrational number.
© Pocket Tutor 2022
Squaring both sides
Squaring 2𝑐𝑐
As it has a factor of two
As it cannot be written as a
fraction with no common
factors.
84
Question 7
a) i)
Step
Step1
Step 2
Step 3
Step 4
Step 5
Length
Removed
1
3
2
9
4
27
8
81
16
243
For the first three steps we can gather the length of the gap from the diagram and then multiply this by the
number of gaps in the line.
2
For steps 4,5 we follow the constant multiple of , which we’ve seen in the first three steps
3
𝟐𝟐 𝒏𝒏
ii) 𝟏𝟏 − � �
𝟑𝟑
𝑆𝑆𝑛𝑛 =
𝑎𝑎 =
𝑟𝑟 =
1
3
𝑎𝑎(1 − 𝑟𝑟 𝑛𝑛 )
1 − 𝑟𝑟
2 1 2
÷ =
9 3 3
2 𝑛𝑛
2 𝑛𝑛
1
1
�1 − � � �
�1 − � � �
3
3
3
3
𝑆𝑆𝑛𝑛 =
→
1
2
1−
3
3
3
1
2 𝑛𝑛
�1 − � � � ×
1
3
3
𝟐𝟐 𝒏𝒏
→ 𝟏𝟏 − � �
𝟑𝟑
© Pocket Tutor 2022
The formula for the sum of a geometric
series can be found on page 22 of The
Maths Tables Book.
𝑎𝑎 = The first term
𝑟𝑟 is the common ratio which we can get
by dividing the second term by the first
term.
Now plugging in our values for 𝑎𝑎 and 𝑟𝑟
To divide by a fraction, we invert it and
multiply by it.
This is the expression we are left with.
85
iii)
𝑆𝑆∞ =
𝑎𝑎
The equation for the sum of an infinite geometric series can
be found on page 22 of The Maths Tables Book.
1−𝑟𝑟
𝑆𝑆∞ =
1
3
1−
2
3
= 𝟏𝟏
We can fill this in with the values of 𝑎𝑎 and 𝑟𝑟 from part ii).
b) i)
Label
A
B
C
D
E
F
End-Point
2
3
2
9
7
9
8
9
7
27
25
27
ii)
It is the end point of a segment.
𝒊𝒊𝒊𝒊𝒊𝒊)
𝟏𝟏
𝟒𝟒
𝑆𝑆∞ =
𝑎𝑎 =
1
3
𝑎𝑎
1 − 𝑟𝑟
The equation for the sum of an infinite geometric
series can be found on page 22 of The Maths
Tables Book.
1
1 1
𝑟𝑟 = − ÷ = −
3
9 3
𝑆𝑆∞ =
1
3
1
1 − (− )
3
=
𝑎𝑎 = the first term, which we can see is
𝟏𝟏
𝟒𝟒
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1
3
𝑟𝑟 is the common ratio which we can get by
dividing the second term by the first term.
Now plugging our values for 𝑎𝑎 and 𝑟𝑟 into the
equation.
86
Question 8
a) €𝟐𝟐𝟐𝟐, 𝟔𝟔𝟔𝟔𝟔𝟔
𝜋𝜋
𝑡𝑡� + 37 500
26
𝜋𝜋
𝑟𝑟(20) = 22500 cos( (20)) + 37500
26
𝑟𝑟(𝑡𝑡) = 22500 cos �
= 20658.5 → €𝟐𝟐𝟐𝟐, 𝟔𝟔𝟔𝟔𝟔𝟔
𝒃𝒃)
To find the revenue after 20 weeks we
sub 20 in for 𝑡𝑡 and then plug the whole
thing into the calculator.
Rounding to the nearest euro.
𝟓𝟓𝟓𝟓
𝟏𝟏𝟏𝟏𝟏𝟏
= 𝒕𝒕 𝐚𝐚𝐚𝐚𝐚𝐚 𝒕𝒕 =
𝟑𝟑
𝟑𝟑
𝜋𝜋
𝑡𝑡� + 37 500 = 26 250
26
𝜋𝜋
22500 cos � 𝑡𝑡� = 26250 − 37500
26
22500 cos �
𝜋𝜋
11250
𝑡𝑡) = −
26
22500
cos(
To find the time when the revenue is 26,250, we
let the expression equal 26,250 and solve for 𝑡𝑡.
Taking 37,500 from both sides.
Dividing across by 22,500.
11250
11250
𝜋𝜋
�
𝑡𝑡 = cos −1 �−
22500
26
Now finding the cos inverse of −
𝜋𝜋
2
𝜋𝜋 ÷
= 𝑡𝑡
26
3
quadrants, we get a second answer by taking
2
𝜋𝜋
𝜋𝜋 =
𝑡𝑡
3
26
4𝜋𝜋
𝜋𝜋
and
=
𝑡𝑡
3
26
4𝜋𝜋 𝜋𝜋
÷
= 𝑡𝑡
3
26
𝟏𝟏𝟏𝟏𝟏𝟏
𝟓𝟓𝟓𝟓
= 𝒕𝒕 𝐚𝐚𝐚𝐚𝐚𝐚 𝒕𝒕 =
𝟑𝟑
𝟑𝟑
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The calculator gives
2𝜋𝜋
3
for this.
22500
However, as cos is positive in the first and 4th
away from 2𝜋𝜋. (You can revise this in the
Trigonometry section.)
2𝜋𝜋
3
87
c) 𝒓𝒓′ (𝒕𝒕) = −
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
𝐬𝐬𝐬𝐬𝐬𝐬 �
𝝅𝝅
𝟐𝟐𝟐𝟐
𝒕𝒕�
𝜋𝜋
𝑡𝑡� + 37 500
26
𝜋𝜋
𝜋𝜋
𝑟𝑟 ′ (𝑡𝑡) = −22500 sin � 𝑡𝑡� ×
26
26
𝑟𝑟(𝑡𝑡) = 22500 cos �
𝒓𝒓′ (𝒕𝒕) = −
𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝝅𝝅
𝐬𝐬𝐬𝐬𝐬𝐬 � 𝒕𝒕�
𝟏𝟏𝟏𝟏
𝟐𝟐𝟐𝟐
To differentiate this expression, we can look to page 25 of
the Maths Tables Book. We can see that −𝑠𝑠𝑠𝑠𝑠𝑠 is the
derivative of 𝑐𝑐𝑐𝑐𝑐𝑐.
We also need to multiply this by the derivative of what’s
𝜋𝜋
inside the cos bracket, i.e. the 𝑡𝑡.
26
Tidying up by multiplying 22500 by the
𝜋𝜋
26
d)
𝑟𝑟 ′ (𝑡𝑡) = −
11250𝜋𝜋
𝜋𝜋
sin � 𝑡𝑡�
13
26
𝑟𝑟 ′ (30) = −
= 1263.44
𝜋𝜋
11250𝜋𝜋
sin � (30)�
26
13
1263.44 > 0
To see if a function is increasing at a certain value, we plug
the value into the derivative. If the result is > 0 then the
function is increasing.
So, we plug 30 in for 𝑡𝑡 in our derivative, and we can see that
the revenue is increasing at this time.
∴ Increasing
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88
e) 𝟐𝟐𝟐𝟐 = 𝐌𝐌𝐌𝐌𝐌𝐌
𝑟𝑟 ′ (𝑡𝑡) = −
11250𝜋𝜋
𝜋𝜋
sin � 𝑡𝑡�
13
26
11250𝜋𝜋
𝜋𝜋
sin � 𝑡𝑡� = 0
13
26
𝜋𝜋
sin � 𝑡𝑡� = 0
26
𝜋𝜋
𝑠𝑠𝑠𝑠𝑛𝑛−1 (0) =
𝑡𝑡
26
π
𝜋𝜋
𝑡𝑡 = 0 and t = π
26
26
−
𝑡𝑡 = 0
𝑡𝑡 = 𝜋𝜋 ÷
𝑡𝑡 = 26
𝜋𝜋
26
𝑟𝑟
′′ (𝑡𝑡)
𝜋𝜋
11250𝜋𝜋
𝜋𝜋
cos � 𝑡𝑡� ×
=−
26
13
26
𝑟𝑟
′′ (0)
𝜋𝜋
11250𝜋𝜋
𝜋𝜋
cos � (0)� ×
=−
26
13
26
= −328.5 < 0
∴ 0 = Max
𝑟𝑟 ′′ (26) = −
328.5 > 0
𝜋𝜋
11250𝜋𝜋
𝜋𝜋
cos � (26)� ×
26
13
26
To find the maximum point of a function, we let its
derivative equal 0 and solve.
Dividing across by −
11250𝜋𝜋
13
Finding the sin inverse of 0
The calculator gives 0 for this but as in part b) there are
two solutions.
As 𝑠𝑠𝑠𝑠𝑠𝑠 is positive in the first and second quadrants, we
find the second value by taking 0 from 𝜋𝜋.
0 and 26 are our two values for 𝑡𝑡
To find which is the max we need to find the second
derivative.
𝑐𝑐𝑐𝑐𝑐𝑐 is the derivative of 𝑠𝑠𝑠𝑠𝑠𝑠, and again we multiply by
𝜋𝜋
the derivative of 𝑡𝑡.
26
Now plugging in 0 for 𝑡𝑡.
As this gives a number less than 0, we know that this is
the maximum point
Now plugging in 26 for 𝑡𝑡.
As this gives a number greater than 0, we know that
this is the minimum point.
∴ 𝟐𝟐𝟐𝟐 = 𝐌𝐌𝐌𝐌𝐌𝐌
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89
Question 9
a) i) 𝝅𝝅𝝅𝝅 + 𝟐𝟐𝟐𝟐 + 𝟐𝟐𝟐𝟐
Perimeter = Length of semicircle +2𝑦𝑦 + 2𝑥𝑥
Length of a semicircle =
→
1
(2𝜋𝜋𝜋𝜋)
2
The perimeter is the sum of all the sides.
1
Length of a circle
2
Taking the length of a circle (2𝜋𝜋𝜋𝜋) from
page 8 of The Maths Tables Book.
1
�2𝜋𝜋(𝑥𝑥)� = 𝜋𝜋𝜋𝜋
2
Subbing in 𝑥𝑥 for the radius.
Adding the length of the semicircle to the
length of the 2 sides and the bottom
𝝅𝝅𝝅𝝅 + 𝟐𝟐𝟐𝟐 + 𝟐𝟐𝟐𝟐
ii)
𝜋𝜋𝜋𝜋 + 2𝑦𝑦 + 2𝑥𝑥 = 12
Letting the expression, we have for the
perimeter equal 12.
2𝑦𝑦 = 12 − 𝜋𝜋𝜋𝜋 − 2𝑥𝑥
𝑦𝑦 =
12 − 𝜋𝜋𝜋𝜋 − 2𝑥𝑥
2
Taking 𝜋𝜋𝜋𝜋 and 2𝑥𝑥 from both sides.
𝑦𝑦 =
12 − (2 + 𝜋𝜋)𝑥𝑥
2
Factorising our the 𝑥𝑥
Dividing across by 2
12 − (𝜋𝜋𝜋𝜋 + 2𝑥𝑥)
𝑦𝑦 =
2
Factorising out the minus 1
b) i)
0
𝒙𝒙
𝑦𝑦 =
12 − (2 + 𝜋𝜋)𝑥𝑥
2
𝟔𝟔
𝟏𝟏𝟏𝟏
𝟐𝟐 + 𝝅𝝅
𝟎𝟎
Plugging in 0 for 𝑥𝑥 gives us 6.
Subbing in
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12
2+𝜋𝜋
for 𝑥𝑥 and plugging it into the calculator gives us 0.
90
ii)
iii)
(0,6),
𝑦𝑦2 − 𝑦𝑦1
𝑥𝑥2 − 𝑥𝑥1
(2.334,0)
0−6
= −𝟐𝟐. 𝟓𝟓𝟓𝟓
2.334 − 0
For every 1m increase in the length of the
radius of the semicircle, the height of the
rectangle decreases by 2.57𝑚𝑚.
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We can use the coordinates from the previous part to
calculate the slope.
�
12
= 2.334�
2 + 𝜋𝜋
Remember that 𝑥𝑥 is the radius of the semicircle and 𝑦𝑦 is the
height of the rectangle. So, we can see from the slope that
when we increase 𝑥𝑥 by one, 𝑦𝑦 decrease by 2.57
91
c) i)
Area = Area of semicircle + Area of rectangle
Area of rectangle: 𝑦𝑦 × 2𝑥𝑥 = 2𝑥𝑥𝑥𝑥
Area of semicircle =
→
=
1
Area of Circle
2
1
1
(𝜋𝜋𝑟𝑟 2 ) → (𝜋𝜋(𝑥𝑥)2 )
2
2
𝜋𝜋𝑥𝑥 2
2
Adding these two areas gives us the area of
the window.
12 − (2 + 𝜋𝜋)𝑥𝑥
𝑦𝑦 =
2
12 − (2 + 𝜋𝜋)𝑥𝑥
𝜋𝜋𝑥𝑥 2
+ 2𝑥𝑥 �
�
2
2
𝜋𝜋𝑥𝑥 2 24𝑥𝑥 − 2𝑥𝑥(2𝑥𝑥 + 𝜋𝜋𝜋𝜋)
+
2
2
𝜋𝜋𝑥𝑥 2 24𝑥𝑥 − 4𝑥𝑥 2 − 2𝜋𝜋𝑥𝑥 2
+
2
2
24𝑥𝑥 − 4𝑥𝑥 2 − 𝜋𝜋𝑥𝑥 2
2
𝑎𝑎(𝑥𝑥) =
24𝑥𝑥 − (𝜋𝜋 + 4)𝑥𝑥
2
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Taking the area of a circle (𝜋𝜋𝑟𝑟 2 ) from page
8 of The Maths Tables Book.
Subbing in 𝑥𝑥 for the radius.
𝜋𝜋𝑥𝑥 2
Area =
+ 2𝑥𝑥𝑥𝑥
2
→
The area of a rectangle = length ×width
2
Remember (from part c ii) and given in q):
12 − (2 + 𝜋𝜋)𝑥𝑥
𝑦𝑦 =
2
So, we sub this in for 𝑦𝑦 in our expression for
the area.
Multiplying out the brackets.
Adding the fractions
Factorising out the 𝑥𝑥 2
92
ii) 𝟏𝟏𝟏𝟏 − (𝝅𝝅 + 𝟒𝟒)𝒙𝒙
24𝑥𝑥 − (𝜋𝜋 + 4)𝑥𝑥 2
𝑎𝑎(𝑥𝑥) =
2
𝑎𝑎(𝑥𝑥) =
𝑎𝑎(𝑥𝑥) =
1
�24𝑥𝑥 − (𝜋𝜋 + 4)(𝑥𝑥 2 )�
2
1
(24𝑥𝑥 − 𝜋𝜋𝑥𝑥 2 − 4𝑥𝑥 2 )
2
𝑎𝑎′ (𝑥𝑥) =
1
(24 − 2𝜋𝜋𝜋𝜋 − 8𝑥𝑥)
2
= 12 − 2𝜋𝜋𝜋𝜋 − 4𝑥𝑥
Rewriting the fraction by factorising out the
half to make it easier to differentiate, also
multiplying out the brackets for the same
reason.
Differentiating and then multiplying by the
half.
Factorising to tidy up (not essential)
𝟏𝟏𝟏𝟏 − (𝝅𝝅 + 𝟒𝟒)𝒙𝒙
iii)
𝑎𝑎′ (𝑥𝑥) = 0
12 − (𝜋𝜋 + 4)𝑥𝑥 = 0
To find when the area is a maximum, we let the
derivative equal 0.
12 = (𝜋𝜋 + 4)𝑥𝑥
12
= 𝑥𝑥
𝜋𝜋 + 4
1.68 = 𝑥𝑥
𝑦𝑦 =
𝑦𝑦 =
12 − (2 + 𝜋𝜋)𝑥𝑥
2
12 − (2 + 𝜋𝜋)(1.68)
= 1.68
2
At maximum 𝒙𝒙 = 𝒚𝒚
→ The area is a maximum when the height of
the rectangle equals the length of the radius.
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Solving for 𝑥𝑥
Subbing our 𝑥𝑥 value into our expression for 𝑦𝑦 to find
the corresponding 𝑦𝑦 value.
We see that when the area is at its maximum 𝑥𝑥 = 𝑦𝑦.
This means that the radius of the semicircle (𝑥𝑥) equals
the height of the rectangle (𝑦𝑦), when the area is at its
maximum.
93
2018 Paper 1
Question 1
a) 𝒙𝒙 = 𝟐𝟐, 𝒚𝒚 = −𝟏𝟏, 𝒛𝒛 = 𝟓𝟓
2𝑥𝑥 + 3𝑦𝑦 − 𝑧𝑧 = −4
3𝑥𝑥 + 2𝑦𝑦 + 2𝑧𝑧 = 14
𝑥𝑥
− 3𝑧𝑧 = −13
Equation 1
Equation 2
Equation 3
Labelling the equations makes it easier to keep
track of what we’re doing.
We want to get rid of the ‘𝑦𝑦’s in equations 1&2
so we can use equation 3 to find one of the
variables.
Equation 1 × 2
Multiplying equation 1 by 2.
Equation 2 × −3
Multiplying equation 2 by −3
Adding these:
Adding the results of this
→ 4𝑥𝑥 + 6𝑦𝑦 − 2𝑧𝑧 = −8
→ −9𝑥𝑥 − 6𝑦𝑦 − 6𝑧𝑧 = −42
4𝑥𝑥 + 6𝑦𝑦 − 2𝑧𝑧 = −8
−9𝑥𝑥 − 6𝑦𝑦 − 6𝑧𝑧 = −42
−5𝑥𝑥
− 8𝑧𝑧 = −50
Equation 3 × 5
→ 5𝑥𝑥 − 15𝑧𝑧 = −65
Equation 4
Now we are going to use equations 3&4 to
solve for 𝑧𝑧
Multiplying equation 3 by 5
Adding Equations 3 & 4
5𝑥𝑥 − 15𝑧𝑧 = −65
Adding the result of this to equation 4
−5𝑥𝑥 − 8𝑧𝑧 = −50
−23𝑧𝑧 = −115
23𝑧𝑧 = 115
𝒛𝒛 = 𝟓𝟓
𝑥𝑥 − 3(5) = −13
𝑥𝑥 − 15 = −13
𝑥𝑥 = −13 + 15
Plugging 5 in for 𝑧𝑧 in equation 3 to find 𝑥𝑥
𝒙𝒙 = 𝟐𝟐
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94
2(2) + 3𝑦𝑦 − (5) = −4
4 + 3𝑦𝑦 − 5 = −4
3𝑦𝑦 − 1 = −4
Plugging our values for 𝑥𝑥 and 𝑧𝑧 into
equation 1 to solve for 𝑦𝑦
3𝑦𝑦 = −4 + 1
3𝑦𝑦 = −3
𝒚𝒚 = −𝟏𝟏
𝒙𝒙 = 𝟐𝟐, 𝒚𝒚 = −𝟏𝟏, 𝒛𝒛 = 𝟓𝟓
b) −𝟗𝟗 ≤ 𝒙𝒙 ≤ −𝟐𝟐
2𝑥𝑥 − 3
≥3
𝑥𝑥 + 2
We multiply across by (𝑥𝑥 + 2)2 to make sure we don’t
have to flip the inequality sign. (It is impossible to know
whether or not you have to do this if multiplying across by
an unknown)
2𝑥𝑥 − 3
(𝑥𝑥 + 2)2 �
� ≥ 3(𝑥𝑥 + 2)2
𝑥𝑥 + 2
The (𝑥𝑥 + 2) divides in to leave us with (𝑥𝑥 + 2)(2𝑥𝑥 − 3) on
the left.
2𝑥𝑥 − 3
(𝑥𝑥 + 2)2 �
� ≥ 3(𝑥𝑥 + 2)2
𝑥𝑥 + 2
2𝑥𝑥 2 − 3𝑥𝑥 + 4𝑥𝑥 − 6 ≥ 3(𝑥𝑥 2 + 4𝑥𝑥 + 4)
2𝑥𝑥 2 + 𝑥𝑥 − 6 ≥ 3𝑥𝑥 2 + 12𝑥𝑥 + 12
2
0 ≥ 𝑥𝑥 + 11𝑥𝑥 + 18
(𝑥𝑥 + 2)(𝑥𝑥 + 9) = 0
𝑥𝑥 = −2, 𝑥𝑥 = −9
Taking 2𝑥𝑥 2 + 𝑥𝑥 − 6 from both sides.
We let the expression = 0 to find the x-values
Solving the quadratic.
From the diagram we can see that the graph is negative
between −9 and −2
−𝟗𝟗 ≤ 𝒙𝒙 ≤ −
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95
Question 2
a)
𝑥𝑥 2 , 5𝑥𝑥 − 8, 𝑥𝑥 + 8
𝑟𝑟 =
𝑟𝑟 =
𝑡𝑡2 5𝑥𝑥 − 8
=
𝑥𝑥 2
𝑡𝑡1
𝑟𝑟 of a geometric series =
𝑡𝑡3
𝑥𝑥 + 8
=
𝑡𝑡2 5𝑥𝑥 − 8
𝑡𝑡𝑛𝑛
𝑡𝑡𝑛𝑛−1
Letting 𝑟𝑟 = 𝑟𝑟
𝑟𝑟 = 𝑟𝑟
𝑥𝑥 + 8
5𝑥𝑥 − 8
=
2
5𝑥𝑥 − 8
𝑥𝑥
Multiplying across by 𝑥𝑥 2 and (5𝑥𝑥 − 8)
(5𝑥𝑥 − 8)(5𝑥𝑥 − 8) = 𝑥𝑥 2 (𝑥𝑥 + 8)
Taking 25𝑥𝑥 2 − 80𝑥𝑥 + 64 from both
sides
25𝑥𝑥 2 − 80𝑥𝑥 + 64 = 𝑥𝑥 3 + 8𝑥𝑥 2
𝒙𝒙𝟑𝟑 − 𝟏𝟏𝟏𝟏𝒙𝒙𝟐𝟐 + 𝟖𝟖𝟖𝟖𝟖𝟖 − 𝟔𝟔𝟔𝟔 = 𝟎𝟎
b) 𝒙𝒙 = 𝟖𝟖
𝑓𝑓(𝑥𝑥) = 𝑥𝑥 3 − 17𝑥𝑥 2 + 80𝑥𝑥 − 64
𝑓𝑓(1) = (1)3 − 17(1)2 + 80(1) − 64
1 − 17 + 80 − 64 = 0
𝑥𝑥 = 1 → 𝑥𝑥 − 1 is a factor
𝑥𝑥 2 − 16𝑥𝑥 + 64
3
If we divide the cubic by a factor, we get
its other factors. (Alternative method is
subbing in (𝑥𝑥 − 1) for 𝑥𝑥).
2
𝑥𝑥 − 1 𝑥𝑥 − 17𝑥𝑥 + 80𝑥𝑥 − 64
𝑥𝑥 3 − 𝑥𝑥 2
Subbing in 1 for 𝑥𝑥.
−16𝑥𝑥 2 + 80𝑥𝑥
−16𝑥𝑥 2 + 16𝑥𝑥
64𝑥𝑥 − 64
64𝑥𝑥 − 64
𝑥𝑥 2 − 16𝑥𝑥 + 64 = 0
(𝑥𝑥 − 8)(𝑥𝑥 − 8) = 8
𝒙𝒙 = 𝟖𝟖
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0
We let 𝑥𝑥 2 − 16𝑥𝑥 + 64 = 0 so we can
solve for the other factors.
96
c) 128
𝑥𝑥 2 , 5𝑥𝑥 − 8, 𝑥𝑥 + 8
Subbing 𝑥𝑥 = 1 into the series
(1)2 , 5(1) − 8, (1) + 8
1, −3,9
Subbing 𝑥𝑥 = 8 into the series
(8)2 , 5(8) − 8, (8) + 8
64,32,16
𝑎𝑎 = 64,
32 1
=
64 2
𝑎𝑎
𝑆𝑆∞ =
1 − 𝑟𝑟
64
1
1−� �
2
𝑎𝑎 = The first term = 64
𝑟𝑟 =
𝑡𝑡2
𝑡𝑡1
From page 22 of The Maths Tables Book
𝑟𝑟 =
𝑆𝑆∞ =
This does not have a sum to infinity as |𝑟𝑟| > 1
= 𝟏𝟏𝟏𝟏𝟏𝟏
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Subbing in for 𝑎𝑎 and 𝑟𝑟
97
Question 3
a) 120°
ℎ(𝑥𝑥) = cos(2𝑥𝑥)
ℎ
′ (𝑥𝑥)
= −2sin(2𝑥𝑥)
At 𝑥𝑥 =
𝜋𝜋
𝜋𝜋
→ −2 sin �2 � ��
3
3
√3
� = −√3
2
−2 �
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 = −√3
𝜃𝜃 = 𝑇𝑇𝑇𝑇𝑛𝑛−1 (√3)
Differentiating cos 𝑎𝑎𝑎𝑎 → − asin 𝑎𝑎𝑎𝑎
We are trying to find the angle it makes
𝜋𝜋
at so we plug this in for 𝑥𝑥
3
If the slope, = −√3 then the angle
made by the line is the tan inverse of
this.
As the slope =
𝜃𝜃 = −60°
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
𝑟𝑟𝑟𝑟𝑟𝑟
=
𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
This value for theta is in the second quadrant, we want the angle the line makes with the positive
sense of the x-axis, (i.e. in the first quadrant)
180 − 60 = 120°
𝒃𝒃)
𝟐𝟐
𝝅𝝅
Average Value =
Interval: 0 ≤ 𝑥𝑥 ≤
1
𝜋𝜋
4
b
1
� 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑
b−a a
𝜋𝜋
4
� cos 2𝑥𝑥 𝑑𝑑𝑑𝑑
𝜋𝜋
−0 0
4
𝜋𝜋
1 sin 2𝑥𝑥 4
𝜋𝜋 � 2 �
0
4
𝜋𝜋
4 sin 2 �4 �
sin 2(0)
�−�
��
��
𝜋𝜋
2
2
4 1
� − 0�
𝜋𝜋 2
To find the average value of
something we integrate it between
certain limits and multiply it by
Using page 26 of the Maths Tables
Book to integrate cos 2𝑥𝑥
1
𝜋𝜋 = 4
4 𝜋𝜋
𝜋𝜋
Plugging and 0 in for 𝑥𝑥
4
4
𝟐𝟐
4 1
� �=
=
2𝜋𝜋 𝝅𝝅
𝜋𝜋 2
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1
𝑏𝑏−𝑎𝑎
98
Question 4
(cos 𝜃𝜃 + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖)𝑛𝑛 = cos(𝑛𝑛𝑛𝑛) + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 (𝑛𝑛𝑛𝑛)
1. Showing it is true for 𝒏𝒏 = 𝟏𝟏
→ (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖)1 = cos(1𝜃𝜃) + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖(1𝜃𝜃)
Subbing in 1 for 𝑛𝑛
2. Assuming it is true for 𝒏𝒏 = 𝒌𝒌
Subbing in 𝑘𝑘 for 𝑛𝑛
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖
𝑘𝑘
→ (𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃 + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝜃𝜃) = cos(𝑘𝑘𝑘𝑘) + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖(𝑘𝑘𝑘𝑘)
3. Proving it is true for 𝒏𝒏 = 𝒌𝒌 + 𝟏𝟏
(𝑐𝑐𝑐𝑐𝑐𝑐 𝜃𝜃 + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝜃𝜃)(𝑘𝑘+1) = cos((𝑘𝑘 + 1)𝜃𝜃) + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖((𝑘𝑘 + 1)𝜃𝜃)
Subbing in (𝑘𝑘 + 1) for 𝑛𝑛
(cos(𝑘𝑘𝑘𝑘) + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖(𝑘𝑘𝑘𝑘)) (cos 𝜃𝜃 + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖) = cos((𝑘𝑘 + 1)𝜃𝜃) + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖((𝑘𝑘 + 1)𝜃𝜃)
Subbing in our assumption
for 𝑛𝑛 = 𝑘𝑘
(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖)𝑘𝑘 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖)1 = cos((𝑘𝑘 + 1)𝜃𝜃) + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖((𝑘𝑘 + 1)𝜃𝜃)
cos(𝑘𝑘𝑘𝑘) . cos 𝜃𝜃 + cos 𝑘𝑘𝑘𝑘 . 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 + 𝑖𝑖𝑖𝑖𝑖𝑖𝑛𝑛𝑘𝑘𝑘𝑘. 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖. 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
= cos((𝑘𝑘 + 1)𝜃𝜃) + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖((𝑘𝑘 + 1)𝜃𝜃)
cos(𝑘𝑘𝑘𝑘) . cos 𝜃𝜃 − 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑖𝑖 cos 𝑘𝑘𝑘𝑘 . 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠+ 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖. 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = cos((𝑘𝑘 + 1)𝜃𝜃) +
𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖((𝑘𝑘 + 1)𝜃𝜃)
cos(𝑘𝑘𝑘𝑘 + 𝜃𝜃) + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 (𝑘𝑘𝑘𝑘 + 𝜃𝜃) = cos((𝑘𝑘 + 1)𝜃𝜃) + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖((𝑘𝑘 + 1)𝜃𝜃)
cos((𝑘𝑘 + 1) 𝜃𝜃) + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖((𝑘𝑘 + 1)𝜃𝜃) = cos((𝑘𝑘 + 1)𝜃𝜃) + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖((𝑘𝑘 + 1)𝜃𝜃)
Multiplying out the
brackets.
Using the cos(𝐴𝐴 + 𝐵𝐵) and
sin(𝐴𝐴 + 𝐵𝐵) identities.
Factorising out 𝜃𝜃
4. Thus, the proposition is true for 𝑛𝑛 = 𝑘𝑘 + 1 provided it is true for 𝑛𝑛 = 𝑘𝑘 but it is true for 𝑛𝑛 = 1 and
therefore true for all positive integers.
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99
b) 1
3
1 √3
𝑖𝑖�
�− +
2
2
𝑟𝑟 = �(𝑎𝑎)2 + (𝑏𝑏)2
1 2
√3
𝑟𝑟 = ��− � + � �
2
2
1 3
𝑟𝑟 = � + = √1 = 1
4 4
𝑏𝑏
𝜃𝜃 = 𝑇𝑇𝑇𝑇𝑛𝑛−1 � �
𝑎𝑎
√3
�
2
𝜋𝜋
−1
𝜃𝜃 = 𝑇𝑇𝑇𝑇𝑛𝑛
= 𝑇𝑇𝑇𝑇𝑛𝑛−1 �√3� = −
1
3
�− �
2
𝜋𝜋 2𝜋𝜋
𝜋𝜋 − =
3
3
2𝜋𝜋
2𝜋𝜋 3
1 �cos
+ 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 �
3
3
�
(cos 𝜃𝜃 + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖)𝑛𝑛 = cos(𝑛𝑛𝑛𝑛) + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 (𝑛𝑛𝑛𝑛)
�cos 3 �
To solve this, we need to write it in
polar form.
First, we need to find the modulus (𝑟𝑟)
Next, we need to find the argument.
𝜋𝜋
1
𝑇𝑇𝑇𝑇𝑛𝑛−1 gives us − , but − +
3
2
√3
2
the second quadrant, so we take
from 𝜋𝜋, ignoring the sign.
is in
𝜋𝜋
3
Using the modulus and the argument
to express it in polar form.
Using De Moivre’s theorem.
2𝜋𝜋
2𝜋𝜋
� + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 3 � ��
3
3
cos(2𝜋𝜋) + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖(2𝜋𝜋)
1 + 0𝑖𝑖
= 𝟏𝟏
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100
Question 5
a) i)
2115
Sum of first 45 terms in row 1:
𝑛𝑛
𝑆𝑆𝑛𝑛 = [2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑]
2
𝑎𝑎 = 4, 𝑑𝑑 = 7 − 4 = 3, 𝑛𝑛 = 45
𝑆𝑆45 =
𝑆𝑆45 =
45
[2(4) + (45 − 1)(3)]
2
45
45
[8 + 132] =
[140]
2
2
To find the difference we need to
find the sum of the first 45 terms in
row 1 and the sum of the first 45
terms in row 2 and subtract them
from each other.
Using the formula for an arithmetic
series on page 22 of the Maths
Tables Book
𝑆𝑆45 = 3150
Sum of first 45 terms in row 2:
𝑆𝑆𝑛𝑛 =
𝑛𝑛
[2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑]
2
𝑎𝑎 = 7, 𝑑𝑑 = 12 − 7 = 5, 𝑛𝑛 = 45
𝑆𝑆45 =
𝑆𝑆45 =
45
[2(7) + (45 − 1)(5)]
2
45
45
[14 + 220] =
[234]
2
2
𝑆𝑆45 = 5265
Difference:
5265 − 3150 = 2115
Subtracting the two figures.
a) ii)
8350
𝑇𝑇60 = The number at the start of the 60th row.
𝑇𝑇𝑛𝑛 = 𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑
𝑛𝑛 = 60, 𝑎𝑎 = 4, 𝑑𝑑 = 3
𝑇𝑇60 = 4 + (60 − 1)3
𝑇𝑇60 = 4 + 177 = 181
181 = 𝑇𝑇1 of the column going across from row 60
To find the number in the 60th row
and 70th column, we need to find
the first number in the 60th row and
then go to the 70th column.
Using the formula for an arithmetic
sequence on page 22 of the Maths
Tables Book
𝑇𝑇2 = The 60th term down the second column
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101
𝑇𝑇𝑛𝑛 = 𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑
We now need to find 𝑇𝑇2 of this
column in order to find the
common difference and then the
70th term.
𝑛𝑛 = 60, 𝑎𝑎 = 7, 𝑑𝑑 = 5
𝑇𝑇60 = 7 + (60 − 1)5
𝑇𝑇60 = 7 + 295 = 302
We do this by finding the 60th term
going down the 2nd column.
𝑇𝑇2 = 302
Now we need to find the 70th column across:
𝑇𝑇𝑛𝑛 = 𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑
𝑛𝑛 = 70, 𝑎𝑎 = 181,
𝑑𝑑 = 𝑇𝑇2 − 𝑇𝑇1 = 302 − 181 = 121
Finally, we find the 70th term in this
sequence.
𝑇𝑇70 = 181 + (70 − 1)(121)
𝑇𝑇70 = 181 + 8349 = 8350
b)
−𝟐𝟐
𝑎𝑎1 = 4, 𝑎𝑎2 = 2
𝑎𝑎𝑛𝑛 = 𝑎𝑎𝑛𝑛−1 − 𝑎𝑎𝑛𝑛−2
𝑎𝑎3 = 𝑎𝑎2 − 𝑎𝑎1 = 2 − 4 = −2
𝑎𝑎4 = 𝑎𝑎3 − 𝑎𝑎2 = −2 − 2 = −4
𝑎𝑎5 = 𝑎𝑎4 − 𝑎𝑎3 = −4 − (−2) = −2
𝑎𝑎6 = 𝑎𝑎5 − 𝑎𝑎4 = −2 − (−4) = 2
𝑎𝑎7 = 𝑎𝑎6 − 𝑎𝑎5 = 2 − (−2) = 4
𝑎𝑎8 = 𝑎𝑎7 − 𝑎𝑎6 = 4 − 2 = 2
Repeating pattern of 4, 2, −2, −4, −2, 2
2 is every 6th number:
2016 = 2 as it is divisible of 6
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→ 2017 = 4, 2018 = 2, 2019 = −2
102
Question 6
a)
(0,0), (1,1), (−1, −1)
𝑥𝑥 = 𝑥𝑥 3
To find where they intersect we let the
functions equal each other.
𝑥𝑥(𝑥𝑥 2 − 1) = 0
Taking 𝑥𝑥 from both sides.
𝑥𝑥 2 = 1
Solving for 𝑥𝑥
𝑥𝑥 3 − 𝑥𝑥 = 0
Factorising out the 𝑥𝑥
𝑥𝑥 = 0
When we square root to find an unknown it
can always equal ± the value.
𝑥𝑥 = √1 → 𝑥𝑥 = ±1
Subbing our values for 𝑥𝑥 back into the equation
for the line to find the corresponding y-values.
ℎ(𝑥𝑥) = 𝑥𝑥
ℎ(0) = 0
(0,0)
ℎ(1) = 1
(1,1)
ℎ(−1) = −1
(−1, −1)
b) i)
1
sq. units
2
1
1
0
0
Area = �� 𝑥𝑥 − � 𝑥𝑥 3 𝑑𝑑𝑑𝑑� + �� 𝑥𝑥 − � 𝑥𝑥 3 𝑑𝑑𝑑𝑑 �
1
0
1
0
0
−1
0
−1
𝑥𝑥 4
𝑥𝑥 2
𝑥𝑥 4
𝑥𝑥 2
�� � − � � � + � � � − � � �
2 0
4 0
2 −1
4 −1
��
(1)2 (0)2
(1)4 (0)4
(0)2 (1)2
(0)4 (−1)4
−
�−�
−
�� + ��
−
�−�
−
��
2
2
4
4
2
2
4
4
1
1
1 1
� − � + �− − �− ��
2
4
2 4
1
1
� � + [− ]
4
4
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To find the area between them we
find the area beneath the curve and
take it from the area beneath the
line. We find the area between 1 and
−1 as this is where they meet. We
add the value for this from the
positive and negative parts of the
graph.
Subbing in 1 and 0 for 𝑥𝑥
1
Forcing − to be positive as it is
4
representing area.
103
1 1 1
+ = sq. units
4 4 2
𝑖𝑖𝑖𝑖)
Using algebra:
𝑘𝑘(𝑥𝑥) = 𝑥𝑥 3
𝑘𝑘 −1 →
𝑦𝑦 = 𝑥𝑥 3
1
𝑦𝑦 3 = 𝑥𝑥
1
𝑥𝑥 3 = 𝑘𝑘 −1 (𝑥𝑥)
1
Plot 𝑥𝑥 3
Or use symmetry:
𝑘𝑘 −1 (𝑥𝑥)
𝑘𝑘(𝑥𝑥)
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104
Question 7
a)
𝑡𝑡(𝑥𝑥) = 𝑘𝑘[ln(1 −
𝑥𝑥
)]
80
𝑡𝑡 = 35.96, 𝑥𝑥 = 35
35.96 = 𝑘𝑘[ln(1 −
35.96
= 𝑘𝑘
35
[ln(1 − )]
80
35
)]
80
−62.499 = 𝑘𝑘
To find 𝑘𝑘 we simply sub in the given values for 𝑡𝑡
and 𝑥𝑥.
Dividing across by [ln(1 −
35
)]
80
To one decimal place
−62.5 = 𝑘𝑘
b)
64wpm
𝑡𝑡 = 100
100 = −62.5[ln(1 −
𝑥𝑥
100
= ln(1 − )
80
−62.5
100
𝑒𝑒 −6.25
𝑥𝑥
)]
80
𝑥𝑥
=1−
80
100
(𝑒𝑒 −6.25
𝑥𝑥
− 1) = −
80
100
−80 �𝑒𝑒 −62.5 − 1� = 𝑥𝑥
63.84 = 𝑥𝑥
64𝑤𝑤𝑤𝑤𝑤𝑤 = 𝑥𝑥
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Plugging in 100 for 𝑡𝑡
Dividing across by −62.5
From page 21 of the Maths Tables Book
𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥 In this case 𝑎𝑎 = 𝑒𝑒
→ 𝑒𝑒 𝑥𝑥 = 𝑦𝑦 ↔ ln𝑎𝑎 𝑦𝑦 = 𝑥𝑥
Multiplying across by −80
To the nearest whole number
105
c)
𝑥𝑥
0
10
20
30
40
50
60
70
𝑡𝑡(𝑥𝑥)
0
8
18
29
43
61
87
130
(wpm)
(days)
𝑡𝑡 = 𝑡𝑡 = −62.5[ln(1 −
0
)] = 0
80
𝑡𝑡 = −62.5[ln(1 −
10
)] = 8
80
𝑡𝑡 = −62.5[ln(1 −
30
)] = 29
80
𝑡𝑡 = −62.5[ln(1 −
𝑡𝑡 = −62.5[ln(1 −
𝑡𝑡 = −62.5[ln(1 −
𝑡𝑡 = −62.5[ln(1 −
𝑡𝑡 = −62.5[ln(1 −
20
)] = 18
80
40
)] = 43
80
50
)] = 61
80
60
)] = 87
80
70
)] 130
80
d)
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106
e) i)
62wpm
ℎ(𝑥𝑥) = 𝑝𝑝(𝑥𝑥) − 𝑡𝑡(𝑥𝑥)
𝑝𝑝(𝑥𝑥) − 𝑡𝑡(𝑥𝑥) = 0
Where 𝑝𝑝(𝑥𝑥) = 𝑡𝑡(𝑥𝑥)
Approximately 62wpm
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107
ii)
17 days
ℎ(𝑥𝑥) = 1.5𝑥𝑥 − (−62.5[ln(1 −
ℎ(𝑥𝑥) = 1.5𝑥𝑥 + 62.5[ln(1 −
ℎ′ (𝑥𝑥) = 1.5 +
Let ℎ′ (𝑥𝑥) = 0
→ 1.5 + 62.5 �
𝑥𝑥
)] )
80
To find the max we differentiate the function and
let it equal 0.
𝑥𝑥
)]
80
1
𝑥𝑥 × − 80
1−
80
1
ln 1 −
𝑥𝑥
80
→
𝑥𝑥
1
1−
𝑥𝑥
80
We also have to multiply this by the derivative of
𝑥𝑥
, (chain rule).
80
1
𝑥𝑥 × − 80� = 0
1−
80
1
Taking 1.5 from both sides and multiplying the
fractions
62.5
= −1.5
𝑥𝑥
�1 − � − 80
80
62.5
= −1.5
−80 + 𝑥𝑥
Multiplying across by (−80 + 𝑥𝑥)
62.5 = −1.5(−80 + 𝑥𝑥)
Dividing across by (−1.5)
62.5
= −80 + 𝑥𝑥
−1.5
−
1
From pg 25 of the Maths Tables Book, ln 𝑥𝑥 → . So
Adding 80 to both sides
125
+ 80 = 𝑥𝑥
3
115
= 𝑥𝑥
3
We now plug this value for 𝑥𝑥 into the original
equation to find the maximum number of days.
𝑥𝑥 = 38.33
ℎ(38.33) = 1.5(38.33) + 62.5 ln �1 −
= 17 days
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38.33
� = 16.73
80
108
Question 8
a)
�0,
1
√2𝜋𝜋
�
The graph intersects the y-axis when 𝑥𝑥 = 0
→ 𝑓𝑓(0) =
1
√2𝜋𝜋
�0,
1
√2𝜋𝜋
(1) =
1
√2𝜋𝜋
�
1
1
𝑒𝑒 −2
(0)2
√2𝜋𝜋
b)
2
√2𝜋𝜋𝜋𝜋
Area of rectangle = length × width
Length = |𝐵𝐵𝐵𝐵|
As the function is symmetric 𝐶𝐶 = �1,
∴ |𝐵𝐵𝐵𝐵| = From − 1 to + 1 = 2
Width =
√2𝜋𝜋𝜋𝜋
�
√2𝜋𝜋𝜋𝜋
Area = 2 ×
2
1
1
√2𝜋𝜋𝜋𝜋
1
√2𝜋𝜋𝜋𝜋
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109
c)
𝑓𝑓(𝑥𝑥) =
1
√2𝜋𝜋
1 2
𝑒𝑒 −2𝑥𝑥
To see if a function is decreasing at a certain point, we
differentiate it and then plug int the x coordinate.
1 1 −1𝑥𝑥 2
𝑒𝑒 2 � × 2(𝑥𝑥)
𝑓𝑓 ′ (𝑥𝑥) = − �
2 √2𝜋𝜋
1
√2𝜋𝜋
From page 25 of The Maths Tables Book 𝑒𝑒 𝑎𝑎𝑎𝑎 → 𝑎𝑎𝑒𝑒 𝑎𝑎𝑎𝑎
1 2
1 1 2
∴ 𝑒𝑒 −2𝑥𝑥 → − 𝑒𝑒 −2𝑥𝑥
2
1 2
𝑒𝑒 −2𝑥𝑥 × −
𝐶𝐶 = (1, 1√2𝜋𝜋)
𝑓𝑓 ′ (1) =
1
√2𝜋𝜋
= −0.24197
1
We then need to multiply by the derivative of the 𝑥𝑥 2
2
𝑒𝑒 −2(1) × −(1)
Now we plug in the x-value of the coordinates of 𝐶𝐶
As the slope is negative the graph is increasing.
→Decreasing
d)
Point of inflection
𝑑𝑑 2 𝑦𝑦
𝑑𝑑𝑥𝑥 2
=0
1 −1𝑥𝑥 2
𝑑𝑑𝑑𝑑
=
𝑒𝑒 2 × −𝑥𝑥
𝑑𝑑𝑑𝑑 √2𝜋𝜋
At a point of inflection
1 −1𝑥𝑥 2
1 −1𝑥𝑥 2
𝑑𝑑 2 𝑦𝑦
=
𝑒𝑒 2 × (−1) + �−𝑥𝑥 ×
𝑒𝑒 2 (−𝑥𝑥)�
2
𝑑𝑑𝑥𝑥
√2𝜋𝜋
√2𝜋𝜋
−
1
√2𝜋𝜋
1 2
𝑒𝑒 −2𝑥𝑥 + �𝑥𝑥 2 ×
1
√2𝜋𝜋
𝑓𝑓 ′′ (−1) = −
1
√2𝜋𝜋
1
2
1
√2𝜋𝜋
�
𝑒𝑒 −2(−1) + �(−1)2 ×
= 0 ∴ Point of inflection
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𝑑𝑑𝑥𝑥 2
=0
Using the product rule:
1 2
𝑒𝑒 −2𝑥𝑥 �
Point of inflection at 𝐵𝐵 �−1,
𝑑𝑑 2 𝑦𝑦
1
√2𝜋𝜋
1
Plugging in the x -value of the B
coordinate.
2
𝑒𝑒 −2(−1) �
110
Question 9
a)
Step
0
1
2
3
Triangles
remaining
1
3
9
27
Fraction of
original triangle
remaining
1
3
4
9
16
27
64
b) i)
3𝑛𝑛
𝑇𝑇𝑛𝑛 = 𝑎𝑎𝑟𝑟 𝑛𝑛
𝑎𝑎 = 1, 𝑟𝑟 =
(1)3𝑛𝑛
𝑇𝑇2 3
= =3
𝑇𝑇1 1
3𝑛𝑛
Using the expression for a geometric
sequence from page 22 of the
Maths Tables Book.
We use 𝑟𝑟 𝑛𝑛 instead of 𝑟𝑟 𝑛𝑛−1 as the
steps begin at 0 and not 1.
ii)
19th step
3𝑘𝑘 = 1 × 109
log 3 1 × 109 = 𝑘𝑘
18.86 = 𝑘𝑘
∴ It exceeds 1 × 109 for the first time on the 19th step.
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Using the laws of logs from page 21
of the Maths Tables Book.
𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥
111
c) i)
17
Sequence for the fraction remaining:
Creating a geometric sequence for the fraction of
the triangle remaining.
𝑇𝑇𝑛𝑛 = 𝑎𝑎𝑟𝑟 𝑛𝑛
Using the expression for a geometric sequence
from page 22 of the Maths Tables Book.
3
𝑇𝑇2 4 3
𝑎𝑎 = 1, 𝑟𝑟 =
= =
𝑇𝑇1 1 4
We use 𝑟𝑟 𝑛𝑛 instead of 𝑟𝑟 𝑛𝑛−1 as the steps begin at
0 and not 1.
𝑛𝑛
3
(1) � �
4
3 𝑛𝑛
� �
4
1
3 𝑛𝑛
� � =
100
4
Using the laws of logs from page 21 of the Maths
Tables Book.
1
= 𝑛𝑛
4 100
𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥
log 3
16.0078 = 𝑛𝑛
∴ 17 is the first time the fraction is less than
1
100
ii)
0
3 𝑛𝑛
lim � � = 0
𝑛𝑛→∞ 4
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As 𝑛𝑛 increases the fraction gets approaches 0
At 𝑛𝑛 = ∞ the limit of the expression is 0
112
d) i)
Step
0
1
2
3
4
Perimeter
3
9
2
27
4
81
8
243
16
1
In Step 1 the triangles split each side in half. As the triangles are equilateral each side is of length .
1
1
1
The perimeter of each triangle therefore is, + + =
There are 3 triangles → Total perimeter =
9
2
2
2
2
3
2
2
𝑆𝑆𝑛𝑛 = 𝑎𝑎𝑟𝑟 𝑛𝑛
9
3
𝑎𝑎 = 3, 𝑟𝑟 = 2 =
3 2
3 𝑛𝑛
𝑠𝑠𝑛𝑛 = 3 � �
2
3 3 81
𝑆𝑆3 = 3 � � =
8
2
3 4 243
𝑆𝑆4 = 3 � � =
16
2
Creating a geometric sequence to find
the perimeters in the 3rd and 4th steps
Plugging in 3 for 𝑛𝑛
Plugging in 4 for 𝑛𝑛
ii)
4368329 units
3 𝑛𝑛
𝑠𝑠𝑛𝑛 = 3 � �
2
𝑆𝑆35
3 35
= 3 � � = 4368329 units
2
iii)
Area = 0
Using the expression for the
geometric sequence we made in
part i).
Subbing in 35 for 𝑛𝑛
From part c ii)
𝑛𝑛
3
lim 3 � � = ∞
𝑛𝑛→∞
2
The perimeter tends to infinity and the area tends to 0
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As 𝑛𝑛 increases so does the
perimeter, as 𝑛𝑛 approaches ∞ so
does the perimeter.
113
2017 Paper 1
Question 1
a)
7 2 129
2 �𝑥𝑥 − � −
8
4
2𝑥𝑥 2 − 7𝑥𝑥 − 10
7
2 �𝑥𝑥 2 − 𝑥𝑥 − 5�
2
7
7 2
7 2
2 �𝑥𝑥 2 − 𝑥𝑥 + � � − � � − 5�
2
4
4
7
49 49
−
− 5�
2 �𝑥𝑥 2 − 𝑥𝑥 +
2
16 16
7
49
7
2 ��𝑥𝑥 − � �𝑥𝑥 − � −
− 5�
4
16
4
To complete the square the coefficient of
the 𝑥𝑥 2 must be 1 so we start by factorising
out the 2.
We then half the coefficient of the 𝑥𝑥. Then,
we square this and add it and take it from
the expression.
We can now factorise the quadratic of
7
𝑥𝑥 2 − 𝑥𝑥 +
2
49
16
7 2 129
�
2 ��𝑥𝑥 − � −
16
4
7 2 129
2 �𝑥𝑥 − � −
4
8
Multiplying −
129
16
by 2
b)
7 129
�
� ,−
8
4
If a quadratic is in the form (𝑥𝑥 − 𝑝𝑝)2 + 𝑞𝑞
The minimum point is (𝑝𝑝, 𝑞𝑞)
c) i)
If a function has two real roots 𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎 > 0
2𝑥𝑥 2 − 7𝑥𝑥 − 10
(−7)2 − 4(2)(−10)
𝑎𝑎 = 2, 𝑏𝑏 = −7, 𝑐𝑐 = −10
49 + 80
129 > 0
∴ Two real roots
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114
ii)
−𝑏𝑏 ± �(𝑏𝑏)2 − 4𝑎𝑎𝑎𝑎
2𝑎𝑎
2𝑥𝑥 2 − 7𝑥𝑥 − 10
Using the minus b formula to find the
roots of the graph. (page 20 of the Maths
Tables Book)
−(−7) ± �(7)2 − 4(2)(−10)
2(2)
7 ± √49 + 80
4
7 √129
±
4
4
129
7
±�
16
4
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Separating the fraction
Squaring 4 to write it as part of the square
root.
115
Question 2
a)
−8 − 8√3𝑖𝑖
−√3 + 𝑖𝑖, 𝑎𝑎 = −√3, 𝑏𝑏 = 1
𝑟𝑟(cos 𝜃𝜃 + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖)
𝑟𝑟 = �𝑎𝑎2 + 𝑏𝑏 2
2
𝑟𝑟 = ��−√3� + (1)2
𝑟𝑟 = √3 + 1 = √4 = 2
θ = tan−1
1
√3
=
𝜃𝜃 = tan−1
𝜋𝜋
6
−√3 + 𝑖𝑖 is in the second quadrant. So, we take it from 𝜋𝜋
𝜋𝜋 −
𝜋𝜋 5𝜋𝜋
=
6
6
𝑧𝑧 = 2 �cos
5𝜋𝜋
5𝜋𝜋
+ 𝑖𝑖 sin �
6
6
5𝜋𝜋
5𝜋𝜋
𝑧𝑧 = �2 �cos
+ 𝑖𝑖 sin ��
6
6
4
𝑧𝑧 4 = 24 �cos 4 �
This is the argument.
4
5𝜋𝜋
5𝜋𝜋
� + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖4 � ��
6
6
20𝜋𝜋
20𝜋𝜋
� + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 �
�)
16(cos �
6
6
𝑏𝑏
𝑎𝑎
Subbing cos and sin into the calculator.
1 √3
16 �− −
𝑖𝑖�
2
2
−8 − 8√3𝑖𝑖
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116
b)
−6
𝑟𝑟 = 3
|𝑤𝑤| = 3, means that the
modulus = 3
𝜃𝜃 = 30
𝑡𝑡 = 𝑧𝑧𝑧𝑧
𝑤𝑤 = 3(cos 30 + 𝑖𝑖 sin 30)
𝜋𝜋
𝜋𝜋
𝑤𝑤 = 3(cos + 𝑖𝑖 sin )
6
6
𝑧𝑧𝑧𝑧 = 2 �cos
𝜋𝜋
𝜋𝜋
5𝜋𝜋
5𝜋𝜋
+ 𝑖𝑖 sin � × 3 �cos + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 �
6
6
6
6
𝜋𝜋
5𝜋𝜋
𝜋𝜋
𝜋𝜋
5𝜋𝜋
5𝜋𝜋
𝜋𝜋
5𝜋𝜋
6 �cos . cos
+ cos . 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖
+ 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 . cos
+ 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 . 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 �
6
6
6
6
6
6
6
6
𝜋𝜋
𝜋𝜋
5𝜋𝜋
𝜋𝜋
5𝜋𝜋
5𝜋𝜋
𝜋𝜋
5𝜋𝜋
6 �cos cos
+ icos 𝑠𝑠𝑠𝑠𝑠𝑠
+ 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 cos
− 𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠 �
6
6
6
6
6
6
6
6
𝜋𝜋
5𝜋𝜋
𝜋𝜋
5𝜋𝜋
𝜋𝜋
5𝜋𝜋
𝜋𝜋
5𝜋𝜋
6 �cos cos
− 𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠
+ icos 𝑠𝑠𝑠𝑠𝑠𝑠
+ 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 cos �
6
6
6
6
6
6
6
6
5𝜋𝜋 𝜋𝜋
𝜋𝜋 5𝜋𝜋
6 �cos � + � + 𝑖𝑖 sin � + ��
6
6
6
6
6(cos 𝜋𝜋 + 𝑖𝑖 sin 𝜋𝜋)
6(−1 + 0𝑖𝑖)
Converting 30 degrees to
radians
Multiplying 𝑧𝑧 by 𝑤𝑤
Multiplying the brackets:
𝑖𝑖 × 𝑖𝑖 = −1
Rearranging so the real and
imaginary numbers are
together.
Using the identities
cos(𝐴𝐴 + 𝐵𝐵) and sin(𝐴𝐴 + 𝐵𝐵)
(page 14 Maths Tables Book)
−6
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117
Question 3
a)
𝟐𝟐
𝒙𝒙 − 𝟏𝟏
𝟑𝟑
𝑓𝑓(𝑥𝑥) =
1 2
𝑥𝑥 − 𝑥𝑥 + 3
3
1
𝑓𝑓(𝑥𝑥 + ℎ) = (𝑥𝑥 + ℎ)2 − (𝑥𝑥 + ℎ) + 3
3
𝑓𝑓(𝑥𝑥 + ℎ) =
𝑓𝑓(𝑥𝑥 + ℎ) =
First, we sub in (𝑥𝑥 + ℎ) for 𝑥𝑥
1 2
(𝑥𝑥 + 2ℎ𝑥𝑥 + ℎ2 ) − 𝑥𝑥 − ℎ + 3
3
1
1 2 2
𝑥𝑥 + ℎ𝑥𝑥 + ℎ2 − 𝑥𝑥 − ℎ + 3
3
3
3
𝑓𝑓(𝑥𝑥 + ℎ) − 𝑓𝑓(𝑥𝑥) =
𝑓𝑓(𝑥𝑥 + ℎ) − 𝑓𝑓(𝑥𝑥) =
=
Tip: Differentiate normally first
so that you can check your
answer.
1
2
ℎ𝑥𝑥 + ℎ2 − ℎ
3
3
1
1 2 2
1
𝑥𝑥 + ℎ𝑥𝑥 + ℎ2 − 𝑥𝑥 − ℎ + 3 − � 𝑥𝑥 2 − 𝑥𝑥 + 3�
3
3
3
3
1
1 2 2
1
𝑥𝑥 + ℎ𝑥𝑥 + ℎ2 − 𝑥𝑥 − ℎ + 3 − 𝑥𝑥 2 + 𝑥𝑥 − 3
3
3
3
3
2
1 2
𝑓𝑓(𝑥𝑥 + ℎ) − 𝑓𝑓(𝑥𝑥) 3 ℎ𝑥𝑥 + 3 ℎ − ℎ
=
ℎ
ℎ
1
2
= 𝑥𝑥 + ℎ − 1
3
3
𝑓𝑓(𝑥𝑥 + ℎ) − 𝑓𝑓(𝑥𝑥) 𝟐𝟐
→ 𝒙𝒙 − 𝟏𝟏
lim
ℎ→0
𝟑𝟑
ℎ
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Then we take away 𝑓𝑓(𝑥𝑥) (the
original function) from the result
of this.
Then we divide by ℎ
Finally, we limit ℎ to 0, so we
plug in 0 for ℎ
Revise differentiation from first
principles in section 2 of
differentiation.
118
b)𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑
𝑓𝑓(𝑥𝑥) = ln(3𝑥𝑥 2 + 2) , 𝑔𝑔(𝑥𝑥) = 𝑥𝑥 + 5
𝑓𝑓(𝑔𝑔(𝑥𝑥)) means we sub in the function 𝑔𝑔(𝑥𝑥) for 𝑥𝑥
in the function 𝑓𝑓(𝑥𝑥)
𝑓𝑓�𝑔𝑔(𝑥𝑥)� = ln(3(𝑥𝑥 + 5)2 + 2)
→ Plugging in 𝑥𝑥 + 5 for 𝑥𝑥
ln(3(𝑥𝑥 2 + 10𝑥𝑥 + 25) + 2)
Multiplying it out.
ln(3𝑥𝑥 2 + 30𝑥𝑥 + 77)
𝑓𝑓 ′ �𝑔𝑔(𝑥𝑥)� =
1
× (6𝑥𝑥 + 30)
2
3𝑥𝑥 + 30𝑥𝑥 + 77
1
𝑓𝑓 �𝑔𝑔 � �� =
4
′
=
1
1 2
1
3 � � + 30 � � + 77
4
4
504
= 0.372
1355
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1
× �6 � � + 30�
4
To differentiate this, we first differentiate 𝑙𝑙𝑙𝑙
(look at page 25 of the Maths Tables Book)
And then multiply this by the derivative of
(3𝑥𝑥 2 + 30𝑥𝑥 + 77)
1
Subbing in for 𝑥𝑥
4
119
Question 4
a)
12th day
𝑇𝑇𝑛𝑛 = 𝑎𝑎𝑟𝑟 𝑛𝑛−1
𝑎𝑎 = 95
𝑟𝑟 =
9
𝑡𝑡2 42.75
=
=
5
20
𝑡𝑡1
𝑇𝑇𝑛𝑛 = 95 �
9 𝑛𝑛−1
�
20
9 𝑛𝑛−1
0.01 = 95 � �
20
0.01
9 𝑛𝑛−1
=� �
95
20
0.01
= 𝑛𝑛 − 1
log 9
20 95
11.47 = 𝑛𝑛 − 1
Making an expression for a
geometric sequence using the
formula on page 22 of The Maths
Tables Book.
Letting this expression = 0.01 to
solve for 𝑛𝑛 (the number of days).
Dividing across by 95.
Using the rule of logs:
𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥
Page 21 of The Maths Tables Book.
12.47 = 𝑛𝑛
12th day
b)
16 + 8√2𝑐𝑐𝑐𝑐
First perimeter = 4 × 2 = 8𝑐𝑐𝑐𝑐
Second Perimeter:
Side2 = 12 + 12
Side = √2
Perimeter = 4 × √2 = 4√2cm
𝑆𝑆∞ =
𝑎𝑎
1 − 𝑟𝑟
1cm
√2
1cm
𝑎𝑎 = 8𝑐𝑐𝑐𝑐
𝑟𝑟 =
4√2 √2
=
8
2
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120
8
√2
1−
2
= 16 + 8√2𝑐𝑐𝑐𝑐
Question 5
a)
𝟏𝟏
𝒙𝒙 = − , 𝒙𝒙 = 𝟏𝟏
𝟐𝟐
𝑓𝑓(𝑥𝑥) = 2𝑥𝑥 3 + 5𝑥𝑥 2 − 4𝑥𝑥 − 3
𝑓𝑓(−3) = 2(−3)3 + 5(−3)2 − 4(−3) − 3
−54 + 45 + 12 − 3
=0
∴ 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
If 𝑥𝑥 = −3 is a root, 𝑥𝑥 + 3 is a factor of the
equation.
→ 𝑥𝑥 + 3 is a factor
2𝑥𝑥 2 − 𝑥𝑥 − 1
𝑥𝑥 + 3 2𝑥𝑥 3 + 5𝑥𝑥 2 − 4𝑥𝑥 − 3
3
2𝑥𝑥 + 6𝑥𝑥
To show that a value is a root of a
function we sub it in for 𝑥𝑥 and show that
it equals 0
Dividing the factor in.
2
−𝑥𝑥 2 − 4𝑥𝑥
−𝑥𝑥 2 − 3𝑥𝑥
−𝑥𝑥 − 3
−𝑥𝑥 − 3
let 2x 2 − 𝑥𝑥 − 1 = 0
0
This leaves us with a quadratic which we
can let = 0 and solve for the remaining
two roots.
(2𝑥𝑥 + 1)(𝑥𝑥 − 1) = 0
2𝑥𝑥 + 1 = 0
𝒙𝒙 = −
𝟏𝟏
𝟐𝟐
𝑥𝑥 − 1 = 0
𝒙𝒙 = 𝟏𝟏
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121
b)
1 100
)
𝑀𝑀𝑀𝑀𝑀𝑀 = (−2,9), 𝑀𝑀𝑀𝑀𝑀𝑀 = ( , −
27
3
𝑦𝑦 = 2𝑥𝑥 3 + 5𝑥𝑥 2 − 4𝑥𝑥 − 3
𝑑𝑑𝑑𝑑
= 6𝑥𝑥 2 + 10𝑥𝑥 − 4
𝑑𝑑𝑑𝑑
6𝑥𝑥 2 + 10𝑥𝑥 − 4 = 0
To find the maximum and minimum values
of a function we differentiate it and let it
equal 0.
Solving the quadratic.
3𝑥𝑥 3 + 5𝑥𝑥 − 2 = 0
(3𝑥𝑥 − 1)(𝑥𝑥 + 2) = 0
𝑥𝑥 =
1
3
=−
100
27
𝑥𝑥 = −2
3
2
1
1
1
1
𝑓𝑓 � � = 2 � � + 5 � � − 4 � � − 3
3
3
3
3
1 100
�
� ,−
27
3
𝑓𝑓(−2) = 2(−2)3 + 5(−2)2 − 4(−2) − 3
=9
(−2,9)
2
𝑑𝑑 𝑦𝑦
= 12𝑥𝑥 + 10
𝑑𝑑𝑥𝑥 2
12(−2) + 10 = −14
−14 < 0 ∴ 𝑚𝑚𝑚𝑚𝑚𝑚
1
Plugging back into the function to find the
3
corresponding y coordinate.
Plugging −2 back into the function to find
the corresponding y coordinate.
To find which is the min and which is the
max we find the second derivative and plug
in the 𝑥𝑥 values.
As −2 gives us a value < 0 (−2,9) is the
1
maximum value, therefore ( , −
3
minimum value.
100
27
) is the
1 100
)
𝑀𝑀𝑀𝑀𝑀𝑀 = (−2,9), 𝑀𝑀𝑀𝑀𝑀𝑀 = ( , −
27
3
c)
𝑎𝑎 >
Adding a constant moves the graph up or down.
100
or 𝑎𝑎 < −9
27
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If we add a number greater than
100
27
two the
minimum point will move above the x-axis
leaving only one root. Similarly, if we subtract a
number < −9 the maximum point will move
below the x-axis leaving only one root.
122
Question 6
a)
𝑔𝑔(𝑥𝑥)
ℎ(𝑥𝑥)
ℎ(𝑥𝑥) = 𝑒𝑒 −𝑥𝑥
ℎ(0) = 𝑒𝑒 −0 = 1
ℎ(0.2) = 𝑒𝑒 −0.2 = 0.82
ℎ(0.4) = 𝑒𝑒 −0.4 = 0.67
ℎ(0.6) = 𝑒𝑒 −0.6 = 0.55
ℎ(0.8) = 𝑒𝑒 −0.8 = 0.45
ℎ(1) = 𝑒𝑒 −1 = 0.37
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123
b)
0.5894
Area = �
0.75
0
𝑒𝑒 𝑥𝑥 𝑑𝑑𝑑𝑑 − �
[𝑒𝑒 𝑥𝑥 ]0.75
− [−1𝑒𝑒 −𝑥𝑥 ]0.75
0
0
0.75
0
𝑒𝑒 −𝑥𝑥 𝑑𝑑𝑑𝑑
[(𝑒𝑒 0.75 ) − (𝑒𝑒 0 )] − [(−1𝑒𝑒 −0.75 ) − (−1𝑒𝑒 −0 )]
[2.117 − 1] − [−0.4724 + 1]
1.117 − 0.5276
= 0.5894
𝑔𝑔(𝑥𝑥)
ℎ(𝑥𝑥)
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124
Question 7
a) 1.1
𝑝𝑝(𝑡𝑡) = 𝑆𝑆𝑒𝑒 0.1𝑡𝑡 × 106
𝑝𝑝(0) = 𝑆𝑆𝑒𝑒 0.1(0) × 106 = 1,100,000
6
𝑆𝑆(1) × 10 = 1,100,000
𝑆𝑆 =
1,100,000
= 𝟏𝟏. 𝟏𝟏
106
𝑡𝑡 = 0 in 2010
So, we plug in 0 for 𝑡𝑡 and let
it equal to the population
(1,100,000).
Dividing across by 106
b) 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 people
2015 → 𝑡𝑡 = 5
𝑝𝑝(𝑡𝑡) = 1.1𝑒𝑒 0.1𝑡𝑡 × 106
𝑡𝑡 = 5 in 2015
= 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 people
Plugging in 5 for 𝑡𝑡
𝑝𝑝(5) = 1.1𝑒𝑒 0.1(5) × 106
c) 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 people
𝑝𝑝(6) = 1.1𝑒𝑒
= 2004330
0.1(6)
6
× 10
𝑝𝑝(6) − 𝑝𝑝(5)
= 2004330 − 1813593 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 people
© Pocket Tutor 2022
Plugging 1.1 in for 𝑆𝑆
Change in population during 2015 = The
population at the start of 2016 – The
population at the start of 2015.
Or 𝑝𝑝(6) − 𝑝𝑝(5)
So, we sub in 6 for 𝑡𝑡 to find 𝑝𝑝(6) and
then take our answer from the previous
question away from it.
125
d)
𝑞𝑞(𝑡𝑡) = 3.9𝑒𝑒 𝑘𝑘𝑘𝑘 × 106
𝑞𝑞(1) = 3.9𝑒𝑒 𝑘𝑘(1) × 106 = 3,709,795
3.9𝑒𝑒 𝑘𝑘 =
3,709,795
106
3.9𝑒𝑒 𝑘𝑘 = 3.709795
𝑒𝑒 𝑘𝑘 =
3.709795
3.9
3.709795
= 𝑘𝑘
log 𝑒𝑒
3.9
−0.0499 = 𝑘𝑘
𝑘𝑘 = −0.05
𝑡𝑡 = 1 in 2011.
Subbing in 1 for 𝑡𝑡 and letting it
equal the population in 2011.
Dividing by 106
Dividing by 3.9
Using the law of logs from page
21 of the Maths Tables Book
𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥
To two decimal places
𝒆𝒆)
2018
𝑝𝑝(𝑡𝑡) = 𝑞𝑞(𝑡𝑡)
Letting the equations equal each other to
see when the populations are equal.
1.1𝑒𝑒 0.1𝑡𝑡 = 3.9𝑒𝑒 −0.05𝑡𝑡
Dividing across by 106
1.1𝑒𝑒 0.1𝑡𝑡 × 106 = 3.9𝑒𝑒 −0.05𝑡𝑡 × 106
3.9
𝑒𝑒 0.1𝑡𝑡
=
−0.05𝑡𝑡
1.1
𝑒𝑒
𝑒𝑒 0.1𝑡𝑡−(−0.05𝑡𝑡)
𝑒𝑒 0.15𝑡𝑡 =
log 𝑒𝑒
3.9
1.1
3.9
=
1.1
3.9
= 0.15𝑡𝑡
1.1
1.2657 = 0.15𝑡𝑡
1.2657
= 𝑡𝑡
0.15
𝑡𝑡 = 8.44 years
Dividing across by 1.1 and 𝑒𝑒 −0.05
When dividing indices, we subtract the
powers (pg 21 Maths Tables Book)
Using the law of logs from page 21 of the
Maths Tables Book
𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥
Dividing by 0.15
If the populations are equal 8.44 years
after 2010, they will be equal in 2018.
The populations will be equal in 2018
© Pocket Tutor 2022
126
f)
2743694
2025 − 2010 = 15
→ Average value between years 0 and 15
To find the average value of an
equation we integrate it within
the given limits.
15
1
� 3.9𝑒𝑒 −0.05𝑡𝑡 × 106 𝑑𝑑𝑑𝑑
15 − 0 0
15
1 3.9 −0.05𝑡𝑡
�
𝑒𝑒
× 106 �
15 −0.05
0
Integrating 𝑒𝑒 −0.05𝑡𝑡 following the
rule on page 26 of The Maths
1
3.9 −0.05(15)
3.9 −0.05(0)
��
𝑒𝑒
× 106 � − �
𝑒𝑒
× 106 ��
15 −0.05
−0.05
1
[(−36844591.11) − (−78000000)]
15
1
Tables Book: 𝑒𝑒 𝑎𝑎𝑎𝑎 → 𝑒𝑒 𝑎𝑎𝑎𝑎
𝑎𝑎
Subbing in 15 and 0 for 𝑡𝑡
1
[41155408.89]
15
= 2743693.9 = 2743694
g)
−130712
𝑑𝑑𝑑𝑑
= −0.05(3.9)𝑒𝑒 −0.05𝑡𝑡 × 106
𝑑𝑑𝑑𝑑
𝑞𝑞 ′ (8) = −0.05(3.9)𝑒𝑒 −0.05(8) × 106
= −130712
© Pocket Tutor 2022
To find the rate of change of something at a given
point we differentiate and plug in the point.
The derivative of 𝑒𝑒 𝑎𝑎𝑎𝑎 = 𝑎𝑎𝑒𝑒 𝑎𝑎𝑎𝑎 (page 25 of The Maths
Tables Book)
Plugging in 8 for 𝑡𝑡
127
Question 8
a)
𝑃𝑃 =
𝐴𝐴
𝐴𝐴
𝐴𝐴
+
…
+
(1 + 𝑖𝑖)𝑡𝑡
1 + 𝑖𝑖 (1 + 𝑖𝑖)2
𝑎𝑎 =
𝐴𝐴
, 𝑛𝑛 = 𝑡𝑡
1 + 𝑖𝑖
𝑆𝑆𝑛𝑛 =
𝑎𝑎(1 − 𝑟𝑟
1 − 𝑟𝑟
𝑛𝑛 )
𝐴𝐴
𝐴𝐴
1 + 𝑖𝑖
1
(1 + 𝑖𝑖)2
→
×
=
𝑟𝑟 =
2
𝐴𝐴
(1 + 𝑖𝑖)
1 + 𝑖𝑖
𝐴𝐴
1 + 𝑖𝑖
1 𝑡𝑡
𝐴𝐴
� �1 − �
��
1 + 𝑖𝑖
1 + 𝑖𝑖
𝑃𝑃 =
1
�
1−�
1 + 𝑖𝑖
�
1 𝑡𝑡
��
𝐴𝐴
1 + 𝑖𝑖
×
𝑃𝑃 =
1
1 + 𝑖𝑖
�
1−�
1 + 𝑖𝑖
�1 − �
𝑃𝑃 =
𝑃𝑃 =
𝑃𝑃 =
𝐴𝐴 �1 −
𝐴𝐴 �
(1)𝑡𝑡
(1 + 𝑖𝑖)𝑡𝑡
1 + 𝑖𝑖 − 1
�
(1 + 𝑖𝑖)𝑡𝑡 − 1
�
(1 + 𝑖𝑖)𝑡𝑡
𝑖𝑖
𝐴𝐴 (1 + 𝑖𝑖)𝑡𝑡 − 1
×
(1 + 𝑖𝑖)𝑡𝑡
𝑖𝑖
𝑡𝑡
𝐴𝐴((1 + 𝑖𝑖) − 1)
𝑃𝑃 =
𝑖𝑖(1 + 𝑖𝑖)𝑡𝑡
𝑃𝑃(𝑖𝑖(1 + 𝑖𝑖)𝑡𝑡 ) = 𝐴𝐴((1 + 𝑖𝑖)𝑡𝑡 − 1)
𝑃𝑃(𝑖𝑖)(1 + 𝑖𝑖)𝑡𝑡
= 𝐴𝐴
(1 + 𝑖𝑖)𝑡𝑡 − 1
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The Principal amount is equal to the sum of
the present values of the repayments.
𝑟𝑟 = the second term divided by the first.
To divide by a fraction, we invert it and
multiply.
Subbing 𝑎𝑎, 𝑟𝑟 and 𝑛𝑛 into 𝑆𝑆𝑛𝑛 =
𝑎𝑎(1−𝑟𝑟 𝑛𝑛 )
1−𝑟𝑟
This is the same as the last line, just rewritten
to show we are going to multiply fraction by
𝐴𝐴
1+𝑖𝑖
Multiplying the top by the top and the bottom
by the bottom.
Multiplying the 1 × (1 + 𝑖𝑖)𝑡𝑡 so it can be
written as one fraction. Also 1 to the power of
anything is 1.
Taking the bracket out to multiply it by the
fraction
𝐴𝐴
𝑖𝑖
Multiplying top by top and bottom by bottom.
Multiplying across by the bottom of the
fraction.
Dividing across by the bracket beside A.
128
b)
i)
2.5% = 0.025
0.025 × 5000 = €𝟏𝟏𝟏𝟏𝟏𝟏
Finding 2.5% of 5000
ii)
21.75% = 0.2175
(1 + 0.2175) = (1 + 𝑟𝑟)12
12
√1.2175 = 1 + 𝑟𝑟
1.01654 = 1 + 𝑟𝑟
𝑟𝑟 = 0.01654
= 𝟏𝟏. 𝟔𝟔𝟔𝟔%
© Pocket Tutor 2022
129
iii)
Payment
Number
Fixed monthly
payment, €A
New balance of
debt (€)
€A
Interest
Previous balance
reduced by (€)
0
5000
1
125
82.50
42.50
4957.50
2
125
81.80
43.20
4914.30
3
125
81.09
43.91
4870.39
Payment 1:
Interest: 5000 × 0.0165 = 82.50
New Balance of Debt: 5082.5 − 125 = 4957.50
Previous Balance Reduced by: 5000 − 4957.50 = 42.50
The original debt + interest gives us 5082.5
Payment 2:
Interest: 4957.50 × 0.0165 = 81.80
New Balance of Debt: 5039.3 − 125 = 4914.30
Previous Balance Reduced by: 4957.50 − 4914.30 = 43.20
Payment 3:
Interest: 4914.30 × 0.0165 = 81.09
New Balance of Debt: 4995.39 − 125 = 4870.39
Previous Balance Reduced by: 4914.30 − 4870.39 = 43.91
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130
iv)
66 months
𝐴𝐴 =
Taking the formula we derived in part 𝑎𝑎. (It
can also be found on page 31 of the Maths
Tables Book)
𝑃𝑃(𝑖𝑖)(1 + 𝑖𝑖)𝑡𝑡
(1 + 𝑖𝑖)𝑡𝑡 − 1
𝐴𝐴 = 125 𝑃𝑃 = 5,000, 𝑖𝑖 = 0.0165
125 =
(5,000)((0.0165)(1 + 0.0165)𝑡𝑡
(1 + 0.0165)𝑡𝑡 − 1
82.5(1.0165)𝑡𝑡
125 =
(1.0165)𝑡𝑡 − 1
𝑡𝑡
125((1.0165) − 1) = 82.5(1.0165)
𝑡𝑡
Subbing in our values for 𝐴𝐴, 𝑃𝑃 and 𝑖𝑖
Multiplying across by the bottom of the
fraction.
125(1.0165)𝑡𝑡 − 125 = 82.5(1.0165)𝑡𝑡
Getting the unknowns on one side.
42.5(1.0165)𝑡𝑡 = 125
Dividing across by 42.5
125(1.0165)𝑡𝑡 − 82.5(1.0165)𝑡𝑡 = 125
(1.0165)𝑡𝑡 =
125
42.5
125
� = 𝑡𝑡
log1.0165 �
42.5
𝑡𝑡 = 65.92
𝑡𝑡 = 66 months
Using the law of logs from page 21 of the
Maths Tables Book
𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥
v)
€36.16
𝑖𝑖 → (1 + 0.085) = (1 + 𝑟𝑟)52
52
√1.085 = 1 + 𝑟𝑟
1.00157 = 1 + 𝑟𝑟
𝑟𝑟 = 0.00157
𝐴𝐴 =
𝑃𝑃(𝑖𝑖)(1 + 𝑖𝑖)𝑡𝑡
(1 + 𝑖𝑖)𝑡𝑡 − 1
𝐴𝐴 =
5000(0.00157)(1 + 0.00157)156
(1 + 0.00157)156 − 1
First of all we need to find the weekly
interest rate.
We then use this in the formula from
page 31 of the Maths Tables Book to find
the amount of each repayment.
𝐴𝐴 = 𝐴𝐴, 𝑃𝑃 = 5000, 𝑖𝑖 = 0.00157, 𝑡𝑡 = 156
𝐴𝐴 = €36.16
© Pocket Tutor 2022
131
vi)
€2609.04
125 × 66 − 36.16 × 156
8250 − 5640.96 = €2609.04
© Pocket Tutor 2022
Paying the weekly credit union
repayments will cost him 5640.96.
Paying the credit card debt will
cost him 8250.
So he saves 8250 − 5640.96
132
Question 9
a)
Putting high tides at 2 and 14: 34
Putting low tide halfway between these at a height of 1.7
b) i)
𝑎𝑎 = 3.6, 𝑏𝑏 = 1.9
Range: [(𝑎𝑎 + 𝑏𝑏), (𝑎𝑎 − 𝑏𝑏)]
𝑎𝑎 + 𝑏𝑏 = 5.5
𝑎𝑎 = 5.5 − 𝑏𝑏
𝑎𝑎 − 𝑏𝑏 = 1.7
(5.5 − 𝑏𝑏) − 𝑏𝑏 = 1.7
5.5 − 1.7 = 2𝑏𝑏
3.8 = 2𝑏𝑏
𝑏𝑏 = 1.9
𝑎𝑎 = 5.5 − 1.9
𝑎𝑎 = 3.6
© Pocket Tutor 2022
The range of a trigonometric function is found
by looking at the values of a and b when the
trig function is in this format:
𝑓𝑓(𝑡𝑡) = 𝑎𝑎 + 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏
Range: [(𝑎𝑎 + 𝑏𝑏), (𝑎𝑎 − 𝑏𝑏)]
Subbing in (5.5 − 𝑏𝑏) for 𝑎𝑎
Dividing by 2
Subbing in 1.9 for 𝑏𝑏
133
ii)
2𝜋𝜋
= The period of the graph
𝑐𝑐
Time between two high tides:
If a function = cos 𝑎𝑎𝑎𝑎, the period of
34
14: 34 − 2: 00 = 12: 34 → 12
hours
60
The period of a function is the time
taken for it to repeat, so in this case
we can find the period by finding the
time between high tides.
Period = 12
12
34 2𝜋𝜋
=
𝑐𝑐
60
34
60
2𝜋𝜋
= 𝑐𝑐 = 0.4999 = 0.5
34
12
60
it =
2𝜋𝜋
𝑎𝑎
Multiplying across by 𝑐𝑐 and then
dividing across by 12
34
60
c)
13: 26 and 15: 42
𝑓𝑓(𝑡𝑡) = 3.6 + 1.9cos(0.5𝑡𝑡)
Subbing in our values for 𝑎𝑎, 𝑏𝑏 and 𝑐𝑐
5.2 − 3.6 = 1.9 cos(0.5𝑡𝑡)
Taking 3.6 from both sides.
5.2 = 3.6 + 1.9cos(0.5𝑡𝑡)
1.6
= cos(0.5𝑡𝑡)
1.9
cos −1
1.6
= 0.5𝑡𝑡
1.9
Letting the equation equal 5.2
Dividing across by 1.9
0.5696 = 0.5𝑡𝑡
0.5696
= 𝑡𝑡
0.5
𝑡𝑡 = 1.139 hours
0.139 × 60 = 8 minutes
→ 1 hour 8 minutes
14: 34 ± 1: 08
= 13: 26 and 15: 42
© Pocket Tutor 2022
Multiplying 60 by 0.139 to find how
many minutes there are in 0.139
hours.
As this is a cos graph the height of
5.2 metres is reached on either side
of high tide. So, it is reached 1 hour
and 8 minutes before high tide as
well as 1 hour and 8 minutes after
high tide.
134
2016 Paper 1
Question 1
a)
(−4 − 3𝑖𝑖)
If a complex number is the root of an equation its conjugate is also a root. We
get the conjugate by changing the sign of the imaginary part of the number
b)
𝟏𝟏𝟏𝟏 + 𝟎𝟎𝟎𝟎
Modulus = 𝑟𝑟 = �𝑎𝑎2 + 𝑏𝑏 2
𝑟𝑟 = �(1)2 + (1)2 = √2
tan 𝜃𝜃 =
tan−1
𝜃𝜃 =
𝜋𝜋
4
First, we need to write the number in polar
form, to do this we find the modulus and the
argument.
1
1
1
= 𝜃𝜃
1
√2 �cos
𝜋𝜋
𝜋𝜋 8
+ 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 �
4
4
𝜋𝜋
𝜋𝜋
8
�√2� �cos 8 � � + 𝑖𝑖 sin 8 � ��
4
4
16(1 + 0𝑖𝑖)
𝟏𝟏𝟏𝟏 + 𝟎𝟎𝟎𝟎
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Writing the number in polar form,
Now we multiply the argument by the power
and put the modulus to the power.
Plugging the cos and sin into the calculator.
135
c)
𝟏𝟏 − 𝟐𝟐𝟐𝟐
We’re asked to find the roots of
2
𝑧𝑧 + (−2 + 𝑖𝑖)𝑧𝑧 + 3 − 𝑖𝑖
𝑧𝑧 =
𝑧𝑧 =
𝑧𝑧 =
𝑧𝑧 =
𝑧𝑧 =
𝑧𝑧 =
𝑧𝑧 =
−𝑏𝑏 ± √𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎
2𝑎𝑎
𝑧𝑧 2 + (−2 + 𝑖𝑖)𝑧𝑧 + 3 − 𝑖𝑖
We can use the minus b formula to
find the roots of the equation.
−(−2 + 𝑖𝑖) ± �(−2 + 𝑖𝑖)2 − 4(1)(3 − 𝑖𝑖)
2(1)
𝑎𝑎 = 1 𝑏𝑏 = −2 + 𝑖𝑖 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐 = 3 − 𝑖𝑖
2 − 𝑖𝑖 ± √−9
2
Separating out the √−9 to √9√−1
allows us to write it as an imaginary
number.
2 − 𝑖𝑖 ± 3𝑖𝑖
2
Solving the minus b formula gives
us
𝑧𝑧 = 1 + 𝑖𝑖 𝑧𝑧 = 1 − 2𝑖𝑖
2 − 𝑖𝑖 ± √4 − 4𝑖𝑖 − 1 − 12 + 4𝑖𝑖
2
2 − 𝑖𝑖 ± √9√−1
2
2 + 2𝑖𝑖
2
𝑧𝑧 = 1 + 𝑖𝑖
1 − 2𝑖𝑖
𝑧𝑧 =
2 − 4𝑖𝑖
2
𝑧𝑧 = 1 − 2𝑖𝑖
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So the other root is 1 − 2𝑖𝑖
136
Question 2
a)
𝑥𝑥 ≤ 2, 𝑥𝑥 ≥ 6
|𝑥𝑥 − 4| ≥ 2
(𝑥𝑥 − 4)2 ≥ (2)2
𝑥𝑥 2 − 8𝑥𝑥 + 16 ≥ 4
𝑥𝑥 2 − 8𝑥𝑥 + 12 ≥ 0
(𝑥𝑥 − 6)(𝑥𝑥 − 2) = 0
𝑥𝑥 = 6, 𝑥𝑥 = 2
𝑥𝑥 ≤ 2, 𝑥𝑥 ≥ 6
© Pocket Tutor 2022
Squaring both sides to get rid
of the modulus bars.
Taking 4 from both sides.
Solving the quadratic.
As the quadratic has a positive
𝑥𝑥 2 we know it is U shaped and
therefore will be greater than
2 outside these numbers.
137
b)
𝑥𝑥 =
17
11
𝑦𝑦 = −
𝑜𝑜𝑜𝑜 𝑥𝑥 = −2
15
𝑜𝑜𝑜𝑜 𝑦𝑦 = 1
11
𝑥𝑥 2 + 𝑥𝑥𝑥𝑥 + 2𝑦𝑦 2 = 4
2𝑥𝑥 + 3𝑦𝑦 = −1
𝑥𝑥 =
−1 − 3𝑦𝑦
2
Getting 𝑥𝑥 in terms of 𝑦𝑦
−1 − 3𝑦𝑦 2
−1 − 3𝑦𝑦
�
� +�
� 𝑦𝑦 + 2𝑦𝑦 2 = 4
2
2
2
2
−𝑦𝑦 − 3𝑦𝑦
1 + 6𝑦𝑦 + 9𝑦𝑦
�+
+ 2𝑦𝑦 2 − 4 = 0
�
4
2
−𝑦𝑦 − 3𝑦𝑦 2
� + 4(2𝑦𝑦 2 ) − 4(4) = 0
1 − 6𝑦𝑦 + 9𝑦𝑦 2 + 4 �
2
Subbing this in for 𝑥𝑥 in the other equation.
Squaring the bracket
Multiplying across by 4 to get rid of the
fractions.
1 + 6𝑦𝑦 + 9𝑦𝑦 2 − 2𝑦𝑦 − 6𝑦𝑦 2 + 8𝑦𝑦 2 − 16
11𝑦𝑦 2 + 4𝑦𝑦 − 15 = 0
Solving the quadratic.
(11𝑦𝑦 + 15)(𝑦𝑦 − 1) = 0
𝑦𝑦 = −
𝑥𝑥 = �
−1 − 3𝑦𝑦
�
2
𝑥𝑥 = �
𝑥𝑥 =
15
𝑜𝑜𝑜𝑜 𝑦𝑦 = 1
11
−1 − 3 �−
17
11
2
15
�
11 �
−1 − 3(1)
�
𝑜𝑜𝑜𝑜 𝑥𝑥 = �
2
Subbing our values for 𝑦𝑦 into our expression for
𝑥𝑥.
𝑜𝑜𝑜𝑜 𝑥𝑥 = −2
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138
Question 3
a) i)
𝑥𝑥
𝑓𝑓(𝑥𝑥) =
2
𝑒𝑒 𝑥𝑥
𝑔𝑔(𝑥𝑥) = 𝑒𝑒 𝑥𝑥 − 1
𝟎𝟎
𝟎𝟎. 𝟓𝟓
𝟏𝟏
𝒍𝒍𝒍𝒍(𝟒𝟒)
3
2
1.21
0.74
0
0.65
1.72
0.5
ii)
𝑓𝑓(𝑥𝑥)
𝑔𝑔(𝑥𝑥)
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139
iii)
0.7
𝑓𝑓(𝑥𝑥)
𝑔𝑔(𝑥𝑥)
𝑥𝑥 = 0.7
b) 𝐥𝐥𝐥𝐥 𝟐𝟐 = 𝒙𝒙
2
= 𝑒𝑒 𝑥𝑥 − 1
𝑒𝑒 𝑥𝑥
2 = 𝑒𝑒 𝑥𝑥 (𝑒𝑒 𝑥𝑥 − 1)
2 = 𝑒𝑒 𝑥𝑥+𝑥𝑥 − 𝑒𝑒 𝑥𝑥
2 = 𝑒𝑒 2𝑥𝑥 − 𝑒𝑒 𝑥𝑥
0 = (𝑒𝑒 𝑥𝑥 )2 − 𝑒𝑒 𝑥𝑥 − 2
(𝑒𝑒 𝑥𝑥
𝑥𝑥
− 2)(𝑒𝑒 + 1) = 0
𝑒𝑒 𝑥𝑥 = 2, 𝑒𝑒 𝑥𝑥 = −1
log 𝑒𝑒 2 = 𝑥𝑥
𝐥𝐥𝐥𝐥 𝟐𝟐 = 𝒙𝒙
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Letting the functions equal each other.
Multiplying across by 𝑒𝑒 𝑥𝑥
When multiplying indices, we add the powers.
Rewriting 𝑒𝑒 2𝑥𝑥
Factorising the quadratic.
𝑒𝑒 𝑥𝑥 = −1 is not defined so we discard this answer.
Using the law of logs on page 21 of the Maths Tables
Book.
𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥
𝑙𝑙𝑙𝑙𝑙𝑙 with base 𝑒𝑒 can be rewritten as 𝑙𝑙𝑙𝑙
140
Question 4
a)
8𝑛𝑛 − 1
Show for 𝑛𝑛 = 1
81 − 1 = 7… Divisible by 7
Assume for 𝑛𝑛 = 𝑘𝑘
8𝑘𝑘 − 1 = 7𝑀𝑀
8𝑘𝑘 = 7𝑀𝑀 + 1
Prove 𝑛𝑛 = 𝑘𝑘 + 1
8𝑘𝑘+1 − 1
8𝑘𝑘 . 81 − 1
(7𝑀𝑀 + 1)(8) − 1
56𝑀𝑀 + 8 − 1
1. To prove by induction, we show that the
statement is true for 𝑛𝑛 = 1.
2. We assume that it is true for 𝑛𝑛 = 𝑘𝑘
7𝑀𝑀 represents any number of which 7 is a
factor.
3. We use our assumption to prove that it is
true for 𝑛𝑛 = 𝑘𝑘 + 1
Subbing in our result of assuming 𝑛𝑛 = 𝑘𝑘 is
true.
56𝑀𝑀 + 7
56𝑀𝑀 + 7 is divisible by 7 → 7(8𝑀𝑀 + 1)
𝑃𝑃1 is true
If 𝑃𝑃𝑘𝑘 is true then 𝑃𝑃𝑘𝑘+1 is true.
Since 𝑃𝑃1 is true, then by induction 𝑃𝑃𝑘𝑘 is true for all natural numbers ≥ 1
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141
b) i) 𝟑𝟑𝟑𝟑 − 𝒒𝒒
log 𝑎𝑎
8
3
→ log 𝑎𝑎 8 − log 𝑎𝑎 3
→ log 2 (2)3 − log 𝑎𝑎 3
→ 3 log 𝑎𝑎 2 − log 𝑎𝑎 3
𝟑𝟑𝟑𝟑 − 𝒒𝒒
Using the rules from page 21 of The Maths
𝑥𝑥
Tables Book: log 𝑎𝑎 � � = log 𝑎𝑎 𝑥𝑥 − log 𝑎𝑎 𝑦𝑦
Rewriting 8 as 23
𝑦𝑦
log 𝑎𝑎 𝑥𝑥 𝑞𝑞 = 𝑞𝑞𝑞𝑞𝑞𝑞𝑔𝑔𝑎𝑎 𝑥𝑥
Subbing in 𝑝𝑝 and 𝑞𝑞
ii) 𝟐𝟐𝟐𝟐 − 𝟒𝟒𝟒𝟒 + 𝟐𝟐
𝑙𝑙𝑙𝑙𝑔𝑔𝑎𝑎
9𝑎𝑎2
16
𝑙𝑙𝑙𝑙𝑔𝑔𝑎𝑎 9𝑎𝑎2 − 𝑙𝑙𝑙𝑙𝑔𝑔𝑎𝑎 16
log 𝑎𝑎 (3𝑎𝑎)2 − log 𝑎𝑎 (2)4
2 log 𝑎𝑎 3𝑎𝑎 − 4log 𝑎𝑎 2
2 log 𝑎𝑎 3 + 2 log 𝑎𝑎 𝑎𝑎 − 4 log 𝑎𝑎 2
2𝑞𝑞 + 2(1) − 4𝑝𝑝
𝟐𝟐𝟐𝟐 − 𝟒𝟒𝟒𝟒 + 𝟐𝟐
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Using the rules from page 21 of The Maths
𝑥𝑥
Tables Book: log 𝑎𝑎 � � = log 𝑎𝑎 𝑥𝑥 − log 𝑎𝑎 𝑦𝑦
𝑦𝑦
log 𝑎𝑎 𝑥𝑥 𝑞𝑞 = 𝑞𝑞𝑞𝑞𝑞𝑞𝑔𝑔𝑎𝑎 𝑥𝑥
log 𝑎𝑎 (𝑥𝑥𝑥𝑥) = log 𝑎𝑎 𝑥𝑥 + log 𝑎𝑎 𝑦𝑦
log 𝑎𝑎 𝑎𝑎 = 1
Subbing in 𝑝𝑝 and 𝑞𝑞
142
Question 5
a) i)
𝑥𝑥 = 10
Pythagoras’ theorem.
(5𝑥𝑥 − 9)2 = (𝑥𝑥 − 1)2 + (4𝑥𝑥)2
2
2
25𝑥𝑥 − 90𝑥𝑥 + 81 = 𝑥𝑥 − 2𝑥𝑥 + 1 + 16𝑥𝑥
8𝑥𝑥 2 − 88𝑥𝑥 + 80 = 0
𝑥𝑥 2 − 11𝑥𝑥 + 10 = 0
2
Squaring out the brackets.
Dividing across by 8
Solving the quadratic.
(𝑥𝑥 − 10)(𝑥𝑥 − 1) = 0
𝑥𝑥 = 10 𝑜𝑜𝑜𝑜 𝑥𝑥 = 1
As 𝑥𝑥 = 1 gives us a negative
length when plugged into the
side (5𝑥𝑥 − 9), it cannot be the
answer, ∴ 𝑥𝑥 = 10
5(1) − 9 = −4
∴ 𝑥𝑥 = 10
ii)
2
(5(10) − 9)2 = (10 − 1)2 + �4(10)�
1681 = 81 + 1600
Subbing in 10 for 𝑥𝑥 in each of the sides and
showing that the hypotenuse squared equals the
sum of the squares of the other 2 sides.
1681 = 1681
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143
b) i)
You can also show it is injective by
showing the inverse of the function
exists, as if a function has an inverse
it is bijective.
A bijective function is one which is
both injective and surjective.
Note: As the horizontal line crosses
exactly once this shows that the
function is also bijective.
Horizontal line crosses the function at least once, therefore it is injective
𝒊𝒊𝒊𝒊) 𝒇𝒇−𝟏𝟏 (𝒙𝒙) =
𝒙𝒙 + 𝟐𝟐
𝟑𝟑
𝑓𝑓(𝑥𝑥) = 3𝑥𝑥 − 2
𝑦𝑦 = 3𝑥𝑥 − 2
𝑦𝑦 + 2 = 3𝑥𝑥
𝑦𝑦 + 2
= 𝑥𝑥
3
𝒙𝒙 + 𝟐𝟐
𝒇𝒇−𝟏𝟏 (𝒙𝒙) =
𝟑𝟑
© Pocket Tutor 2022
To find the inverse of a function we
get the function in terms of 𝑥𝑥 and
then sub in 𝑥𝑥 where 𝑦𝑦 is.
Adding 2 to both sides
Dividing by 3
Subbing in 𝑥𝑥
144
Question 6
a) 𝟖𝟖𝟖𝟖 + 𝟏𝟏𝟏𝟏
(2𝑥𝑥 + 4)2
𝑓𝑓(𝑥𝑥 + ℎ) = (2(𝑥𝑥 + ℎ) + 4)2
(2𝑥𝑥 + 2ℎ + 4)2
4𝑥𝑥 2 + 8𝑥𝑥ℎ + 16𝑥𝑥 + 4ℎ2 + 16ℎ + 16
Subbing in (𝑥𝑥 + ℎ) for 𝑥𝑥
Squaring out the bracket.
4𝑥𝑥 2 + 8𝑥𝑥ℎ + 16𝑥𝑥 + 4ℎ2 + 16ℎ + 16 − (4𝑥𝑥 2 + 16𝑥𝑥 + 16)
Taking the original function
away from what we got when
we plugged in (𝑥𝑥 + ℎ)
𝑓𝑓(𝑥𝑥 + ℎ) − 𝑓𝑓(𝑥𝑥) 8𝑥𝑥ℎ + 4ℎ2 + 16ℎ
=
ℎ
ℎ
Dividing the result by ℎ
𝑓𝑓(𝑥𝑥 + ℎ) − 𝑓𝑓(𝑥𝑥)
= 8𝑥𝑥 + 4(0) + 16
lim
ℎ→0
ℎ
Limiting ℎ to 0 which
effectively means plugging in
0 for any remaining ℎ
𝑓𝑓(𝑥𝑥 + ℎ) − 𝑓𝑓(𝑥𝑥) = 4𝑥𝑥 2 + 8𝑥𝑥ℎ + 16𝑥𝑥 + 4ℎ2 + 16ℎ + 16 − (2𝑥𝑥 + 4)2
8𝑥𝑥ℎ + 4ℎ2 + 16ℎ
= 8𝑥𝑥 + 4ℎ + 16
𝟖𝟖𝟖𝟖 + 𝟏𝟏𝟏𝟏
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145
b) i)
1
𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 � �
𝑥𝑥
→ 𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 𝑥𝑥 −1
𝑑𝑑𝑑𝑑
= 1. sin 𝑥𝑥 −1 + 𝑥𝑥(cos(𝑥𝑥 −1 ) × −𝑥𝑥 −2 )
𝑑𝑑𝑑𝑑
1
1
1
sin + 𝑥𝑥 �− 2 cos �
𝑥𝑥
𝑥𝑥
𝑥𝑥
1 1
1
sin − cos
𝑥𝑥 𝑥𝑥
𝑥𝑥
1
𝑥𝑥
can be rewritten as 𝑥𝑥 −1 , this can make it
easier to differentiate.
Using the product rule we differentiate 𝑥𝑥
and multiply it by sin 𝑥𝑥 −1 , we then
differentiate, sin 𝑥𝑥 −1 , by differentiating sin
and then multiplying it by the derivative of
𝑥𝑥 −1 . We then multiply this derivative by 𝑥𝑥
ii) 𝟎𝟎. 𝟏𝟏𝟏𝟏
𝑥𝑥 =
4
𝜋𝜋
1 1
1
sin − cos
4 4
4
𝜋𝜋 𝜋𝜋
𝜋𝜋
= 0.151 = 𝟎𝟎. 𝟏𝟏𝟏𝟏
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To find the slope at a point we plug the 𝑥𝑥 value into the
derivative of the function.
4
Plugging in for 𝑥𝑥 and then putting it into the
calculator.
𝜋𝜋
146
Question 7
a) i)
𝟓𝟓
𝒄𝒄𝒄𝒄/𝒔𝒔
𝟑𝟑𝟑𝟑𝟑𝟑
Find = Given × Need
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
=
×
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝑉𝑉 =
4 3
𝜋𝜋𝑟𝑟
3
𝑑𝑑𝑑𝑑
4
= 3 � � 𝜋𝜋𝑟𝑟 2 → 4𝜋𝜋𝑟𝑟 2
𝑑𝑑𝑑𝑑
3
1
1
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
=
=
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 4𝜋𝜋𝑟𝑟 2
1
𝑑𝑑𝑑𝑑
= 250 ×
4𝜋𝜋𝑟𝑟 2
𝑑𝑑𝑑𝑑
1
𝑑𝑑𝑑𝑑
= 250 ×
4𝜋𝜋(20)2
𝑑𝑑𝑑𝑑
=
𝟓𝟓
𝒄𝒄𝒄𝒄/𝒔𝒔
𝟑𝟑𝟑𝟑𝟑𝟑
© Pocket Tutor 2022
Volume of a sphere from a page 10 of The
Maths Tables Book.
Differentiating this to find
We need
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
which is equal to one over
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
Subbing back into our 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 = 𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺𝐺 ×
𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁
Subbing in 20 for 𝑟𝑟 to find the rate of
increase when the radius is 20
147
ii) 𝟐𝟐𝟐𝟐𝟐𝟐𝒎𝒎𝟐𝟐 /𝒔𝒔
Find = Given × Need
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
=
×
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝐴𝐴 = 4𝜋𝜋𝑟𝑟 2
𝑑𝑑𝑑𝑑
= 8𝜋𝜋𝜋𝜋
𝑑𝑑𝑑𝑑
5
𝑑𝑑𝑑𝑑
=
× 8𝜋𝜋𝜋𝜋
𝑑𝑑𝑑𝑑 32𝜋𝜋
5
𝑑𝑑𝑑𝑑
=
× 8𝜋𝜋(20) = 𝟐𝟐𝟐𝟐𝟐𝟐𝒎𝒎𝟐𝟐 /𝒔𝒔
𝑑𝑑𝑑𝑑 32𝜋𝜋
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Surface Area of a sphere from page
10 The Maths Tables Book
Differentiating this to get
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
Going back to our Find = Given ×
need
Subbing in 20 for 𝑟𝑟
148
b) i) 𝒙𝒙 = 𝟎𝟎, 𝒙𝒙 = 𝟏𝟏𝟏𝟏
0 = −𝑥𝑥 2 + 10𝑥𝑥
To find when the ball is on the ground we let
𝑓𝑓(𝑥𝑥) = 0 as the equation is for the height of
the ball.
2
𝑥𝑥 − 10𝑥𝑥 = 0
𝑥𝑥(𝑥𝑥 − 10) = 0
Factorising out the 𝑥𝑥 and solving for 𝑥𝑥
𝒙𝒙 = 𝟎𝟎, 𝒙𝒙 = 𝟏𝟏𝟏𝟏
𝐢𝐢𝐢𝐢)
𝟓𝟓𝟓𝟓
𝒎𝒎
𝟑𝟑
10
1
� −𝑥𝑥 2 + 10𝑥𝑥 𝑑𝑑𝑑𝑑
10 − 0 0
10
𝑥𝑥 3 10𝑥𝑥 2
1
�− +
�
3
2 0
10
10
𝑥𝑥 3
1
�− + 5𝑥𝑥 2 �
10
3
0
(10)3
(0)3
1
��−
+ 5(10)2 � − �−
+ 5(0)2 ��
3
3
10
1 500
�
− 0�
10 3
=
To find the average height of the ball above
the ground we integrate the function
between the 𝑥𝑥 values when it is on the
ground.
Integrating the expression
Subbing in 10 and 0 for 𝑥𝑥
𝟓𝟓𝟓𝟓
𝒎𝒎
𝟑𝟑
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149
Question 8
a) i)
4.529m
𝑓𝑓(𝑥𝑥) = −0.274𝑥𝑥 2 + 1.193𝑥𝑥 + 3.23
𝑑𝑑𝑑𝑑
= −(2)0.274𝑥𝑥 + 1.193
𝑑𝑑𝑑𝑑
−0.548𝑥𝑥 + 1.193 = 0
1.193 = 0.548𝑥𝑥
1.193
= 𝑥𝑥
0.548
𝑥𝑥 = 2.177
𝑓𝑓(2.177) = −0.274(2.177)2 + 1.193(2.177) + 3.23
= 4.529m
ii)
We find the slope at the basket by
plugging the x-coordinate into the
derivative of the function.
𝟓𝟓𝟓𝟓°
The slope of the curve at the basket:
𝑓𝑓 ′ (4.5) = −0.548(4.5) + 1.193 = −1.273
tan 𝜃𝜃 = −1.273
𝜃𝜃 = tan
−1
1.273 = 51.8 = 52°
tan 𝜃𝜃 = The slope at this point as the
slope is the rise over run, which is the
Rise
Run
𝜃𝜃
same as
opposite
adjacent
iii)
(2.677,3.964)
A(−0.5,2.565) → 𝐶𝐶(0,2)
Translation 𝑥𝑥 + 0.5, 𝑦𝑦 − 0.565
2.177 + 0.5 = 2.677
4.529 − 0.565 = 3.964
(2.677,3.964)
© Pocket Tutor 2022
A is translated to 𝐶𝐶 by adding 0.5 to the 𝑥𝑥
coordinate and taking 0.565 away rom the 𝑦𝑦
coordinate.
We do the same to the coordinates for the
maximum value of 𝑓𝑓(𝑥𝑥) to find the
coordinates for the maximum value of 𝑔𝑔(𝑥𝑥)
150
iv) 𝒈𝒈(𝒙𝒙) = −𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐𝒙𝒙𝟐𝟐 + 𝟏𝟏. 𝟔𝟔𝟔𝟔𝟔𝟔 + 𝟐𝟐
𝑔𝑔(𝑥𝑥) = 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐
𝐶𝐶(0,2)
→ 𝑔𝑔(0) = 𝑎𝑎(0)2 + 𝑏𝑏(0) + 𝑐𝑐 = 2
𝑐𝑐 = 2
𝐵𝐵(4.5,3.05)
→ 𝑔𝑔(4.5) = 𝑎𝑎(4.5)2 + 𝑏𝑏(4.5) + 2 = 3.05
20.25𝑎𝑎 + 4.5𝑏𝑏 = 1.05
4.5𝑏𝑏 = 1.05 − 20.25𝑎𝑎
𝑏𝑏 =
𝑏𝑏 =
Letting 𝑔𝑔(𝑥𝑥) = The general formula for a
quadratic.
Plugging in 0 for 𝑥𝑥 and letting it equal 2 as (0,2)
is a point on the graph.
Plugging in 4.5 for 𝑥𝑥 and letting it equal 3.05 as
(4.5,3.05) is a point on the graph. Also subbing
in 2 for c.
Getting b in terms of a
1.05 − 20.25𝑎𝑎
4.5
7
− 4.5𝑎𝑎
30
Max Coordinates = 𝑎𝑎(2.677)2 + 𝑏𝑏(2.677) + 2 = 3.964
7.166𝑎𝑎 + 2.677𝑏𝑏 = 1.964
7.166𝑎𝑎 + 2.677 �
7
− 4.5𝑎𝑎� = 1.964
30
7.166𝑎𝑎 + 0.6246 − 12.0465𝑎𝑎 = 1.964
−4.8805𝑎𝑎 = 1.3354
𝑎𝑎 = −
1.3394
4.8805
Plugging in 2.677 for 𝑥𝑥 and letting it equal 3.964
as (2.677,3.964) is a point on the graph
Subbing in the expression we found when
writing b in terms of a.
Solving for 𝑎𝑎
𝑎𝑎 = −0.274
𝑏𝑏 =
7
− 4.5𝑎𝑎
30
→ 𝑏𝑏 =
7
− 4.5(−0.274)
30
𝑏𝑏 = 1.467
Subbing our value for 𝑎𝑎 into our expression for 𝑏𝑏
Subbing our values for a, b and c into the original
equation
𝒈𝒈(𝒙𝒙) = −𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐𝒙𝒙𝟐𝟐 + 𝟏𝟏. 𝟒𝟒𝟒𝟒𝟒𝟒𝟒𝟒 + 𝟐𝟐
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151
b) i)
1000points, 1020points
200m:
𝑦𝑦 = 𝑎𝑎(𝑏𝑏 − 𝑥𝑥)𝑐𝑐
𝑦𝑦 = (4.99087)(42.5 − 23.8)1.81
𝑦𝑦 = 1000.48 = 1000points
Javelin:
𝑦𝑦 = 𝑎𝑎(𝑥𝑥 − 𝑏𝑏)𝑐𝑐
𝑦𝑦 = (15.9803)(58.2 − 3.8)1.04
𝑦𝑦 = 1020.01 = 1020points
ii)
72.23m
1295 = (15.9803)(𝑥𝑥 − 3.8)1.04
1295
= (𝑥𝑥 − 3.8)1.04
15.9803
1.04
1295
= 𝑥𝑥 − 3.8
15.9803
�
68.4343 = 𝑥𝑥 − 3.8
72.23𝑚𝑚 = 𝑥𝑥
© Pocket Tutor 2022
Letting the equation = 1295
Dividing across by 15.9803
Finding the 1.04 root of both sides
Adding 3.8 to both sides.
Note this can be solved using logs.
152
iii) 𝟏𝟏. 𝟖𝟖𝟖𝟖
𝑦𝑦 = 𝑎𝑎(𝑏𝑏 − 𝑥𝑥)𝑐𝑐
Converting two minutes and 1.84 seconds into
seconds
1087 = (0.11193)(254 − 121.84)𝑐𝑐
Subbing the new values for 𝑎𝑎, 𝑏𝑏 and 𝑥𝑥 into the
formula for the 200m race and letting it equal
the points Jesse got. (1087)
𝑥𝑥 = 2 × 60 = 120 + 1.84 = 121.84
1087
= (132.16)𝑐𝑐
0.11193
1087
= 𝑐𝑐
log132.16
0.11193
𝑐𝑐 = 1.8798 = 𝟏𝟏. 𝟖𝟖𝟖𝟖
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Dividing across by (0.11193)
Using the law of logs from pg. 21of The Maths
Tables Books:
𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥
153
Question 9
a) i)
7
𝑆𝑆𝑛𝑛 =
𝑎𝑎(1 − 𝑟𝑟 𝑛𝑛 )
1 − 𝑟𝑟
𝑎𝑎 = 4, 𝑟𝑟 =
1
2
1 𝑛𝑛
4(1 − � � )
2
7.9375 =
1
1−
2
1 𝑛𝑛
4 �1 − � � �
2
7.9375 =
1
2
1
1 𝑛𝑛
(7.9375) = 4 − 4 � �
2
2
Geometric series formula from page 22 of The Maths
Tables Book.
Setting up the equation with 𝑎𝑎 and 𝑟𝑟 and letting it
equal the length.
Multiplying across by
1
2
1 𝑛𝑛
3.96875 − 4 = −4 � �
2
Taking 4 from both sides.
1
1 𝑛𝑛
=� �
128
2
Using the law of logs from pg. 21of The Maths Tables
Books:
−0.03125
1 𝑛𝑛
=� �
−4
2
1
= 𝑛𝑛
2 128
log 1
𝑛𝑛 = 7
Dividing by −4
𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥
ii)
8units
𝑆𝑆∞ =
𝑆𝑆∞ =
𝑎𝑎
1 − 𝑟𝑟
4
1−
1
2
= 8units
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Infinite Geometric series formula from page 22 of
The Maths Tables Book.
Subbing in 𝑎𝑎 and 𝑟𝑟
154
iii)
(3.2,1.6)
Sum to infinity of the changes in the x − coordinate
𝑎𝑎
𝑆𝑆∞ =
1 − 𝑟𝑟
𝑎𝑎 = 4, 𝑟𝑟 = −
𝑆𝑆∞ =
4
1
4
1
1 − �− �
4
Ignoring the times each doesn’t change:
= 3.2
Sum to infinity of the changes in the y − coordinates
𝑎𝑎 = 2, 𝑟𝑟 = −
𝑆𝑆∞ =
2
1
4
1
1 − �− �
4
As the pattern starts at the origin the
sum to infinity in the changes to each
coordinate will give us the final
coordinates.
We can see 𝑟𝑟 = −
1
4
Subbing 𝑎𝑎 and 𝑟𝑟 into the formula for a
series to infinity on page 22 of the
Maths Tables Book.
Repeating this process with 𝑎𝑎 = 2 for
the y-coordinate.
= 1.6
(3.2,1.6)
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155
b) i)
Female
Male
Female
Female
Male
ii)
8,13
𝑛𝑛 + 2 = 𝑛𝑛 + 1 + 𝑛𝑛
6=5+4
𝐺𝐺6 = 𝐺𝐺5 + 𝐺𝐺4
If we are putting 𝐺𝐺6 on the left it is
equal to 𝐺𝐺5 + 𝐺𝐺4 (one term before
plus two terms before).
𝐺𝐺6 = 5 + 3 = 8
𝐺𝐺7 = 𝐺𝐺6 + 𝐺𝐺5 = 8 + 5 = 13
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156
iii)
𝐺𝐺𝑛𝑛 =
𝐺𝐺3 =
𝑛𝑛
𝑛𝑛
�1 + √5� − �1 − √5�
2𝑛𝑛 √5
(3)
�1 + √5�
3
− �1 − √5�
2(3) √5
3
(3)
Subbing in 3 for 𝑛𝑛 and letting it = 2
Multiplying across by the bottom of the
fraction.
=2
Separating out the brackets so we can
square them.
�1 + √5� − �1 − √5� = 2(23 √5)
3
3
�1 + √5� − �1 − √5� = 2(23 √5)
2
2
�1 + √5��1 + √5� − �1 − √5��1 − √5� = 2(23 √5)
2
2
�1 + √5� �1 + 2√5 + �√5� � − �1 − √5��1 − 2√5 + (√5� ) = 2(23 √5)
2
2
3
2
2
3
�1 + 2√5 + �√5� + √5 + 2�√5� + �√5� � − �1 − 2√5 + �√5� − √5 + 2�√5� − �√5� �
= 2(23 √5)
2
3
2
3
�1 + 3√5 + 3�√5� + �√5� � − �1 − 3√5 + 3�√5� − �√5� � = 2(23 √5)
3
6√5 + 2�√5� = 16√5
2
2√5 �3 + �√5� � = 8(2√5)
2
3 + �√5� = 8
3+5=8
8 = 8 𝑄𝑄. 𝐸𝐸. 𝐷𝐷
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Squaring the brackets and then multiplying
out the bracket we factorised out.
Subtracting the left hand side and
multiplying out the right hand side.
Dividing across by 2√5
157
2015 Paper 1
Question 1
a)
3
Multiplying the previous number by each time.
4
b)
3 9 27 81
�
2 + 2� + +
+
2 8 32 128
=
𝟔𝟔𝟔𝟔𝟔𝟔
𝒎𝒎
𝟔𝟔𝟔𝟔
c) 14m
𝑆𝑆∞ =
𝑎𝑎 =
𝑎𝑎
1 − 𝑟𝑟
3
3
, 𝑟𝑟 =
2
4
3
𝑆𝑆∞ = 2 + 2 � 2 �
3
1−
4
= 𝟏𝟏𝟏𝟏
© Pocket Tutor 2022
3
3
The ball fell 2m. It then bounced up m and fell back down 𝑚𝑚
2
2
and so on. Therefore, we add the 2m it fell to two times the height
of each bounce.
Taking the equation for the sum to infinity of a geometric series
from page 22 of The Maths Tables Book.
3
We start on as the ball only fell from 2m it did not bounce up
2
2m and then fall.
We multiply the distance of the bounces by 2 as they bounced
up and then fell back down the same distance
We add the original 2m the ball travelled when it was dropped.
158
Question 2
𝒙𝒙 = 𝟏𝟏, 𝟏𝟏 + 𝟐𝟐√𝟑𝟑, 𝟏𝟏 − 𝟐𝟐√𝟑𝟑
𝑥𝑥 3 − 3𝑥𝑥 2 − 9𝑥𝑥 + 11 = 0
𝑓𝑓(1) = (1)3 − 3(1)2 − 9(1) + 11 = 0
1 − 3 − 9 + 11 = 0
0=0
𝑥𝑥 = 1 is a root
When we want to solve a cubic, we plug in 1 for
𝑥𝑥, then we plug in −1, then 2, then −2 until we find a
root. In this case 𝑥𝑥 = 1 is a root as when we plug in
one for 𝑥𝑥 we get 0.
This means that 𝑥𝑥 − 1 is a factor of the equation.
∴ 𝑥𝑥 − 1 is a factor
𝑥𝑥 2 − 2𝑥𝑥 − 11
𝑥𝑥 − 1 𝑥𝑥 3 − 3𝑥𝑥 2 − 9𝑥𝑥 + 11
So, we divide the cubic by 𝑥𝑥 − 1
𝑥𝑥 3 − 𝑥𝑥 2
−2𝑥𝑥 2 − 9𝑥𝑥
−2𝑥𝑥 2 + 2𝑥𝑥
−11𝑥𝑥 + 11
−11𝑥𝑥 + 11
0
2
𝑥𝑥 − 2𝑥𝑥 − 11 = 0
𝑥𝑥 =
𝑥𝑥 =
−𝑏𝑏 ± √𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎
2𝑎𝑎
−(−2) ± �(−2)2 − 4(1)(−11)
2(1)
𝒙𝒙 = 𝟏𝟏 + 𝟐𝟐√𝟑𝟑, 𝟏𝟏 − 𝟐𝟐√𝟑𝟑 and 𝒙𝒙 = 𝟏𝟏
© Pocket Tutor 2022
We now use the −𝑏𝑏 formula to solve the quadratic
we got from the division.
The −𝑏𝑏 formula can be found on page 20 of The
Maths Tables Book
Plugging this into our calculator (once with a plus and
then with a minus in front of the square root), gives
us our other two roots.
159
Question 3
a) i)
By plugging in each of these values for 𝑥𝑥, we get the corresponding values for 𝑓𝑓(𝑥𝑥)
ii) 𝟑𝟑𝟑𝟑 square units
𝐴𝐴 =
ℎ
[𝑦𝑦 + 𝑦𝑦𝑛𝑛 + 2(𝑦𝑦2 + 𝑦𝑦3 + 𝑦𝑦4 + 𝑦𝑦5 + 𝑦𝑦6 )]
2 1
ℎ = 1, 𝑦𝑦1 = 0, 𝑦𝑦𝑛𝑛 = 0
1
𝐴𝐴 = [0 + 0 + 2(5 + 8 + 9 + 8 + 5)]
2
1
[2(35)]
2
=
The formula for the trapezoidal rule can be found on page
12 of The Maths Tables Book.
𝑦𝑦1 = the first term. yn = the last term and h = 1 as the
width of each section is 1. We know this as the 𝑥𝑥 values
are going up in ones.
𝐴𝐴 = 𝟑𝟑𝟑𝟑 square units
b) i) 𝟑𝟑𝟑𝟑
9
� −𝑥𝑥 2 + 12𝑥𝑥 − 27 𝑑𝑑𝑑𝑑
3
9
𝑥𝑥 3 12𝑥𝑥 2 27𝑥𝑥
�− +
�
−
1 3
3
2
��−
(3)3 12(3)2
(9)3 12(9)2
+
− 27(9)� − �−
+
− 27(3)��
3
2
3
2
(−243 + 486 − 243) − (−9 + 54 − 81)
Integrating the expression
Subbing in 9 for 𝑥𝑥, and then subbing in 3
for 𝑥𝑥
Taking one bracket from the other.
(0) − (−36)
= 𝟑𝟑𝟑𝟑
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160
ii) 2.8%
36 − 35 = 1
1 100
×
= 𝟐𝟐. 𝟖𝟖%
1
36
© Pocket Tutor 2022
The error between the part ii and iii is 1 square unit.
Putting this over the total answer to get a
percentage.
161
Question 4
a) 𝟓𝟓 + 𝒊𝒊
1
1
2
= +
𝑧𝑧1 𝑧𝑧2 𝑧𝑧3
1
1
2
=
+
𝑧𝑧1 2 + 3𝑖𝑖 3 − 2𝑖𝑖
1
1
2 + 3𝑖𝑖
2 3 − 2𝑖𝑖
�
�
=
×
+
𝑧𝑧1 3 − 2𝑖𝑖 2 + 3𝑖𝑖 3 − 2𝑖𝑖 2 + 3𝑖𝑖
2
3 − 2𝑖𝑖
2 + 3𝑖𝑖
=
+
𝑧𝑧1 (3 − 2𝑖𝑖)(2 + 3𝑖𝑖) (3 − 2𝑖𝑖)(2 + 3𝑖𝑖)
3 − 2𝑖𝑖 + 2 + 3𝑖𝑖
2
=
𝑧𝑧1 (3 − 2𝑖𝑖)(2 + 3𝑖𝑖)
5 + 𝑖𝑖
5 + 𝑖𝑖
2
=
=
𝑧𝑧1 6 + 9𝑖𝑖 − 4𝑖𝑖 − 6(−1) 12 + 5𝑖𝑖
𝑧𝑧1 12 + 5𝑖𝑖
=
5 + 𝑖𝑖
2
𝑧𝑧1 12 + 5𝑖𝑖 5 − 𝑖𝑖
=
×
5 + 𝑖𝑖
5 − 𝑖𝑖
2
𝑧𝑧1 60 + 25𝑖𝑖 − 12𝑖𝑖 − 5(−1)
=
25 − 5𝑖𝑖 + 5𝑖𝑖 − (−1)
2
𝑧𝑧1 65 + 13𝑖𝑖
=
26
2
𝑧𝑧1 = 2 ×
65 + 13𝑖𝑖
26
Subbing in for 𝑧𝑧2 and 𝑧𝑧3 .
To write as one fraction we multiply each
fraction by the bottom of the other fraction
over itself. This is the same as multiplying by 1
so we do not have to do it across the whole
equation.
Adding the two fractions.
Adding the top and multiplying out the bottom.
Inverting both sides
Multiplying across by the conjugate of the
bottom to get rid of the complex number on
the bottom.
Multiplying across by the 2 and then dividing in
gives us our answer
𝑧𝑧1 = 𝟓𝟓 + 𝒊𝒊
b)
𝑎𝑎(1 − 𝑟𝑟 𝑛𝑛 )
1 − 𝑟𝑟
𝑤𝑤
𝑎𝑎 = 1, 𝑟𝑟 = = 𝑤𝑤
1
𝑆𝑆𝑛𝑛 =
1(1 − 𝑤𝑤 0 )
𝑆𝑆0 =
1 − 𝑤𝑤
𝑆𝑆 =
=
1(1 − 1)
1 − 𝑤𝑤
0
= 𝟎𝟎
1 − 𝑤𝑤
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Taking the formula for the sum of a geometric
series form page 22 of The Maths Tables Book.
𝑎𝑎 =The first term, 𝑟𝑟 = The second term divided
by the first term.
Subbing this into the equation.
0 divided by any number = 0
162
Question 5
a) 𝒙𝒙 = 𝟑𝟑
𝑥𝑥 = √𝑥𝑥 + 6
𝑥𝑥 2 = 𝑥𝑥 + 6
Squaring both sides.
(𝑥𝑥 + 2)(𝑥𝑥 − 3) = 0
Factorising and solving for 𝑥𝑥
𝑥𝑥 2 − 𝑥𝑥 − 6 = 0
Putting all the terms on one side.
𝑥𝑥 = 3 𝑜𝑜𝑜𝑜 𝑥𝑥 = −2
Now subbing our 𝑥𝑥 values in for 𝑥𝑥 in
the original equation.
𝑥𝑥 = √𝑥𝑥 + 6
As we can see the −2 does not work
so it is not a solution for 𝑥𝑥
𝑥𝑥 = −2 → (−2) = �(−2) + 6
−2 ≠ √4
(3) = √3 + 6
3 = √9
3 does work so it is a solution for 𝑥𝑥
∴ 𝒙𝒙 = 𝟑𝟑
𝒃𝒃) 𝟏𝟏 −
𝟏𝟏
𝟐𝟐√𝒙𝒙 + 𝟔𝟔
𝑥𝑥 − √𝑥𝑥 + 6
1
→ 𝑥𝑥−(𝑥𝑥 + 6)2
1
1
𝑑𝑑𝑑𝑑
= 1 − (𝑥𝑥 + 6)−2 × 1
2
𝑑𝑑𝑑𝑑
= 𝟏𝟏 −
𝟏𝟏
𝟐𝟐√𝒙𝒙 + 𝟔𝟔
© Pocket Tutor 2022
It can help to rewrite the square root as being to the power of a
half before we differentiate.
To differentiate this, we use the chain rule. (Multiplying the
bracket by the power and taking one from the power, then
multiplying it by the derivative of whatever is inside the bracket)
1
𝑎𝑎−2 =
1
1
1
using this to rewrite (𝑥𝑥 + 6)−2
2
√𝑎𝑎
163
c) (−𝟓𝟓. 𝟕𝟕𝟕𝟕, −𝟔𝟔. 𝟐𝟐𝟐𝟐)
Turning point
1−
1=
1
2√𝑥𝑥 + 6
1
𝑑𝑑𝑑𝑑
=0
𝑑𝑑𝑑𝑑
=0
2√𝑥𝑥 + 6
The turning point of the function is when
the slope is equal to 0. So, we let the
derivative equal 0.
Adding the fraction to both sides.
2√𝑥𝑥 + 6 = 1
Multiplying across by the bottom of the
fraction.
1 2
𝑥𝑥 + 6 = � �
2
Squaring both sides
√𝑥𝑥 + 6 =
𝑥𝑥 + 6 =
1
4
𝑥𝑥 = −5
3
4
𝑥𝑥 =
1
2
1
−6
4
𝑓𝑓(𝑥𝑥) = 𝑥𝑥 − √𝑥𝑥 + 6
3
3
3
𝑓𝑓 �−5 � = �−5 � − ��−5 � + 6 = −6.25
4
4
4
Dividing across by 2
Taking 6 from both sides
Plugging our 𝑥𝑥-value back into the original
equation to find the y – coordinate.
(−𝟓𝟓. 𝟕𝟕𝟕𝟕, −𝟔𝟔. 𝟐𝟐𝟐𝟐)
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164
Question 6
a) i) 𝟒𝟒. 𝟐𝟐𝟐𝟐%
(1 + 𝑖𝑖) = (1 + 𝑟𝑟)12
(1 + 𝑖𝑖) = (1 + 0.0035)12
(1 + 𝑖𝑖) = 1.042818
𝑖𝑖 = 0.042818
= 𝟒𝟒. 𝟐𝟐𝟐𝟐%
ii) 𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑%
(1 + 0.045) = (1 + 𝑟𝑟)12
1.045 = (1 + 𝑟𝑟)12
Using the expression (1 + 𝑖𝑖) = (1 + 𝑟𝑟)12 to convert from
monthly interest rate to an annual rate.
Writing 0.35% in as a decimal
Taking one from both sides and then multiplying by 100
to write it as a percentage.
Using the expression (1 + 𝑖𝑖) = (1 + 𝑟𝑟)12 to convert from
annual interest rate to a monthly rate.
Writing 4.5% in as a decimal.
12
Putting both sides to the root 12
1.00367 = 1 + 𝑟𝑟
Taking one from both sides and then multiplying by 100 to
write it as a percentage.
√1.045 = 1 + 𝑟𝑟
0.00367 = 𝑟𝑟
𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑% = 𝑟𝑟
b) €𝟖𝟖𝟖𝟖𝟖𝟖
𝐴𝐴 = 𝑃𝑃
𝑖𝑖(1 + 𝑖𝑖)𝑡𝑡
(1 + 𝑖𝑖)𝑡𝑡 − 1
𝑃𝑃 = 80000, 𝑡𝑡 = 10 × 12 = 120, 𝑖𝑖 = 0.0035
0.0035(1 + 0.0035)120
�
(1 + 0.0035)120 − 1
𝐴𝐴 = 80000 �
𝐴𝐴 = €817.59
= €𝟖𝟖𝟖𝟖𝟖𝟖
© Pocket Tutor 2022
Taking the formula from page 31 of the Maths Tables
Book. Remembering to convert 𝑖𝑖 from a percentage to
a decimal.
Plugging in the values
Plugging the expression into the calculator and
rounding to the nearest euro.
165
Question 7
a) i)
𝑓𝑓(𝑥𝑥) = 0.0024𝑥𝑥 3 + 0.018𝑥𝑥 2 + 𝑐𝑐𝑐𝑐 + 𝑑𝑑
𝑓𝑓(0) = 0.0024(0)2 + 0.018(0)2 + 𝑐𝑐(0) + 𝑑𝑑 = 0
𝑑𝑑 = 0
As the plane lands at the origin (O), we
know that when 𝑥𝑥 = 0, 𝑦𝑦 is also equal to 0
so we plug in 0 for 𝑥𝑥 and let the equation
equal 0.
ii)
0.0024(−5)3 + 0.018(−5)2 + 𝑐𝑐(−5) + 0 = 0.15
9
−0.3 +
− 5𝑐𝑐 = 0.15
20
−0.3 +
0 = 5𝑐𝑐
9
− 0.15 = 5𝑐𝑐
20
As the plane passes through the point
(−5, 0.15), we know that when 𝑥𝑥 = −5,
𝑦𝑦 is equal to 0.15.
So, we plug in −5 for 𝑥𝑥 and let the
equation equal 0.15. We also plug in 0 for
d
𝑐𝑐 = 0
b) i) −𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
𝑓𝑓(𝑥𝑥) = 0.0024𝑥𝑥 3 + 0.018𝑥𝑥 2
𝑓𝑓 ′ (𝑥𝑥) = 3(0.0024)𝑥𝑥 2 + 2(0.018)𝑥𝑥
𝑓𝑓 ′ (−4) = 3(0.0024)(−4)2 + 2(0.018)(−4)
= −𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
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As 𝑑𝑑 and 𝑐𝑐 = 0 we are left with this as our
equation.
Differentiating by rule
Plugging in −4 for 𝑥𝑥
166
ii) 𝜽𝜽 = 𝟐𝟐°
𝜃𝜃
Think of this as looking at the
slope of the plane’s path –
it’s going down. What is the
angle of descent? We found
the slope in part (i) so let the
slope equal the tan function
to find the angle.
tan 𝜃𝜃 = −0.0288
𝜃𝜃 = tan−1 −0.0288
𝜃𝜃 = −1.6497 = 𝟐𝟐°
tan 𝜃𝜃 = −0.0288
c)
𝑓𝑓 ′ (𝑥𝑥) = 3(0.0024)𝑥𝑥 2 + 2(0.018)𝑥𝑥
𝑓𝑓
′ (𝑥𝑥)
2
= 0.0072𝑥𝑥 + 0.036𝑥𝑥
𝑓𝑓 ′′ (𝑥𝑥) = 2(0.0072)𝑥𝑥 + 0.036
𝑓𝑓 ′′ (𝑥𝑥) = 0.0144𝑥𝑥 + 0.036
0.0144𝑥𝑥 + 0.036 = 0
0.0144𝑥𝑥 = −0.036
𝑥𝑥 = −2.5
𝑓𝑓(𝑥𝑥) = 0.0024𝑥𝑥 3 + 0.018𝑥𝑥 2
𝑓𝑓(−2.5) = 0.0024(−2.5)3 + 0.018(−2.5)2
The point of inflection is found by getting the
second derivative and letting it equal 0
Taking the derivative from b) i) above and
tidying it up.
Finding the second derivative
Letting it equal 0
Dividing across by 0.0144 to find 𝑥𝑥
Plugging our 𝑥𝑥 −value into the original equation
to find the corresponding 𝑦𝑦 −coordinate.
−0.0375 + 0.1125 = 0.075
0.075 = 0.075
Point of Inflection = (−2.5,0.075)
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167
d) i)
𝑓𝑓(𝑥𝑥) = 0.0024𝑥𝑥 3 + 0.018𝑥𝑥 2
Subbing in −𝑥𝑥 − 5 for 𝑥𝑥
𝑓𝑓(−𝑥𝑥 − 5) = 0.0024(−𝑥𝑥 − 5)3 + 0.018(−𝑥𝑥 − 5)2
Separating the bracket to be
cubed into a bracket to be
squared times itself.
0.0024(−𝑥𝑥 − 5)(−𝑥𝑥 − 5)2 + 0.018(𝑥𝑥 2 + 10𝑥𝑥 + 25)
0.0024(−𝑥𝑥 − 5)(𝑥𝑥 2 + 10𝑥𝑥 + 25) + 0.018𝑥𝑥 2 + 0.18𝑥𝑥 + 0.45
3
2
2
2
0.0024(−𝑥𝑥 − 10𝑥𝑥 − 25𝑥𝑥 − 5𝑥𝑥 − 50𝑥𝑥 − 125) + 0.018𝑥𝑥 + 0.18𝑥𝑥 + 0.45
0.0024(−𝑥𝑥 3 − 15𝑥𝑥 2 − 75𝑥𝑥 − 125) + 0.018𝑥𝑥 2 + 0.18𝑥𝑥 + 0.45
Squaring this bracket and the one
on the right.
Multiplying out the brackets.
−0.0024𝑥𝑥 3 − 0.036𝑥𝑥 2 − 0.18𝑥𝑥 − 0.3 + 0.018𝑥𝑥 2 + 0.18𝑥𝑥 + 0.45
−0.0244𝑥𝑥 3 − 0.018𝑥𝑥 2 + 0𝑥𝑥 + 0.15
𝑦𝑦 = 0.0024𝑥𝑥 3 + 0.018𝑥𝑥 2
−𝑦𝑦 + 0.15 = −0.0244𝑥𝑥 3 − 0.018𝑥𝑥 2 + 0.15
𝑦𝑦 = 𝑓𝑓(𝑥𝑥) = 0.0024𝑥𝑥 3 + 0.018𝑥𝑥 2
If we multiply this by −1 and add
0.15, it equals the equation.
ii) (𝒙𝒙, 𝒚𝒚)
Point: (−𝑥𝑥 − 5, −𝑦𝑦 + 0.15)
Point of inflection: (−2.5, 0.075)
Change in 𝑥𝑥 −value = (−2.5) − (−𝑥𝑥 − 5) = −𝑥𝑥 + 2.5
Change in 𝑦𝑦 − value = (0.075) − (−𝑦𝑦 + 0.15) = 𝑦𝑦 − 0.075
Image:
𝑥𝑥 −value = −2.5 + (−𝑥𝑥 + 2.5) = 𝑥𝑥
𝑦𝑦 −value = 0.075 + (𝑦𝑦 − 0.75) = 𝑦𝑦
To find the distance between this point
and the point of inflection we take the 𝑥𝑥value from the 𝑥𝑥-value of the point of
inflection. We repeat this for the y-value.
We then add these values to the 𝑥𝑥 and 𝑦𝑦
values of the point of inflection to find
the coordinates of the image through the
other side of the point of inflection.
(𝑥𝑥, 𝑦𝑦)
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168
Question 8
a) i)
We get this by multiplying 4 × 106 by the number of minutes
ii)
iii)
𝑉𝑉 = (4 × 106 )𝑡𝑡
The volume can be calculated by multiplying 4 × 106 by
the number of minutes (𝑡𝑡)
b) i) 𝟎𝟎. 𝟏𝟏𝟏𝟏𝒓𝒓𝟐𝟐 cm3
2
Volume of a cylinder = 𝜋𝜋𝑟𝑟 ℎ
𝜋𝜋𝑟𝑟 2 (0.1) = 𝟎𝟎. 𝟏𝟏𝟏𝟏𝒓𝒓𝟐𝟐 cm3
© Pocket Tutor 2022
If the oil forms a circle on the water and has a height, it has
formed a cylinder.
The formula for the volume of a cylinder can be found on
page 10 of The Maths Tables Book
169
ii) 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏. 𝟑𝟑 cm per minute
Find = Given × Need
Figuring out how we are going to get the change in the
length of the radius with respect to time.
𝑣𝑣 = (4 × 106 )𝑡𝑡
Differentiating our equation for the volume of oil to get
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
=
×
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
= 4 × 106
𝑑𝑑𝑑𝑑
𝑣𝑣 = 0.1𝜋𝜋𝑟𝑟 2
𝑑𝑑𝑑𝑑
= 0.2𝜋𝜋𝜋𝜋
𝑑𝑑𝑑𝑑
1
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
1
→
=
=
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 0.2𝜋𝜋𝜋𝜋
1
𝑑𝑑𝑑𝑑
= (4 × 106 ) ×
0.2𝜋𝜋𝜋𝜋
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
1
= (4 × 106 ) ×
𝑑𝑑𝑑𝑑
0.2𝜋𝜋(5000)
= 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏. 𝟑𝟑 cm per minute
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
Differentiating our equation for the volume of the oil
slick to find
We need
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
so we put
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
under 1
Subbing back into our original expression
Subbing in 5000 for 𝑟𝑟, (having converted metres to
centimetres)
c)
Area of a circle = 𝜋𝜋𝑟𝑟 2
From page 8 of The Maths Tables Book
Find = Given × Need
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
=
×
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
= 2𝜋𝜋𝜋𝜋
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
1
� × 2𝜋𝜋𝜋𝜋
= (4 × 106 ) �
𝑑𝑑𝑑𝑑
0.2𝜋𝜋𝜋𝜋
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
= 4 × 107 cm2 per minute
© Pocket Tutor 2022
Differentiating the area of a circle
Plugging in our value for
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
The 𝑟𝑟’s cancel leaving us with 4 × 107
170
d) 13 hours
𝐴𝐴 = 𝜋𝜋𝑟𝑟 2
𝐴𝐴 = 𝜋𝜋(100,000)
1 × 1010 𝜋𝜋 𝑐𝑐𝑚𝑚2
2
1 × 1010 𝜋𝜋
= 785.398 minutes
4 × 107
785.398 ÷ 60 = 𝟏𝟏𝟏𝟏 hours
© Pocket Tutor 2022
The area of the slick (a circle) is increasing at a constant rate.
So, if we find the area of the slick at a radius of 1km and
divide this by the rate we can find the time.
1km = 100,000𝑐𝑐𝑐𝑐
Dividing the area by the rate at which the area is increasing.
Dividing by 60 to give the answer to the nearest hour.
171
Question 9
a) 𝟏𝟏𝟏𝟏 𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡 𝟓𝟓𝟓𝟓 𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦
𝑓𝑓(𝑡𝑡) = 12.25 + 4.75 sin �
2𝜋𝜋
𝑡𝑡�
365
2𝜋𝜋
(76)�
𝑓𝑓(76) = 12.25 + 4.75 sin �
365
= 12.25 + 4.5873
= 16.837 hours
0.837 × 60 = 50 minutes
Plugging in 76 for 𝑡𝑡.
Multiplying the decimal by 60 minutes to
convert it to minutes.
→ 𝟏𝟏𝟏𝟏 𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡 𝟓𝟓𝟓𝟓 𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦
b) 26th of April
12.25 + 4.75 sin �
4.75 sin �
2𝜋𝜋
𝑡𝑡� = 15
365
2𝜋𝜋
𝑡𝑡� = 15 − 12.25
365
2.75
2𝜋𝜋
𝑡𝑡� =
sin �
4.75
365
�
�
2𝜋𝜋
𝑡𝑡� = sin−1 (0.5789)
365
2𝜋𝜋
𝑡𝑡� = 0.6174
365
𝑡𝑡 = 0.6174 ÷
𝑡𝑡 = 35.87
2𝜋𝜋
365
= 36 days
36 days after March 21st is the 26th of April
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Letting the equation equal 15 hours.
Taking 12.25 from both sides
Dividing across by 4.75
Getting the sin inverse
Dividing across by the fraction
Rounding to the nearest day
There are 31 days in March
172
𝒄𝒄) 𝒇𝒇′ (𝒕𝒕) =
𝟐𝟐𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏
𝐜𝐜𝐜𝐜𝐜𝐜(
𝒕𝒕)
𝟑𝟑𝟑𝟑𝟑𝟑
𝟕𝟕𝟕𝟕𝟕𝟕
𝑓𝑓(𝑡𝑡) = 12.25 + 4.75 sin �
𝑓𝑓 ′ (𝑡𝑡) = 4.75 cos �
′ (𝒕𝒕)
𝒇𝒇
2𝜋𝜋
𝑡𝑡�
365
2𝜋𝜋
2𝜋𝜋
�
𝑡𝑡� × �
365
365
𝟐𝟐𝟐𝟐
𝟏𝟏𝟏𝟏𝟏𝟏
𝐜𝐜𝐜𝐜𝐜𝐜(
𝒕𝒕)
=
𝟑𝟑𝟑𝟑𝟑𝟑
𝟕𝟕𝟕𝟕𝟕𝟕
From page 25 of The Maths Tables Book we can see that
cos is the derivative of sin, we then need to multiply this
by the derivative of the brackets.
Multiplying (
2𝜋𝜋
365
) by 4.75
d) 17 hours
𝑓𝑓 ′ (𝑡𝑡) =
2𝜋𝜋
19𝜋𝜋
cos(
𝑡𝑡) = 0
365
730
The longest day is when the function is at its
maximum. When a function is at its maximum the
derivative equals 0.
2𝜋𝜋
𝑡𝑡) = 0
365
cos(
Dividing across by
2𝜋𝜋
�
𝑡𝑡� = cos −1 0
365
𝑡𝑡 = 91.25
2𝜋𝜋
𝑡𝑡�
365
𝑓𝑓(91.25) = 12.25 + 4.75 sin �
= 𝟏𝟏𝟏𝟏 hours
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. (0 over anything is 0)
Dividing across by the fraction
𝜋𝜋 2𝜋𝜋
÷
2 365
𝑓𝑓(𝑡𝑡) = 12.25 + 4.75 sin �
730
Getting the cos inverse
𝜋𝜋
2𝜋𝜋
�
𝑡𝑡� =
2
365
𝑡𝑡 =
19𝜋𝜋
2𝜋𝜋
(91.25)�
365
The longest day occurs when 𝑡𝑡 = 91.25
Plugging this back into the original equation to find
the length of this day.
Plugging the expression into the calculator
173
e) 𝟏𝟏𝟏𝟏 hours 𝟏𝟏𝟏𝟏 minutes
𝑏𝑏
1
� 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑
𝑏𝑏 − 𝑎𝑎 𝑎𝑎
→
184
1
2𝜋𝜋
� �12.25 + 4.75 sin �
𝑡𝑡�� 𝑑𝑑𝑑𝑑
184 − 0 0
365
184
1
1
2𝜋𝜋
�12.25𝑡𝑡 + 4.75 �− 2𝜋𝜋 cos �
𝑡𝑡���
184
365
365
0
To find the average length we integrate within
the limits of 0 days and 184 days.
1
Integrating sin 𝑎𝑎𝑎𝑎 goes to − cos 𝑥𝑥
𝑎𝑎
Putting a fraction under 1 inverts it, and
multiplying the minus out gives us −4.75
2𝜋𝜋
365
2𝜋𝜋
1
365
(184)��� − �12.25(0) − 4.75 �
(0)���
� cos �
� �12.25(184) − 4.75 �
cos �
365
2𝜋𝜋
365
184
2𝜋𝜋
1
[(2529.843) − (−275.9349)]
184
Plugging each bracket into the calculator.
= 15.24879 hours
Multiplying the decimal by 60 to convert it to
minutes
1
[2805.7779]
184
. 24879 × 60 = 14.9 = 15 minutes
→ 𝟏𝟏𝟏𝟏 hours 𝟏𝟏𝟏𝟏 minutes
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174
2014 Paper 1
Question 1
a)
𝑥𝑥 = −3, 𝑥𝑥 = −1, 𝑥𝑥 = 2
If these are the three roots of the equation, then we
can write out the factors as so.
(𝑥𝑥 + 3)(𝑥𝑥 + 1)(𝑥𝑥 − 2)
To find the equation we need to multiply the factors
together.
(𝑥𝑥 2 + 4𝑥𝑥 + 3)(𝑥𝑥 − 2)
Multiplying the brackets (𝑥𝑥 + 3)(𝑥𝑥 + 1) first.
Factors: 𝑥𝑥 + 3, 𝑥𝑥 + 1, 𝑥𝑥 − 2
2
(𝑥𝑥 + 𝑥𝑥 + 3𝑥𝑥 + 3)(𝑥𝑥 − 2)
𝑥𝑥 3 + 4𝑥𝑥 2 + 3𝑥𝑥 − 2𝑥𝑥 2 − 8𝑥𝑥 − 6
𝑥𝑥 3 + 2𝑥𝑥 2 − 5𝑥𝑥 − 6
b) i) (𝟎𝟎, −𝟔𝟔), (𝟏𝟏 − 𝟖𝟖), (−𝟑𝟑, 𝟎𝟎)
𝑓𝑓(𝑥𝑥) = 𝑔𝑔(𝑥𝑥)
𝑥𝑥 3 + 2𝑥𝑥 2 − 5𝑥𝑥 − 6 = −2𝑥𝑥 − 6
Letting the two equations equal each
other so we can find the points of
intersection.
𝑥𝑥(𝑥𝑥 2 + 2𝑥𝑥 − 3) = 0
Factorising out the 𝑥𝑥
𝑥𝑥 3 + 2𝑥𝑥 2 − 3𝑥𝑥 = 0
𝑥𝑥�(𝑥𝑥 − 1)(𝑥𝑥 + 3)� = 0
𝑥𝑥 = 0,
(𝑥𝑥 − 1) = 0,
𝑥𝑥 = 1,
𝑔𝑔(𝑥𝑥) = −2𝑥𝑥 − 6
𝑔𝑔(0) = −2(0) − 6 = −6
(𝟎𝟎, −𝟔𝟔)
𝑔𝑔(1) = −2(1) − 6 = −8
(𝑥𝑥 + 3) = 0
𝑥𝑥 = −3
Factorising the quadratic inside the
bracket.
Letting each part = 0 and solving for 𝑥𝑥.
Plugging in each 𝑥𝑥 value we found into
the equation for 𝑔𝑔(𝑥𝑥). This lets us find
the corresponding y-value for each 𝑥𝑥
value.
Writing out each of the coordinates
where the two functions intersect.
(𝟏𝟏, −𝟖𝟖)
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175
𝑔𝑔(−3) = −2(−3) − 6 = 0
(−𝟑𝟑, 𝟎𝟎)
ii)
Plotting the three points of intersection we found above gives us enough points to draw the graph
𝑔𝑔(𝑥𝑥).
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176
Question 2
a) (𝟏𝟏 + 𝟐𝟐𝟐𝟐),
𝟑𝟑
𝟐𝟐
𝑧𝑧1 = 1 − 2𝑖𝑖
If a complex number is the root of an equation, its
conjugate is also a root. To find a conjugate we change
the sign in front of the imaginary part.
𝑧𝑧̅1 = 𝟏𝟏 + 𝟐𝟐𝟐𝟐
(𝑧𝑧 − 1 + 2𝑖𝑖)(𝑧𝑧 − 1 − 2𝑖𝑖)
𝑧𝑧 2 − 𝑧𝑧 − 2𝑖𝑖𝑖𝑖 − 𝑧𝑧 + 1 + 2𝑖𝑖 + 2𝑖𝑖𝑖𝑖 − 2𝑖𝑖 − 4𝑖𝑖 2
𝑧𝑧 2 − 2𝑧𝑧 + 1 − 4𝑖𝑖 2
𝑧𝑧 2 − 2𝑧𝑧 + 5
If 𝑧𝑧 = 1 − 2𝑖𝑖 is a root, then 𝑧𝑧 − 1 + 2𝑖𝑖 is a factor.
Similarly, 𝑧𝑧 − 1 − 2𝑖𝑖 is a factor from the root 1 + 2𝑖𝑖.
Multiplying these two factors together gives us a factor
without any imaginary numbers.
Remember that 𝑖𝑖 2 = −1
2𝑧𝑧 − 3
𝑧𝑧 2 − 2𝑧𝑧 + 5 |2𝑧𝑧 3 − 7𝑧𝑧 2 + 16𝑧𝑧 − 15
2𝑧𝑧 3 − 4𝑧𝑧 2 + 10𝑧𝑧
Dividing our factor into the cubic allows us to find the
third factor.
−3𝑧𝑧 2 + 6𝑧𝑧 − 15
−3𝑧𝑧 2 + 6𝑧𝑧 − 15
0
2𝑧𝑧 − 3 is a factor
2𝑧𝑧 = 3
𝑧𝑧 =
Solving the third factor to find the third root of the
equation.
𝟑𝟑
𝟐𝟐
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177
b) i)
𝑤𝑤 = (1 − 2𝑖𝑖)(1 + 2𝑖𝑖)
1 + 2𝑖𝑖 − 2𝑖𝑖 − 4𝑖𝑖
1 − 4(−1) = 5
2
Multiplying 𝑧𝑧1 and its conjugate together to find
𝑤𝑤. Then plotting the three points with the real
numbers on the x-axis and the imaginary ones on
the y-axis.
𝑤𝑤 = 5 + 0𝑖𝑖
ii) 𝟓𝟓𝟓𝟓°
By drawing a line across the x -axis we get the
1
right-angled triangle shown. The angle is 𝜃𝜃 as it
is half the angle we are looking for.
opposite 2
1
=
tan 𝜃𝜃 =
adjacent 4
2
2
1
𝜃𝜃 = tan−1
4
2
1
𝜃𝜃 = 26.57
2
𝜃𝜃 = 26.57 × 2 = 𝟓𝟓𝟓𝟓°
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2
1
𝜃𝜃
2
We can see from the axes that the
opposite side measures 2 and the
adjacent one measures (5 − 1) =4.
178
Question 3
a)
To Prove: P(n) = 1 + 2 + 3 + ⋯ + 𝑛𝑛 =
𝑛𝑛(𝑛𝑛 + 1)
2
Show true for 𝑛𝑛 = 1:
When proving something by induction we
follow these steps:
1. Show that it is true for 𝑛𝑛 = 1.
1(1 + 1)
1=
2
1=1
2. Assume that it is true for 𝑛𝑛 = 𝑘𝑘.
Assume true for 𝑛𝑛 = 𝑘𝑘
1 + 2 + 3 + ⋯ + 𝑘𝑘 =
Plugging in 𝑘𝑘 for 𝑛𝑛 on both sides.
𝑘𝑘(𝑘𝑘 + 1)
2
Prove true for 𝑛𝑛 = 𝑘𝑘 + 1
1 + 2 + 3 + ⋯ + 𝑘𝑘 + (𝑘𝑘 + 1) =
(𝑘𝑘 + 1)�(𝑘𝑘 + 1) + 1�
2
(𝑘𝑘 + 1)�(𝑘𝑘 + 1) + 1�
𝑘𝑘(𝑘𝑘 + 1)
+ (𝑘𝑘 + 1) =
2
2
𝑘𝑘(𝑘𝑘 + 1) 2(𝑘𝑘 + 1) (𝑘𝑘 + 1)(𝑘𝑘 + 2)
+
=
2
2
2
(𝑘𝑘 + 2)(𝑘𝑘 + 1) (𝑘𝑘 + 1)(𝑘𝑘 + 2)
=
2
2
But 𝑃𝑃(1) is true, so 𝑃𝑃(2) is true etc.
Hence, 𝑃𝑃(𝑛𝑛) is true for all 𝑛𝑛.
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3. Prove that it is true for 𝑛𝑛 = 𝑘𝑘 + 1
Plugging in 𝑘𝑘 + 1 on both sides.
Plugging in our assumption for the sum up to
𝑘𝑘 on the left-hand side.
Expressing 𝑘𝑘 + 1 as a fraction on the left-hand
side.
Factorising the left hand side to write the top
of the fraction as (𝑘𝑘 + 2)(𝑘𝑘 + 1). This is the
same as the right hand side, so we have
proven it is true for 𝑘𝑘 + 1 and hence all values
of 𝑛𝑛.
179
b)
𝑆𝑆𝑛𝑛 =
𝑛𝑛
(2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑)
2
Taking the equation for the sum of an arithmetic series
from page 22 of the Maths Tables Book
𝑎𝑎 = 2, 𝑛𝑛 = 𝑛𝑛, 𝑑𝑑 = (4 − 2) = 2
𝑎𝑎 = the first term, d = the second term − the first term.
𝑛𝑛
�2(2) + (𝑛𝑛 − 1)(2)�
2
𝑛𝑛
𝑆𝑆𝑛𝑛 = (4 + 2𝑛𝑛 − 2)
2
𝑛𝑛
𝑆𝑆𝑛𝑛 = (2 + 2𝑛𝑛) → 𝑛𝑛(1 + 𝑛𝑛)
2
Subbing in our values for 𝑎𝑎, 𝑑𝑑 and 𝑛𝑛
𝑆𝑆𝑛𝑛 =
Simplifying it gives us our answer.
= 𝒏𝒏 + 𝒏𝒏𝟐𝟐
c) 𝒏𝒏𝟐𝟐
Sum of 𝑛𝑛 natural numbers =
𝑛𝑛(𝑛𝑛 + 1)
2
Sum of 2𝑛𝑛 natural numbers →
(2𝑛𝑛)((2𝑛𝑛) + 1)
= 2𝑛𝑛2 + 𝑛𝑛
2
Sum of 𝑛𝑛 even numbers = (n2 + 𝑛𝑛)
Sum of 𝑛𝑛 odd numbers = Sum of 2𝑛𝑛 natural numbers – Sum of 𝑛𝑛 even
numbers:
𝑛𝑛 even numbers plus 𝑛𝑛 odd numbers
= 2𝑛𝑛 natural numbers.
So, to find an expression for odd
numbers we need to find an
expression for 2𝑛𝑛 natural numbers.
We then subtract the expression for
𝑛𝑛 even numbers from this.
2𝑛𝑛2 + 𝑛𝑛 − (n2 + 𝑛𝑛)
= 𝒏𝒏𝟐𝟐
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180
Question 4
a) 𝟒𝟒𝟒𝟒 − 𝟑𝟑
𝑓𝑓(𝑥𝑥) = 2𝑥𝑥 2 − 3𝑥𝑥 − 6
𝑓𝑓(𝑥𝑥 + ℎ) = 2(𝑥𝑥 + ℎ)2 − 3(𝑥𝑥 + ℎ) − 6
2(𝑥𝑥 2 + 2ℎ𝑥𝑥 + ℎ2 ) − 3𝑥𝑥 − 3ℎ − 6
First, we plug in (𝑥𝑥 + ℎ) for 𝑥𝑥
𝑓𝑓(𝑥𝑥 + ℎ) − 𝑓𝑓(𝑥𝑥) = 2𝑥𝑥 2 + 4𝑥𝑥ℎ + 2ℎ2 − 3𝑥𝑥 − 3ℎ − 6 − (2𝑥𝑥 2 − 3𝑥𝑥 − 6)
Then we take 𝑓𝑓(𝑥𝑥), the original
function, away from 𝑓𝑓(𝑥𝑥 + ℎ).
2𝑥𝑥 2 + 4𝑥𝑥ℎ + 2ℎ2 − 3𝑥𝑥 − 3ℎ − 6
4𝑥𝑥ℎ + 2ℎ2 − 3ℎ
𝑓𝑓(𝑥𝑥 + ℎ) − 𝑓𝑓(𝑥𝑥)
= 4𝑥𝑥 + 2ℎ − 3
ℎ
Lim �
ℎ→0
𝑓𝑓(𝑥𝑥 + ℎ) − 𝑓𝑓(𝑥𝑥)
� = 4𝑥𝑥 + 2(0) − 3
ℎ
𝟒𝟒𝟒𝟒 − 𝟑𝟑
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Multiplying out the brackets.
Then we divide by ℎ.
Then we limit ℎ to 0, which effectively
means plug in 0 for any remaining ℎ.
181
b) (𝟐𝟐, 𝟏𝟏), (−𝟔𝟔, 𝟑𝟑)
𝑓𝑓(𝑥𝑥) =
2𝑥𝑥
𝑥𝑥 + 2
To find the slope of the tangent to a curve we
differentiate the equation of the curve.
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 𝑣𝑣 𝑑𝑑𝑑𝑑 − 𝑢𝑢 𝑑𝑑𝑑𝑑
=
𝑑𝑑𝑑𝑑
𝑣𝑣 2
Taking the quotient rule from page 25 of the Maths Tables
𝑢𝑢
Book, where 𝑦𝑦 = .
𝑣𝑣
𝑑𝑑𝑑𝑑 (𝑥𝑥 + 2)2 − 2𝑥𝑥(1)
=
(𝑥𝑥 + 2)2
𝑑𝑑𝑑𝑑
So, 𝑢𝑢 = 2𝑥𝑥 and 𝑣𝑣 = 𝑥𝑥 + 2
Plugging these and their derivatives into the expression.
4
2𝑥𝑥 + 4 − 2𝑥𝑥
=
2
(𝑥𝑥 + 2)2
(𝑥𝑥 + 2)
1
4
=
(𝑥𝑥 + 2)2 4
4(4)
=1
(𝑥𝑥 + 2)2
4(4) = (𝑥𝑥 + 2)
2
Now letting the derivative equal the given slope of the
tangent.
Multiplying across by 4.
2
Multiplying across by the bottom of the fraction.
16 = 𝑥𝑥 + 4𝑥𝑥 + 4
2
Squaring out the bracket.
𝑥𝑥 + 4𝑥𝑥 − 12 = 0
Solving the quadratic gives us the 𝑥𝑥 − values at which the
1
(𝑥𝑥 − 2)(𝑥𝑥 + 6) = 0
slope of the tangent is .
4
𝑥𝑥 = 2, 𝑥𝑥 = −6
𝑓𝑓(2) =
2(2)
= 1,
(2) + 2
(𝟐𝟐, 𝟏𝟏), (−𝟔𝟔, 𝟑𝟑)
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𝑓𝑓(6) =
2(−6)
=3
(−6) + 2
Plugging the 𝑥𝑥 values we found, into the equation of the
curve to find their corresponding y values.
Listing the coordinates.
182
Question 5
a)
� 5 cos 3𝑥𝑥 𝑑𝑑𝑑𝑑
1
5 � � sin 3𝑥𝑥 + 𝑐𝑐
3
𝟓𝟓
𝐬𝐬𝐬𝐬𝐬𝐬 𝟑𝟑𝟑𝟑 + 𝒄𝒄
𝟑𝟑
1
The integral of cos 𝑎𝑎𝑎𝑎 is sin 𝑎𝑎𝑎𝑎.
1
𝑎𝑎
So, multiplying by and integrating cos to
become sin.
3
Don’t forget the + 𝐶𝐶!
b) i) 𝒚𝒚 = 𝒙𝒙𝟐𝟐 − 𝟐𝟐𝟐𝟐 − 𝟖𝟖
� 2𝑥𝑥 − 2 𝑑𝑑𝑑𝑑
2 2
𝑥𝑥 − 2𝑥𝑥 + 𝑐𝑐 → 𝑥𝑥 2 − 2𝑥𝑥 + 𝑐𝑐
2
𝑓𝑓(−2) = 0
→ 𝑓𝑓(−2) = (−2)2 − 2(−2) + 𝑐𝑐 = 0
To find the equation of a curve when we have an
expression for the slope of a tangent, we integrate
the expression.
To integrate 2𝑥𝑥 we add 1 to the power and then
divide by the new power.
Now we use the point we were given to find the
value of 𝑐𝑐.
4 + 4 + 𝑐𝑐 = 0
Plugging in −2 for 𝑥𝑥 and letting it equal 0 as we
know the point (−2,0) is on the curve.
𝒚𝒚 = 𝒙𝒙𝟐𝟐 − 𝟐𝟐𝟐𝟐 − 𝟖𝟖
Writing out the full equation.
𝑐𝑐 = −8
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183
ii) −𝟖𝟖
Average value:
𝑏𝑏
1
� 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑
𝑏𝑏 − 𝑎𝑎 𝑎𝑎
𝑏𝑏 = 3, 𝑎𝑎 = 0, 𝑓𝑓(𝑥𝑥) = 𝑥𝑥 2 − 2𝑥𝑥 − 8
3
1
� 𝑥𝑥 2 − 2𝑥𝑥 − 8 𝑑𝑑𝑑𝑑
3−0 0
3
1 𝑥𝑥 3 2𝑥𝑥 2
� −
− 8𝑥𝑥�
2
3 3
0
(0)3
1 (3)3
��
− (3)2 − 8(3)� − �
− (0)2 − 8(0)��
3
3
3
1
[(9 − 9 − 24) − 0]
3
1
[−24] = −𝟖𝟖
3
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To find the average value of a function
between certain limits we integrate the
function between these limits and
multiply this by 1 over the upper limit
minus the lower limit.
Integrating the expression by adding
one to the power of each 𝑥𝑥 and
dividing by the new power.
Plugging in our two limits separately
and subtracting the results.
Multiplying in by the
1
3
184
Question 6
a) i)
𝑇𝑇𝑛𝑛 = ln 𝑎𝑎𝑛𝑛
𝑇𝑇1 = ln 𝑎𝑎1
𝑇𝑇2 = ln 𝑎𝑎2
𝑇𝑇3 = ln 𝑎𝑎3
𝑇𝑇2 − 𝑇𝑇1 = 2ln 𝑎𝑎 − ln 𝑎𝑎 = ln 𝑎𝑎
𝑇𝑇3 − 𝑇𝑇2 = 3 ln 𝑎𝑎 − 2 ln 𝑎𝑎 = ln 𝑎𝑎
𝑇𝑇3 − 𝑇𝑇2 = 𝑇𝑇2 − 𝑇𝑇1 ∴ Arithmetic Sequence
Plugging in 1,2 and 3 for 𝑛𝑛 to find the first 3 terms.
To show that they are in an arithmetic sequence we need to
show that there is a constant difference between each term.
Note ln 𝑎𝑎2 = 2 ln 𝑎𝑎 , and ln 𝑎𝑎3 = 3 ln 𝑎𝑎 and so on. (Page 21 of
the Maths Tables Book).
Subtracting term 1 from term 2 and then subtracting term 2
from term 3.
We can see that the difference between the 3rd and 2nd terms
and the difference between the 2nd and 1st terms are equal.
ii)
𝑇𝑇𝑛𝑛 = ln 𝑎𝑎𝑛𝑛
→ 𝑛𝑛 ln 𝑎𝑎
𝑇𝑇𝑛𝑛−1 = ln 𝑎𝑎(𝑛𝑛−1)
→ (𝑛𝑛 − 1) ln 𝑎𝑎
𝑇𝑇𝑛𝑛 − 𝑇𝑇𝑛𝑛−1 = 𝑛𝑛 ln 𝑎𝑎 − (𝑛𝑛 − 1) ln 𝑎𝑎
= 𝑛𝑛 ln 𝑎𝑎 − (𝑛𝑛 ln 𝑎𝑎 − 1 ln 𝑎𝑎)
= ln 𝑎𝑎
This is a constant ∴ sequence is arithmetic
Common difference = 𝑇𝑇𝑛𝑛 − 𝑇𝑇𝑛𝑛−1 = 𝐥𝐥𝐥𝐥 𝒂𝒂
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To prove that a sequence is arithmetic we
need to prove that there is a constant
difference for all values of 𝑛𝑛
To do this we find the difference between
𝑇𝑇𝑛𝑛 and 𝑇𝑇𝑛𝑛−1 .
Again, using the fact that log 𝑎𝑎 𝑥𝑥 𝑞𝑞 = 𝑞𝑞𝑞𝑞𝑞𝑞𝑔𝑔𝑎𝑎 𝑥𝑥
(page 21 of the Maths Tables Book).
Subtracting 𝑇𝑇𝑛𝑛−1 from 𝑇𝑇𝑛𝑛 to find the
difference.
The difference is ln 𝑎𝑎 which is a constant.
185
b) 𝟕𝟕. 𝟑𝟑𝟑𝟑𝟑𝟑
𝑇𝑇1 + 𝑇𝑇2 + 𝑇𝑇3 + ⋯ + 𝑇𝑇98 + 𝑇𝑇99 + 𝑇𝑇100 = 10100
Taking the equation for the sum of an arithmetic series
from page 22 of the Maths Tables Book.
𝑎𝑎 = ln 𝑎𝑎 , 𝑑𝑑 = ln 𝑎𝑎 , 𝑛𝑛 = 100
We know that the first term (a) is ln 𝑎𝑎 from part 𝑎𝑎 𝑖𝑖).
𝑆𝑆𝑛𝑛 =
𝑛𝑛
(2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑)
2
10100 =
100
(2(ln 𝑎𝑎) + (100 − 1) ln 𝑎𝑎)
2
10100 = 50(2 ln 𝑎𝑎 + 99 ln 𝑎𝑎)
10100
= 101 ln 𝑎𝑎
50
202 = 101 ln 𝑎𝑎
202
= ln 𝑎𝑎
101
ln 𝑎𝑎 = 2
𝑎𝑎 = 𝑒𝑒 2
= 𝟕𝟕. 𝟑𝟑𝟑𝟑𝟑𝟑
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We know that the constant difference (d) = ln 𝑎𝑎 from
the last part.
Subbing in our values for 𝑎𝑎, 𝑛𝑛 and 𝑑𝑑 and letting the
equation equal 10,100.
Dividing across by 50.
Dividing across by 101
ln 𝑎𝑎 is the same as log 𝑒𝑒 𝑎𝑎
𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥 (page 21 of the Maths Tables
Book)
So 𝑒𝑒 2 = 𝑎𝑎
186
c)
(𝑇𝑇1 + 𝑇𝑇2 + 𝑇𝑇3 + ⋯ + 𝑇𝑇10 ) + 100𝑑𝑑
(ln 𝑎𝑎 + 2 ln 𝑎𝑎 + 3 ln 𝑎𝑎 + ⋯ + 10 ln 𝑎𝑎) + 100(ln 𝑎𝑎)
𝑛𝑛
𝑆𝑆𝑛𝑛 = (2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑)
2
𝑎𝑎 = ln 𝑎𝑎 , 𝑑𝑑 = ln 𝑎𝑎 , 𝑛𝑛 = 10
10
(2(ln 𝑎𝑎) + (10 − 1) ln 𝑎𝑎) + 100 ln 𝑎𝑎
2
5(11 ln 𝑎𝑎) + 100 ln 𝑎𝑎
To verify that this is true we need to find the sum
of both sides and show that they are equal.
(Plugging in ln 𝑎𝑎 for 𝑑𝑑, as we found earlier).
Finding the sum of the left-hand side using the
formula for the sum of an arithmetic series on
page 22 of the Maths Table Book and then adding
the 100 ln 𝑎𝑎
= 155 ln 𝑎𝑎
(𝑇𝑇11 + 𝑇𝑇12 + 𝑇𝑇13 + ⋯ + 𝑇𝑇20 )
(11 ln 𝑎𝑎 + 12 ln 𝑎𝑎 + 13 ln 𝑎𝑎 + ⋯ + 20 ln 𝑎𝑎)
𝑆𝑆𝑛𝑛 =
Now, finding the sum of the right-hand side using
the formula for the sum of an arithmetic series on
page 22 of the Maths Table Book.
𝑛𝑛
(2𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑)
2
𝑎𝑎 = 11 ln 𝑎𝑎 , 𝑑𝑑 = ln 𝑎𝑎 , 𝑛𝑛 = 10
10
(2(11ln 𝑎𝑎) + (10 − 1) ln 𝑎𝑎)
2
5(31 ln 𝑎𝑎)
= 155 ln 𝑎𝑎
We can see that they are equal.
155 ln 𝑎𝑎 = 155 ln 𝑎𝑎
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187
Question 7
a) i)
𝑛𝑛 = 1
We can pick 1 to be the natural
number we sub in for 𝑛𝑛.
𝑎𝑎 = 2𝑛𝑛 + 1 → 2(1) + 1 = 3
𝑏𝑏 = 2𝑛𝑛2 + 2𝑛𝑛 → 2(1)2 + 2(1) = 4
2
Subbing in one for 𝑛𝑛 in each of the
three equations.
2
𝑐𝑐 = 2𝑛𝑛 + 2𝑛𝑛 + 1 → 2(1) + 2(1) + 1 = 5
Subbing each of the values we got
into 𝑎𝑎2 + 𝑏𝑏 2 = 𝑐𝑐 2
(3)2 + (4)2 = (5)2
9 + 16 = 25
We can see that both sides are equal,
so it is a Pythagorean triple.
25 = 25
ii)
𝑎𝑎2 + 𝑏𝑏 2 = 𝑐𝑐 2
(2𝑛𝑛 + 1)2 + (2𝑛𝑛2 + 2𝑛𝑛)2 =
(2𝑛𝑛2 + 2𝑛𝑛 + 1)2
(4𝑛𝑛2 + 4𝑛𝑛 + 1) + (4𝑛𝑛4 + 4𝑛𝑛3 + 4𝑛𝑛3 + 4𝑛𝑛2 )
= 4𝑛𝑛4 + 4𝑛𝑛3 + 2𝑛𝑛2 + 4𝑛𝑛3 + 4𝑛𝑛2 + 2𝑛𝑛 + 2𝑛𝑛2 + 2𝑛𝑛 + 1
4𝑛𝑛4 + 8𝑛𝑛3 + 8𝑛𝑛2 + 4𝑛𝑛 + 1 = 4𝑛𝑛4 + 8𝑛𝑛3 + 8𝑛𝑛2 + 4𝑛𝑛 + 1
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Subbing in the expressions for 𝑎𝑎, 𝑏𝑏 and 𝑐𝑐
respectively. Squaring out each of the brackets.
As both sides are equal, we have proven they will
always form a Pythagorean triple where 𝑛𝑛 ∈ 𝑁𝑁.
188
b) i)
|𝑃𝑃𝑃𝑃| = |𝑃𝑃𝑃𝑃| − |𝑀𝑀𝑀𝑀|
|𝑃𝑃𝑃𝑃| = (7 − 𝑥𝑥) − (2)
|𝑃𝑃𝑃𝑃| = 5 − 𝑥𝑥
|𝑃𝑃𝑃𝑃|2 = |𝐵𝐵𝐵𝐵|2 + |𝑃𝑃𝑃𝑃|2
|𝑃𝑃𝑃𝑃|2 = (2)2 + (5 − 𝑥𝑥)2 = 𝑥𝑥 2 − 10𝑥𝑥 + 29
|𝑃𝑃𝑃𝑃|2 = |𝐴𝐴𝐴𝐴|2 + |𝑃𝑃𝑃𝑃|2
As |𝐴𝐴𝐴𝐴| = 7 |𝐷𝐷𝐷𝐷| is also equal to 7. Therefore we
can say that |𝑃𝑃𝑃𝑃| = |𝐷𝐷𝐷𝐷| − |𝐷𝐷𝐷𝐷|, so 7 − 𝑥𝑥.
Using this to find |𝑃𝑃𝑃𝑃|.
If we draw a line from 𝑃𝑃 to 𝐵𝐵, we have a right-angled
triangle with hypotenuse |𝑃𝑃𝑃𝑃| so we can use
Pythagoras’ theorem to find an expression for |𝑃𝑃𝑃𝑃|2 .
|𝑃𝑃𝑃𝑃|2 = 22 + 𝑥𝑥 2 = 4 + 𝑥𝑥 2
Drawing a line from P to A, allows us to do the same
to find an expression for |𝑃𝑃𝑃𝑃|2 .
|𝑃𝑃𝑃𝑃|2 = |𝑃𝑃𝑃𝑃|2 + |𝐶𝐶𝐶𝐶|2
And doing the same for |𝑃𝑃𝑃𝑃|.
𝑓𝑓(𝑥𝑥) = |𝑃𝑃𝑃𝑃|2 + |𝑃𝑃𝑃𝑃|2 + |𝑃𝑃𝑃𝑃|2
Bringing the expressions we found for each line
together and adding them gives us the equation
𝑓𝑓(𝑥𝑥).
|𝑃𝑃𝑃𝑃|2 = (7 − 𝑥𝑥)2 + (2)2 = 𝑥𝑥 2 − 14𝑥𝑥 + 53
(4 + 𝑥𝑥 2 ) + (𝑥𝑥 2 − 10𝑥𝑥 + 29) + (𝑥𝑥 2 − 14𝑥𝑥 + 53)
𝑓𝑓(𝑥𝑥) = 3𝑥𝑥 2 − 24𝑥𝑥 + 86
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189
ii) 𝒌𝒌 = 𝟒𝟒, 𝐌𝐌𝐌𝐌𝐌𝐌 = 𝟑𝟑𝟑𝟑
𝑓𝑓(𝑥𝑥) = 3𝑥𝑥 2 − 24𝑥𝑥 + 86
To find a minimum value of a function we differentiate
it and let it equal 0.
6𝑥𝑥 − 24 = 0
Letting the function equal 0 and solving for 𝑥𝑥.
𝑓𝑓
′ (𝑥𝑥)
= 6𝑥𝑥 − 24
6𝑥𝑥 = 24
𝒙𝒙 = 𝟒𝟒 = 𝒌𝒌
𝑓𝑓 ′′ (𝑥𝑥) = 6
6 > 0 ∴ 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀
𝑓𝑓(4) = 3(4)2 − 24(4) + 86
= 𝟑𝟑𝟑𝟑
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Finding the second derivative allows us to confirm that
this is a minimum value. As 6 > 0 we know it’s a
minimum.
Subbing 4 back in for 𝑥𝑥 in the original equation to find
the minimum value.
190
Question 8
a) 𝑪𝑪(𝟐𝟐𝟐𝟐, 𝟕𝟕. 𝟒𝟒𝟒𝟒𝟒𝟒)
𝑦𝑦 = −0.013𝑥𝑥 2 + 0.624𝑥𝑥
𝑑𝑑𝑑𝑑
= −0.026𝑥𝑥 + 0.624
𝑑𝑑𝑑𝑑
−0.026𝑥𝑥 + 0.624 = 0
0.624 = 0.026𝑥𝑥
𝑥𝑥 =
0.624
= 24
0.026
𝑦𝑦 = −0.013(24)2 + 0.624(24)
𝑦𝑦 = 7.488
𝑪𝑪(𝟐𝟐𝟐𝟐, 𝟕𝟕. 𝟒𝟒𝟒𝟒𝟒𝟒)
To find the maximum point of an equation we
differentiate it and let it equal 0.
Letting the derivative equal 0 and solving for 𝑥𝑥.
Subbing this 𝑥𝑥 value back into the original
equation to find the corresponding 𝑦𝑦 value.
Writing out the coordinate.
b) 𝑫𝑫(𝟏𝟏𝟏𝟏, 𝟓𝟓), 𝑬𝑬(𝟑𝟑𝟑𝟑, 𝟓𝟓)
Equation of 𝐷𝐷𝐷𝐷: 𝑦𝑦 = 5
5 = −0.013𝑥𝑥 2 + 0.624𝑥𝑥
2
−0.013𝑥𝑥 + 0.624𝑥𝑥 − 5 = 0
−𝑏𝑏 ± √𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎
2𝑎𝑎
−0.624 ± �(0.624)2 − 4(−0.013)(−5)
2(−0.013)
𝑥𝑥 = 10 𝑜𝑜𝑜𝑜 𝑥𝑥 = 38
𝑫𝑫(𝟏𝟏𝟏𝟏, 𝟓𝟓), 𝑬𝑬(𝟑𝟑𝟑𝟑, 𝟓𝟓)
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𝐷𝐷𝐷𝐷 is a line straight across at 𝑦𝑦 = 5.
Letting the equation for 𝐷𝐷𝐷𝐷 equal the
equation for the parabola to find the points
where the line and the curve intersect, as
they intersect at the points 𝐷𝐷 and 𝐸𝐸.
Using the minus b formula which is listed
on page 20 of the Maths Tables Book to
solve for 𝑥𝑥.
Plugging this expression into the calculator
with a plus and then a minus finds us our
two values of 𝑥𝑥.
Listing the coordinates of 𝐷𝐷 and 𝐸𝐸.
191
c) 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 = Area under curve from 0 to 10 + Area of rectangle +Area under curve from 38 to 48.
10
48
� (−0.013𝑥𝑥 2 + 0.624𝑥𝑥)𝑑𝑑𝑑𝑑 = � (−0.013𝑥𝑥 2 + 0.624𝑥𝑥)𝑑𝑑𝑑𝑑
0
38
10
→ 2 � (−0.013𝑥𝑥 2 + 0.624𝑥𝑥)𝑑𝑑𝑑𝑑
2 �−
0
0.013 3 0.624 2 10
𝑥𝑥 +
𝑥𝑥 �
2
3
0
2 ��−
0.013
0.624
0.624
0.013
(10)3 +
(10)2 � − �−
(0)3 +
(0)2 ��
3
2
2
3
2[26.8667 − 0]
= 53.73
Area of rectangle:
To find the area under a curve we
integrate the equation of the curve
between certain limits.
The area on both sides of the
rectangle are equal so we can say
that the area is two times the area
between 0 and 10.
Plugging 10 and 0 in for 𝑥𝑥
respectively.
This is the area under the curve
from 0 to 10 plus the area under
the curve from 38 to 48
The rectangle has height 5 and
length 38 − 10 = 28. Multiplying
length by height gives us the area.
5 × 28 = 140
53.73 + 140 = 193.73
Adding the areas and rounding to
the nearest whole number.
= 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
d) 𝒚𝒚 − 𝟕𝟕. 𝟒𝟒𝟒𝟒𝟒𝟒 = −𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎(𝒙𝒙 − 𝟐𝟐𝟐𝟐)𝟐𝟐
𝑦𝑦 = −0.013𝑥𝑥 2 + 0.624𝑥𝑥
2
= −0.013(𝑥𝑥 − 48𝑥𝑥)
= −0.013(𝑥𝑥 2 − 48𝑥𝑥 + (−24)2 − (−24)2 )
= −0.013((𝑥𝑥 − 24)2 − 576)
𝑦𝑦 = −0.013(𝑥𝑥 − 24)2 + 7.488
𝒚𝒚 − 𝟕𝟕. 𝟒𝟒𝟒𝟒𝟒𝟒 = −𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎(𝒙𝒙 − 𝟐𝟐𝟐𝟐)𝟐𝟐
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To write the equation in the form given in the question we
need to complete the square. In order to complete the square,
we first need to factorise out the number in front of the 𝑥𝑥 2 .
Then we add and subtract half of the coefficient of the 𝑥𝑥,
squared.
Factorising.
Multiplying −576 by the −0.013
Writing it in the form given in the question.
192
e) 𝒚𝒚 + 𝟒𝟒 = −𝟐𝟐(𝒙𝒙 − 𝟑𝟑)𝟐𝟐
Given function: coefficient of 𝑥𝑥 2 ; −0.013;
2
New function: coefficient of 𝑥𝑥 ; −2;
maximum point (24,7.488)
maximum point (3, −4)
Given function: 𝑦𝑦 − 𝟕𝟕. 𝟒𝟒𝟒𝟒𝟒𝟒 = −𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎(𝑥𝑥 − 𝟐𝟐𝟐𝟐)2
New function: 𝑦𝑦 − (−𝟒𝟒) = −𝟐𝟐(𝑥𝑥 − 𝟑𝟑)2
𝒚𝒚 + 𝟒𝟒 = −𝟐𝟐(𝒙𝒙 − 𝟑𝟑)
𝟐𝟐
© Pocket Tutor 2022
By looking at the coefficient of the
𝑥𝑥 2 and the maximum point of the
function given in the question we
can see where these pieces of
information slot into the form we
just wrote the equation in.
Taking the information given to us
about the new function we can slot
its maximum point and coefficient
of the 𝑥𝑥 2 into the same places in
this form.
193
Question 9
a)
𝑦𝑦 = 100 − 23 = 77
𝑦𝑦 = 𝐴𝐴𝑒𝑒 𝑘𝑘𝑘𝑘
77 = 𝐴𝐴𝑒𝑒
𝟕𝟕𝟕𝟕 = 𝑨𝑨
𝑘𝑘0
Difference between water and room temp = 100 − 23
Subbing this in for 𝑦𝑦 and letting 𝑡𝑡 = 0 as this is the
temperature difference before any time has passed.
𝑒𝑒 0 = 1
b) 𝒌𝒌 = −𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
𝑦𝑦 = 77𝑒𝑒 𝑘𝑘𝑘𝑘
Subbing 77 in for 𝐴𝐴 in the equation.
65 = 77𝑒𝑒 𝑘𝑘(5)
Subbing in 5 for 𝑡𝑡 and the temperature difference we found
for 𝑦𝑦.
𝑡𝑡 = 5, 𝑦𝑦 = 88 − 23 = 65
65
= 𝑒𝑒 5𝑘𝑘
77
ln
65
= 5𝑘𝑘
77
−0.16942 = 5𝑘𝑘
𝑘𝑘 = −
0.16942
5
Finding the temperature difference after 5 minutes.
Dividing across by 77.
Using the law of logs on page 21 of The Maths Tables Book.
𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥, note log e 𝑥𝑥 = ln 𝑥𝑥.
Subbing this into the calculator and then dividing by 5.
To 3 significant figures.
𝑘𝑘 = −𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
c) 𝟑𝟑𝟑𝟑 𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦
𝑦𝑦 = 50 − 23 = 27
Finding the temperature difference when the water is at 50
degrees.
27
= 𝑒𝑒 −0.0339𝑡𝑡
77
Subbing this in for 𝑦𝑦 and subbing in the value we just found for 𝑘𝑘.
27
ln � � ÷ −0.0339 = 𝑡𝑡
77
𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥, noting that log e 𝑥𝑥 = ln 𝑥𝑥.
27 = 77𝑒𝑒 −0.0339𝑡𝑡
27
ln
= −0.0339𝑡𝑡
77
𝑡𝑡 = 30.9
𝑡𝑡 = 𝟑𝟑𝟑𝟑 𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦𝐦
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Dividing across by 77.
Again, Using the law of logs on page 21 of The Maths Tables Book.
Dividing across by −0.0339 and plugging it all into the calculator.
Rounding to the nearest minute.
194
d)
Part e) i)
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195
e) ii)
𝑚𝑚 = −𝟎𝟎. 𝟎𝟎𝟎𝟎
Any value for 𝑚𝑚 which is less than 𝑘𝑘 gives a faster rate of decay.
f) i) −𝟐𝟐. 𝟓𝟓𝟓𝟓, −𝟏𝟏. 𝟖𝟖𝟖𝟖
𝑦𝑦 = 77𝑒𝑒 −0.0339𝑡𝑡
𝑑𝑑𝑑𝑑
= (−0.0339)(77)𝑒𝑒 −0.0339𝑡𝑡
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
= −2.61𝑒𝑒 −0.0339𝑡𝑡
𝑑𝑑𝑑𝑑
𝑡𝑡 = 1 → −2.61𝑒𝑒 −0.0339(1)
= −𝟐𝟐. 𝟓𝟓𝟓𝟓
𝑡𝑡 = 10 → −2.61𝑒𝑒 −0.0339(10)
= −𝟏𝟏. 𝟖𝟖𝟖𝟖
To find the rate of change of a function we
differentiate the function.
The derivative of 𝑒𝑒 𝑎𝑎𝑎𝑎 is 𝑎𝑎𝑒𝑒 𝑎𝑎𝑎𝑎 . This is listed
on page 25 of the Maths Tables Book.
Now subbing in 1 for 𝑡𝑡 to find the rate of
change after 1 minute.
Subbing in 10 for 𝑡𝑡 to find the rate of
change after 10 minutes.
ii)
𝑑𝑑𝑑𝑑
= −2.61𝑒𝑒 −0.0339𝑡𝑡
𝑑𝑑𝑑𝑑
2
𝑑𝑑 𝑦𝑦
= −0.0339(−2.61)𝑒𝑒 −0.0339𝑡𝑡
𝑑𝑑𝑡𝑡 2
0.088𝑒𝑒
∴
−0.0339𝑡𝑡
>0
𝑑𝑑𝑦𝑦
is increasing.
𝑑𝑑𝑑𝑑
© Pocket Tutor 2022
If we find the second derivative, we can find
the rate of change of the rate of change.
Differentiating as we did above.
The second derivative is greater than 0 and
therefore the rate of change is increasing.
196
2013 Paper 1
Question 1
a)
𝑧𝑧 =
𝑧𝑧 =
4
1 + √3𝑖𝑖
4
1 + √3𝑖𝑖
×
4 − 4√3𝑖𝑖
1 − √3𝑖𝑖
1 − √3𝑖𝑖
1 + √3𝑖𝑖 − √3𝑖𝑖 − 3𝑖𝑖 2
4 − 4√3𝑖𝑖
1 − 3(−1)
4 − 4√3𝑖𝑖
= 1 − √3𝑖𝑖
4
𝐛𝐛) 𝟐𝟐(𝐜𝐜𝐜𝐜𝐜𝐜
1 − √3𝑖𝑖
We get the conjugate by changing the sign in the middle of the
complex number, i.e. 1 + √3𝑖𝑖 → 1 − √3𝑖𝑖.
Multiplying the top by the top and the bottom by the bottom.
𝑖𝑖 2 = −1
Dividing in by the 4.
𝟓𝟓𝟓𝟓
𝟓𝟓𝟓𝟓
+ 𝒊𝒊 𝐬𝐬𝐬𝐬𝐬𝐬 )
𝟑𝟑
𝟑𝟑
𝑟𝑟 = �𝑎𝑎2 + 𝑏𝑏2
2
𝑟𝑟 = �(1)2 + �−√3� = 2
𝑏𝑏
𝑎𝑎
𝜋𝜋
−1 √3
=
𝜃𝜃 = tan
1
3
𝜃𝜃 = tan−1
𝜋𝜋 5𝜋𝜋
𝜃𝜃 = 2𝜋𝜋 − =
3
3
𝟐𝟐(𝐜𝐜𝐜𝐜𝐜𝐜
To rewrite a complex number that is on the bottom of a fraction we
multiply the fraction by the conjugate of the complex number over
itself. We can do this as it is the same as multiplying by one.
𝟓𝟓𝟓𝟓
𝟓𝟓𝟓𝟓
+ 𝒊𝒊 𝐬𝐬𝐬𝐬𝐬𝐬 )
𝟑𝟑
𝟑𝟑
© Pocket Tutor 2022
To write a complex number in polar form we need to find the modulus (r)
and the argument (𝜃𝜃) and then plug them into the expression 𝑟𝑟(cos 𝜃𝜃 +
𝑖𝑖 sin 𝜃𝜃).
Remember that the complex number is currently written in the form 𝑎𝑎 +
𝑏𝑏𝑏𝑏.
Finding the modulus and then the reference angle.
We then take the reference angle away from 2𝜋𝜋 because, as we can see in
the diagram, the point is in the 4th quadrant. This gives us the argument.
Plugging the modulus and the argument into 𝑟𝑟(cos 𝜃𝜃 + 𝑖𝑖 sin 𝜃𝜃).
197
c)
𝑧𝑧 10 = �2 �cos
10
5𝜋𝜋
5𝜋𝜋
+ 𝑖𝑖 sin ��
3
3
5𝜋𝜋
5𝜋𝜋
210 �cos � × 10� + 𝑖𝑖 sin � × 10��
3
3
50𝜋𝜋
50𝜋𝜋
210 �cos
+ 𝑖𝑖 sin
�
3
3
1 √3
𝑖𝑖�
210 �− +
2
2
10
−2
1 √3
� −
𝑖𝑖�
2
2
−29 �1 − √3�
© Pocket Tutor 2022
To solve this question, we need to use De
Moivre’s theorem.
Following De Moivre’s theorem we put the
modulus (the 2) to the power of ten and then
multiply the argument (the angle) by 10.
Plugging the cos and sin into the calculator,
remembering to put the calculator into radians.
Factorising out the minus to get it closer to the
form asked for in the question.
Multiplying in by 2 to simplify the bracket and so
reducing the modulus to the power of 9.
198
Question 2
𝐚𝐚) 𝒙𝒙 ≤ −𝟑𝟑, 𝒙𝒙 ≥
𝟓𝟓
𝟐𝟐
2𝑥𝑥 2 + 𝑥𝑥 − 15 ≥ 0
2𝑥𝑥 2 + 𝑥𝑥 − 15 = 0
(2𝑥𝑥 − 5)(𝑥𝑥 + 3) = 0
2𝑥𝑥 = 5 𝑥𝑥 = −3
5
𝑥𝑥 =
2
𝒙𝒙 ≤ −𝟑𝟑, 𝒙𝒙 ≥
𝟓𝟓
𝟐𝟐
© Pocket Tutor 2022
Letting the expression equal 0 and solving for 𝑥𝑥 by factorising the
quadratic.
Sketching a graph with the two roots we found can show us for
what values of 𝑥𝑥 the graph will be greater than 0. (We know the
graph is U shaped as the equation has a positive 𝑥𝑥 2 ).
We can see from the graph that the function is greater than 0
when 𝑥𝑥 is less than −3 and greater than
5
2
199
b) 𝒙𝒙 = 𝟒𝟒, 𝒚𝒚 = 𝟏𝟏𝟏𝟏, 𝒛𝒛 = 𝟐𝟐
𝑥𝑥 + 𝑦𝑦 + 𝑧𝑧 = 16
5
𝑥𝑥 + 𝑦𝑦 + 10𝑧𝑧 = 40
2
1
2𝑥𝑥 + 𝑦𝑦 + 4𝑧𝑧 = 21
2
𝑒𝑒𝑒𝑒. 1
𝑒𝑒𝑒𝑒. 2 × 2 → 5𝑥𝑥 + 2𝑦𝑦 + 20𝑧𝑧 = 80
𝑒𝑒𝑒𝑒. 3
𝑒𝑒𝑒𝑒. 1 × −2 → −2𝑥𝑥 − 2𝑦𝑦 − 2𝑧𝑧 = −32
5𝑥𝑥 + 2𝑦𝑦 + 20𝑧𝑧 = 80
−2𝑥𝑥 − 2𝑦𝑦 − 2𝑧𝑧 = −32
3𝑥𝑥 + 18𝑧𝑧 = 48
𝑒𝑒𝑒𝑒. 4
𝑒𝑒𝑒𝑒. 3 × −2 → −4𝑥𝑥 − 𝑦𝑦 − 8𝑧𝑧 = −42
−4𝑥𝑥 − 𝑦𝑦 − 8𝑧𝑧 = −42
𝑥𝑥 + 𝑦𝑦 + 𝑧𝑧 = 16
−3𝑥𝑥 − 7𝑧𝑧 = −26
−3𝑥𝑥 − 7𝑧𝑧 = −26
3𝑥𝑥 + 18𝑧𝑧 = 48
11𝑧𝑧 = 22
𝒛𝒛 = 𝟐𝟐
𝑒𝑒𝑒𝑒. 5: − 3𝑥𝑥 − 7𝑧𝑧 = −26
−3𝑥𝑥 − 7(2) = −26
𝒙𝒙 = 𝟒𝟒
𝑒𝑒𝑒𝑒. 1: 𝑥𝑥 + 𝑦𝑦 + 𝑧𝑧 = 16
4 + 𝑦𝑦 + 2 = 16
𝒚𝒚 = 𝟏𝟏𝟏𝟏
Labelling the equations 1,2 and 3.
𝑒𝑒𝑒𝑒. 2
𝑒𝑒𝑒𝑒. 5
Multiplying equation 2 by 2 to simplify it.
Multiplying equation 1 by −2.
Adding equations 2 and 1.
This gives us a new equation in 𝑥𝑥 and 𝑧𝑧.
Multiplying equation 3 by −2 to simplify it.
Adding this to equation 1.
This gives us a second equation in 𝑥𝑥 and 𝑧𝑧.
Adding equations 4 and 5 together.
This gives us a value for 𝑧𝑧.
Plugging our value for 𝑧𝑧 into equation 5 to find
a value for 𝑥𝑥.
Plugging our values for 𝑥𝑥 and 𝑧𝑧 into equation 1
to find a value for 𝑦𝑦.
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200
Question 3
a)
0.693𝑡𝑡
𝑄𝑄 = 𝑒𝑒 − 5730
𝑄𝑄 = 𝑒𝑒 −
0.693(2000)
5730
𝑄𝑄 = 𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕
To find the proportion of carbon-14 in the
item we plug in 2000 for 𝑡𝑡.
b) 𝟖𝟖, 𝟗𝟗𝟗𝟗𝟗𝟗 years
0.693𝑡𝑡
0.3402 = 𝑒𝑒 − 5730
0.693𝑡𝑡
𝑙𝑙𝑙𝑙 0.3402 = −
5730
5730(ln 0.3402) = −0.693𝑡𝑡
5730(ln 0.3402)
= 𝑡𝑡
−0.693
𝑡𝑡 = 8915 → 𝟖𝟖, 𝟗𝟗𝟗𝟗𝟗𝟗 years
Plugging in the given proportion of carbon-14 for Q.
Using the rule: 𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥, found on page
21 of the Maths Tables Book.
log 𝑒𝑒 0.3402 is the same as ln 0.3402
Multiplying across by the bottom of the fraction.
Dividing across by −0.693
Rounding our answer to two significant figures.
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201
Question 4
a) i)
(1 + 𝑖𝑖)12 = (1 + 0.04)
12
(1 + 𝑖𝑖) = √1.04
1 + 𝑖𝑖 = 1.00327
𝑖𝑖 = 0.00327
𝑖𝑖 = 0.327%
ii) €𝟑𝟑𝟑𝟑𝟑𝟑
12 rooting both sides.
Taking 1 from both sides.
Multiplying by 100 to find the percentage.
𝐹𝐹 = 𝑃𝑃(1 + 𝑖𝑖)
15,000 = 𝑃𝑃(1 + 0.00327) + 𝑃𝑃(1 + 0.00327)2 + ⋯ + 𝑃𝑃(1 + 0.00327)36
𝑆𝑆𝑛𝑛 =
𝑎𝑎(1 − 𝑟𝑟 𝑛𝑛 )
1 − 𝑟𝑟
𝑃𝑃(1.00327)2
𝑎𝑎 = 𝑃𝑃(1.00327), 𝑟𝑟 =
= 1.00327, 𝑛𝑛 = 36
𝑃𝑃(1.00327)
15,000 =
𝑃𝑃(1.00327)(1 − (1.00327)36 )
1 − 1.00327
15,000(1 − 1.00327) = 𝑃𝑃(1.00327)(1 − (1.00327)36 )
15,000(1 − 1.00327)
= 𝑃𝑃
(1.00327)(1 − (1.00327)36 )
𝑃𝑃 = €392.02 = €𝟑𝟑𝟑𝟑𝟑𝟑
The equation for the future value of
an investment can be found on page
30 of the Maths Tables Book.
The value of the investment is the sum of the
future values of the monthly payments. This
means that we can find the sum of payments of
€P and let it equal 15,000. The equation for the
sum of a series can be found on page 22 of the
Maths Tables Book.
𝑎𝑎 = The first term. 𝑟𝑟 = the second term divided
by the first term. 𝑛𝑛 = the total number of
payments.
Multiplying across by the bottom of the fraction
and then dividing across by (1.00327)(1 −
(1.00327)36 ).
Plugging it into the calculator and rounding to the
nearest euro.
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202
b) €𝟒𝟒𝟒𝟒𝟒𝟒
𝐴𝐴 = 𝑃𝑃
𝑖𝑖(1 + 𝑖𝑖)𝑡𝑡
(1 + 𝑖𝑖)𝑡𝑡 − 1
𝑃𝑃 = 15000, 𝑖𝑖 = 0.00866, 𝑡𝑡 = 36
0.00866(1 + 0.00866)36
𝐴𝐴 = 15000 �
�
(1 + 0.00866)36 − 1
𝐴𝐴 = 486.77 → €𝟒𝟒𝟒𝟒𝟒𝟒
Taking the amortisation formula from page 31 of the
Maths Tables Book, where 𝐴𝐴 is the monthly
repayment amount and 𝑃𝑃 is the principal.
Plugging in the given values for 𝑃𝑃,𝑖𝑖 and 𝑡𝑡.
Plugging the expression into the calculator and
rounding it to the nearest euro.
Question 5
a)
b)
A quadratic differentiates to a line which differentiates to a constant.
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203
Question 6
a)
sin(0) = 0 ,
sin
5𝜋𝜋
= 0.5,
6
sin
𝜋𝜋
= 0.5,
6
𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 0
sin
𝜋𝜋
= 0.866,
3
sin
𝜋𝜋
= 1,
2
sin
2𝜋𝜋
= 0.866
3
b) 𝟏𝟏. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 𝟐𝟐
𝐴𝐴 =
ℎ
[𝑦𝑦 + 𝑦𝑦𝑛𝑛 + 2(𝑦𝑦2 + 𝑦𝑦3 + 𝑦𝑦4 + ⋯ + 𝑦𝑦𝑛𝑛−1 )]
2 1
𝜋𝜋
ℎ=
6
The trapezoidal rule can be found on
page 12 of the Maths Tables Book.
𝜋𝜋
𝐴𝐴 = 6 [0 + 0 + 2(0.5 + 0.866 + 1 + 0.866 + 0.5)]
2
𝐴𝐴 =
c) 𝟐𝟐
𝜋𝜋
[7.464] = 𝟏𝟏. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮 𝟐𝟐
12
𝜋𝜋
� sin 𝑥𝑥 𝑑𝑑𝑑𝑑
0
[− cos 𝑥𝑥]𝜋𝜋0
(− cos 𝜋𝜋) − (− cos 0)
1 − (−1) = 𝟐𝟐
ℎ = The gap between each height
measurement.
Filling in the heights from the table in
part a).
Multiplying it out.
The integral of sin 𝑥𝑥 is listed on page 26 of
the Maths Tables Book.
Plugging in 𝜋𝜋 and 0 for 𝑥𝑥 and subtracting
the results.
d)
Difference in results: 2 − 1.95407 = 0.04593
0.04593
× 100 = 𝟐𝟐. 𝟑𝟑%
2
© Pocket Tutor 2022
Putting the difference between
the two answers over the
correct answer (the one from
integrating).
204
Question 7
a)
€20 → 12,000
€19 → 13,000
€18 → 𝟏𝟏𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎
For each euro the price decreases we add 1,000 to the
attendance. So, if we decrease the price by €2, we
increase the attendance by 2,000
b)
12,000 + (20 − 𝑥𝑥)1000
12,000 + 20,000 − 1000𝑥𝑥
𝟑𝟑𝟑𝟑, 𝟎𝟎𝟎𝟎𝟎𝟎 − 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
12,000 is the starting attendance. We then add 1,000 for every
euro under €20 the ticket costs. So, if we take the new ticket
price (x) from €20 we get the number of thousands of people
we need to add.
c)
𝒇𝒇(𝒙𝒙) = (𝟑𝟑𝟑𝟑, 𝟎𝟎𝟎𝟎𝟎𝟎 − 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏)𝒙𝒙
d) €𝟏𝟏𝟏𝟏
𝑓𝑓(𝑥𝑥) = (32,000 − 1000𝑥𝑥)𝑥𝑥
𝑓𝑓(𝑥𝑥) = 32,000𝑥𝑥 − 1000𝑥𝑥 2
𝑓𝑓 ′ (𝑥𝑥) = 32,000 − 2000𝑥𝑥
32,000 − 2000𝑥𝑥 = 0
2000𝑥𝑥 = 32,000
𝑥𝑥 = €𝟏𝟏𝟏𝟏
e) €𝟐𝟐𝟐𝟐𝟐𝟐, 𝟎𝟎𝟎𝟎𝟎𝟎
𝑓𝑓(𝑥𝑥) = (32,000 − 1000𝑥𝑥)𝑥𝑥
𝑓𝑓(16) = �32,000 − 1000(16)�(16)
𝑓𝑓(16) = €𝟐𝟐𝟐𝟐𝟐𝟐, 𝟎𝟎𝟎𝟎𝟎𝟎
© Pocket Tutor 2022
To find the expected income we multiply the expected
attendance by the ticket price. So, we multiply the expression
from part b) by 𝑥𝑥.
To find the price which gives a maximum expected
income, we differentiate the function and let it
equal 0. We then solve for 𝑥𝑥.
Plugging our value for 𝑥𝑥 from the previous part into
the function gives us the maximum expected
income.
205
f) €𝟖𝟖𝟖𝟖, 𝟎𝟎𝟎𝟎𝟎𝟎
32,000 − 1000𝑥𝑥 = 25,000
To find the ticket price when the stadium is full, we take our
expression for the attendance, from part b), and we let it equal
the capacity of the stadium (25,000).
−1000𝑥𝑥 = −7,000
−7000
= €7
𝑥𝑥 =
−1000
Solving for 𝑥𝑥 gives us the ticket price.
€7 × 25,000 = €175,000
We then multiply this price by the capacity to find the income.
€256,000 − 175,000 = €𝟖𝟖𝟖𝟖, 𝟎𝟎𝟎𝟎𝟎𝟎
Now we take the income, from when the stadium is full, away
from the maximum income we found in part e).
g) 𝟐𝟐, 𝟓𝟓𝟓𝟓𝟓𝟓 tickets
𝑥𝑥 = number of single tickets
𝑓𝑓 = number of family tickets
𝑦𝑦 = cost of family ticket
Labelling the three unknowns.
𝑥𝑥 + 4𝑓𝑓 = 25,000
16𝑥𝑥 + 𝑓𝑓𝑓𝑓 = 365,000
𝑒𝑒𝑒𝑒. 1
𝑒𝑒𝑒𝑒. 2
16(𝑥𝑥 − 4,000) + (𝑓𝑓 + 1,000)𝑦𝑦 = 351,000
16𝑥𝑥 − 64,000 + 𝑓𝑓𝑓𝑓 + 1000𝑦𝑦 = 351,000
16𝑥𝑥 + 𝑓𝑓𝑓𝑓 + 1000𝑦𝑦 = 415,000
𝑒𝑒𝑒𝑒. 3
𝑒𝑒𝑒𝑒. 2 × −1 → −16𝑥𝑥 − 𝑓𝑓𝑓𝑓 = −365,000
16𝑥𝑥 + 𝑓𝑓𝑓𝑓 + 1000𝑦𝑦 = 415,000
−16𝑥𝑥 − 𝑓𝑓𝑓𝑓
= −365,000
1000𝑦𝑦 = 50,000
𝑦𝑦 = 50
16𝑥𝑥 + 𝑓𝑓(50) = 365,000
16𝑥𝑥 + 50𝑓𝑓 = 365,000
𝑒𝑒𝑒𝑒. 1 × −16 → −16𝑥𝑥 − 64𝑓𝑓 = −400,000
16𝑥𝑥 + 50𝑓𝑓 = 365,000
−16𝑥𝑥 − 64𝑓𝑓 = −400,000
−14𝑓𝑓 = −35,000
𝑓𝑓 = 𝟐𝟐, 𝟓𝟓𝟓𝟓𝟓𝟓 tickets
The total attendance is equal to the number of single
tickets sold plus the number of family tickets times 4,
as each family ticket admits four people. So, we can
say 𝑥𝑥 + 4𝑓𝑓 = 25,000.
The total income is equal to the number of each type
of ticket sold, times the price of that ticket. A single
ticket costs €16 so we can multiply 𝑥𝑥 by €16. We have
let the cost of a family ticket = 𝑦𝑦 so we can multiply 𝑓𝑓
by 𝑦𝑦. So, we can say that 16𝑥𝑥 + 𝑓𝑓𝑓𝑓 = 365,000
Finally, if we add 1000 to the number of family tickets
sold, we can take 14,000 from the total income and
4,000 from the number of single tickets sold. By
plugging this information into equation 2 and
simplifying we get equation 3.
Multiplying equation 2 by minus one and then adding
it to equation 3.
Subbing the value we found for 𝑦𝑦 into equation 2.
Multiplying equation 1 by −16 and then adding it to
equation 2.
This allows us to solve for 𝑓𝑓, the number of family
tickets.
© Pocket Tutor 2022
206
Question 8
a)
𝑠𝑠(𝑡𝑡) = 6𝑡𝑡 + 0.3𝑡𝑡 2 − 0.01𝑡𝑡 3
2
𝑠𝑠(10) = 6(10) + 0.3(10) − 0.01(10)
𝟖𝟖𝟖𝟖𝟖𝟖
b) 𝟓𝟓𝟓𝟓
𝑠𝑠(𝑡𝑡) = 6𝑡𝑡 + 0.3𝑡𝑡 2 − 0.01𝑡𝑡 3
𝑑𝑑𝑑𝑑
= 6 + (2)0.3𝑡𝑡 − (3)0.01𝑡𝑡 2
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
= 6 + 0.6𝑡𝑡 − 0.03𝑡𝑡 2
𝑑𝑑𝑑𝑑
6 + 0.6𝑡𝑡 − 0.03𝑡𝑡 2 = 8.25
0.03𝑡𝑡 2 − 0.6𝑡𝑡 − 6 = −8.25
0.03𝑡𝑡 2 − 0.6𝑡𝑡 + 2.25 = 0
3𝑡𝑡 2 − 60𝑡𝑡 + 225
𝑡𝑡 2 − 20𝑡𝑡 + 75 = 0
(𝑡𝑡 − 5)(𝑡𝑡 − 15) = 0
𝑡𝑡 = 5, 𝑡𝑡 = 15
0 ≤ 𝑡𝑡 ≤ 10 ∴ 𝑡𝑡 = 𝟓𝟓𝟓𝟓
© Pocket Tutor 2022
3
Taking the first equation as it is correct for when 𝑡𝑡 is less
than or equal to 10.
Subbing in 10 for 𝑡𝑡 and plugging it into the calculator.
We have an equation for displacement. To
find an equation for velocity we need to
differentiate this equation.
Letting this equation equal the given
velocity.
Multiplying across by −1
Multiplying across by 100
Dividing across by 3.
Factorising and solving for 𝑡𝑡. (If you struggle
to find the factors, the minus b formula can
also be used).
As our equation is only correct when t ≤ 10
we take 5 seconds as our answer.
207
c)
𝟗𝟗. 𝟗𝟗𝟗𝟗
𝑑𝑑𝑑𝑑
= 6 + 0.6𝑡𝑡 − 0.03𝑡𝑡 2
𝑑𝑑𝑑𝑑
𝑑𝑑 2 𝑠𝑠
= 0.6 − 0.06𝑡𝑡
𝑑𝑑𝑡𝑡 2
0.6 − 0.06𝑡𝑡 = 0.006
0.6 − 0.006 = 0.06𝑡𝑡
0.594 = 0.06𝑡𝑡
Differentiating the equation for velocity
gives us an equation for acceleration.
Letting this equation equal the acceleration
given.
Solving for 𝑡𝑡
𝑡𝑡 = 𝟗𝟗. 𝟗𝟗𝟗𝟗
d) 𝟕𝟕𝟕𝟕 seconds
𝑡𝑡 = 10 → 80𝑚𝑚
𝑡𝑡 > 10 → 𝑠𝑠 = 620 − 80 = 540𝑚𝑚
𝑑𝑑𝑑𝑑
= 6 + 0.6𝑡𝑡 − 0.03𝑡𝑡 2
𝑑𝑑𝑑𝑑
6 + 0.6(10) − 0.03(10)2
9𝑚𝑚𝑠𝑠 −1
540
= 60𝑠𝑠
9
10 + 60 = 𝟕𝟕𝟕𝟕 seconds
© Pocket Tutor 2022
We know from part a) that the raindrop falls
80m in the first ten seconds. We need to find
how long it takes to fall the 540m after that.
By taking our equation for the velocity of the
drop from part b) we can figure out the terminal
velocity it reaches at 10s.
We can then divide the remaining distance by
this velocity to find the time it will take to cover
this distance.
Adding the 10 seconds it takes to reach terminal
velocity gives us our answer.
208
e) 𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐/𝒔𝒔
Find = Given × Need
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
=
×
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
4
𝑉𝑉 = 𝜋𝜋𝑟𝑟 3
3
𝑑𝑑𝑑𝑑
= 4𝜋𝜋𝑟𝑟 2
𝑑𝑑𝑑𝑑
1
𝑑𝑑𝑑𝑑
=
𝑑𝑑𝑑𝑑 4𝜋𝜋𝑟𝑟 2
1
𝑑𝑑𝑑𝑑
=6×
4𝜋𝜋𝑟𝑟 2
𝑑𝑑𝑑𝑑
1
𝑑𝑑𝑑𝑑
=6×
4𝜋𝜋(1.5)2
𝑑𝑑𝑑𝑑
When tackling questions involving related rates
of change the word equation Find = Given x
Need can be helpful. You can revise this in the
‘rates of change’ section under differentiation.
The volume of a sphere from page 10 of the
Maths Tables Book.
Differentiating this to find
But we need
put 1 over
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
for our equation above, so we
to get this.
Plugging 6 into our equation as this was given
in the question as
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
and then plugging in
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
.
Putting in the given radius for 𝑟𝑟 and multiplying
out to get our answer.
𝑑𝑑𝑑𝑑
= 𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐/𝒔𝒔
𝑑𝑑𝑑𝑑
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209
Question 9
a) i)
ii)
b)
𝒏𝒏𝟐𝟐
We can see from part a) ii) that the number
of triangles is equal to the number of the
pattern squared.
c)
𝟑𝟑𝟑𝟑
© Pocket Tutor 2022
We can see from part a) ii) that the number of
extra match sticks needed for each patter is equal
to the number of the pattern times 3.
210
𝒅𝒅) 𝒂𝒂 =
𝟑𝟑
𝟑𝟑
, 𝒃𝒃 =
𝟐𝟐
𝟐𝟐
𝑢𝑢𝑛𝑛 = 𝑎𝑎𝑛𝑛2 + 𝑏𝑏𝑏𝑏
𝑢𝑢1 = 𝑎𝑎(1)2 + 𝑏𝑏(1) = 3
Plugging in 1 for 𝑛𝑛 and letting it equal
three, the number of matches in the first
pattern.
𝑎𝑎 = 3 − 𝑏𝑏
Rewriting to get 𝑎𝑎 in terms of 𝑏𝑏
𝑢𝑢2 = 𝑎𝑎(2)2 + 𝑏𝑏(2) = 9
Plugging in 2 for 𝑛𝑛 and letting it equal
nine, the number of matches in the
second pattern.
𝑎𝑎 + 𝑏𝑏 = 3
4𝑎𝑎 + 2𝑏𝑏 = 9
4(3 − 𝑏𝑏) + 2𝑏𝑏 = 9
12 − 4𝑏𝑏 + 2𝑏𝑏 = 9
−2𝑏𝑏 = −3
𝒃𝒃 =
𝟑𝟑
𝟐𝟐
𝑎𝑎 = 3 − 𝑏𝑏
3
𝑎𝑎 = 3 − � �
2
𝒂𝒂 =
Subbing in (3 − 𝑏𝑏) for 𝑎𝑎.
Solving for 𝑏𝑏
Plugging this value into our expression for
𝑎𝑎.
𝟑𝟑
𝟐𝟐
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211
e) 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 triangles
𝑢𝑢𝑛𝑛 = 𝑎𝑎𝑛𝑛2 + 𝑏𝑏𝑛𝑛
𝑢𝑢𝑛𝑛 =
3 2 3
𝑛𝑛 + 𝑛𝑛
2
2
4134 =
3 2 3
𝑛𝑛 + 𝑛𝑛
2
2
Subbing in the values for 𝑎𝑎 and 𝑏𝑏 from above.
Letting the equation equal the given number of
matchsticks.
Multiplying across by 2
8268 = 3𝑛𝑛2 + 3𝑛𝑛
3𝑛𝑛2 + 3𝑛𝑛 − 8268 = 0
𝑛𝑛2 + 𝑛𝑛 − 2756
(𝑛𝑛 + 53)(𝑛𝑛 − 52)
𝑛𝑛 = −53, 𝑛𝑛 = 52
𝑛𝑛2 = (52)2 = 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐 triangles
© Pocket Tutor 2022
Dividing across by 3
Factorising (the minus b formula could be used
here). Solving for n.
Taking the equation for the number of triangles we
found in part b) and subbing in the pattern number
we just found for 𝑛𝑛.
212
2012 Paper 1
Question 1
𝐚𝐚) 𝒂𝒂 = −
𝟏𝟏𝟏𝟏
𝟐𝟐
, 𝒃𝒃 =
𝟕𝟕
𝟕𝟕
𝒂𝒂 = 𝟏𝟏, 𝒃𝒃 = −𝟏𝟏
𝑎𝑎2 − 𝑎𝑎𝑎𝑎 + 𝑏𝑏 2 = 3
𝑎𝑎 + 2𝑏𝑏 + 1 = 0
Using the second equation to write 𝑎𝑎 in
terms of 𝑏𝑏.
𝑎𝑎 = −2𝑏𝑏 − 1
(−2𝑏𝑏 − 1)2 − (−2𝑏𝑏 − 1)(𝑏𝑏) + 𝑏𝑏 2 = 3
2
2
2
4𝑏𝑏 + 4𝑏𝑏 + 1 + 2𝑏𝑏 + 𝑏𝑏 + 𝑏𝑏 = 3
7𝑏𝑏 2 + 5𝑏𝑏 − 2 = 0
(7𝑏𝑏 − 2)(𝑏𝑏 + 1) = 0
𝒃𝒃 =
𝟐𝟐
, 𝒃𝒃 = −𝟏𝟏
𝟕𝟕
𝑎𝑎 = −2𝑏𝑏 − 1
2
𝑎𝑎 = −2 � � − 1
7
𝒂𝒂 = −
𝟏𝟏𝟏𝟏
𝟕𝟕
𝑎𝑎 = −2(−1) − 1
𝒂𝒂 = 𝟏𝟏
© Pocket Tutor 2022
Subbing the result in for 𝑎𝑎 in the other
equation.
Squaring out the brackets.
Factorising and solving the quadratic.
Subbing our values for 𝑏𝑏 into the
expression we found for 𝑎𝑎.
213
b) 𝒙𝒙 ≥ 𝟓𝟓, 𝒙𝒙 < 𝟑𝟑
2𝑥𝑥 − 5 5
≤
2
𝑥𝑥 − 3
(𝑥𝑥 − 3)2
2𝑥𝑥 − 5 5
≤ (𝑥𝑥 − 3)2
2
𝑥𝑥 − 3
5
(𝑥𝑥 − 3)(2𝑥𝑥 − 5) ≤ (𝑥𝑥 2 − 6𝑥𝑥 + 9)
2
2(2𝑥𝑥 2 − 11𝑥𝑥 + 15) ≤ 5(𝑥𝑥 2 − 6𝑥𝑥 + 9)
4𝑥𝑥 2 − 22𝑥𝑥 + 30 ≤ 5𝑥𝑥 2 − 30𝑥𝑥 + 45
0 ≤ 𝑥𝑥 2 − 8𝑥𝑥 + 15
0 = (𝑥𝑥 − 5)(𝑥𝑥 − 3)
We multiply across by the bottom of the
fraction squared. We do this because if we
multiply across by an unknown, we do not know
if we have to flip the inequality sign or not,
squaring the expression solves this issue.
Multiplying across by two and multiplying out
the brackets.
Getting everything on one side.
Letting the expression equal 0 and solving.
𝑥𝑥 = 5, 𝑥𝑥 = 3
2(4) − 5 5
≤
2
(4) − 3
3 5
≤
1 2
∴
𝒙𝒙 ≥ 𝟓𝟓, 𝒙𝒙 < 𝟑𝟑
© Pocket Tutor 2022
Subbing in a number between 3 and 5 to see if
the expression is true for this value.
It is not true, so we know that 𝑥𝑥 lies outside
these values → 𝑥𝑥 ≥ 5 and 𝑥𝑥 < 3.
𝑥𝑥 is not equal to 3 as this would give us an
undefined fraction
214
Question 2
a) & b)
c)
We are told that 𝐺𝐺 contains imaginary numbers made up of integers, we also know that 𝑄𝑄 only
contains rational numbers. Therefore 𝑎𝑎𝑎𝑎 cannot be an irrational number as we are multiplying two
rational numbers together.
© Pocket Tutor 2022
215
Question 3
a)
𝑟𝑟 = 5
𝑧𝑧 = 5
4𝜋𝜋
1
, 𝜃𝜃 =
9
16
1
4𝜋𝜋
4𝜋𝜋
�cos
+ 𝑖𝑖 sin �
16
9
9
If 𝑤𝑤 to the power of
𝑤𝑤 4 = 𝑧𝑧
𝑤𝑤 =
1
𝑧𝑧 4
=
1
𝑧𝑧 4
Using the modulus and
argument to write 𝑧𝑧 in
polar form.
1
4𝜋𝜋 1
1 4
4𝜋𝜋 1
� � + 𝑖𝑖 sin
� ��
= �5 � �cos
9 4
16
9 4
𝜋𝜋
3
𝜋𝜋
�cos + 𝑖𝑖 sin �
9
2
9
3
𝜋𝜋 𝑛𝑛𝑛𝑛
𝜋𝜋 𝑛𝑛𝑛𝑛
�cos � + � + 𝑖𝑖 sin � + ��
2
2
2
9
9
3
𝟑𝟑
𝝅𝝅
𝝅𝝅
𝜋𝜋 0𝜋𝜋
𝜋𝜋 0𝜋𝜋
𝑛𝑛 = 0 → �cos � + � + 𝑖𝑖 sin � + �� = �𝐜𝐜𝐜𝐜𝐜𝐜 + 𝒊𝒊 𝐬𝐬𝐬𝐬𝐬𝐬 �
2
𝟐𝟐
𝟗𝟗
𝟗𝟗
2
2
9
9
𝟏𝟏𝟏𝟏𝟏𝟏
𝟑𝟑
𝟏𝟏𝟏𝟏𝟏𝟏
3
𝜋𝜋 (1)𝜋𝜋
𝜋𝜋 (1)𝜋𝜋
� + 𝑖𝑖 sin � +
�� = �𝐜𝐜𝐜𝐜𝐜𝐜
�
+ 𝒊𝒊 𝐬𝐬𝐬𝐬𝐬𝐬
𝑛𝑛 = 1 → �cos � +
𝟐𝟐
𝟏𝟏𝟏𝟏
2
2
𝟏𝟏𝟏𝟏
2
9
9
1
find 𝑧𝑧 4 .
Following De Moivre’s
theorem by putting the
modulus to the power
1
of and multiplying the
4
1
argument by .
4
To find the four roots
𝑛𝑛𝑛𝑛
we add to the
2
argument. We then sub
in 0,1,2 and 3 for 𝑛𝑛 to
find the four roots in
polar form.
𝟑𝟑
𝟏𝟏𝟏𝟏𝟏𝟏
3
𝟏𝟏𝟏𝟏𝟏𝟏
𝜋𝜋 (2)𝜋𝜋
𝜋𝜋 (2)𝜋𝜋
� + 𝑖𝑖 sin � +
�� = �𝐜𝐜𝐜𝐜𝐜𝐜
�
𝑛𝑛 = 2 → �cos � +
+ 𝒊𝒊 𝐬𝐬𝐬𝐬𝐬𝐬
𝟐𝟐
𝟗𝟗
2
2
2
𝟗𝟗
9
9
𝟑𝟑
𝟐𝟐𝟐𝟐𝟐𝟐
3
𝟐𝟐𝟐𝟐𝟐𝟐
𝜋𝜋 (3)𝜋𝜋
𝜋𝜋 (3)𝜋𝜋
� + 𝑖𝑖 sin � +
�� = �𝐜𝐜𝐜𝐜𝐜𝐜
�
𝑛𝑛 = 3 → �cos � +
+ 𝒊𝒊 𝐬𝐬𝐬𝐬𝐬𝐬
𝟐𝟐
𝟏𝟏𝟏𝟏
2
2
2
𝟏𝟏𝟏𝟏
9
9
© Pocket Tutor 2022
1
four equals 𝑧𝑧. 𝑤𝑤 = 𝑧𝑧 4 .
So, we have to use De
Moivre’s theorem to
216
b)
𝑤𝑤0 =
𝑤𝑤1 =
𝜋𝜋
3
𝜋𝜋
�cos + 𝑖𝑖 sin � = 1.4 + 0.5𝑖𝑖
9
2
9
11𝜋𝜋
3
11𝜋𝜋
�cos
� = −0.5 + 1.4𝑖𝑖
+ 𝑖𝑖 sin
18
2
18
10𝜋𝜋
3
10𝜋𝜋
� = −1.4 − 0.5𝑖𝑖
𝑤𝑤2 = �cos
+ 𝑖𝑖 sin
9
2
9
29𝜋𝜋
3
29𝜋𝜋
� = 0.5 − 1.4𝑖𝑖
𝑤𝑤3 = �cos
+ 𝑖𝑖 sin
18
2
18
© Pocket Tutor 2022
To plot each of the roots we have to convert
them from polar form to rectangular form.
We do this by plugging each cos and sin into
3
the calculator and multiplying in the .
2
(Remember to change your calculator to
radians).
217
Question 4
a)
𝑎𝑎 + 𝑎𝑎𝑎𝑎 + 𝑎𝑎𝑟𝑟 2 + ⋯ + 𝑎𝑎𝑟𝑟 𝑛𝑛−1 =
𝑎𝑎(1 − 𝑟𝑟 𝑛𝑛 )
1 − 𝑟𝑟
To prove by induction, we first show that the
statement is true for 𝑛𝑛 = 1, by plugging in 1 for 𝑛𝑛.
Show to be true for 𝑛𝑛 = 1
𝑎𝑎𝑟𝑟1−1 =
𝑎𝑎(1) =
𝑎𝑎 = 𝑎𝑎
𝑎𝑎(1 − 𝑟𝑟1 )
1 − 𝑟𝑟
𝑎𝑎(1 − 𝑟𝑟)
(1 − 𝑟𝑟)
Assume it is true for 𝑛𝑛 = 𝑘𝑘
We then plug in 𝑘𝑘 for 𝑛𝑛 and assume that this is
true.
Prove that it is true for 𝑛𝑛 = 𝑘𝑘 + 1
We then plug in 𝑘𝑘 + 1 for 𝑛𝑛 and prove that this is
true using the assumption we made for 𝑛𝑛 = 𝑘𝑘.
𝑎𝑎 + 𝑎𝑎𝑎𝑎 + ⋯ + 𝑎𝑎𝑟𝑟 𝑘𝑘−1 =
𝑎𝑎�1 − 𝑟𝑟 𝑘𝑘 �
1 − 𝑟𝑟
𝑎𝑎 + 𝑎𝑎𝑎𝑎 + ⋯ + 𝑎𝑎𝑟𝑟 𝑘𝑘−1 + 𝑎𝑎𝑟𝑟 (𝑘𝑘+1)−1 =
𝑎𝑎 + 𝑎𝑎𝑎𝑎 + ⋯ + 𝑎𝑎𝑟𝑟 𝑘𝑘−1 + 𝑎𝑎𝑟𝑟 𝑘𝑘 =
𝑎𝑎�1 − 𝑟𝑟
𝑎𝑎(1 − 𝑟𝑟 )
+ 𝑎𝑎𝑟𝑟 𝑘𝑘 =
1 − 𝑟𝑟
1 − 𝑟𝑟
𝑘𝑘
𝑎𝑎�1 − 𝑟𝑟 𝑘𝑘+1 �
1 − 𝑟𝑟
𝑘𝑘+1
𝑘𝑘
𝑘𝑘
�
𝑘𝑘+1
𝑎𝑎�1 − 𝑟𝑟 � + 𝑎𝑎𝑟𝑟 (1 − 𝑟𝑟) 𝑎𝑎�1 − 𝑟𝑟
=
1 − 𝑟𝑟
1 − 𝑟𝑟
𝑘𝑘
𝑘𝑘
𝑎𝑎�1 − 𝑟𝑟 � + 𝑎𝑎(𝑟𝑟 − 𝑟𝑟
1 − 𝑟𝑟
𝑘𝑘+1
)
=
𝑎𝑎�1 − 𝑟𝑟 𝑘𝑘+1 �
1 − 𝑟𝑟
�
𝑘𝑘+1
𝑎𝑎�1 − 𝑟𝑟
1 − 𝑟𝑟
𝑎𝑎(1 − 𝑟𝑟 𝑘𝑘 + 𝑟𝑟 𝑘𝑘 − 𝑟𝑟 𝑘𝑘+1 ) 𝑎𝑎�1 − 𝑟𝑟 𝑘𝑘+1 �
=
1 − 𝑟𝑟
1 − 𝑟𝑟
Subbing in 𝑘𝑘 + 1 for 𝑛𝑛.
Focusing on the left-hand side. Subbing in our
assumption for the sum of the terms up to 𝑎𝑎𝑟𝑟 𝑘𝑘−1
Multiplying 𝑎𝑎𝑟𝑟 𝑘𝑘 by (1 − 𝑟𝑟) in order to write this
side as a single fraction.
�
Multiplying in the 𝑟𝑟 𝑘𝑘
Factorising out the 𝑎𝑎’s
Tidying up the left-hand side.
𝑎𝑎�1 − 𝑟𝑟 𝑘𝑘+1 � 𝑎𝑎(1 − 𝑟𝑟 𝑘𝑘+1 )
=
1 − 𝑟𝑟
1 − 𝑟𝑟
Thus, the proposition is true for 𝑛𝑛 = 𝑘𝑘 + 1 if it is true for 𝑛𝑛 = 𝑘𝑘, but it is true for 𝑛𝑛 = 1 and
therefore true for all values of 𝑛𝑛.
© Pocket Tutor 2022
218
𝐛𝐛)
𝟏𝟏𝟏𝟏𝟏𝟏
𝟑𝟑𝟑𝟑
5.21212121….
5+
21
21
21
+
+
+ ⋯.
100 10000 1000000
𝑆𝑆∞ =
𝑎𝑎
21
21
21
1
, 𝑎𝑎 =
, 𝑟𝑟 =
÷
=
1 − 𝑟𝑟
100
10000 100 100
21
100
5+
1
1−
100
=
𝟏𝟏𝟏𝟏𝟏𝟏
𝟑𝟑𝟑𝟑
© Pocket Tutor 2022
Recurring decimals can be expressed as
infinite geometric series.
We exclude the ‘5’ from the series as it is
not repeating. So, we construct a series of
the repeating 0.212121.
Taking the formula for an infinite
geometric series form page 22 of The
Maths Tables Book. Filling in 𝑎𝑎 and 𝑟𝑟 and
adding the 5.
219
Question 5
a) (−𝟏𝟏, 𝟕𝟕), (𝟓𝟓, 𝟑𝟑𝟑𝟑)
𝑓𝑓(𝑥𝑥) = 2𝑥𝑥 2 − 3𝑥𝑥 + 2
To find where the two curves meet, we let the
two equations equal each other.
2𝑥𝑥 2 − 3𝑥𝑥 + 2 = 𝑥𝑥 2 + 𝑥𝑥 + 7
Getting all the terms on one side.
𝑔𝑔(𝑥𝑥) = 𝑥𝑥 2 + 𝑥𝑥 + 7
𝑥𝑥 2 − 4𝑥𝑥 − 5 = 0
(𝑥𝑥 + 1)(𝑥𝑥 − 5) = 0
Solving the quadratic.
𝑓𝑓(𝑥𝑥) = 2𝑥𝑥 2 − 3𝑥𝑥 + 2
Plugging each 𝑥𝑥 value into the equation for 𝑓𝑓(𝑥𝑥)
to find the corresponding 𝑦𝑦 values.
𝑥𝑥 = −1, 𝑥𝑥 = 5
2
𝑓𝑓(−1) = 2(−1) − 3(−1) + 2 = 7
(−𝟏𝟏, 𝟕𝟕)
𝑓𝑓(5) = 2(5)2 − 3(5) + 2 = 37
(𝟓𝟓, 𝟑𝟑𝟑𝟑)
b) 𝟑𝟑𝟑𝟑 square units
5
5
𝐴𝐴 = � 𝑥𝑥 + 𝑥𝑥 + 7 𝑑𝑑𝑑𝑑 − � 2𝑥𝑥 2 − 3𝑥𝑥 + 2 𝑑𝑑𝑑𝑑
−1
2
5
−1
5
𝑥𝑥 3 𝑥𝑥 2
2𝑥𝑥 3 3𝑥𝑥 2
� + + 7𝑥𝑥� − �
−
+ 2𝑥𝑥�
3
2
3
2
−1
−1
��
(5)3 (5)2
(−1)3 (−1)2
+
+ 7(5)� − �
+
+ 7(−1)��
3
2
3
2
− ��
2(−1)3 3(−1)2
2(5)3 3(5)2
−
+ 2(5)� − �
−
+ 2(−1)��
3
2
3
2
41
25
535
335
�
− �− �� − �
− �− ��
6
6
6
6
= 96 − 60
= 𝟑𝟑𝟔𝟔 square units
© Pocket Tutor 2022
The area under a curve can
be found by integrating the
equation for the curve
between certain limits. The
area between the curves can
be found by taking the area
under 𝑓𝑓(𝑥𝑥) away from the
area under 𝑔𝑔(𝑥𝑥).
Integrating both curves and
plugging in the limits.
Subtracting the areas.
You can revise finding area
like this in the Integration
section.
220
Question 6
a)
1 2
𝑓𝑓(𝑥𝑥) = 𝑒𝑒 −2𝑥𝑥
1 2
𝑓𝑓 ′ (𝑥𝑥) = 𝑒𝑒 −2𝑥𝑥 × −𝑥𝑥
1
2
1 2
𝑓𝑓 ′′ (𝑥𝑥) = −𝑥𝑥 × �𝑒𝑒 −2𝑥𝑥 × −𝑥𝑥� + 𝑒𝑒 −2𝑥𝑥 × (−1)
1 2
𝑥𝑥 2 𝑒𝑒 −2𝑥𝑥
−
1 2
𝑒𝑒 −2𝑥𝑥
1 2
(𝑥𝑥 2 − 1)𝑒𝑒 −2𝑥𝑥
© Pocket Tutor 2022
1 2
derivative of − 𝑥𝑥 2 .
1 2
−𝑥𝑥𝑒𝑒 −2𝑥𝑥
1 2
1 2
To differentiate 𝑒𝑒 −2𝑥𝑥 , multiply 𝑒𝑒 −2𝑥𝑥 by the
To find the second derivative we need to use
the product rule. To do this we multiply (−𝑥𝑥)
1 2
by the derivative of 𝑒𝑒 −2𝑥𝑥 , we then multiply
1 2
𝑒𝑒 −2𝑥𝑥 by the derivative of (−𝑥𝑥) and add the
results.
1 2
Factorising out 𝑒𝑒 −2𝑥𝑥 to display the result in
the form given in the question.
221
b)
𝑓𝑓 ′′ (𝑥𝑥) = 0
(𝑥𝑥 2 −
1 2
1)𝑒𝑒 −2𝑥𝑥
We need to find the point of inflection so that we can find
and equation for the line P is on, then we can find the point
where this line crosses the 𝑥𝑥 − axis.
=0
(𝑥𝑥 2 − 1) = 0
The point of inflection can be found by letting the second
derivative equal 0 and solving for 𝑥𝑥. Dividing across by
𝑥𝑥 2 = 1
12
𝑥𝑥 = ±1
𝑒𝑒 −2𝑥𝑥 leaves us with (𝑥𝑥 2 − 1) = 0
1 2
We take +1 as our 𝑥𝑥 value as we are told in the question
that P is in the first quadrant, where 𝑥𝑥 is positive.
𝑓𝑓(𝑥𝑥) = 𝑒𝑒 −2𝑥𝑥
1
𝑓𝑓(1) = 𝑒𝑒 −2
=
1
𝑒𝑒 −2
(1)2
We plug this value into the equation for the graph to find
the 𝑦𝑦 −coordinate of 𝑃𝑃.
1
𝑓𝑓 ′ (𝑥𝑥) = −𝑒𝑒 −2𝑥𝑥
𝑓𝑓 ′ (1) =
1
(1)
−(1)𝑒𝑒 −2
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
𝑥𝑥1 = 1, 𝑦𝑦1 = 𝑒𝑒
1
−
1
2 , 𝑚𝑚
= −𝑒𝑒
= −𝑒𝑒
−
−
1
2
1
2
1
𝑦𝑦 − 𝑒𝑒 −2 = −𝑒𝑒 −2 (𝑥𝑥 − 1)
1
1
𝑦𝑦 = 0 → (0) − 𝑒𝑒 −2 = −𝑒𝑒 −2 (𝑥𝑥 − 1)
1 = 𝑥𝑥 − 1
𝑥𝑥 = 2
© Pocket Tutor 2022
We can find the slope of the tangent at the point P by
plugging the 𝑥𝑥 −coordinate into the first derivative of the
graph.
Subbing the slope and coordinates we found into the
equation of a line, from page 18 of the Maths Tables Book.
Letting 𝑦𝑦 = 0 as 𝑦𝑦 equals 0 all along the 𝑥𝑥 − axis.
1
Dividing across by 𝑒𝑒 −2
Solving for 𝑥𝑥
222
Question 7
a)
ℎ = �10 −
ℎ = �10 −
𝑡𝑡 2
�
200
2
0
� = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
200
Plugging in 0 for 𝑡𝑡.
b) 𝟒𝟒𝟒𝟒𝟒𝟒 seconds
64 = �10 −
𝑡𝑡 2
�
200
𝑡𝑡
√64 = 10 −
200
8 − 10 = −
𝑡𝑡
200
(−2)200 = −𝑡𝑡
Plugging in 64 for h.
Square rooting both sides.
Taking 10 from both sides.
Multiplying across by 200
−400 = −𝑡𝑡
𝑡𝑡 = 𝟒𝟒𝟒𝟒𝟒𝟒 seconds
© Pocket Tutor 2022
223
c) 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝒎𝒎𝟑𝟑 /𝒔𝒔
Find = Given × Need
To find the rate the volume is decreasing with respect
to time we need to use related rates of change.
𝑉𝑉 = 𝜋𝜋𝑟𝑟 2 ℎ
The equation for the volume of a cylinder is given on
page 10 of the Maths Tables Book.
𝑑𝑑𝑑𝑑 𝑑𝑑ℎ 𝑑𝑑𝑑𝑑
=
×
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑ℎ
𝑑𝑑𝑑𝑑
= 𝜋𝜋𝑟𝑟 2
𝑑𝑑ℎ
𝑑𝑑𝑑𝑑
= 𝜋𝜋(52)2
𝑑𝑑ℎ
𝑑𝑑𝑑𝑑
= 2704𝜋𝜋
𝑑𝑑ℎ
ℎ = �10 −
𝑡𝑡 2
�
200
−1
𝑡𝑡
𝑑𝑑ℎ
�×
= 2 �10 −
200
200
𝑑𝑑𝑑𝑑
−1
400
𝑑𝑑ℎ
�×
= 2 �10 −
200
200
𝑑𝑑𝑑𝑑
Differentiating to find
𝑑𝑑𝑑𝑑
𝑑𝑑ℎ
Plugging in 52 for 𝑟𝑟 as this is given in the question.
Using the chain rule to differentiate the equation
given for the height of the water to find
𝑑𝑑ℎ
𝑑𝑑𝑑𝑑
Plugging in 400 for 𝑡𝑡 as we found in part 𝑏𝑏 that 𝑡𝑡 =
400 when the height is 64.
2
𝑑𝑑ℎ
=−
25
𝑑𝑑𝑑𝑑
2
𝑑𝑑𝑑𝑑
= − × 2704𝜋𝜋
25
𝑑𝑑𝑑𝑑
= −679.589322
→ 𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯 𝐢𝐢𝐢𝐢 𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝 𝐚𝐚𝐚𝐚 𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝟔𝒎𝒎𝟑𝟑 /𝒔𝒔
Plugging our results into:
𝑑𝑑𝑑𝑑 𝑑𝑑ℎ 𝑑𝑑𝑑𝑑
=
×
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑ℎ
d)
𝑑𝑑𝑑𝑑
= 𝐴𝐴𝐴𝐴
𝑑𝑑𝑑𝑑
679.589322 = 𝜋𝜋𝑟𝑟 2 𝑣𝑣
679.589322 = 𝜋𝜋(1)2 𝑣𝑣
679.589322
= 𝑣𝑣
𝜋𝜋
𝑣𝑣 = 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑/𝒔𝒔
© Pocket Tutor 2022
Letting the rate at which the volume is decreasing equal
the area of the hole times the speed of the water coming
out of the hole.
Area of a circle = 𝜋𝜋𝑟𝑟 2 (pg. 8 of the Maths Tables Book).
Subbing in 1 for 𝑟𝑟 as given at the start of the question.
Dividing across by 𝜋𝜋 gives us our answer.
(We subbed in 679.589322 for the rate, instead of 680, to
be more accurate).
224
e)
𝐴𝐴𝐴𝐴 = −
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
The speed of the water coming out of the hole
𝑑𝑑ℎ 𝑑𝑑𝑑𝑑
(𝜋𝜋)𝑣𝑣 = − � �
𝑑𝑑𝑑𝑑 𝑑𝑑ℎ
𝜋𝜋𝜋𝜋 = −2 �10 −
−1
𝑡𝑡
�×
× 2704𝜋𝜋
200
200
𝑡𝑡
1
�10 −
� × 2704𝜋𝜋
𝜋𝜋𝜋𝜋 =
200
100
𝑡𝑡
�
𝑣𝑣 = 27.04 �10 −
200
multiplied by the Area equals
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
(from the
previous part, we inserted the minus because the
volume is decreasing).
Plugging in 𝜋𝜋 for the area (previous part) and
𝑑𝑑ℎ
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
× � � for
𝑑𝑑ℎ
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
from part ‘c’.
𝑑𝑑ℎ
Subbing in the values for
𝑑𝑑𝑑𝑑
found in part ‘c’.
and
𝑑𝑑𝑑𝑑
𝑑𝑑ℎ
Dividing across by 𝜋𝜋.
𝑡𝑡 2
�
ℎ = �10 −
200
We know from the question that
𝑡𝑡 2
�
ℎ = �10 −
200
→ 𝑣𝑣 = 27.04√ℎ
Subbing in √ℎ for �10 −
√ℎ = �10 −
𝑡𝑡
�
200
Which is a constant multiple of √ℎ
which we
Square rooting both sides
𝑡𝑡
200
�
f) 𝒄𝒄 = 𝟎𝟎. 𝟔𝟔
𝑣𝑣 = 27.04√ℎ
𝑣𝑣 = 27.04�(1)
𝑣𝑣 = 27.04
𝑣𝑣 = 𝑐𝑐√1962ℎ
27.04 = 𝑐𝑐�1962(1)
𝑐𝑐 =
27.04
Using the equation, we just found for the speed
of the water with regards to ℎ.
Plugging in one for ℎ to find the speed when the
water is at this height.
Taking the equation from the question and
plugging in 1 for ℎ and 27.04 for 𝑣𝑣.
Dividing across by the square root.
√1962
𝒄𝒄 = 𝟎𝟎. 𝟔𝟔
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225
Question 8
a)
𝑟𝑟 2 + ℎ2 = (9)2
We know that the slant length is 9 as the cone is
made from a circular piece of paper of radius 9cm.
𝑟𝑟 2 = 81 − ℎ2
Using Pythagoras theorem, we can express 𝑟𝑟 2 in
terms of ℎ.
2
2
𝑟𝑟 + ℎ = 81
𝑉𝑉 =
1 2
𝜋𝜋𝑟𝑟 ℎ
3
1
𝜋𝜋(81 − ℎ2 )ℎ
3
𝜋𝜋
𝑉𝑉 = ℎ(81 − ℎ2 )
3
𝑉𝑉 =
© Pocket Tutor 2022
Taking the equation for the volume of a cone from
page 10 of The Maths Tables Book.
Subbing in (81 − ℎ2 ) for 𝑟𝑟 2 .
Writing the equation in the form given in the
question.
226
b) 𝒉𝒉 = 𝟐𝟐, 𝒉𝒉 = 𝟕𝟕. 𝟖𝟖𝟖𝟖
𝑉𝑉 =
𝜋𝜋
ℎ(81 − ℎ2 )
3
Plugging in the given value for the volume.
154𝜋𝜋 𝜋𝜋
= ℎ(81 − ℎ2 )
3
3
154𝜋𝜋 = 𝜋𝜋ℎ(81 − ℎ2 )
Multiplying across by 3 then dividing across by 𝜋𝜋.
154 = 81ℎ − ℎ3
Multiplying out the bracket and then writing out
the cubic.
let ℎ = 1 → (1)3 − 81(1) + 154 ≠ 0
As we know there is one positive integer value,
we let 𝑥𝑥 = 1,2 etc. to find a root of the equation.
154 = ℎ(81 − ℎ2 )
ℎ3 − 81ℎ + 154 = 0
𝑙𝑙𝑙𝑙𝑙𝑙 ℎ = 2 → (2)3 − 81(2) + 154 = 0
0=0
ℎ2 + 2ℎ − 77
ℎ = 2 is a root.
Therefore ℎ − 2 is a factor.
We divide the cubic by this to find the other
factors of the equation.
ℎ − 2 ) ℎ3 + 0ℎ2 − 81ℎ + 154
ℎ3 − 2ℎ2
2ℎ2 − 81ℎ
2ℎ2 − 4ℎ
−77ℎ + 154
−77ℎ + 154
0
ℎ2 + 2ℎ − 77 = 0
−𝑏𝑏 ±
√𝑏𝑏 2
2𝑎𝑎
− 4𝑎𝑎𝑎𝑎
−2 ± �(2)2 − 4(1)(−77)
2(1)
𝒉𝒉 = 𝟕𝟕. 𝟖𝟖𝟖𝟖 ℎ = −9.83
𝒉𝒉 = 𝟐𝟐
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We can use the −𝑏𝑏 formula to solve for the other
two roots of the equation.
As the question is looking for the positive integer
7.83 is our answer.
227
c) 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝒎𝒎𝟑𝟑
𝜋𝜋
ℎ(81 − ℎ2 )
3
𝜋𝜋
𝑉𝑉 = (81ℎ − ℎ3 )
3
𝑉𝑉 =
𝑑𝑑𝑑𝑑 𝜋𝜋
= (81 − 3ℎ2 )
𝑑𝑑ℎ 3
𝜋𝜋
(81 − 3ℎ2 ) = 0
3
81 − 3ℎ2 = 0
To find the maximum volume of the cup we
differentiate the expression for the volume and
then let the derivative equal 0. This will give us the
height when the volume is a maximum. We then
sub this back into the equation to find the volume
at this height.
Multiplying in by the ℎ to make differentiating
easier.
Letting the derivative equal 0
81 = 3ℎ2
Dividing across by
𝜋𝜋
3
27 = ℎ2
Dividing across by 3
√27 = ℎ
Square rooting both sides.
𝜋𝜋
ℎ(81 − ℎ2 )
3
𝜋𝜋
2
𝑉𝑉 = �√27�(81 − �√27� )
3
Now plugging the value, we found for ℎ into the
equation for the volume of the cup.
𝑉𝑉 =
293.84𝑐𝑐𝑚𝑚3 → 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝒎𝒎𝟑𝟑
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Plugging the expression into the calculator gives us
our final answer.
228
d)
Cup in part (c)
Cups in part (b)
Radius (𝑟𝑟)
8.77cm
Height (h)
2cm
Capacity (𝑉𝑉)
154𝜋𝜋
= 161𝑐𝑐𝑚𝑚3
3
4.43 cm
7.35cm
7.83 cm
5.20 cm
154
= 161𝑐𝑐𝑚𝑚3
3
294𝑐𝑐𝑚𝑚3
Cups in part b
ℎ=2
ℎ = 7.83
𝑟𝑟 2 = 81 − ℎ2
𝑟𝑟 2 = 81 − ℎ2
𝑟𝑟 2 = 77
𝑟𝑟 2 = 19.69
𝑟𝑟 2 = 81 − (2)2
𝑟𝑟 2 = 81 − (7.83)2
𝑟𝑟 = 8.77𝑐𝑐𝑐𝑐
𝑟𝑟 = 4.43𝑐𝑐𝑐𝑐
𝑉𝑉 = Given in question
Cup in part c
ℎ = √27 = 5.2, 𝑉𝑉 = 294
𝑟𝑟 2 = 81 − ℎ2
2
𝑟𝑟 2 = 81 − �√27�
𝑟𝑟 2 = 81 − 27
𝑟𝑟 2 = 54
𝑟𝑟 = 7.35𝑐𝑐𝑐𝑐
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229
e)
The cup with the radius of 4.43𝑐𝑐𝑐𝑐 and height 7.83𝑐𝑐𝑐𝑐 as the other two cups are too wide and
shallow to hold.
f) 𝟏𝟏𝟏𝟏𝟏𝟏°
𝜃𝜃
�
𝑙𝑙 = 2𝜋𝜋𝜋𝜋 �
360
𝑙𝑙
× 360 = 𝜃𝜃
2𝜋𝜋𝜋𝜋
𝑟𝑟 = 9
𝑙𝑙 = Circumference of the top of the cup
Circumference = 2𝜋𝜋𝜋𝜋
2𝜋𝜋(4.43) = 8.86𝜋𝜋
8.86𝜋𝜋
× 360 = 𝜃𝜃
2𝜋𝜋(9)
𝜃𝜃 = 𝟏𝟏𝟏𝟏𝟏𝟏°
© Pocket Tutor 2022
Taking the equation for the length of an arc
from page 9 of the Maths Tables Book.
Rearranging the equation to get 𝜃𝜃 by itself.
The radius of the circle is given as 9 at the
start of the question.
The length of the arc missing from the circle is
the same as the circumference of the top of
the cup.
Using the equation for the circumference of a
circle from page 8. Using the radius of the top
of the cup we chose in part e).
Subbing our values for 𝑟𝑟 and 𝑙𝑙 into the
equation and rounding to the nearest degree.
230
Question 9
a) i)
Altitude (km)
Pressure
(kPa)
0
101.3
1
89.4
2
79.0
3
69.7
4
61.6
5
54.4
101.3 × 0.883 = 89.4
89.4 × 0.883 = 79.0
79.0 × 0.883 = 69.7
69.7 × 0.883 = 61.6
61.6 × 0.883 = 54.4
ii) 1%
89.9 − 89.4
× 100 = 0.56%
89.9
79.5 − 79.0
× 100 = 0.63%
79.5
70.1 − 69.7
× 100 = 0.57%
70.1
Finding the percentage error of each one by
putting the difference between the values in
the two tables over the original value and
multiplying by 100.
61.6 − 61.6
= 0%
61.6
54.0 − 54.4
× 100 = 0.74%
54.0
Hannah’s model is accurate to within 1%
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231
b) i)
Thomas’ model at altitude = 1
𝑝𝑝 = 101.3 × 𝑒𝑒 −0.1244ℎ
Using each of their models to find the
pressure at an altitude of 1km.
Letting ℎ = 1
101.3 × 𝑒𝑒 −0.1244(1) = 89.4506
Hannah’s model at altitude = 1
→ 101.3 × 0.883 = 89.4479
89.450 − 89.448 = 0.0027
0.0027 < 0.01
Subtracting the two answers gives us a
difference less than 0.01𝑘𝑘𝑘𝑘𝑘𝑘
ii)
He might have assumed it was of the form 101.3𝑒𝑒 −𝑘𝑘𝑘𝑘 and then used one of the observations to find
𝑘𝑘.
He might have put various values of 𝑡𝑡 into 𝑝𝑝(𝑡𝑡) = 101.3−𝑘𝑘𝑘𝑘 , and found an average of the resulting
values of 𝑘𝑘.
He might have got the natural log of the ratio of consecutive terms.
He might have plotted the log of the pressure against the altitude and used the slope of the best-fit
line to find 𝑘𝑘.
c) i)
Hannah’s model gives values for the pressure at separate (whole number) values for the altitude.
Thomas’s model gives a value for the pressure at any real value of the altitude, whether it’s a whole
number or not.
ii)
You are not restricted to the specific discrete values of the independent variable; you can also work
with values between any two given values – any value you like.
© Pocket Tutor 2022
232
d)
𝑝𝑝 = 101.3 × 𝑒𝑒 −0.1244ℎ
ℎ = 8.848 kilometres
𝑝𝑝 = 101.3 × 𝑒𝑒 −0.1244(8.848)
= 𝟑𝟑𝟑𝟑. 𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕
Converting the altitude to kilometres and
then plugging it in for ℎ in Thomas’ model.
e) 𝟓𝟓. 𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓
𝑝𝑝(ℎ) =
1
𝑝𝑝(0)
2
101.3 × 𝑒𝑒 −0.1244ℎ =
Sea level is an altitude of 0km.
1
�101.3 × 𝑒𝑒 −0,1244(0) �
2
101.3 × 𝑒𝑒 −0.1244ℎ = 50.65
𝑒𝑒 −0.1244ℎ =
𝑒𝑒 −0.1244ℎ =
50.65
101.3
1
2
1
ln = −0.1244ℎ
2
1
2 =ℎ
−0.1244
ln
ℎ = 𝟓𝟓. 𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓
© Pocket Tutor 2022
So, we let the equation for the pressure at a certain
height equal half the pressure at sea level. We find
the pressure at sea level by plugging 0 in for ℎ.
Dividing across by 101.3
Using the law of logs from page 21 of The Maths
Tables Book: 𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥
(Note, ln 𝑥𝑥 is the same as log 𝑒𝑒 𝑥𝑥)
Dividing across by −0.1244 and plugging the
expression into the calculator.
233
s
f) 𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀 𝟐𝟐𝟐𝟐 𝐟𝐟𝐟𝐟𝐟𝐟𝐟𝐟𝐟𝐟𝐟𝐟
101.3 − 1 = 100.3
𝑝𝑝 = 101.3 × 𝑒𝑒 −0.1244ℎ
100.3 = 101.3 × 𝑒𝑒 −0.1244ℎ
100.3
= 𝑒𝑒 −0.1244ℎ
101.3
ln
ln
100.3
= −0.1244ℎ
101.3
100.3
÷ (−0.1244) = ℎ
101.3
0.8𝑘𝑘𝑘𝑘 = ℎ
80 metres = ℎ
80
= 𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚 𝟐𝟐𝟐𝟐 𝐟𝐟𝐟𝐟𝐟𝐟𝐟𝐟𝐟𝐟𝐟𝐟
3
© Pocket Tutor 2022
Taking 1 from the pressure at sea level as we are looking for
the height at which the pressure has changed by 1 kilopascal
from the pressure at sea level.
Letting this pressure equal the equation for pressure at a given
height.
Dividing across by 101.3
Using the law of logs from page 21 of the Maths Tables Book:
𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥
(Note, ln 𝑥𝑥 is the same as log 𝑒𝑒 𝑥𝑥)
Dividing across by (−0.1244) and plugging the expression into
the calculator.
Estimating 3m as the distance between each floor, and
dividing this into our answer to get the number of floors.
234
Paper 2
Usual Paper 2 topics
•
•
•
Probability
Statistics
Geometry (Incl.
constructions/theor
ems)
© Pocket Tutor 2022
•
•
•
•
The Line
The Circle
Trigonometry
Length, Area,
Volume
•
Transformation &
Enlargements
235
2021 Paper 2
Question 1
a)
(1 − 0.15)10 × 0.15 × 11 = 𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑
To find the probability that 1 player is left footed we multiply the
probability of ten players being right footed (1 minus the probability
of a player being left footed, to the power of ten) by the probability
of one player being left footed. We then multiply this by 11 as this
can happen in any one of 11 ways.
b)
𝟎𝟎. 𝟕𝟕𝟕𝟕
P(Less than 3) = P(0 left footed) + P(1 left footed) + P(2 left footed)
𝑃𝑃(0 left footed):
(1 − 0.15)11 = 0.167
𝑃𝑃(1 left footed) = 0.325
𝑃𝑃(2 left footed):
�
11
� (1 − 0.15)9 × (0.15)2 = 0.287
2
𝑃𝑃(Less than 3) = 0.167 + 0.325 + 0.287
To find the probability that fewer
than three players are left footed
we add the probabilities of there
being 0, 1 or 2 left footed players
on the team.
Calculating the probability 0 being left footed by
putting the probability of a player being right footed
to the power of 11.
Taking P(1 left footed) from part a).
Remember that we need to multiply by 11-choose-2
to find the probability of 2 people being left footed
as this can happen in numerous different ways.
Adding the three probabilities gives us our answer.
= 0.779 → 𝟎𝟎. 𝟕𝟕𝟕𝟕
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236
c)
𝟎𝟎. 𝟖𝟖𝟖𝟖
𝑃𝑃(at least 8 right footed) = 𝑃𝑃(8 right footed) + 𝑃𝑃(9 right footed) + 𝑃𝑃(10 right footed)
𝑃𝑃(8 right footed) = �
10
� (1 − 0.15)8 × (0.15)2 = 0.2758
8
10
𝑃𝑃(9 right footed) = � � (1 − 0.15)9 × (0.15)1 = 0.3474
9
𝑃𝑃(10 right footed) = (1 − 0.15)10 = 0.1968
𝑃𝑃(at least 8 right footed) = 0.2758 + 0.3474 + 0.1968 = 𝟎𝟎. 𝟖𝟖𝟖𝟖
To find the probability of at least 8 being
right footed, we add the probabilities of 8,
9 or all 10 of the outfield players being right
footed.
Calculating the probability of 8 being right
footed by putting the probability of a player
being right footed to the power of 8 and
then the probability of a player being left
footed to the power of 2 as there are 10
players outfield. Remember that we need
to multiply by 10-choose-8 as this can
happen in numerous different ways.
Repeating this process for the probability of
9 being right footed.
Finally adding the three probabilities.
© Pocket Tutor 2022
237
Question 2
a)
−𝟐𝟐 = 𝒌𝒌
3𝑥𝑥 − 6𝑦𝑦 + 2 = 0
�𝑘𝑘,
2𝑘𝑘 + 2
�
3
To find the value of 𝑘𝑘, we sub the 𝑥𝑥 coordinate in
for 𝑥𝑥 and the 𝑦𝑦 coordinate in for 𝑦𝑦 in the equation
and let it equal 0.
We can then solve for 𝑘𝑘.
2𝑘𝑘 + 2
�+2=0
3(𝑘𝑘) − 6 �
3
Subbing in the coordinates.
3𝑘𝑘 − 4𝑘𝑘 − 4 + 2 = 0
Multiplying out the bracket.
3𝑘𝑘 − 2(2𝑘𝑘 + 2) + 2 = 0
Multiplying out the brackets, the 3 divides into
the 6 on the outside of the bracket, leaving us
with −2(2𝑘𝑘 + 2).
−𝑘𝑘 − 2 = 0
Adding and subtracting like terms.
−𝟐𝟐 = 𝒌𝒌
Adding 𝑘𝑘 to both sides.
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238
b)
𝑡𝑡 = −𝟑𝟑, 𝑠𝑠 = 𝟐𝟐 𝒐𝒐𝒐𝒐 𝒕𝒕 = −
𝟒𝟒𝟒𝟒
𝟐𝟐
, 𝒔𝒔 =
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
𝑥𝑥 − 2𝑦𝑦 − 8 = 0
(𝑠𝑠, 𝑡𝑡)
(𝑠𝑠) − 2(𝑡𝑡) − 8 = 0
𝑠𝑠 = 2𝑡𝑡 + 8
4𝑥𝑥 + 3𝑦𝑦 + 6 = 0
distance =
|𝑎𝑎𝑥𝑥1 + 𝑏𝑏𝑦𝑦1 + 𝑐𝑐|
√𝑎𝑎2 + 𝑏𝑏 2
distance = 1, 𝑎𝑎 = 4, 𝑏𝑏 = 3, 𝑐𝑐 = 6, 𝑥𝑥1 = 2𝑡𝑡 + 8, 𝑦𝑦1 = 𝑡𝑡
1=
|4(2𝑡𝑡 + 8) + 3(𝑡𝑡) + 6|
�(4)2 + (3)2
|4(2𝑡𝑡 + 8) + 3(𝑡𝑡) + 6|
1=
5
5 = |4(2𝑡𝑡 + 8) + 3(𝑡𝑡) + 6|
5 = |8𝑡𝑡 + 32 + 3𝑡𝑡 + 6|
5 = |11𝑡𝑡 + 38|
© Pocket Tutor 2022
We can get an equation in 𝑠𝑠 and 𝑡𝑡 by subbing,
𝑠𝑠 in for 𝑥𝑥 and 𝑡𝑡 in for 𝑦𝑦 in the equation of the
line and letting it equal 0.
We can rearrange the equation to get 𝑠𝑠 in
terms of 𝑡𝑡.
Now we can find another expression by using
the distance from a point to a line formula
from page 9 of the Maths Tables Book.
We can sub in the given distance and the
values from the equation of the line.
The coordinates (𝑠𝑠, 𝑡𝑡) are 𝑥𝑥1 , 𝑦𝑦1 , but as we
found earlier, we can write 𝑠𝑠 as 2𝑡𝑡 + 8. So,
we sub this in for 𝑥𝑥1 .
Plugging the square root on the bottom into
the calculator.
Multiplying across by 5.
Multiplying out the brackets.
Adding like terms.
239
11𝑡𝑡 + 38 = ±5
As there are modulus bars on the right hand
side, we say that whatever is in the modulus
bars is equal to plus or minus 5.
11𝑡𝑡 + 38 = 5
11𝑡𝑡 = −33
As we are asked to find one value of 𝑠𝑠 and 𝑡𝑡,
we can just take the positive value of 5 and
solve for 𝑡𝑡.
𝑡𝑡 = −𝟑𝟑
𝑠𝑠 = 2𝑡𝑡 + 8
𝑠𝑠 = 2(−3) + 8
Subbing the value, we found for 𝑡𝑡 into our
expression for 𝑠𝑠.
𝑠𝑠 = 𝟐𝟐
Other answer:
𝒕𝒕 = −
𝟐𝟐
𝟒𝟒𝟒𝟒
, 𝒔𝒔 =
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
c) i)
𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮
𝐴𝐴(4,2), 𝐶𝐶(16,11)
𝑥𝑥1 = 4, 𝑦𝑦1 = 2,
𝑥𝑥2 = 16, 𝑦𝑦2 = 11
|𝐴𝐴𝐴𝐴| = �(𝑥𝑥2 − 𝑥𝑥1 )2 + (𝑦𝑦2 − 𝑦𝑦1 )2
|𝐴𝐴𝐴𝐴| = �(16 − 4)2 + (11 − 2)2
|𝐴𝐴𝐴𝐴| = 15
|𝐴𝐴𝐴𝐴| ∶ |𝐷𝐷𝐷𝐷| = 2 ∶ 1
15 ÷ 3 = 5
|𝐴𝐴𝐴𝐴| = 5 × 2 = 𝟏𝟏𝟏𝟏 𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮𝐮
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As we know the ratio of the distances |𝐴𝐴𝐴𝐴| and
|𝐷𝐷𝐷𝐷| which are both on the line 𝐴𝐴𝐴𝐴, we can find
the distance |𝐴𝐴𝐴𝐴| by finding the distance |𝐴𝐴𝐴𝐴|.
Taking the expression for the distance between
two points from page 18 of the Maths Tables
Book.
Subbing in the coordinates of 𝐴𝐴 and 𝐶𝐶 and
plugging into the calculator.
We know that the distances |𝐴𝐴𝐴𝐴| and |𝐷𝐷𝐷𝐷| are in
the ratio 2: 1, so we can find |𝐴𝐴𝐴𝐴| by dividing the
length of |𝐴𝐴𝐴𝐴| by 3 and multiplying the result by
2.
240
ii)
𝑬𝑬(𝟐𝟐𝟐𝟐, 𝟖𝟖)
|𝐴𝐴𝐴𝐴| = 33 → 𝐵𝐵 is (37,2)
The translation 𝐶𝐶 → 𝐵𝐵: 𝑥𝑥 increase by 21.
1
→ 𝑥𝑥 = 16 + (21) = 23
3
The translation 𝐶𝐶 → 𝐵𝐵: 𝑦𝑦 decreases by 9.
1
→ 𝑦𝑦 = 11 − (9) = 8
3
𝑬𝑬(𝟐𝟐𝟐𝟐, 𝟖𝟖)
© Pocket Tutor 2022
The question tells us that the lines AB and DE are
horizontal. This means that they are on the same
level and therefore have the same 𝑦𝑦 coordinates.
We also know that the distance from A to B is 33.
So, we can add 33 to the coordinates of 𝐴𝐴 to find
the coordinates of 𝐵𝐵.
We can find the coordinates of 𝐸𝐸 by looking at the
translation from 𝐶𝐶 to 𝐵𝐵.
The 𝑥𝑥 coordinate increases by 21 from 𝐶𝐶 to 𝐵𝐵. We
know that |𝐶𝐶𝐶𝐶| and |𝐵𝐵𝐵𝐵| are in the same ratio as
|𝐴𝐴𝐴𝐴| and |𝐷𝐷𝐷𝐷| is |𝐷𝐷𝐷𝐷| and |𝐴𝐴𝐵𝐵| are horizontal.
So, we multiply the translations from 𝐶𝐶 to 𝐵𝐵 by
to find the translations from 𝐶𝐶 to 𝐸𝐸.
241
1
3
Question 3
a)
2√3
𝐴𝐴
𝟒𝟒√𝟐𝟐
𝐷𝐷(3,2)
|𝐷𝐷𝐷𝐷| = �(𝑥𝑥2 − 𝑥𝑥1 )2 + (𝑦𝑦2 − 𝑦𝑦1 )2
2
|𝐷𝐷𝐷𝐷| = �(3 − 1)2 + �2 − (−2)�
|𝐷𝐷𝐷𝐷| = 2√5
|𝐴𝐴𝐴𝐴|2 = |𝐴𝐴𝐴𝐴|2 + |𝐶𝐶𝐶𝐶|2
2
|𝐴𝐴𝐴𝐴|2 = �2√3� + �2√5�
|𝐴𝐴𝐴𝐴|2 = 32
|𝐴𝐴𝐴𝐴| = 𝟒𝟒√𝟐𝟐
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𝐶𝐶(1, −2)
2
To help us find the radius we have
drawn out the right-angled triangle
𝐴𝐴𝐴𝐴𝐴𝐴. We know the length 𝐴𝐴𝐴𝐴 is 4√3.
As the point 𝐷𝐷 is the midpoint, we
can find the distance 𝐴𝐴𝐴𝐴 by halving
this.
The distance 𝐴𝐴𝐴𝐴 is the radius length,
we can find it using Pythagoras’
theorem but first we need to find
|𝐶𝐶𝐶𝐶|. We do this using the distance
formula from page 18 of the Maths
Tables Book.
Finally, using Pythagoras’ theorem to
find the radius length.
242
b) i)
2
2
𝑐𝑐: 𝑥𝑥 + 𝑦𝑦 + 4𝑥𝑥 − 2𝑦𝑦 − 95 = 0
Centre: (−𝑔𝑔, −𝑓𝑓)
→ (−2,1)
Radius: �𝑔𝑔2 + 𝑓𝑓 2 − 𝑐𝑐
→ �(2)2 + (−1)2 − (−95) = 10
𝑠𝑠: (𝑥𝑥 − 7)2 + (𝑦𝑦 − 13)2 = 25
centre: (h, k)
→ (7,13)
Radius: 𝑟𝑟
→ √25 = 5
Distance from centre 𝑐𝑐 (−2,1), to centre 𝑠𝑠 (7,13):
𝑑𝑑 = �(𝑥𝑥2 − 𝑥𝑥1 )2 + (𝑦𝑦2 − 𝑦𝑦1 )2
𝑑𝑑 = �(7 − (−2))2 + (13 − 1)2
𝑑𝑑 = 15
𝑟𝑟1 + 𝑟𝑟2 = distance between centres
10 + 5 = 15
15 = 15
© Pocket Tutor 2022
To show that two circles touch externally we
need to show that the distance between their
centres is equal to the sum of their radii.
First, we write down the centre and the radius
of circle 𝑐𝑐.
We use page 19 of the Maths Tables Book to
help us.
Remember the circle, 𝑐𝑐, is written in the form
𝑥𝑥 2 + 𝑦𝑦 2 + 2𝑔𝑔𝑔𝑔 + 2𝑓𝑓𝑓𝑓 + 𝑐𝑐 = 0
We now do the same for circle 𝑠𝑠, also using
page 19 but the part for circles written in the
from: (𝑥𝑥 − ℎ)2 + (𝑦𝑦 − 𝑘𝑘)2 = 𝑟𝑟 2 .
𝑟𝑟 = the radius.
Now we take the distance formula from page
18 of the Maths Tables Book, and sub in the
coordinates (−2,1) for 𝑥𝑥1 , 𝑦𝑦1 and (7,13) for
𝑥𝑥2 , 𝑦𝑦2 .
Plugging into the calculator gives us the
distance between the centres.
We can see that this equals the sum of the
radii, so the circles touch externally.
243
ii)
Example answer:
Slope from centre 𝑐𝑐 to centre 𝑠𝑠:
𝑦𝑦2 − 𝑦𝑦1
𝑥𝑥2 − 𝑥𝑥1
13 − 1 4
=
3
7+2
𝑦𝑦 − 𝑦𝑦1 =
𝑦𝑦 − 1 =
4
(𝑥𝑥 − 𝑥𝑥1 )
3
4
�𝑥𝑥 − (−2)�
3
3𝑦𝑦 − 3 = 4(𝑥𝑥 + 2)
3𝑦𝑦 − 3 = 4𝑥𝑥 + 8
3𝑦𝑦 = 4𝑥𝑥 + 11
Let 𝑥𝑥 = 10
3𝑦𝑦 = 4(10) + 11
3𝑦𝑦 = 51
𝑦𝑦 =
Any circle that touches 𝑐𝑐 at that point must be
on 𝑙𝑙 (i.e. line through the two centres).
We can find the equation of a line through the
two centres by finding the slope between the
two centres.
We then sub this slope and the coordinates of
one of the centres into the equation of a line
from page 18 of the Maths Tables Book.
Multiplying across by 3.
Multiplying out the brackets.
Now that we have a line through the centres,
we can find the centre of another circle by
letting 𝑥𝑥 = 10 and finding the corresponding 𝑦𝑦
coordinate.
51
= 17
3
(𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏)
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244
Question 4
a) i)
cos(𝐴𝐴 + 𝐵𝐵) = cos 𝐴𝐴 . cos 𝐵𝐵 − sin 𝐴𝐴 . sin 𝐵𝐵
cos(𝐴𝐴 + 𝐴𝐴) = cos 𝐴𝐴 . cos 𝐴𝐴 − sin 𝐴𝐴 . sin 𝐴𝐴
cos(2𝐴𝐴) = cos 2 𝐴𝐴 − sin2 𝐴𝐴
Starting with the identity for cos(𝐴𝐴 + 𝐵𝐵) from page 14
of the Maths Tables Book. Subbing in 𝐴𝐴, for each 𝐵𝐵.
Tidying up.
ii)
𝟑𝟑
𝟓𝟓
sin
1
𝜃𝜃
=
2 √5
cos(2𝐴𝐴) = cos 2 𝐴𝐴 − sin2 𝐴𝐴
𝜃𝜃
𝜃𝜃
→ cos 𝜃𝜃 = cos 2 − sin2
2
2
cos 𝜃𝜃 = �1 − sin2
𝜃𝜃
𝜃𝜃
� − sin2
2
2
𝜃𝜃
cos 𝜃𝜃 = 1 − 2 sin2
2
sin
1
𝜃𝜃
=
2 √5
2
From page 13 of the Maths Tables Book, we
know that cos 2 𝜃𝜃 + sin2 𝜃𝜃 = 1.
𝜃𝜃
So, cos 2 𝜃𝜃 = 1 − sin2 𝜃𝜃. Equally cos 2 = 1 −
2
𝜃𝜃
sin2 ,
2
So, subbing this in.
We know this from the question.
1
cos 𝜃𝜃 = 1 − 2 � �
√5
𝟑𝟑
cos 𝜃𝜃 =
𝟓𝟓
𝜃𝜃
Subbing in 𝜃𝜃 for 2𝐴𝐴 and for 𝐴𝐴.
2
© Pocket Tutor 2022
So, we can sub
1
√5
𝜃𝜃
in for sin .
2
245
b)
𝑩𝑩 = 𝟏𝟏𝟏𝟏𝟏𝟏°, 𝟑𝟑𝟑𝟑𝟑𝟑°
tan(𝐵𝐵 + 150°) = −√3,
0° ≤ 𝜃𝜃 ≤ 360°
𝐵𝐵 + 150° = tan−1 �√3�
60° = the reference angle
Tan is negative in the 2nd and 4th quadrants:
→ 180 − 60 = 120°, 360 − 60 = 300°
𝐵𝐵 + 150° = 120° + 𝑛𝑛360
𝐵𝐵 = −30 + 𝑛𝑛360,
or 𝐵𝐵 + 150 = 300 + 𝑛𝑛360
or 𝐵𝐵 = 150 + 𝑛𝑛360
let 𝑛𝑛 = 0:
𝐵𝐵 = −30 + (0)360, 𝐵𝐵 = 150 + (0)360
𝐵𝐵 = −30
𝐵𝐵 = 150
let 𝑛𝑛 = 1:
𝐵𝐵 = −30 + (1)360 𝐵𝐵 = 150 + (1)360
𝐵𝐵 = 330,
𝑩𝑩 = 𝟏𝟏𝟏𝟏𝟏𝟏°, 𝟑𝟑𝟑𝟑𝟑𝟑°
© Pocket Tutor 2022
𝐵𝐵 = 510
To solve this equation, we first find the tan
inverse of both sides, ignoring the signs.
Doing this gives us the reference angle of 𝜃𝜃.
As −√3 is negative, we use the CAST acronym to
see where tan is negative. Tan is negative in the
2nd and 4th quadrants, so we take the reference
angle away from 180 and we take it away from
360.
Now, we write out the general solution (𝜃𝜃 +
𝑛𝑛360).
Taking 150° from both sides.
Now, subbing in 0 for 𝑛𝑛.
We ignore the value −30 as we are told we are
looking for theta in the range 0° ≤ 𝜃𝜃 ≤ 360°.
Now we sub in 1 for 𝑛𝑛.
Again, ignoring the value outside of the given
range. We can see that any value of 𝑛𝑛 higher
than 1 will only give us values greater than 360°
so we stop there.
246
Question 5
a) i)
Volume of sphere:
4
𝑉𝑉 = 𝜋𝜋𝑟𝑟 3
3
Volume of 1 cone:
𝑉𝑉 =
1 2
𝜋𝜋𝑟𝑟 ℎ
3
ℎ = 𝑟𝑟
→ 𝑉𝑉 =
𝑉𝑉 =
1 2
𝜋𝜋𝑟𝑟 (𝑟𝑟)
3
1 3
𝜋𝜋𝑟𝑟
3
If we let the radius of the sphere equal 𝑟𝑟, we can write the
volume of the sphere using the equation on page 10 of the
Maths Tables Book.
Similarly, we can write the volume of one of the cones
using the formula from page 10 of the Maths Tables Book.
We can rewrite ℎ as 𝑟𝑟, as the height of the cone is the
distance from the centre of the sphere to its edge, which is
also the radius length.
Plugging in 𝑟𝑟 for the ℎ in the formula for the volume of a
cone.
Multiplying out.
Volume of 2 cones:
1 3
𝜋𝜋𝑟𝑟 × 2
3
2 3
𝜋𝜋𝑟𝑟
3
=
Multiplying the result by 2 gives us the volume of 2 cones.
This is half the volume of the sphere, so we know that
exactly half the volume of the sphere is remaining.
ii)
𝟕𝟕 𝒄𝒄𝒄𝒄 = 𝒓𝒓
2
686
𝜋𝜋 = 𝜋𝜋𝑟𝑟 3
3
3
686𝜋𝜋 = 2𝜋𝜋𝑟𝑟 3
686𝜋𝜋
= 𝑟𝑟 3
2𝜋𝜋
3
686𝜋𝜋
= 𝑟𝑟
2𝜋𝜋
�
We can find the radius of one of the cones by letting the expression
we found for the volume of the two cones combined in the last part,
equal the given volume.
Multiplying across by 3.
Dividing across by 2𝜋𝜋.
Cube rooting both sides gives us our answer.
𝟕𝟕 𝒄𝒄𝒄𝒄 = 𝒓𝒓
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247
b)
𝟏𝟏𝟏𝟏: 𝟒𝟒𝟒𝟒
𝑑𝑑
𝑑𝑑
− 1.75 =
95
60
𝑑𝑑
𝑑𝑑
(95)
95 � � − 1.75(95) =
95
60
19𝑑𝑑
− 166.25 = 𝑑𝑑
12
19𝑑𝑑
− 𝑑𝑑 = 166.25
12
7
𝑑𝑑 = 166.25
12
𝑑𝑑 = 285
285 ÷ 60
= 4.75 hours → 4hrs 45 mins
9 + 4hrs 45 mins
We know that time is equal to distance divided by speed.
We also know that the second van took 1 hour and 45
minutes (1.75 hours) less time than the second van.
So, we can say that the distance, 𝑑𝑑, divided by the first
van’s speed minus 1.75 hours is equal to the distance
divided by the second van’s speed.
Multiplying across by 95 to get rid of the fractions.
Taking 𝑑𝑑 from both sides and adding 166.25 to both sides.
Dividing across by
vans.
7
12
, gives us the distance travelled by the
Now, we divide the distance by the speed of the van that
left at 9am, to find how long it took.
We add this time to 9am to see what time the vans arrived
at.
= 𝟏𝟏𝟏𝟏: 𝟒𝟒𝟒𝟒
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248
Question 6
a)
Given:
2 Similar Triangles ABC and DEF
To Prove:
|𝐴𝐴𝐴𝐴| |𝐴𝐴𝐴𝐴| |𝐵𝐵𝐵𝐵|
=
=
|𝐷𝐷𝐷𝐷| |𝐷𝐷𝐷𝐷| |𝐸𝐸𝐸𝐸|
Construction:
Pick a point G on the line segment AB so that |AG|=|DE|and a point H on the line segment AC so
that |AH|=|DF|
© Pocket Tutor 2022
249
Proof:
Triangle AGH is congruent to the triangle DEF (SAS)
|<AGH| = |<ABC|
The angles are equal as | < 𝐴𝐴𝐴𝐴𝐴𝐴| equals the angle | < 𝐷𝐷𝐷𝐷𝐷𝐷|
which we have been given is equal to | < 𝐴𝐴𝐴𝐴𝐴𝐴|.
Therefore, GH is parallel to BC
Therefore,
|AG| |AH|
=
|AB| |AC|
Therefore,
|DE| |DF|
=
|AB| |AC|
Therefore,
|AB| |AC|
=
,
|DE| |DF|
Q. E. D.
similarly
theorem 12
since |AG| = |DE| and |AH| = |DF|
|𝐴𝐴𝐴𝐴| |𝐵𝐵𝐵𝐵|
=
|𝐷𝐷𝐷𝐷| |𝐸𝐸𝐸𝐸|
b)
|< 𝐻𝐻𝐻𝐻𝐻𝐻| = |< 𝐻𝐻𝐻𝐻𝐻𝐻|
|< 𝑄𝑄𝑄𝑄𝑄𝑄| = |< 𝑃𝑃𝑃𝑃𝑃𝑃|
Alternate angles.
Vertically opposite angles.
So, triangles are similar
So
|𝐴𝐴𝐴𝐴| |𝐴𝐴𝐴𝐴|
=
|𝐻𝐻𝐻𝐻| |𝑄𝑄𝑄𝑄|
So |𝐴𝐴𝐴𝐴| × |𝑄𝑄𝑄𝑄| = |𝐴𝐴𝐴𝐴| × |𝐻𝐻𝐻𝐻|
© Pocket Tutor 2022
Multiplying across by the bottom of each fraction.
250
Question 7
a)
Distance = Speed × Time
12
= 0.2 → time = 1.2 hours
60
Distance = 25 × 1.2 = 30km
Total distance:
30 + 28 + 4 = 62𝑘𝑘𝑘𝑘
We can calculate the length of the cycle by
multiplying Mary’s speed by the time it takes her to
cycle from A to B.
First, we convert the 12 minutes into hours by
dividing them by 60. This gives Mary a time of 1.2
hours.
We multiply this by her speed to get the distance.
Finally, we add this to the distances given to find
the total distance.
b)
𝟐𝟐. 𝟓𝟓𝟓𝟓𝟓𝟓/𝒉𝒉
4.8 − 1.2 = 3.6
Swim: 4km, run: 28km
28
4
+
= 3.6
𝑥𝑥 5.6𝑥𝑥
28
4
� = 5.6𝑥𝑥(3.6)
5.6𝑥𝑥 � � + 5.6𝑥𝑥 �
5.6𝑥𝑥
𝑥𝑥
5.6(4) + 28 = 20.16𝑥𝑥
50.4 = 20.16𝑥𝑥
50.4
= 𝑥𝑥
20.16
𝑥𝑥 = 𝟐𝟐. 𝟓𝟓𝟓𝟓𝟓𝟓/𝒉𝒉
© Pocket Tutor 2022
First of all, we take away the cycling time, we found in
the last part, from the total time to find how long Mary
spent swimming and running.
If we let Mary’s swimming speed equal 𝑥𝑥, we can call her
running speed 5.6𝑥𝑥 as she runs 5.6 times faster than she
swims.
Now, we can divide the distance she swims by 𝑥𝑥 and the
distance she runs by 5.6𝑥𝑥 and let this equal her time, as
time is equal to distance over speed.
Now, multiplying across by 5.6𝑥𝑥 to get rid of the
fractions.
Dividing across by 20.16 gives us the value of 𝑥𝑥, Mary’s
swimming speed.
251
c)
𝑐𝑐 2 = 𝑎𝑎2 + 𝑏𝑏 2 − 2𝑎𝑎𝑎𝑎 cos 𝐶𝐶
𝑎𝑎 = 4, 𝑏𝑏 = 28, 𝑐𝑐 = 30
(30)2
=
(4)2
+
(28)2
− 2(4)(28) cos 𝐶𝐶
900 − 800 = −2(4)(28) cos 𝐶𝐶
100 = −2(4)(28) cos 𝐶𝐶
100
= cos 𝐶𝐶
−2(4)(28)
−
25
= cos 𝐶𝐶
56
cos −1 �−
25
� = 𝐶𝐶
56
28𝑘𝑘𝑘𝑘
30𝑘𝑘𝑘𝑘
4𝑘𝑘𝑘𝑘
As we have the lengths of the three sides of
the triangle, we can use the cosine rule from
page 16 of the Maths Tables Book to find the
angle.
Writing out the equation so that the side, 𝑐𝑐,
opposite the angle is on the left hand side.
Subbing in the lengths given and the one we
found in part a).
Rearranging and then finding the cos inverse
gives us the measure of the angle.
𝐶𝐶 = 116.5°
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252
d)
𝟓𝟓𝟓𝟓. 𝟏𝟏𝐤𝐤𝐤𝐤𝟐𝟐
Area =
1
𝑎𝑎𝑎𝑎 sin 𝐶𝐶
2
𝑎𝑎 = 4, 𝑏𝑏 = 28, 𝐶𝐶 = 116.5°
Area =
1
(4)(28) sin 116.5
2
Taking the equation for the area of a triangle from page 16 of
the Maths Tables Book.
Subbing in the values for 𝑎𝑎 and 𝑏𝑏 we had in the last part and
subbing in the angle we just found for 𝐶𝐶.
Plugging into the calculator gives us the answer.
Area = 𝟓𝟓𝟓𝟓. 𝟏𝟏𝐤𝐤𝐤𝐤𝟐𝟐
e)
𝟑𝟑. 𝟑𝟑𝟑𝟑𝟑𝟑
Area = 50.1km2
Area =
1
base ×⊥ height
2
base = 30, height = h
50.1 =
1
(30)ℎ
2
1
50.1 ÷ (30) = ℎ
2
𝟑𝟑. 𝟑𝟑𝟑𝟑𝟑𝟑 = ℎ
© Pocket Tutor 2022
As we know the area of the triangle, we can find
the distance from 𝐶𝐶 to 𝐴𝐴𝐴𝐴, using the other
formula for the area of a triangle from page 9 of
the Maths Tables Book.
The length from 𝐶𝐶 to 𝐴𝐴𝐴𝐴 is the perpendicular
height, so we let that equal ℎ, we then fill in the
length of 𝐴𝐴𝐴𝐴 as the base and let the expression
equal the area we found in the last part.
Solving for ℎ.
253
f)
𝟐𝟐𝟐𝟐𝟐𝟐
tan 𝜃𝜃 =
𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
tan 0.05 =
|𝐴𝐴𝐴𝐴|
30
30 tan 0.05 = |𝐴𝐴𝐴𝐴|
|𝐴𝐴𝐴𝐴| = 0.026𝑘𝑘𝑘𝑘
0.026𝑘𝑘𝑘𝑘 × 1000 = 𝟐𝟐𝟐𝟐𝟐𝟐
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𝑇𝑇
𝐴𝐴
We drew out the right angled
triangle shown with base AB
and the angle of elevation from
𝑇𝑇 to 𝐵𝐵 given.
30
0.05°
𝐵𝐵
We can now use the tan ratio
to find the length of the side
|𝐴𝐴𝐴𝐴|.
The answer we get is in
kilometres, so we multiply it by
1000 to convert it to metres.
254
Question 8
a) i)
𝟐𝟐𝟐𝟐𝟐𝟐
90% → 0.9
0.9/(0.8997) → 1.28
𝑥𝑥 − 𝜇𝜇
= 1.28
𝜎𝜎
𝑥𝑥 = 176, 𝜎𝜎 = 36
𝑥𝑥 − 176
= 1.28
36
The top 10% means that there’s 90% below that
score. So, we go to page 37 of the Maths Tables
Book to find the z-score corresponding to 0.9.
Now taking the equation for the 𝑧𝑧 −score and letting
it equal this value.
Subbing in the mean score and the standard
deviation given.
𝑥𝑥 − 176 = 1.28(36)
Multiplying across by 36.
𝑥𝑥 = 222.08
We round up as the minimum mark to be above 90%
of people is the next mark up.
𝑥𝑥 = 1.28(36) + 176
Minimum mark of 223
© Pocket Tutor 2022
Adding 176 to both sides.
255
ii)
𝟒𝟒𝟒𝟒. 𝟖𝟖𝟖𝟖%
𝑥𝑥 − 𝜇𝜇
𝜎𝜎
𝑥𝑥 = 165, 𝜇𝜇 = 176, 𝜎𝜎 = 36
165 − 176
= −0.31
36
0.31 → 0.6217
1 − 0.6217 = 0.3783
210 − 176
= 0.94
36
0.94 → 0.8264
0.8264 − 0.3783 = 0.4481
0.4481 × 100 = 𝟒𝟒𝟒𝟒. 𝟖𝟖𝟖𝟖%
© Pocket Tutor 2022
To calculate the percentage of students within the given
range, we calculate the proportion of people who scored
below 165 and take this away from the proportion who
scored below 210.
So, we use the z-score formula to find the z-score of 165.
Now, we go to the table on page 36 of the Maths Tables
Book, to find the corresponding proportion.
This figure represents the proportion of students who scored
more than 165, so we take this away from 1 to find the
proportion who scored less than 165.
Now we find the z-score corresponding to a score of 210.
Again, we go to page 36 to find the corresponding
proportion.
This is the proportion of people who scored less than 210. So,
we take the proportion who scored less than 165 away from
this to find the proportion who scored within the range
165 − 210. Finally, we multiply by 100 to get a percentage.
256
b) i)
−𝟏𝟏. 𝟕𝟕𝟕𝟕
𝑧𝑧 =
𝑥𝑥 − 𝜇𝜇
𝜎𝜎
√𝑛𝑛
𝑥𝑥 = 19.8, 𝜇𝜇 = 21, 𝜎𝜎 = 5.2, 𝑛𝑛 = 60.
19.8 − 21
= −𝟏𝟏. 𝟕𝟕𝟕𝟕
5.2
√60
To find the test statistic, we take the
formula from page 35 of the Maths
Tables Book, where 𝜎𝜎 = the standard
deviation and 𝑛𝑛 = the sample size.
Subbing in the values and plugging into
the calculator gives us the answer.
ii)
1.79 → 0.9633
Probability > 1.79 = 1 − 0.9633 = 0.0367
Probability < −1.79 = 1 − 0.9633 = 0.0367
𝑝𝑝 − value = 0.0367 + 0.0367
= 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
0.0734 > 0.05
∴ 𝐖𝐖𝐖𝐖 𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫 𝐭𝐭𝐭𝐭𝐭𝐭 𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚 𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡𝐡.
We go to the tables on pages 36 & 37 of
the Maths Tables Book to convert the zscore.
We then calculate the 𝑝𝑝 − value by
finding the probability of the z-score
being greater than 1.79 or less than
−1.79.
Adding these values together gives us
the 𝑝𝑝 − value.
As the result is greater than 0.05 we
reject the alternative hypothesis.
Therefore, there is not enough evidence to say that news report is incorrect in its claims that
students in Ireland study an average of 21 hours per week.
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257
c) i)
𝟎𝟎. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
12 classroom keys, 6 lab keys, 5 office keys, total = 23.
Not office keys = 12 + 6 = 18
18 17 16 5
×
×
×
= 𝟎𝟎. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
23 22 21 20
To calculate the probability that the first office key is
drawn on the fourth draw, we multiply the
probability of a non-office key being draw on each of
the previous draws by the probability of getting the
office key.
The number of keys and non-office keys goes down
with each draw as the keys are not replaced.
Note: This question could have been answered in two ways as it did not specify whether the keys
were replaced between each draw. If you assume that the keys are replaced the answer is: 0.1042.
ii)
𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
Probability of classroom key, lab, office:
5
60
12 6
×
×
=
23 22 21 1771
60
× 3! = 𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
1771
© Pocket Tutor 2022
First, we calculate the probability of a classroom key, a
lab key and an office key being drawn in that order.
The keys aren’t replaced between draws so the number
on the bottom of the fraction decreases each time.
We then multiply the result by 3! as the keys could be
drawn in this many different orders.
258
Question 9
a) i)
𝟗𝟗𝟗𝟗 minutes
840𝑘𝑘𝑘𝑘
𝑎𝑎2 = 𝑏𝑏 2 + 𝑐𝑐 2 − 2𝑏𝑏𝑏𝑏 cos 𝐴𝐴
𝑎𝑎 = 𝑎𝑎, 𝑏𝑏 = 1450, 𝑐𝑐 = 840, 𝐴𝐴 = 8.57
𝑎𝑎2 = (1450)2 + (840)2 − 2(1450)(840) cos(8.57)
𝑎𝑎2 = 399299.0556
𝑎𝑎 = √399299.0556
𝑎𝑎 = 631.9𝑘𝑘𝑘𝑘
time =
631.9
= 1.504 hours
420
1.504 × 60 = 𝟗𝟗𝟗𝟗 minutes
© Pocket Tutor 2022
We can find the length |𝐴𝐴𝐴𝐴| by multiplying the
plane’s speed (420) by the number of hours it
flew that direction (2). Filling this into the
diagram shows us that we can solve for the
length of 𝐵𝐵𝐵𝐵 using the cosine rule.
Taking the cosine rule from page 16 of the
Maths Tables Book.
Subbing in the lengths of the other two sides
and the angle opposite the side 𝐵𝐵𝐵𝐵.
Plugging it into the calculator and then square
rooting both sides gives us the length.
Finally, we divide this by the speed of the plane
to find the time it takes.
Multiplying the result by 60 to convert the
answer to minutes.
259
ii)
Total distance:
840 + 631.9 + 1450 = 2921.9
To calculate the amount of fuel used, we first
calculate the total distance travelled by adding up
the three legs.
Total time:
2921.9
= 6.9569 hours
420
6.9569 × 60 = 417.41 minutes
417.41 × 60 = 25044 seconds
We then divide this distance by the speed of the
plane to find the total time taken.
We multiply this by 60 to convert it to minutes and
then by 60 again to convert it to seconds.
Amount of fuel:
25044 × 3.8 = 95167.2 litres
𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗. 𝟐𝟐 < 𝟏𝟏𝟏𝟏𝟏𝟏, 𝟎𝟎𝟎𝟎𝟎𝟎
Finally, we multiply the result by the fuel used per
second.
The result is less than the capacity of the fuel tank.
b) i)
𝑉𝑉(𝑡𝑡) = 110√2 sin(120𝜋𝜋𝜋𝜋)
Period:
𝟏𝟏
2π
=
120π 𝟔𝟔𝟔𝟔
Range: �+𝟏𝟏𝟏𝟏𝟏𝟏√𝟐𝟐, −𝟏𝟏𝟏𝟏𝟏𝟏√𝟐𝟐�
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260
ii)
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261
iii)
𝟏𝟏𝟏𝟏𝟏𝟏. 𝟗𝟗𝟗𝟗 𝐕𝐕𝐕𝐕𝐕𝐕𝐕𝐕𝐕𝐕
𝑉𝑉(𝑡𝑡) = 110√2 sin(120𝜋𝜋𝜋𝜋)
𝑉𝑉(6.67) = 110√2 sin�120𝜋𝜋(6.67)�
𝑉𝑉(6.67) = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟗𝟗𝟗𝟗 𝐕𝐕𝐕𝐕𝐕𝐕𝐕𝐕𝐕𝐕
To answer this question, we simply sub in 6.67 for
𝑡𝑡 and plug the equation into the calculator. Make
sure your calculator is in radians.
iv)
𝒕𝒕 =
𝟏𝟏
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬
𝟒𝟒𝟒𝟒𝟒𝟒
110√2 sin(120𝜋𝜋𝜋𝜋) = 110
sin(120𝜋𝜋𝜋𝜋) =
110√2
120𝜋𝜋𝜋𝜋 = sin−1
120𝜋𝜋𝜋𝜋 =
𝑡𝑡 =
𝒕𝒕 =
𝜋𝜋
4
110
110
110√2
𝜋𝜋
÷ 120𝜋𝜋
4
𝟏𝟏
𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬
𝟒𝟒𝟒𝟒𝟒𝟒
© Pocket Tutor 2022
To find one value of 𝑡𝑡 for which the
voltage is 110 volts, we let 𝑉𝑉(𝑡𝑡) = 110.
Dividing across by 110√2.
Finding the sin inverse of both sides.
Dividing across by 120𝜋𝜋 gives us our
answer.
262
v)
𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓 𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯/𝐬𝐬𝐬𝐬𝐬𝐬
𝑉𝑉(𝑡𝑡) = 110√2 sin(120𝜋𝜋𝜋𝜋)
𝑉𝑉
′ (𝑡𝑡)
= 110√2 cos(120𝜋𝜋𝜋𝜋) × 120𝜋𝜋
𝑉𝑉 ′ (2) = 110√2 cos(120𝜋𝜋(2)) × 120𝜋𝜋
𝑉𝑉 ′ (2) = 58646.05
→ 𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓 𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯𝐯/𝐬𝐬𝐬𝐬𝐬𝐬
© Pocket Tutor 2022
To find the rate of change when 𝑡𝑡 = 2, we need to
differentiate 𝑉𝑉(𝑡𝑡).
To differentiate sin, we change the sin to cos and
multiply by the derivative of the angle.
So, we multiply by the derivative of 120𝜋𝜋𝜋𝜋, which is
120𝜋𝜋.
Now, we sub in 2 for 𝑡𝑡 in the derivative and plug
into the calculator. Make sure your calculator is in
radians.
263
Question 10
a) i)
𝟎𝟎. 𝟏𝟏𝟏𝟏𝟏𝟏
𝑛𝑛
� � 𝑝𝑝𝑟𝑟 𝑞𝑞 𝑛𝑛−𝑟𝑟
𝑟𝑟
𝑛𝑛 = 9, 𝑟𝑟 = 2, 𝑝𝑝 = 0.08, 𝑞𝑞 = 0.92
9
� � (0.08)2 (0.92)(9−2) × 0.08
2
= 𝟎𝟎. 𝟏𝟏𝟏𝟏𝟏𝟏
We can use Bernoulli trials to solve this question.
To answer this question, we find the probability of 2 of the
first 9 having type O-negative blood and then multiply this
by the probability of the tenth person having this type.
Taking the expression from page 33 of the Maths Tables
Book. 𝑛𝑛 = the number of trials, 𝑟𝑟 =the number of
successes (o-type), 𝑝𝑝 = probability of success (o-type, 8%→
0.08), 𝑞𝑞 = probability of unsuccessful outcome (not otype).
Subbing in the values and plugging into the calculator.
ii)
𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑
At least one → 1 − probability of none
To find the probability of at least one having a blood type
of O-negative, we take the probability of none of them
having this blood type away from 1.
Probability of none:
(0.92)5 = 0.6591
1 − 0.6591 = 𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑
© Pocket Tutor 2022
We find the probability of no one having this blood type
by multiplying 0.92 by itself 5 times, which is the same as
putting it to the power of 5.
Taking the result away from 1 gives us our answer.
264
iii)
𝟒𝟒𝟒𝟒
1 − 𝑥𝑥 = 0.97
1 − 0.97 = 𝑥𝑥
𝑥𝑥 = 0.03
(0.92)𝑛𝑛 = 0.03
𝑛𝑛 = log 0.92 0.03
𝑛𝑛 = 42.05
→ 𝟒𝟒𝟒𝟒
© Pocket Tutor 2022
Taking 0.97 away from 1, gives us the probability of none of the
donors having blood type O-negative.
Now, we can let this probability equal the expression we used in
the last part, which is the probability of a person having a different
blood type, to the power of the number of donors, 𝑛𝑛.
So, we can use logs to solve for the number of donors.
From page 21 of the Maths Tables Book:
𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥
𝑎𝑎 = 0.92, 𝑦𝑦 = 0.03, 𝑥𝑥 = 𝑛𝑛
Subbing these values in and plugging into the calculator gives us
our answer.
265
b)
0.8 × 70 = €56
0.2 × (150 + 80) = €46
46 + 56 = €𝟏𝟏𝟏𝟏𝟏𝟏
To find the expected value we multiply the probability of each
outcome by that outcome and add the results.
So, multiplying the cost of the first scenario by 0.8. Then multiplying
the two costs of the second scenario added together, by its
probability which is 1 − 0.8 = 0.2.
Adding the results gives us our answer.
Note: There are two interpretations of this question. In this interpretation the initial €70 is not
charged if the repair is not successful. In the other interpretation the initial €70 is charged
regardless. This gives an answer of €116.
c)
€𝟏𝟏𝟏𝟏𝟏𝟏
18000 × 0.0001 = 1.8
18000 × 0.002 = 36
1.8 × 120000 = €216,000
36 × 40000 = €1,440,000
1,440,000 + 216,000 = €1,656,000
1,656,000 + 900,000 = 2,556,000
2,556,000
= €𝟏𝟏𝟏𝟏𝟏𝟏
18000
© Pocket Tutor 2022
To calculate the premiums, we need to first calculate the
expected amount the insurance company will have to pay
out.
So, we multiply the number of policy holders by the
probability of death and then by the probability of
disability.
We then multiply each of the results by their respective
payout to find the expected payouts the insurance company
will have to make.
We add these together and then we add the profits the
company wants to make.
Finally, we divide by the number of policy holders to find
how much they should pay each.
266
2020 Paper 2
Question 1
a)
𝟎𝟎
𝑚𝑚 =
𝑦𝑦2 − 𝑦𝑦1
𝑥𝑥2 − 𝑥𝑥1
3 − (−12)
𝑚𝑚 =
−4 − 6
𝑚𝑚 = −
3
2
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
3
𝑦𝑦 − (−12) = − �𝑥𝑥 − (6)�
2
3
𝑦𝑦 + 12 = − (𝑥𝑥 − 6)
2
2𝑦𝑦 + 24 = −3(𝑥𝑥 − 6)
2𝑦𝑦 + 24 = −3𝑥𝑥 + 18
We can find the perpendicular distance from 𝐴𝐴 to 𝐵𝐵𝐵𝐵
by using the equation on page 19 of the Maths Tables
Book for the perpendicular distance between a point
and a line. To do this we first need to find the
equation of the line 𝐵𝐵𝐵𝐵.
So, we find the slope between 𝐵𝐵 and 𝐶𝐶 using the slope
formula.
We then plug this and one of the points into the
equation of a line found on page 18 of the Maths
Tables Book.
Multiplying across by 2.
3𝑥𝑥 + 2𝑦𝑦 + 6 = 0
Rearranging to get all the terms on one side.
|𝑎𝑎𝑥𝑥1 + 𝑏𝑏𝑦𝑦1 + 𝑐𝑐|
Taking the perpendicular distance formula from the
Maths Tables Book.
√𝑎𝑎2
+
𝑏𝑏 2
|3(2) + 2(−6) + 6|
�(3)2 + (2)2
= 𝟎𝟎
Remember that the equation of a line is written in the
form 𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐. Subbing in the point 𝐴𝐴(2, −6)
As the distance between A and the line BC is 0 we
know that the three points are on the same line.
The points 𝑨𝑨, 𝑩𝑩 and 𝑪𝑪 are colinear.
© Pocket Tutor 2022
267
b)
𝟑𝟑𝟑𝟑. 𝟒𝟒𝟒𝟒𝟒𝟒°
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = ±
𝑚𝑚1 − 𝑚𝑚2
1 + 𝑚𝑚1 𝑚𝑚2
To find the angle between two lines we use
this formula from page 19 of the Maths
Tables Book.
To use this formula, we need the slope of
each line.
𝑚𝑚 of line 𝑎𝑎:
𝑥𝑥 − 2𝑦𝑦 + 1 = 0
To find the slope of line 𝑎𝑎 we rewrite its
equation in the form 𝑦𝑦 = 𝑚𝑚𝑚𝑚 + 𝑐𝑐, where
𝑚𝑚 = the slope.
2𝑦𝑦 = 𝑥𝑥 + 1
𝑦𝑦 =
𝑚𝑚 =
1
1
𝑥𝑥 +
2
2
1
2
We know that the line 𝑏𝑏 makes an angle of
60° with the 𝑥𝑥-axis.
𝑚𝑚 of line 𝑏𝑏:
𝑚𝑚 = 60°
We also know that 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 =
tan(60) = √3
the same as the slope �
𝜃𝜃 = tan
−1 �
1
√3 − 2
1
1 + (√3) � �
2
�. So, if we find
Now subbing these two slopes into the
equation.
1
√3 − 2
1
1 + (√3) � �
2
𝑟𝑟𝑟𝑟𝑟𝑟
, which is
tan 60 we get the slope of line 𝑏𝑏.
𝑚𝑚1 − 𝑚𝑚2
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = ±
1 + 𝑚𝑚1 𝑚𝑚2
tan 𝜃𝜃 = ±
𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟
𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
�
Plugging the tan inverse into the calculator
gives us the acute angle between the two
lines.
𝜃𝜃 = 𝟑𝟑𝟑𝟑. 𝟒𝟒𝟒𝟒𝟒𝟒°
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268
Question 2
a)
𝟗𝟗 units
𝑥𝑥 2 + 𝑦𝑦 2 − 4𝑥𝑥 + 2𝑦𝑦 − 4 = 0
Centre: (2, −1)
radius: �(−2)2 + (1)2 − (−4) = 3
Distance from centre to 𝐵𝐵:
�(𝑥𝑥2 − 𝑥𝑥1 )2 + (𝑦𝑦2 − 𝑦𝑦1 )2
�(2 − 5)2 + �(−1) − 8�
2
|𝐴𝐴𝐴𝐴| = √90
Pythagoras’ theorem in triangle 𝐴𝐴𝐴𝐴𝐴𝐴:
|𝑇𝑇𝑇𝑇|2
+
|𝐵𝐵𝐵𝐵|2
=
|𝐴𝐴𝐴𝐴|2
2
32 + |𝐵𝐵𝐵𝐵|2 = �√90�
2
|𝐵𝐵𝐵𝐵|2 = �√90� − 32
|𝐵𝐵𝐵𝐵|2 = 81
As the circle is in the form 𝑥𝑥 2 + 𝑦𝑦 2 + 2𝑔𝑔𝑔𝑔 + 2𝑓𝑓𝑓𝑓 + 𝑐𝑐 =
0, the centre is (−𝑔𝑔, −𝑓𝑓). (page 19 Maths Tables Book)
So −4 → 2 and 2 → −1 gives us (2, −1).
We can get the radius using the
equation �𝑔𝑔2 + 𝑓𝑓 2 − 𝑐𝑐, this is the distance from the
centre to point T.
We can calculate the distance from the centre to B
using the distance formula from page 18 of the Maths
Tables Book. Subbing in the centre coordinates and the
point 𝐵𝐵(5,8).
Now, if we draw a triangle from T to A to B, where TA is
at a right angle as it is a radius touching a tangent, we
can use Pythagoras’ theorem to find the distance BT.
Subbing in the radius for |𝑇𝑇𝑇𝑇| and the length of |𝐴𝐴𝐴𝐴|
for the hypotenuse.
Square rooting both sides.
|𝐵𝐵𝐵𝐵| = 𝟗𝟗 units
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269
b)
(𝒙𝒙 − 𝟒𝟒)𝟐𝟐 + 𝒚𝒚𝟐𝟐 = 𝟐𝟐𝟐𝟐, (𝒙𝒙 + 𝟐𝟐)𝟐𝟐 + 𝒚𝒚𝟐𝟐 = 𝟐𝟐𝟐𝟐
centre of a circle: (−𝑔𝑔, −𝑓𝑓).
From page 19 of the Maths Tables Book.
centre: (−g, 0)
radius:
Subbing in 0 for −𝑓𝑓 as the centre is on the 𝑥𝑥
axis, so 𝑦𝑦 = 0.
�𝑔𝑔2 + 𝑓𝑓 2 − 𝑐𝑐
Now subbing these centre coordinates into
the equation for the radius from page 19,
and letting it equal the given radius.
�𝑔𝑔2 + (0)2 − 𝑐𝑐 = 5
𝑔𝑔2 − 𝑐𝑐 = 25
eq. 1
Now taking the general equation for a circle
from page 19 and subbing in 0 for 𝑓𝑓.
𝑥𝑥 2 + 𝑦𝑦 2 + 2𝑔𝑔𝑔𝑔 + 2𝑓𝑓𝑓𝑓 + 𝑐𝑐 = 0
𝑥𝑥 2 + 𝑦𝑦 2 + 2𝑔𝑔𝑔𝑔 + 𝑐𝑐 = 0
(1)2 + (4)2 + 2𝑔𝑔(1) + 𝑐𝑐 = 0
17 + 2𝑔𝑔 + 𝑐𝑐 = 0
eq.2
𝑔𝑔2 − 𝑐𝑐 = 25
eq. 1
𝑔𝑔2 − 25 = 𝑐𝑐
© Pocket Tutor 2022
Squaring both sides.
Subbing in the coordinates (1,4) of the
point given on the circles. We now have two
equations which we can use to solve for 𝑔𝑔.
Rewriting equation one to get 𝑐𝑐 on one side
by itself.
270
17 + 2𝑔𝑔 + 𝑐𝑐 = 0
17 + 2𝑔𝑔 +
(𝑔𝑔2
eq.2
− 25) = 0
Now subbing this expression for 𝑐𝑐 into
equation 2.
𝑔𝑔2 + 2𝑔𝑔 − 8 = 0
(𝑔𝑔 − 2)(𝑔𝑔 + 4) = 0
Factorising the quadratic.
𝑔𝑔 = 2, 𝑔𝑔 = −4
Solving for the two values of 𝑔𝑔.
centre: (−g, 0)
Subbing these values into our expression
for the centre.
→ (−2,0) and (4,0)
Equations:
(𝑥𝑥 − ℎ)2 + (𝑦𝑦 − 𝑘𝑘)2 = 𝑟𝑟 2
(𝑥𝑥 − (−2))2 + (𝑦𝑦 − 0)2 = (5)2
(𝒙𝒙 + 𝟐𝟐)𝟐𝟐 + 𝒚𝒚𝟐𝟐 = 𝟐𝟐𝟐𝟐
Now using these centres and the given
radius to find the equation of each circle
using the formula on page 19 of the Maths
Tables Book.
(𝑥𝑥 − 4)2 + (𝑦𝑦 − 0)2 = (5)2
(𝒙𝒙 − 𝟒𝟒)𝟐𝟐 + 𝒚𝒚𝟐𝟐 = 𝟐𝟐𝟐𝟐
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271
Question 3
a)
To solve this question, we first need to figure out the other
angles in the triangles.
𝟔𝟔. 𝟒𝟒𝟒𝟒𝟒𝟒
17° 33°
128°
95°
85°
We can find | < 𝐸𝐸𝐸𝐸𝐸𝐸| by taking 52 away from 180° as
they’re on a straight line.
We can find the angle at G in the bottom triangle by
subtracting 5° and 90° from 180° as angles in a triangle add
up to 180°.
We can find | < 𝐹𝐹𝐹𝐹𝐹𝐹| by subtracting 85° from 180° as they
are on a straight line.
We can find | < 𝐸𝐸𝐸𝐸𝐸𝐸| by taking 35° and 128° from 180° as
angles in a triangle add up to 180°.
We can find | < 𝐹𝐹𝐹𝐹𝐺𝐺| by taking 52° and 95° from 180 for
the same reason.
sin rule :
𝑎𝑎
𝑏𝑏
=
sin 𝐴𝐴 sin 𝐵𝐵
|𝐻𝐻𝐻𝐻|
6
=
sin 17 sin 35
6
× sin 35 = |𝐻𝐻𝐻𝐻|
sin 17
11.771 = |𝐻𝐻𝐻𝐻|
|𝐹𝐹𝐹𝐹|
11.771
=
sin 95° sin 33
11.771
× sin 33 = |𝐹𝐹𝐹𝐹|
sin 95°
|𝐹𝐹𝐹𝐹| = 𝟔𝟔. 𝟒𝟒𝟒𝟒𝟒𝟒
© Pocket Tutor 2022
Now we can use the sine rule from page 16
of the Maths Tables Book to solve this
question.
We first find |𝐻𝐻𝐻𝐻| with the sine rule
remembering that the angle A is the angle
opposite the side 𝑎𝑎.
Multiplying across by sin 35.
Now that we have the length |𝐻𝐻𝐻𝐻| we can
use the sine rule to find |𝐹𝐹𝐹𝐹| .
Multiplying both sides by sin 33.
272
b)
𝒌𝒌 = 𝟗𝟗
|< 𝐵𝐵𝐵𝐵𝐴𝐴| = 60°
To find the ratio of the areas we need to find a
way of writing |𝑂𝑂𝑂𝑂|, the radius of 𝑠𝑠 in terms of
the radius, 𝑟𝑟, of 𝑐𝑐.
→ |< 𝐶𝐶𝐶𝐶𝐶𝐶| = 30°
| < 𝐶𝐶𝐶𝐶𝐶𝐶| is half | < 𝐵𝐵𝐵𝐵𝐵𝐵| as 𝑂𝑂𝑂𝑂 and 𝑂𝑂𝑂𝑂 are
tangents to 𝑐𝑐 and 𝑂𝑂𝑂𝑂 goes through the centre of
the circle 𝑐𝑐.
𝑜𝑜𝑜𝑜𝑜𝑜
𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
ℎ𝑦𝑦𝑦𝑦
sin 30 =
𝑟𝑟
|𝐷𝐷𝐷𝐷|
We can write |𝐷𝐷𝐷𝐷| in terms of 𝑟𝑟 using the sin
ratio in the triangle drawn in the question.
𝑟𝑟
1
=
2 |𝐷𝐷𝐷𝐷|
Multiplying across by |𝐷𝐷𝐷𝐷| and then multiplying
across by 2.
|𝐷𝐷𝐷𝐷| = 2𝑟𝑟
|𝑂𝑂𝑂𝑂| = |𝐷𝐷𝐷𝐷| + 𝑟𝑟
Now getting |𝑂𝑂𝑂𝑂| in terms of 𝑟𝑟 by subbing in
what we found for |𝐷𝐷𝐷𝐷|.
|𝑂𝑂𝑂𝑂| = 2𝑟𝑟 + 𝑟𝑟
|𝑂𝑂𝑂𝑂| = 3𝑟𝑟
Area c = πr 2
2
Area s = π(3r) = 9𝜋𝜋𝑟𝑟
Area s: Area c = 9: 1
𝒌𝒌 = 𝟗𝟗
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2
Using the equation of the area of a circle from
page 8 of the Maths Tables Book.
Plugging 3𝑟𝑟 in for the radius of the circle 𝑠𝑠 in
place of |𝑂𝑂𝑂𝑂|.
9𝜋𝜋𝑟𝑟 2 : 𝜋𝜋𝑟𝑟 2 → 9: 1.
273
Question 4
a)
𝟓𝟓𝟓𝟓 𝟏𝟏𝟏𝟏𝟏𝟏
,
𝟑𝟑
𝟑𝟑
tan
1
𝜃𝜃
=−
2
√3
𝜃𝜃
1
= tan−1 � �
2
√3
Tan is negative in the second and fourth
quadrants, so we calculate 𝜃𝜃 accordingly.
reference angle =
𝜋𝜋
6
2nd quadrant: π −
π 5𝜋𝜋
=
6
6
𝜃𝜃 5𝜋𝜋
=
+ 2𝑛𝑛𝑛𝑛
6
2
→
𝟓𝟓𝟓𝟓
𝟑𝟑
𝜃𝜃 11𝜋𝜋
=
+ 2𝑛𝑛𝑛𝑛
6
2
𝜃𝜃 =
Subbing in 0 for 𝑛𝑛.
We don’t continue past 0 as anything else
would go outside the limits of 0 ≤ 𝜃𝜃 ≤ 4𝜋𝜋
π 11𝜋𝜋
=
6
6
11𝜋𝜋
+ 4𝑛𝑛𝑛𝑛
3
𝑛𝑛 = 0:
→
Multiplying across by 2 to get 𝜃𝜃 by itself.
5𝜋𝜋
+ 4(0)𝜋𝜋
3
4th quadrant: 2π −
𝟏𝟏𝟏𝟏𝟏𝟏
𝟑𝟑
As tan is negative in the second quadrant,
we take the reference angle away from 𝜋𝜋.
Now writing out the general solution.
5𝜋𝜋
+ 4𝑛𝑛𝑛𝑛
𝜃𝜃 =
3
𝑛𝑛 = 0:
We first, find the reference angle by
ignoring the sign and finding the tan
inverse.
11𝜋𝜋
+ 4(0)𝜋𝜋
3
© Pocket Tutor 2022
Tan is also negative in the 4th quadrant, so
we take the reference angle from 2𝜋𝜋.
Repeating the same process as above to
find the second value of 𝜃𝜃 which is within
the limits of 0 ≤ 𝜃𝜃 ≤ 4𝜋𝜋.
274
b)
𝟒𝟒. 𝟒𝟒𝟒𝟒𝟒𝟒
Area of COA = Area of sector − 21
1 2
𝑟𝑟 𝜃𝜃 − 21
2
1
(7)2 (1.2) − 21 = 8.4
2
Area of COA =
1
𝑎𝑎𝑎𝑎 sin 𝐶𝐶
2
1
|𝐶𝐶𝐶𝐶|(7) sin 1.2 = 8.4
2
|𝐶𝐶𝐶𝐶| =
8.4
1
(7) sin 1.2
2
|𝐶𝐶𝐶𝐶| = 2.57
|𝐵𝐵𝐵𝐵| = 7 − 2.57
|𝐵𝐵𝐵𝐵| = 4.43 → 𝟒𝟒. 𝟒𝟒𝟒𝟒𝟒𝟒
© Pocket Tutor 2022
|𝐵𝐵𝐵𝐵| = 7 − |𝐶𝐶𝐶𝐶| so, we can find |𝐵𝐵𝐵𝐵| by first
finding |𝐶𝐶𝐶𝐶|.
We can find |𝐶𝐶𝐶𝐶| by finding the area of the
triangle COA.
Taking the equation for the area of a sector
from page 9 of the Maths Tables Book.
Subbing in the values and taking away the
shaded area leaves us with the area of the
triangle.
Taking the equation for the area of a triangle
from page 9, subbing in the given side and
angle and letting it equal the area we just
found. (Make sure your calculator is in
radians).
Solving for the length of |𝐶𝐶𝐶𝐶|.
Taking the length of |𝐶𝐶𝐶𝐶| from 7 gives us the
length of |𝐵𝐵𝐵𝐵|.
275
Question 5
a) i)
𝑃𝑃(𝐵𝐵|𝐴𝐴) =
𝑃𝑃(𝐴𝐴 ∩ 𝐵𝐵)
𝑃𝑃(𝐴𝐴)
1
𝑃𝑃(𝐵𝐵|𝐴𝐴) = 2
3
4
𝟐𝟐
=
𝟑𝟑
Taking the equation for conditional
probability.
Subbing in the probability of A intersection
B and the probability of 𝐴𝐴.
Plugging into the calculator.
ii)
𝑃𝑃(𝐴𝐴 ∪ 𝐵𝐵) = 𝑃𝑃(𝐴𝐴) + 𝑃𝑃(𝐵𝐵) − 𝑃𝑃(𝐴𝐴 ∩ 𝐵𝐵)
1
11 3
= + 𝑃𝑃(𝐵𝐵) −
2
12 4
11 3 1
− + = 𝑃𝑃(𝐵𝐵)
12 4 2
2
= 𝑃𝑃(𝐵𝐵)
3
If independent: 𝑃𝑃(𝐴𝐴) × 𝑃𝑃(𝐵𝐵) = 𝑃𝑃(𝐴𝐴 ∩ 𝐵𝐵)
3 2 1
× =
4 3 2
1 1
=
2 2
∴ 𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢.
© Pocket Tutor 2022
If two events are independent, then: 𝑃𝑃(𝐴𝐴) ×
𝑃𝑃(𝐵𝐵) = 𝑃𝑃(𝐴𝐴 ∩ 𝐵𝐵).
So, we need to find 𝑃𝑃(𝐵𝐵). The probability of 𝐴𝐴
union 𝐵𝐵 is equal to the probability of 𝐴𝐴 plus the
probability of 𝐵𝐵 minus the probability of the
intersection.
Subbing the given values in and solving for
𝑃𝑃(𝐵𝐵).
Now using this value to check if the events are
independent.
𝑃𝑃(𝐴𝐴) times 𝑃𝑃(𝐵𝐵) equals the intersection so the
events are independent.
276
b)
1
1
2
3
1
2
2
3
4
1
2
2
3
4
2
3
3
4
5
3
4
4
5
6
No remainder: 5
Remainder of 1: 5
Remainder of 2: 6
→ Not Fair as Lee has 6 chances to win while the others have 5.
© Pocket Tutor 2022
By drawing out a sample space of all
the possible outcomes we can count
how many have no remainder when
divided by 3, how many have a
remainder of 1 and how many a
remainder of 2.
There is not the same amount of
each therefore the game is not fair.
277
Question 6
a)
30 × .7 + 60 × .25 + 10 × .09 = 36.9%
→ 𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑
To find the percentage of cars that are Volkswagen we
multiply the percentage of each category by the
proportion of those which are Volkswagen. Adding the
results gives us the total percentage which are
Volkswagen. We convert this to a decimal to give the
probability.
b) i)
𝑛𝑛
� � 𝑝𝑝𝑟𝑟 𝑞𝑞 𝑛𝑛−𝑟𝑟
𝑟𝑟
𝟏𝟏𝟏𝟏𝟏𝟏
5 1 2 3 3 1
� �� � � � × =
4 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
4
2 4
We can answer this question using Bernoulli trials. Taking the equation
from page 33 of the Maths Tables Book.
Where, 𝑛𝑛 =the total number of trials, 𝑟𝑟 = the number of successful
outcomes, 𝑝𝑝 = the probability of a successful outcome and 𝑞𝑞 = the
probability of an unsuccessful outcome.
As we are looking for the probability that Joe passes with exactly two
others, we find the probability of 2 out of 5 passing using Bernoulli trials
and then multiply this by the probability of Joe passing.
© Pocket Tutor 2022
278
ii)
𝒂𝒂 = 𝟏𝟏, 𝒃𝒃 = 𝟏𝟏, 𝒄𝒄 = 𝟐𝟐
Probability of less than two people or two people passing:
0 pass + 1 pass + 2 pass
1 𝑛𝑛
� �
2
𝑛𝑛 1 1 1 𝑛𝑛−1
� �� � � �
1 2
2
𝑛𝑛 1 2 1 𝑛𝑛−2
� �� � � �
2 2
2
1 𝑛𝑛
𝑛𝑛 1 1 1 𝑛𝑛−1
𝑛𝑛 1 2 1 𝑛𝑛−2
� � + � �� � � �
+ � �� � � �
2
2
2
1 2
2 2
© Pocket Tutor 2022
To find an expression for two people or less
passing we need to add the probabilities of 0
passing, 1 passing and 2 passing.
1
The probability of everyone failing is to the
power of the number of people, 𝑛𝑛.
2
The probability of 1 person passing can be
calculated using Bernoulli trials as in the last
part, with 𝑟𝑟, the number of successful outcomes
= 1 and leaving 𝑛𝑛 unknown.
The probability of 2 people passing can be
calculated the same way just with 𝑟𝑟 = 2.
Adding these three expressions gives us our
answer.
279
1
2𝑛𝑛
1 𝑛𝑛
1
� � = 𝑛𝑛 as 1𝑛𝑛 = 1.
2
2
1
𝑛𝑛
1 𝑛𝑛−1+1
1 𝑛𝑛
𝑛𝑛 × � �
→ 𝑛𝑛 × � � → 𝑛𝑛 × 𝑛𝑛 → 𝑛𝑛
2
2
2
2
Now taking the third part of the sum, the
choose function for example of:
𝑛𝑛(𝑛𝑛 − 1)
𝑛𝑛 × (𝑛𝑛 − 1)
1 𝑛𝑛−2+2 𝑛𝑛(𝑛𝑛 − 1) 1
� �
→
× 𝑛𝑛 →
2
2𝑛𝑛+1
2×1
2
2
𝑛𝑛 𝑛𝑛(𝑛𝑛 − 1)
1
+ 𝑛𝑛 +
𝑛𝑛
2𝑛𝑛+1
2
2
2𝑛𝑛
𝑛𝑛(𝑛𝑛 − 1)
+ 𝑛𝑛+1 +
𝑛𝑛+1
2
2𝑛𝑛+1
2
2
2 + 2𝑛𝑛 + 𝑛𝑛2 − 𝑛𝑛
2𝑛𝑛+1
𝑛𝑛2 + 𝑛𝑛 + 2
2𝑛𝑛+1
Now taking the second part of the sum, we can
rewrite �𝑛𝑛1� as 𝑛𝑛 and we can multiply the
fractions together by adding the powers.
5×4×3
5
� �=
, therefore we can
3×2×1
3
Rewrite, �𝑛𝑛2� as
𝑛𝑛×(𝑛𝑛−1)
2×1
and we can multiply
the fractions by adding the powers. Also,
2𝑛𝑛 × 2 = 2𝑛𝑛+1 (page 21 Maths Tables Book).
Now that we have expressed each part of the
sum as a single fraction, we can add them
together.
To add the fractions, we multiply the top and
the bottom of the first two fractions by two so
that they all have the base 2𝑛𝑛+1
Now adding the fractions.
𝒂𝒂 = 𝟏𝟏, 𝒃𝒃 = 𝟏𝟏, 𝒄𝒄 = 𝟐𝟐
© Pocket Tutor 2022
280
Question 7
a) i)
(9)2 = 𝑥𝑥 2 + (3.3)2
(9)2
−
(3.3)2
70.11 = 𝑥𝑥 2
= 𝑥𝑥
2
8.37𝑐𝑐𝑐𝑐 = 𝑥𝑥
𝑥𝑥 𝑐𝑐𝑐𝑐
9 𝑐𝑐𝑐𝑐
By drawing a line from the centre of
the circle to the tip of the cone we
create a right angled triangle with sides
as shown.
We can then use Pythagoras’ theorem
to calculate 𝑥𝑥, the vertical height.
3.3𝑐𝑐𝑐𝑐
ii)
𝐴𝐴 = 𝜋𝜋𝜋𝜋𝜋𝜋
Taking the equation for the curved surface area of a
cone from page 10 of the Maths Tables Book.
𝐴𝐴 = 𝟗𝟗𝟗𝟗. 𝟑𝟑𝟑𝟑𝟑𝟑𝒎𝒎𝟐𝟐
Subbing in the radius and the slant height 𝑙𝑙.
𝐴𝐴 = 𝜋𝜋(3.3)(9)
© Pocket Tutor 2022
281
iii)
𝟏𝟏𝟏𝟏𝟏𝟏°
circumference of cup = arc length
circumference of cup = 2πr
circumference of cup = 2π(3.3)
𝜃𝜃
�
Arc length = 2πr �
360
θ
�
Arc length = 2π(9) �
360
2π(9) �
θ
� = 2𝜋𝜋(3.3)
360
2𝜋𝜋(9)(𝜃𝜃) = 2𝜋𝜋(3.3) × 360
𝜃𝜃 =
2𝜋𝜋(3.3) × 360
2𝜋𝜋(9)
The length of the arc shown is equal to the
circumference of the cup. So, we can calculate
the circumference of the cup and let this equal
the arc length in order to solve for 𝜃𝜃.
Taking the equation for the circumference of a
circle from page 8 of the Maths Tables Book.
Taking the equation for the arc length when 𝜃𝜃 is
in degrees from page 9 of the Maths Tables Book
and subbing in the radius.
Letting these two equations equal each other.
Multiplying across by 360.
Dividing across by 2𝜋𝜋(9).
Plugging into the calculator.
𝜃𝜃 = 𝟏𝟏𝟏𝟏𝟏𝟏°
© Pocket Tutor 2022
282
b)
𝟔𝟔𝟔𝟔. 𝟐𝟐𝟐𝟐𝒎𝒎𝟑𝟑
𝑉𝑉 =
1 2
𝜋𝜋𝑟𝑟 ℎ
3
We can use the equation from page 10 of the
Maths Tables Book to calculate the volume of the
smaller cone.
𝑟𝑟
3.3
=
8.37 7.37
To find the radius we can use the fact that the two
cones contain two similar triangles. Therefore, we
can put the radius of each cone over the vertical
height of each and solve for the radius of the
smaller circle.
3.3
× 7.37 = 𝑟𝑟
8.37
2.9057 = 𝑟𝑟
𝑣𝑣 =
1
𝜋𝜋(2.9057)2 (7.37)
3
Subbing this radius and the height given into the
equation for the volume of a cone.
𝑣𝑣 = 𝟔𝟔𝟔𝟔. 𝟐𝟐𝟐𝟐𝒎𝒎𝟑𝟑
c)
𝟏𝟏𝟏𝟏 seconds
Volume of water per second:
area of pipe end × 2.5.
area: πr 2
𝐴𝐴 = 𝜋𝜋(0.8)2
Volume: π(0.8)2 × 2.5 =
8
𝜋𝜋𝜋𝜋𝑚𝑚3 /𝑠𝑠𝑠𝑠𝑠𝑠
5
Time taken = Volume up to line 𝐹𝐹 ÷ Volume per second:
8
65.2 ÷ 𝜋𝜋
5
= 𝟏𝟏𝟏𝟏 seconds
© Pocket Tutor 2022
To find how long it will take to fill the cup we need
to find the volume of water flowing through the
pipe per second. We can do this by finding the
area of the circle opening at the end of the pipe
and multiplying it by the flow rate given.
Calculating the area of the circle using the formula
from page 8 of the Maths Tables Book.
Now multiplying the result by the flow rate.
Now to find how long it takes to fill the cup we
take the volume of the cup we found in the last
part and divide it by the volume flowing per
second we just found.
283
d)
𝟏𝟏. 𝟐𝟐𝟐𝟐𝟐𝟐
𝑟𝑟
3.3
=
8.37 ℎ
𝑟𝑟 =
3.3ℎ
8.37
𝑉𝑉 =
1 2
𝜋𝜋𝑟𝑟 ℎ
3
1 3.3ℎ 2
� ℎ = 60
𝑉𝑉 = 𝜋𝜋 �
3 8.37
�
1
3.3ℎ 2
� ℎ = 60 ÷ 𝜋𝜋
3
8.37
180
10.89ℎ2
ℎ=
2
(8.37)
𝜋𝜋
ℎ3 =
3
180 10.89ℎ2
÷
(8.37)2
𝜋𝜋
ℎ=�
180 10.89ℎ2
÷
(8.37)2
𝜋𝜋
ℎ = 7.17
8.37 − 7.17 = 𝟏𝟏. 𝟐𝟐𝟐𝟐𝟐𝟐
© Pocket Tutor 2022
We can let the formula for the volume of a cone equal
the volume given to calculate the height of the cone.
But first we need its radius.
Again, we can find an expression for the radius using
the fact that we have two similar triangles involving
the cones heights and radii. Rearranging to get the
expression in terms of 𝑟𝑟.
Now, taking the equation for the volume of a cone
from page 10 of the Maths Tables Book.
Subbing in the expression we found for 𝑟𝑟 and letting
the equation equal the given volume.
1
Dividing across by 𝜋𝜋.
3
Dividing across by the fraction.
Cube rooting both sides and subbing into the
calculator.
Taking the height away from the full cone height to
find the distance, 𝑥𝑥, below the rim at which the line
should be drawn.
284
Question 8
a) i)
𝟑𝟑𝟑𝟑𝟑𝟑 marks
𝑧𝑧 =
𝑥𝑥 − 𝑥𝑥̅
𝜎𝜎
𝑥𝑥 − 280
= 0.68
90
𝑥𝑥 = 0.68(90) + 280
𝑥𝑥 = 341.2
→ 𝟑𝟑𝟑𝟑𝟑𝟑 marks
To find the minimum mark required to be in the top 25%
we first find the 𝑧𝑧 −score which matches up with 75%. By
going to page 36 of the Maths Tables Book and finding 0.75
in the table we can see that the 𝑧𝑧 value is 0.68.
We can now let the equation for the 𝑧𝑧 − score equal this
value and sub in the mean score and standard deviation.
This allows us to calculate 𝑥𝑥, the minimum mark needed.
Rounding up as a score of 341 won’t be enough to be in the
top 25%.
ii)
𝑧𝑧 =
𝑥𝑥 − 𝑥𝑥̅
𝜎𝜎
260 − 280
= −0.22
90
0.22 → 0.5871
1 − 0.5871 = 0.4129
= 41.29% ∴ above the 40th percentile
→ Can resit
© Pocket Tutor 2022
To figure out whether or not Eileen’s result puts her
above the 40th percentile we first need to calculate her
𝑧𝑧 − score.
Subbing her result, the mean result and the standard
deviation into the formula.
The −0.22 tells us that Eileen’s result was 0.22
standard deviations below the mean. Ignoring the sign
and finding the corresponding value for 0.22 on page
36 of the Maths Tables Book. This shows us the
proportion of people who scored higher than Eileen. To
find the proportion below her and thus her percentile,
we take this away from 1.
Converting the result to a percentage shows that
Eileen’s score puts her above the 40th percentile.
285
b) i)
95% of the data lies in the interval −1.96 ≤ 𝑧𝑧 ≤ 1.96
b) ii)
𝒑𝒑 = 𝟎𝟎. 𝟖𝟖
1.96 �
𝑝𝑝(1 − 𝑝𝑝)
= 95% confidence interval
𝑛𝑛
𝑝𝑝 = 𝑝𝑝, 𝑛𝑛 = 2500, confidence interval = 0.01568
1.96�
𝑝𝑝(1 − 𝑝𝑝)
= 0.01568
2500
𝑝𝑝(1 − 𝑝𝑝) 0.01568
=
1.96
2500
�
𝑝𝑝(1 − 𝑝𝑝)
0.01568
�
=�
2500
1.96
𝑝𝑝(1 − 𝑝𝑝) = 2500 �
𝑝𝑝(1 − 𝑝𝑝) =
𝑝𝑝 − 𝑝𝑝2 =
𝑝𝑝2 − 𝑝𝑝 +
𝑝𝑝 =
𝑝𝑝 =
4
25
As we have the confidence interval and the
number of people surveyed, we can use
this formula from page 34 of the Maths
Tables Book to calculate 𝑝𝑝.
Subbing in the number of people surveyed
for 𝑛𝑛 and letting the equation equal the
confidence interval.
Dividing across by 1.96.
2
Squaring both sides.
0.01568 2
�
1.96
Multiplying across by 2500.
Plugging the right hand side into the
calculator.
4
25
4
=0
25
Rearranging to form a quadratic.
−𝑏𝑏 ± √𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎
2𝑎𝑎
−(−1) ± �(−1)2 − 4(1) �
2(1)
𝑝𝑝 = 0.2, 0.8 → 𝒑𝒑 = 𝟎𝟎. 𝟖𝟖
© Pocket Tutor 2022
4
�
25
There are no obvious factors so we can use
the −𝑏𝑏 formula from page 20 of the Maths
Tables Book to solve for 𝑝𝑝.
0.2 is outside the given range so we take
the answer as 𝑝𝑝 = 0.8.
286
c)
𝐻𝐻0 = The mean weight of the bags has not changed
Stating the null hypothesis and
the alternative hypothesis.
𝐻𝐻1 = The mean weight of the bags has changed
𝑧𝑧 =
𝑥𝑥 − 𝜇𝜇
𝜎𝜎
√𝑛𝑛
Taking the equation for the 𝑧𝑧
score which takes into account
the standard error from page 35
of the Maths Tables Book.
𝑥𝑥 = 13.1, 𝜇𝜇 = 12, 𝜎𝜎 = 4.5, 𝑛𝑛 = 80
Subbing in the sample mean,
the mean, the standard
deviation, and then sample size.
13.1 − 12
= 2.186
4.5
√80
2.186 > 1.96
Plugging into the calculator.
∴ We reject the null hypothesis, the mean weight of the bags has changed.
2.186 is greater than 1.96 and
therefore at the 95% level of
confidence we can say that the
mean weight has changed.
d)
𝟏𝟏𝟏𝟏. 𝟔𝟔𝟔𝟔%
3000
= 75
40
→ Probability that the mean weight > 75
𝑥𝑥 − 𝜇𝜇
𝑧𝑧 = 𝜎𝜎
√𝑛𝑛
75 − 73 √10
=
12
3
√40
= 1.054
1.05 = 0.8531
1 − 0.8531 = 0.1469
→ 𝟏𝟏𝟏𝟏. 𝟔𝟔𝟔𝟔%
© Pocket Tutor 2022
Dividing the maximum weight by the number of
people to find what the average weight would have
to be for the weight to be above the maximum.
Taking the equation for the 𝑧𝑧 score from page 35 of
the Maths Tables Book.
Subbing in the mean weight needed, the mean
weight, the standard deviation and the number of
passengers and plugging into the calculator.
Finding the corresponding value for 1.05 on page
36 of the Maths Tables Book.
This is the probability of the weight being under 75,
so we take this away from 1 to get the probability
of the mean weight being over 75kg.
287
e)
𝟒𝟒, 𝟔𝟔, 𝟗𝟗, 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏, 𝟐𝟐𝟐𝟐, 𝟐𝟐𝟐𝟐
Median = 12.5 → 𝐷𝐷 + 𝐸𝐸 = 25
lower quartile = 7.5 → B + C = 15
upper quartile = 19.5 → 𝐹𝐹 + 𝐺𝐺 = 39
𝐺𝐺 = 23
→ 𝐹𝐹 + 23 = 39
𝐹𝐹 = 16
𝐵𝐵 + 𝐶𝐶 + 𝐷𝐷 + 𝐸𝐸 + 𝐹𝐹 + 𝐺𝐺 = 25 + 15 + 39
𝐵𝐵 + 𝐶𝐶 + 𝐷𝐷 + 𝐸𝐸 + 𝐹𝐹 + 𝐺𝐺 = 79
Total sum = 13.5 × 8 = 108
∴ 𝐴𝐴 + 𝐻𝐻 = 108 − 79
𝐴𝐴 + 𝐻𝐻 = 29
𝐻𝐻 − 𝐴𝐴 = 21
𝐻𝐻 = 21 + 𝐴𝐴
𝐴𝐴 + (21 + 𝐴𝐴) = 29
2𝐴𝐴 = 8
𝐴𝐴 = 4
© Pocket Tutor 2022
As there are 8 terms the median is equal to the 4th
and the 5th numbers added together and divided by
two. So 𝐷𝐷 + 𝐸𝐸 = twice the median.
We can apply the same logic to the 2nd and 3rd
numbers for the lower quartile. and the 6th and 7th
numbers for the upper quartile.
We have been told that 𝐺𝐺 = 23 so we can use this
to calculate 𝐹𝐹.
Now putting together, the sums we figured out
above.
We know that the sum of all the numbers is equal to
8 times the mean.
Taking away the sum of 𝐵𝐵, 𝐶𝐶, 𝐷𝐷, 𝐸𝐸, 𝐹𝐹 and G, leaves
us with 𝐴𝐴 and 𝐻𝐻 adding up to 29.
We know that the range (the last number minus the
first number) is equal to 21.
So, we can get 𝐻𝐻 in terms of 𝐴𝐴 and plug this
expression in for 𝐻𝐻 in 𝐴𝐴 + 𝐻𝐻 = 29.
Solving for 𝐴𝐴.
288
𝐻𝐻 − 4 = 21
Now subbing this value for 𝐴𝐴 into the expression
for the range to find 𝐻𝐻.
𝐷𝐷 + 𝐸𝐸 = 25,
We know that 𝐷𝐷 and 𝐸𝐸 cannot be 12 and 13 as
they must be two or more apart, we also know
that they cannot be 10 and 15 as 𝐸𝐸 would then be
within 1 of 𝐹𝐹.
𝐻𝐻 = 25
so 𝐷𝐷 = 11, 𝐸𝐸 = 14
𝐵𝐵 + 𝐶𝐶 = 15,
so 𝐵𝐵 = 6, 𝐶𝐶 = 9
𝟒𝟒, 𝟔𝟔, 𝟗𝟗, 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏, 𝟏𝟏𝟏𝟏, 𝟐𝟐𝟐𝟐, 𝟐𝟐𝟐𝟐
© Pocket Tutor 2022
We know that 𝐵𝐵 and 𝐶𝐶 cannot be 7 and 8 as the
difference between them would not be 2. We
also know that they can’t be 5 and 10 as 𝐵𝐵 would
then be within 1 of 𝐴𝐴.
289
Question 9
a)
𝟖𝟖𝟖𝟖. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
15km/h × 0.5 = 7.5𝑘𝑘𝑘𝑘
Multiplying each speed by 0.5
to see how far they have
travelled in half an hour.
90 − 7.5 = 82.5
30km/h × 0.5 = 15𝑘𝑘𝑘𝑘
𝑥𝑥
𝑥𝑥 2 = (82.5)2 + (15)2
𝑥𝑥 2 = 7031.25
15
Ship A has travelled 7.5km
towards Port B and ship B has
travelled 15km away from
port B.
We can use Pythagoras’
theorem to find the distance
between the ships.
𝑥𝑥 = √7031.25
𝑥𝑥 = 𝟖𝟖𝟖𝟖. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
b)
(90 − 15𝑡𝑡)2 + (30𝑡𝑡)2 = 𝑠𝑠 2
8100 − 1350𝑡𝑡 − 1350𝑡𝑡 + 225𝑡𝑡 2 + 900𝑡𝑡 2 = 𝑠𝑠 2
1125𝑡𝑡 2 − 2700𝑡𝑡 + 8100 = 𝑠𝑠 2
(1125𝑡𝑡 2
− 2700𝑡𝑡 +
© Pocket Tutor 2022
1
8100)2
= 𝑠𝑠
We know that the distance from ship A to port B = 90
minus the distance travelled by ship A. We can write this
as 90 minus its speed (7.5) times time (t). We know that
the distance of ship B from port B is equal to its speed (30)
times time (t).
We know that we can calculate the distance between the
ships using Pythagoras’ theorem. So, the distance
between the ships squared is equal to each of their
distances from port B squared and added together.
290
c)
𝟖𝟖𝟖𝟖. 𝟓𝟓𝟓𝟓𝟓𝟓
1
𝑠𝑠 = (1125𝑡𝑡 2 − 2700𝑡𝑡 + 8100)2
1
𝑑𝑑𝑑𝑑 1
= (1125𝑡𝑡 2 − 2700𝑡𝑡 + 8100)−2 × (2(1125𝑡𝑡) − 2700)
𝑑𝑑𝑑𝑑 2
1
𝑑𝑑𝑑𝑑 1
= (1125𝑡𝑡 2 − 2700𝑡𝑡 + 8100)−2 × (2250𝑡𝑡 − 2700)
𝑑𝑑𝑑𝑑 2
2250𝑡𝑡 − 2700
𝑑𝑑𝑑𝑑
=
𝑑𝑑𝑑𝑑 2√1125𝑡𝑡 2 − 2700𝑡𝑡 + 8100
2250𝑡𝑡 − 2700
2√1125𝑡𝑡 2 − 2700𝑡𝑡 + 8100
Taking the power down and then
multiplying by the derivative of what’s
inside the bracket.
1
𝑥𝑥 −2 =
1
√𝑥𝑥
, so we can rewrite the bracket
1
by the two under the on the left.
2
Letting it equal 0 to solve for 𝑡𝑡.
Multiplying across by the bottom of the
fraction.
2700
𝑡𝑡 =
2250
Adding 2700 to both sides then dividing
across by 2250.
𝒕𝒕 = 𝟏𝟏. 𝟐𝟐
𝑠𝑠 = (1125𝑡𝑡 2 − 2700𝑡𝑡 +
𝑠𝑠 =
So, we differentiate it using chain rule and
then let it equal 0 and solve for 𝑡𝑡.
as a square root under a fraction multiplied
=0
2250𝑡𝑡 − 2700 = 0
(1125(1.2)2
To find when the ships are closest to each
other we need to find the minimum value
of this equation.
1
8100)2
− 2700(1.2) +
𝑠𝑠 = 80.49 → 𝟖𝟖𝟖𝟖. 𝟓𝟓𝟓𝟓𝟓𝟓
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This is the time at which the distance is a
minimum.
1
8100)2
Subbing this time into the original equation
to find the minimum distance.
291
2019 Paper 2
Question 1
a) i)
𝟒𝟒𝟒𝟒
𝟗𝟗𝟗𝟗
12 8
24
×
=
20 19 95
8 12 24
×
=
20 19 95
24 24 𝟒𝟒𝟒𝟒
+
=
95 95 𝟗𝟗𝟗𝟗
This selection can happen in two ways. 1. A boy is chosen and then a girl
2. A girl is chosen and then a boy
We need to find the probability of each of these and add them to get
the total probability
ii)
𝟖𝟖𝟖𝟖
12 11 10 8
×
×
×
=
20 19 18 17 𝟗𝟗𝟗𝟗𝟗𝟗
With each selection there is one fewer pupil to draw from, so the
number on the bottom goes down. The same happens with the
number of boys.
b)
6
8
� � × � � = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
3
4
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As question 1 of part A must be done, we say there are 3 choices from a
possible 6 in Section A. In Section B you can choose 4 from 8. The total
number of combinations is equal to the number of combinations in each
multiplied together.
292
Question 2
Using the equation formula of a line from page
18 of The Maths Tables Book.
a)
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥`1 )
𝑦𝑦2 − 𝑦𝑦1
𝑚𝑚 =
𝑥𝑥2 − 𝑥𝑥1
𝑚𝑚 =
𝑏𝑏
𝑏𝑏 − 0
=−
𝑎𝑎
0 − 𝑎𝑎
𝑏𝑏
𝑦𝑦 − 𝑏𝑏 = − (𝑥𝑥 − 0)
𝑎𝑎
𝑎𝑎𝑎𝑎 − 𝑎𝑎𝑎𝑎 = −𝑏𝑏𝑏𝑏
𝑎𝑎𝑎𝑎 − 𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏 = 0
𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏 = 𝑎𝑎𝑎𝑎
𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏 𝑎𝑎𝑎𝑎
+
=
𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎
First, we need to find the slope.
Plugging the coordinates into the equation for
the slope of a line also from page 18 of The
Maths Tables Book.
Plugging the point (0, 𝑏𝑏) and the slope into
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥`1 ).
Multiplying across by 𝑎𝑎.
Rearranging
Dividing across by 𝑎𝑎𝑎𝑎
𝑦𝑦 𝑥𝑥
+ =1
𝑏𝑏 𝑎𝑎
b) i)
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥`1 )
𝑦𝑦 − 0 = 𝑚𝑚(𝑥𝑥 − 6)
𝒚𝒚 = 𝒎𝒎𝒎𝒎 − 𝟔𝟔𝟔𝟔
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Plugging the point (6,0) into
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥`1 ) and leaving 𝑚𝑚 as
the slope.
293
𝒎𝒎
𝟐𝟐𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟏𝟏
�
𝒊𝒊𝒊𝒊) �
,
𝟒𝟒 + 𝟑𝟑𝟑𝟑 𝟒𝟒 + 𝟑𝟑𝟑𝟑
Eq 1
4𝑥𝑥 + 3𝑦𝑦 = 25
Eq 2
𝑦𝑦 = 𝑚𝑚𝑚𝑚 − 6𝑚𝑚
4𝑥𝑥 + 3(𝑚𝑚𝑚𝑚 − 6𝑚𝑚) = 25
4𝑥𝑥 + 3𝑚𝑚𝑚𝑚 − 18𝑚𝑚 = 25
𝑥𝑥(4 + 3𝑚𝑚) − 18𝑚𝑚 = 25
𝑥𝑥(4 + 3𝑚𝑚) = 25 + 18𝑚𝑚
𝟐𝟐𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟏𝟏
𝑥𝑥 =
𝟒𝟒 + 𝟑𝟑𝟑𝟑
𝑦𝑦 = 𝑚𝑚𝑚𝑚 − 6𝑚𝑚
25 + 18𝑚𝑚
� − 6𝑚𝑚
𝑦𝑦 = 𝑚𝑚 �
4 + 3𝑚𝑚
25𝑚𝑚 + 18𝑚𝑚2
𝑦𝑦 =
− 6𝑚𝑚
4 + 3𝑚𝑚
𝑦𝑦 =
2
25𝑚𝑚 + 18𝑚𝑚 − 6𝑚𝑚(4 + 3𝑚𝑚)
4 + 3𝑚𝑚
25𝑚𝑚 + 18𝑚𝑚2 − 24𝑚𝑚 + 18𝑚𝑚2
4 + 3𝑚𝑚
𝒎𝒎
𝑦𝑦 =
𝟒𝟒 + 𝟑𝟑𝟑𝟑
𝑦𝑦 =
�
We need to find where these two lines meet.
Subbing (𝑚𝑚𝑚𝑚 − 6𝑚𝑚) in for 𝑦𝑦 in equation 1, as
we can see from equation 2 that 𝑦𝑦 is equal to
this.
Factorising out the 𝑥𝑥
Dividing across by 4 + 3𝑚𝑚 gives us our 𝑥𝑥coordinate
Now subbing our expression for 𝑥𝑥 in for 𝑥𝑥 in
equation 2.
Multiplying out the bracket.
Multiplying 6𝑚𝑚 by the bottom of the fraction
to put it over the fraction.
Tidying up.
Listing the coordinates.
𝒎𝒎
𝟐𝟐𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟏𝟏
�
,
𝟒𝟒 + 𝟑𝟑𝟑𝟑 𝟒𝟒 + 𝟑𝟑𝟑𝟑
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294
Question 3
a) 𝒌𝒌 = 𝟏𝟏𝟏𝟏, 𝒌𝒌 = −𝟒𝟒
(𝑥𝑥 − 2)2 + (𝑦𝑦 − 3)2 = 65
2
2
�(−2) − 2� + �(𝑘𝑘) − 3� = 65
(−4)2 + 𝑘𝑘 2 − 6𝑘𝑘 + 9 = 65
𝑘𝑘 2 − 6𝑘𝑘 + 9 + 16 − 65 = 0
𝑘𝑘 2 − 6𝑘𝑘 − 40 = 0
(𝑘𝑘 − 10)(𝑘𝑘 + 4) = 0
If the point is on the circle, we plug in −2 for 𝑥𝑥
and 𝑘𝑘 for 𝑦𝑦 into the equation of the circle.
We can then solve for 𝑘𝑘
Squaring out the brackets.
Rearranging
Solving the quadratic.
𝒌𝒌 = 𝟏𝟏𝟏𝟏, 𝒌𝒌 = −𝟒𝟒
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295
b) (𝒙𝒙 − 𝟏𝟏)𝟐𝟐 + (𝒚𝒚 − 𝟏𝟏)𝟐𝟐 = 𝟏𝟏
𝑥𝑥 2 + 𝑦𝑦 2 + 2𝑔𝑔𝑔𝑔 + 2𝑓𝑓𝑓𝑓 + 𝑐𝑐 = 0
𝑔𝑔 = 𝑓𝑓
The general equation of a circle from page 19 of The Maths Tables
Book. Page 19 also tells us that the centre is (−𝑔𝑔, −𝑓𝑓)
As the circle touches the 𝑥𝑥 − 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 the radius equals 𝑓𝑓.
Centre: (−𝑔𝑔, −𝑓𝑓) → (−𝑔𝑔, −𝑔𝑔)
3𝑥𝑥 − 4𝑦𝑦 + 6 = 0
As the circle touches the 𝑦𝑦 − 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 the radius equals 𝑔𝑔.
Therefore 𝑓𝑓 = 𝑔𝑔 = The radius
This means that we can right the centre as (−𝑔𝑔, −𝑔𝑔)
The perpendicular distance from the tangent to the centre equals
the radius (𝑔𝑔).
|𝑎𝑎𝑥𝑥1 + 𝑏𝑏𝑦𝑦1 + 𝑐𝑐|
√𝑎𝑎2 + 𝑏𝑏 2
|3(−𝑔𝑔) + (−4)(−𝑔𝑔) + 6|
�(3)2 + (−4)2
| − 3𝑔𝑔 + 4𝑔𝑔 + 6|
= 𝑔𝑔
5
|𝑔𝑔 + 6| = 5𝑔𝑔
(𝑔𝑔 + 6)2 = 25𝑔𝑔2
= 𝑔𝑔
So, we use the equation for the distance from a point to a line to
find the distance from the centre to the tangent and let it equal 𝑔𝑔
Multiplying across by 5
Squaring both sides to get rid of the modulus.
𝑔𝑔2 + 12𝑔𝑔 + 36 = 25𝑔𝑔2
Rearranging
2𝑔𝑔2 − 1𝑔𝑔 − 3 = 0
Solving the quadratic.
24𝑔𝑔2 − 12𝑔𝑔 − 36 = 0
(2𝑔𝑔 − 3)(𝑔𝑔 + 1)
𝑔𝑔 =
3
2
Dividing across by 12
𝑔𝑔 = −1
−𝑔𝑔 = 1 𝑜𝑜𝑜𝑜 − 𝑔𝑔 = −
3
2
Centre = (1,1) 𝑟𝑟 = 1
(𝒙𝒙 − 𝟏𝟏)𝟐𝟐 + (𝒚𝒚 − 𝟏𝟏)𝟐𝟐 = 𝟏𝟏
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As the centre = (−𝑔𝑔, −𝑔𝑔) we find what −𝑔𝑔 equals.
3
We don’t use 𝑔𝑔 = − as a centre coordinate, as we can see from
2
the diagram that the circle is in the first quadrant and thus the
centre coordinates must be positive.
Plugging the centre and the radius into (𝑥𝑥 − ℎ)2 + (𝑦𝑦 − 𝑘𝑘)2 = 𝑟𝑟 2
296
Question 4
a)
cos 2𝜃𝜃 = cos(𝜃𝜃 + 𝜃𝜃)
cos 2𝜃𝜃 = cos 𝜃𝜃𝜃𝜃𝜃𝜃𝜃𝜃𝜃𝜃 − sin 𝜃𝜃 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
cos 2𝜃𝜃 = cos 2 𝜃𝜃 − sin2 𝜃𝜃
cos 2𝜃𝜃 = (1 − sin2 𝜃𝜃) − sin2 𝜃𝜃
cos 2𝜃𝜃 = 1 − 2 sin2 𝜃𝜃
© Pocket Tutor 2022
Pages 13&14 of The Maths Tables Book will help us with this question
Using the identity: cos(𝐴𝐴 + 𝐵𝐵) = cos 𝐴𝐴 cos 𝐵𝐵 − sin 𝐴𝐴 sin 𝐵𝐵
From page 13: cos 2 𝜃𝜃 + sin2 𝜃𝜃 = 1
→ cos 2 𝜃𝜃 = 1 − sin2 𝜃𝜃, plugging this in for cos 2 𝜃𝜃
297
b) 𝐜𝐜𝐜𝐜𝐜𝐜 𝑨𝑨 =
𝟏𝟏
𝟑𝟑
𝑥𝑥 2 + 𝑥𝑥 2 = Hypotenuse2
2
2
2𝑥𝑥 = Hypotenuse
First, we label each side of the cube with length 𝑥𝑥.
𝑥𝑥
√2𝑥𝑥 = Hypotenuse
𝑥𝑥
2
𝑥𝑥 2 + �√2𝑥𝑥� = Hypotenuse2
𝑥𝑥 2 + 2𝑥𝑥 2 = Hypotenuse2
3𝑥𝑥 2 = Hypotenuse2
Now we can use Pythagoras’ theorem to find the
length of the dotted line, where √2𝑥𝑥 is the diagonal
across the base of the cube and 𝑥𝑥 is the height of the
side of the cube.
𝑥𝑥
√3𝑥𝑥 = Hypotenuse
√2𝑥𝑥
𝑎𝑎2 = 𝑏𝑏 2 + 𝑐𝑐 2 − 2𝑏𝑏𝑏𝑏 cos 𝐴𝐴
2
2
√3𝑥𝑥
√3𝑥𝑥
√3𝑥𝑥 √3𝑥𝑥
� +�
� − 2�
��
� cos 𝐴𝐴
𝑥𝑥 = �
2
2
2
2
2
2
𝑥𝑥 2
√3𝑥𝑥 √3𝑥𝑥
2�
��
�
2
2
𝑥𝑥 2
√3𝑥𝑥 √3𝑥𝑥
��
�
2
2
2�
=
−
√3𝑥𝑥 √3𝑥𝑥
2�
��
�
2
2
2
2
√3𝑥𝑥 √3𝑥𝑥
��
�
2
2
2�
√3𝑥𝑥 √3𝑥𝑥
��
�
2
2
2�
cos 𝐴𝐴 =
−𝑥𝑥 2 + �
2
= − cos 𝐴𝐴
2
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= − cos 𝐴𝐴
Now we can use the cosine rule to find the shaded
angle. The cosine rule is on page 16 of The Maths
Tables Book.
𝑎𝑎 = The length of the side of the cube.
𝑏𝑏 and 𝑐𝑐 = half the length of the dotted diagonals, so
√3𝑥𝑥
2
Dividing across by 2 �
√3𝑥𝑥
√3𝑥𝑥
�� �
2
2
Getting cos 𝐴𝐴 by itself
Subtracting the two fractions.
2
√3𝑥𝑥
√3𝑥𝑥
� +�
�
2
2
√3𝑥𝑥 √3𝑥𝑥
2�
��
�
2
2
− cos 𝐴𝐴
2
√3𝑥𝑥
√3𝑥𝑥
� +�
�
2
2
√3𝑥𝑥
√3𝑥𝑥
𝑥𝑥 − ��
� +�
� �
2
2
2
2
√3𝑥𝑥
√3𝑥𝑥
� +�
�
�
2
2
�
In order to find the length of the diagonal from the
top left corner to the bottom right corner (the
dotted line) we first need to find the length of the
diagonal across the base. We can do this using
Pythagoras’ theorem.
Multiplying across by −1
298
cos 𝐴𝐴 =
cos 𝐴𝐴 =
3𝑥𝑥 2 3𝑥𝑥 2
+
4
4
2
3𝑥𝑥
�
2�
4
−𝑥𝑥 2 +
−𝑥𝑥 2 +
6𝑥𝑥 2
4
3𝑥𝑥 2
2
𝑥𝑥 2
cos 𝐴𝐴 = 2 2
3𝑥𝑥
�
�
2
cos 𝐴𝐴 =
cos 𝐴𝐴 =
𝐜𝐜𝐜𝐜𝐜𝐜 𝑨𝑨 =
Squaring out the brackets on the top and
multiplying out the bottom.
To divide by a fraction, we invert it and
multiply.
𝑥𝑥 2
2
× 2
2 3𝑥𝑥
𝑥𝑥 2
3𝑥𝑥 2
𝟏𝟏
𝟑𝟑
Question 5
a)
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299
b) 𝟑𝟑𝟑𝟑°
|𝐷𝐷𝐷𝐷| =
1
|𝐴𝐴𝐴𝐴|
2
Given in the question.
As OB is half AB
→ |𝐷𝐷𝐷𝐷| = |𝑂𝑂𝑂𝑂|
→ Triangle 𝑂𝑂𝑂𝑂𝑂𝑂 = Equilateral
|< 𝑂𝑂𝑂𝑂𝑂𝑂| = 60
As DC = 𝑂𝑂𝑂𝑂 (the radius) and, OD and OC are both radii
All angles in an equilateral triangle are 60
By alternate angles
|< 𝐴𝐴𝐴𝐴𝐴𝐴| = 60
Triangle AOD is isosceles as |𝑂𝑂𝑂𝑂| = |𝑂𝑂𝑂𝑂|
As OA and OD are both radii, an isosceles triangle is one with
two equal sides, the base angles are equal. As we know AOD
is 60, we take this from 180, which leaves 120° between the
other two angles.
|< 𝐴𝐴𝐴𝐴𝐴𝐴| = 90
As BE is a tangent it makes an angle of 90° with the line
through the centre of the circle.
|< 𝐵𝐵𝐵𝐵𝐵𝐵| = 180 − 90 − 60 = 𝟑𝟑𝟑𝟑°
Taking the angles OAD and ABE from 180 leaves us with BEA
|< 𝑂𝑂𝑂𝑂𝑂𝑂| = |< 𝑂𝑂𝑂𝑂𝑂𝑂| =
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120
= 60
2
300
Question 6
𝒂𝒂) 𝒙𝒙 =
𝟏𝟏𝟏𝟏
𝟏𝟏𝟏𝟏
, 𝒚𝒚 =
𝟒𝟒𝟒𝟒
𝟑𝟑𝟑𝟑
𝑃𝑃(𝐹𝐹 ∩ 𝑆𝑆) = 𝑃𝑃(𝐹𝐹) × 𝑃𝑃(𝑆𝑆)
As the events are independent.
9
1
=
× 𝑃𝑃(𝑆𝑆)
5 20
1 9
÷
= 𝑃𝑃(𝑆𝑆)
5 20
4
= 𝑃𝑃(𝑆𝑆)
9
𝑃𝑃(𝑆𝑆\𝐹𝐹) =
4 1 𝟏𝟏𝟏𝟏
− =
= 𝒙𝒙
9 5 𝟒𝟒𝟒𝟒
𝑃𝑃(𝐹𝐹 ∪ 𝑆𝑆)′ = 1 − �
𝑦𝑦 =
𝟏𝟏𝟏𝟏
𝟑𝟑𝟑𝟑
11 1 1
+ + �
45 5 4
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Taking the intersection from our value for P(S)
to find 𝑥𝑥
Taking the sum of the probabilities in the circles
from 1 to find 𝑦𝑦
301
b)
𝟐𝟐5
1
= Probability of German × (Probability of not German)
6
1
𝑥𝑥
2𝑥𝑥 + 10
�
=
�
6 3𝑥𝑥 + 10
3𝑥𝑥 + 9
2𝑥𝑥 2 + 10𝑥𝑥
1
= 2
6 9𝑥𝑥 + 27𝑥𝑥 + 30𝑥𝑥 + 90
2𝑥𝑥 2 + 10𝑥𝑥
1
= 2
6 9𝑥𝑥 + 57𝑥𝑥 + 90
1
(9𝑥𝑥 2 + 57𝑥𝑥 + 90) � � = 2𝑥𝑥 2 + 10𝑥𝑥
6
2
If there are 𝑥𝑥 germans, the probability of one being
selected is 𝑥𝑥 over the total number of students.
The total number is 𝑥𝑥 + 2𝑥𝑥 +10, where 2𝑥𝑥 is the
number of Irish students.
We multiply this by the probability of a nonGerman student being chosen, which is the
number of other students over the total number
−1, as one student has already been selected.
Multiplying across by the bottom of the fraction.
Multiplying across by 6
2
9𝑥𝑥 + 57𝑥𝑥 + 90 = 6(2𝑥𝑥 + 10𝑥𝑥)
9𝑥𝑥 2 + 57𝑥𝑥 + 90 = 12𝑥𝑥 2 + 60𝑥𝑥
3𝑥𝑥 2 + 3𝑥𝑥 − 90
Dividing across by 3
(𝑥𝑥 + 6)(𝑥𝑥 − 5) = 0
There cannot be a negative number of people, so
we discard 6 as an answer
2
𝑥𝑥 + 𝑥𝑥 − 30
𝑥𝑥 = −6, 𝑥𝑥 = 5
German children: 5
Irish children: 5 × 2 = 10
Spanish Children: 10
There are twice as many Irish students as Germans
(given).
Total: 5 + 10 + 10 = 𝟐𝟐𝟐𝟐
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302
Question 7
a) i) 𝟑𝟑𝟑𝟑√𝟓𝟓
𝑂𝑂
|𝐴𝐴𝐴𝐴|2 + |𝑂𝑂𝑂𝑂|2 = |𝐴𝐴𝐴𝐴|2
|𝐴𝐴𝐴𝐴|2 + (60)2 = (90)2
90
|𝐴𝐴𝐴𝐴|2 = 8100 − 3600
|𝐴𝐴𝐴𝐴|2 = 4500
|𝐴𝐴𝐴𝐴| = 𝟑𝟑𝟑𝟑√𝟓𝟓
Taking the triangle ODA.
90 − 30
𝐴𝐴
2
2
𝑎𝑎 = 𝑏𝑏 + 𝑐𝑐 − 2𝑏𝑏𝑏𝑏 cos 𝐴𝐴
2
�30√5� =
(90)2
+
(60)2
𝑂𝑂
− 2(90)(60)𝑐𝑐𝑐𝑐𝑐𝑐 𝑂𝑂
4500 − 8100 − 3600 = − 2(90)(60)𝑐𝑐𝑐𝑐𝑐𝑐 𝑂𝑂
−7200 = −10800𝑐𝑐𝑐𝑐𝑐𝑐 𝑂𝑂
−7200
= 𝑐𝑐𝑐𝑐𝑐𝑐 𝑂𝑂
−10800
𝑐𝑐𝑐𝑐𝑠𝑠 −1
2
= 𝑂𝑂
3
We can find AD using Pythagoras’ theorem.
𝐷𝐷
ii) 𝟎𝟎. 𝟖𝟖𝟖𝟖
2
OD = 𝑂𝑂𝑂𝑂 − 𝐷𝐷𝐷𝐷, 𝑂𝑂𝑂𝑂 is the radius (90) and 𝐷𝐷𝐷𝐷
is given as 30.
𝐴𝐴
90
30√5
60
𝐷𝐷
Taking the cosine rule from page 16 of
The Maths Tables Book.
Filling in the lengths of the sides,
remembering that 𝑎𝑎 has to be the side
opposite the angle.
Dividing across by −10800
|< 𝐷𝐷𝐷𝐷𝐷𝐷| = 𝟎𝟎. 𝟖𝟖𝟖𝟖
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303
iii) 𝟎𝟎. 𝟐𝟐𝟐𝟐𝒎𝒎𝟐𝟐
Area of segment = Area of sector – Area of 2 triangles
1
Area of sector = 𝑟𝑟 2 𝜃𝜃
2
𝑟𝑟 = 0.9𝑚𝑚, 𝜃𝜃 = 2(0.84) = 1.68
→
1
(0.9)2 (1.68)
2
= 0.68𝑚𝑚2
Taking the equation for the sector of a circle from
page 9 of The Maths Tables Book.
Converting the lengths to metres as the question says
give your answer in 𝑚𝑚2
The angle is 2 times 𝐷𝐷𝐷𝐷𝐷𝐷, as we want the angle AOC
1
Area of triangle = 𝑎𝑎𝑎𝑎 sin 𝐶𝐶
2
𝑎𝑎 = 0.9, 𝑏𝑏 = 0.6 𝐶𝐶 = 0.84
1
(0.9)(0.6) sin(0.84)
2
Taking the equation for the area of a triangle from
page 9 of The Maths Tables Book
Again, converting the lengths to metres
= 0.201
Area of segment = Area of sector – Area of 2 triangles
Area of segment = 0.68 − 2(0.201)
= 0.278 → 𝟎𝟎. 𝟐𝟐𝟐𝟐𝒎𝒎𝟐𝟐
Multiplying the area of the triangle we found by 2 as
there are two of them above the segment.
iv)
0.28 × 2.5 = 𝟎𝟎. 𝟕𝟕𝒎𝒎𝟑𝟑
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As we have the area of the cross section of
the trough, we just need to multiply it by the
length of the trough to find the volume
304
b) i) 𝟐𝟐𝟐𝟐. 𝟕𝟕𝟕𝟕𝟕𝟕𝒎𝒎𝟐𝟐
Volume = Volume of hemisphere + Volume of Cylinder + Volume of Cone
Volume of hemisphere = volume of sphere
To find the volume of the top half of
the timer we have to find the volume
of the three different parts and add
them together.
𝑟𝑟 = 1.25
Taking the equation ( 𝜋𝜋𝑟𝑟 3 ) for the
→
2
1
4
× � 𝜋𝜋𝑟𝑟 3 � = 𝜋𝜋𝑟𝑟 3
3
2
3
1
2
4
3
2
𝜋𝜋(1.25)3 = 4.091
3
volume of a sphere from page 10 of
The Maths Tables Book.
Volume of Cylinder = 𝜋𝜋𝑟𝑟 2 ℎ
The volume of a cylinder is also on
page 9.
𝑟𝑟 = 1.25, ℎ = 3.5
𝜋𝜋(1.25)2 (3.5) = 17.181
1
Volume of Cone = 𝜋𝜋𝑟𝑟 2 ℎ
𝑟𝑟 = 1.25, ℎ = 1.5
The volume of a cone is also on page 9.
3
1
𝜋𝜋(1.25)2 (1.5) = 2.454
3
Total Volume = 4.091 + 17.181 + 2.454 = 23.726
= 𝟐𝟐𝟐𝟐. 𝟕𝟕𝟕𝟕𝟕𝟕𝒎𝒎𝟐𝟐
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Adding the three volumes and
rounding to two decimal places.
305
ii) 𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
23.73 × 0.02 = 0.4746𝑐𝑐𝑚𝑚2
0.4746 =
1 2
𝜋𝜋𝑟𝑟 ℎ
3
𝑟𝑟 1.25
=
1.5
ℎ
Getting 𝑟𝑟 in terms of ℎ
5
ℎ
6
0.4746 =
0.4746 =
0.4746 =
0.4746 ÷
Letting this volume equal the equation for the volume
of a cone.
Finding the ratio of radius to height of the cone, using
the measurements of the original cone.
𝑟𝑟 5
=
ℎ 6
𝑟𝑟 =
If 98% of the volume of the sand is in the bottom half,
then 2% is left in the top half. Multiplying the total
volume by 0.02 gives us the amount left in the top
half.
1 5 2
𝜋𝜋 � ℎ� (ℎ)
3 6
Now plugging our expression for 𝑟𝑟 into the equation
for the volume of the cone.
25
𝜋𝜋ℎ3
108
Multiplying by the third
1 25 2
𝜋𝜋 � ℎ � ℎ
3 36
25
𝜋𝜋 = ℎ3
108
0.65262 = ℎ3
3
√0.65262 = ℎ
ℎ = 𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
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Squaring the bracket.
Dividing across by
25
108
𝜋𝜋
Cube rooting both sides and rounding to two decimal
places.
306
Question 8
a) i)
� < 𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
𝟎𝟎. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 < 𝒑𝒑
𝑝𝑝̂ ± 1.96�
𝑝𝑝̂ =
𝑝𝑝(1 − 𝑝𝑝)
𝑛𝑛
174
= 0.2175
800
𝑝𝑝 = 0.2715, 𝑛𝑛 = 800
0.2175 ± 1.96�
(0.2175)�1 − (0.2175)�
800
0.2175 ± 0.02859
0.2175 + 0.02859 = 0.2461
0.2175 − 0.02859 = 0.1889
� < 𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
𝟎𝟎. 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏 < 𝒑𝒑
Taking the equation for the standard error of a
proportion from page 34 of The Maths Tables
Book and remembering to multiply by the 1.96
Finding the proportion of new cars.
Using our value for 𝑝𝑝̂ for 𝑝𝑝 as 𝑝𝑝 was not given.
Plugging these values into the equation.
Adding and subtracting the error from the
proportion.
Constructing our confidence interval
ii) 𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
𝑥𝑥̅ − 𝜇𝜇
(𝜎𝜎)
𝑥𝑥̅ = 95, 𝜇𝜇 = 87.3, 𝜎𝜎 = 12
95 − 87.3
= 0.64
(12)
= 0.7389
1 − 0.7389 = 𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
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Using the 𝑧𝑧 −score formula from page 35 of The Maths
Tables Book, we do not use the √𝑛𝑛, as the data we are
dealing with is not from a sample.
Plugging in our values to get the 𝑧𝑧 −score
Converting this to a proportion by looking at pages 36 & 37
of The Maths Tables Book
Taking this proportion from 1 to find the proportion that
travel over 95 km/h
307
iii) 81km/h
0.7 → −0.52
𝑥𝑥̅ − 𝜇𝜇
= −0.52
(𝜎𝜎)
𝑥𝑥̅ − 87.3
= −0.52
12
𝑥𝑥̅ = −0.52(12) + 87.3
𝑥𝑥̅ = 81.06 → 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖/𝒉𝒉
Finding 70% (0.7) in the tables on pages 36 & 37 of The
Maths Tables Book and converting it to a 𝑧𝑧 − 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠.
We say it is −0.52 as the driver is under the mean speed.
Plugging it into the formula and solving for 𝑥𝑥̅
Rounding to the nearest whole number.
b) i)
The magazine can conclude that the average speed has changed.
0.024 < 0.05 therefore, we reject the null hypothesis that the
average speed is 87.3𝑘𝑘𝑘𝑘/ℎ
If a 𝑝𝑝 −value is < 0.05 we reject the
null hypothesis.
ii) 𝟖𝟖𝟖𝟖. 𝟔𝟔𝟔𝟔𝟔𝟔/𝒉𝒉
0.024 ÷ 2 = 0.012
1 − 0.012 = 0.988
0.988 → −2.26
𝑥𝑥̅ − 𝜇𝜇
= −2.26
𝜎𝜎
� �
√𝑛𝑛
𝑥𝑥̅ − 87.3
= −2.26
12
√100
12
� + 87.3
𝑥𝑥̅ = −2.26 �
√100
𝑥𝑥̅ = 84.588
As we have a 𝑝𝑝 −value we need to work backwards to get a 𝑧𝑧 −
𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
Taking the 𝑝𝑝 −value and dividing it by 2, as the 𝑝𝑝 − value covers both
ends of the distribution, but we only want to deal with the one which
is lower than the mean (that is, on the left side of the normal
distribution curve)
Taking 0.012 from 1.
Finding 0.988 in the tables on pages 36&37 of The Maths Tables
Book.
Making it −2.26 as we are dealing with a speed less than the mean.
Letting this equal our equation for finding 𝑧𝑧 − 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
Filling in the old mean, standard deviation and the size of the sample.
Multiplying across by the bottom of the fraction and adding 87.3 to
both sides.
= 𝟖𝟖𝟖𝟖. 𝟔𝟔𝟔𝟔𝟔𝟔/𝒉𝒉
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308
Question 9
a) 𝟓𝟓𝟓𝟓. 𝟒𝟒𝟒𝟒
𝑎𝑎2 = 𝑏𝑏 2 + 𝑐𝑐 2 − 2𝑏𝑏𝑏𝑏 cos 𝐴𝐴
Taking the cosine rule form page 16 of The
Maths Tables Book.
|𝑆𝑆𝑆𝑆|2 = (30)2 + (58)2 − 2(30)(58) cos(68)
Subbing in our values and plugging it all into
the calculator.
𝑎𝑎 = |𝑆𝑆𝑆𝑆|, 𝑏𝑏 = 30, 𝑐𝑐 = 58, 𝐴𝐴 = 68°
|𝑆𝑆𝑆𝑆|2 = 2960.369
|𝑆𝑆𝑆𝑆| = √2960.369
Square rooting both sides.
= 𝟓𝟓𝟓𝟓. 𝟒𝟒𝟒𝟒
b) 𝟑𝟑𝟑𝟑. 𝟕𝟕𝟕𝟕°
𝑆𝑆𝑆𝑆𝑆𝑆 𝐴𝐴 𝑆𝑆𝑆𝑆𝑆𝑆 𝐵𝐵
=
𝑏𝑏
𝑎𝑎
Taking the Sin rule from page 16 of The
Maths Tables Book.
𝐴𝐴 = |< 𝐻𝐻𝐻𝐻𝐻𝐻|, 𝑎𝑎 = 30, 𝐵𝐵 = 68, 𝑏𝑏 = 54.4
sin 𝐴𝐴 sin 68
=
54.4
30
Subbing in our values.
Multiplying across by 30
sin 68
�
sin 𝐴𝐴 = 30 �
54.4
sin 𝐴𝐴 = 0.5113
sin−1 0.5113 = 𝐴𝐴
Finding the sin inverse of 0.5113
|< 𝐻𝐻𝐻𝐻𝐻𝐻| = 𝟑𝟑𝟑𝟑. 𝟕𝟕𝟕𝟕°
Rounding to two decimal places
c)
Area of a triangle =
1
𝑎𝑎𝑎𝑎 sin 𝐶𝐶
2
𝑎𝑎 = 30, 𝑏𝑏 = 58, 𝐶𝐶 = 68
→
1
(30)(58) sin(68) = 𝟖𝟖𝟖𝟖𝟖𝟖. 𝟔𝟔𝟔𝟔𝒎𝒎𝟐𝟐
2
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Taking the formula for the area of a triangle from
page 16 of The Maths Tables Book.
Plugging it all into the calculator.
309
d) i)
Area of a triangle =
→
1
(58)(𝑟𝑟)
2
1
𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 ×⊥ ℎ𝑒𝑒𝑒𝑒𝑒𝑒ℎ𝑡𝑡
2
𝐻𝐻
= 𝟐𝟐𝟐𝟐𝟐𝟐
Taking area of triangle from page
9 of The Maths Tables Book.
𝑃𝑃
𝑟𝑟
58𝑚𝑚
𝑆𝑆
Plugging in the base and 𝑟𝑟 for the
perpendicular height.
ii)
Area of 𝑆𝑆𝑆𝑆𝑆𝑆 = Area HSP + Area PHG + Area PGS
Area HSP = 29r
Area PHG =
1
(30)𝑟𝑟 = 15𝑟𝑟
2
1
Area PGS = (54.4)𝑟𝑟 = 27.2𝑟𝑟
2
Area SGH = 29r + 15r + 27.2r = 71.2r
The triangle 𝑆𝑆𝑆𝑆𝑆𝑆 can be split into the triangles
𝐻𝐻𝐻𝐻𝐻𝐻, 𝑃𝑃𝑃𝑃𝑃𝑃 and 𝑃𝑃𝑃𝑃𝑃𝑃, each with perpendicular height 𝑟𝑟.
So, the area of SGH equals the sum of the areas of these
three triangles.
Area HSP = 29𝑟𝑟 from previous part
Plugging 𝑟𝑟 in for the perpendicular height in each case
and plugging in the length of the side as the base.
Adding the three areas.
iii) 𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑
Area SGH = 71.2𝑟𝑟
Area SGH = 806.65
71.2𝑟𝑟 = 806.65
806.65
𝑟𝑟 =
71.2
Area of SGH = 806.65 (from part c)
Letting the area of SGH = the area of SGH
Dividing across by 71.2 gives us our answer.
𝑟𝑟 = 𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑
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310
e) 𝟏𝟏𝟏𝟏. 𝟔𝟔𝟔𝟔
𝑇𝑇
𝑃𝑃
𝐻𝐻
11.3
34°
𝑘𝑘
|< 𝑃𝑃𝑃𝑃𝑃𝑃| =
|< 𝑃𝑃𝑃𝑃𝑃𝑃| =
sin 𝜃𝜃 =
58𝑚𝑚
1
|< 𝐺𝐺𝐺𝐺𝐺𝐺|
2
1
68 = 34°
2
opp
ℎ𝑦𝑦𝑦𝑦
sin 15.375 =
11.3
|𝑃𝑃𝑃𝑃|
11.3
|𝑃𝑃𝑃𝑃| =
sin 15.375
|𝑃𝑃𝑃𝑃| = 42.619
tan 𝜃𝜃 =
𝑜𝑜𝑝𝑝𝑝𝑝 |𝑇𝑇𝑇𝑇|
=
𝑎𝑎𝑎𝑎𝑎𝑎 |𝑃𝑃𝑃𝑃|
tan 14 =
|𝑇𝑇𝑇𝑇|
42.619
42.619(tan 14) = |𝑇𝑇𝑇𝑇|
|𝑇𝑇𝑇𝑇| = 10.626 = 𝟏𝟏𝟏𝟏. 𝟔𝟔𝟔𝟔
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15.375°
𝑆𝑆
𝑃𝑃
1
|< 𝑃𝑃𝑃𝑃𝑃𝑃| = | < 𝐻𝐻𝐻𝐻𝐻𝐻|
2
14°
𝑆𝑆
Each of these angles is half the large angle
from the original triangle.
1
|< 𝑃𝑃𝑃𝑃𝑃𝑃| = 30.75 = 15.375°
2
Looking at the right-angled triangle 𝑃𝑃𝑃𝑃𝑃𝑃
Multiplying across by |𝑃𝑃𝑃𝑃| and dividing by
sin 15.375
Now that we know the length of 𝑃𝑃𝑃𝑃 we can
use trig ratios to solve for |𝑇𝑇𝑇𝑇|
Note: If you solve this question by finding the
angle HPS and using the sine rule to find |𝑃𝑃𝑃𝑃|,
you get a slightly different length for |PS|
and the final answer is 10.7𝑚𝑚
311
2018 Paper 2
Question 1
a)
1
1
1
(9000) + (7000) + (3000) = 1900
10
4
20
2000 − 1900 = €𝟏𝟏𝟏𝟏𝟏𝟏
To find her expected loss we find the expected
value of finishing in each position. We then
add these values and take them away from
what she paid to enter.
Mary is expected to win €1900, which is €100
less than she paid so she makes a loss of €100
b) €𝟐𝟐𝟐𝟐𝟐𝟐
1
1
1
(9000 + 𝑥𝑥) + (7000 + 𝑥𝑥) + (3000 + 𝑥𝑥) = 2000
10
4
20
9000 + 𝑥𝑥 + 2(7000 + 𝑥𝑥) + 5(3000 + 𝑥𝑥) = 20(2000)
9000 + 𝑥𝑥 + 14000 + 2𝑥𝑥 + 15000 + 5𝑥𝑥 = 40000
If Mary expects to break even the expected
values of her finishes must add up to €2000. So,
we add 𝑥𝑥 to each prize and let it equal 2000.
Multiplying across by 20
8𝑥𝑥 = 40000 − 38000
8𝑥𝑥 = 2000
𝑥𝑥 = €𝟐𝟐𝟐𝟐𝟐𝟐
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312
Question 2
a)
𝑧𝑧1 = 0.44
b)
Looking at page 36 of the Maths Tables Book. Finding the
corresponding value for a probability of 0.67
i) Maths
Using the formula 𝑧𝑧 =
65 − 70
5
1
Maths:
=−
=−
15
15
3
English:
−
𝑥𝑥−𝜇𝜇
𝜎𝜎
to find
her position relative to the class.
2
68 − 72 4
−
=−
10
5
10
2
1
>−
5
3
∴ Mary did better in Maths relative to the class, as her score was not as far below the mean in
Maths as in English.
ii) 83%
1.04 =
𝑥𝑥 − 72
10
1.04(10) = 𝑥𝑥 − 72
10.4 + 72 = 𝑥𝑥
82.4% = 𝑥𝑥
𝟖𝟖𝟖𝟖%
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1.04 is the z-score which corresponds to the top 15%.
(85% get less than this result, so we find this by
finding 0.85 on page 36 of the Maths Tables Book.)
We sub this into the formula:𝑧𝑧 =
And solve for 𝑥𝑥
𝑥𝑥−𝜇𝜇
𝜎𝜎
313
iii) 𝟖𝟖𝟖𝟖. 𝟓𝟓%
82% = 1 standard deviation above the mean.
→
68
% got between 72 and 82
2
52 = 2 standard deviations under the mean
→
95
% got between 52 and 72
2
68 95
+
= 𝟖𝟖𝟖𝟖. 𝟓𝟓%
2
2
© Pocket Tutor 2022
82% is one standard deviation above the mean. One
standard deviation contains 68% of the population.
Half of the standard deviation is above the mean
and half is below
→
68
2
% is between the mean and 82%
52% is two standard deviations below the mean.
Two standard deviations contain 95% of the
population.
→
95
2
% is between 52% and the mean.
314
Question 3
i)
105
To end with 0 there is only one possibility for the 6th number.
So, we need to find the number of permutations of the first 5
digits. There are 10 possibilities for each digit and 5 digits →
105
ii) 300
A code with 4 digits can start in position A, B or C.
A
B
C
The two remaining digits can be filled in 102 ways, therefore the total number of codes is 102 × the
3 starting points of 2018.
102 × 3 = 𝟑𝟑𝟑𝟑𝟑𝟑
b)
𝒂𝒂 = 𝟏𝟏, 𝒃𝒃 = 𝟕𝟕, 𝒄𝒄 = 𝟏𝟏𝟏𝟏, 𝒅𝒅 = 𝟏𝟏𝟏𝟏
(𝑛𝑛 + 3)! (𝑛𝑛 + 2)!
(𝑛𝑛 + 1)! (𝑛𝑛 + 1)!
(𝑛𝑛 + 3)(𝑛𝑛 + 2)(𝑛𝑛 + 1)! (𝑛𝑛 + 2)(𝑛𝑛 + 1)!
(𝑛𝑛 + 1)! (𝑛𝑛 + 1)!
(𝑛𝑛 + 3)(𝑛𝑛 + 2)(𝑛𝑛 + 1)! (𝑛𝑛 + 2)(𝑛𝑛 + 1)!
(𝑛𝑛 + 1)! (𝑛𝑛 + 1)!
(𝑛𝑛 + 3)(𝑛𝑛 + 2)(𝑛𝑛 + 2)
(𝑛𝑛2 + 5𝑛𝑛 + 6)(𝑛𝑛 + 2)
(𝑛𝑛 + 3)! = (𝑛𝑛 + 3)(𝑛𝑛 + 2)(𝑛𝑛 + 1)!
(𝑛𝑛 + 2)! = (𝑛𝑛 + 2)(𝑛𝑛 + 1)!
The (𝑛𝑛 + 1)! ′𝑠𝑠 cancel.
Multiplying out the brackets
𝑛𝑛3 + 2𝑛𝑛2 + 5𝑛𝑛2 + 10𝑛𝑛 + 6𝑛𝑛 + 12
𝑛𝑛3 + 7𝑛𝑛2 + 16𝑛𝑛 + 12
𝒂𝒂 = 𝟏𝟏, 𝒃𝒃 = 𝟕𝟕, 𝒄𝒄 = 𝟏𝟏𝟏𝟏, 𝒅𝒅 = 𝟏𝟏𝟏𝟏
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315
Question 4
a) 𝟕𝟕𝟕𝟕°, 𝟏𝟏𝟏𝟏𝟏𝟏°, 𝟐𝟐𝟐𝟐𝟐𝟐°, 𝟐𝟐𝟐𝟐𝟐𝟐°
cos 2𝑥𝑥 = −
√3
2
√3
� = 30
2
2𝑥𝑥 = cos −1 �
2𝑥𝑥 = 150 + 360𝑛𝑛
𝑥𝑥 = 75 + 180𝑛𝑛
or
or
𝑛𝑛 = 0 → 𝑥𝑥 = 75 + 180(0)
𝑥𝑥 = 𝟕𝟕𝟕𝟕° or 𝟏𝟏𝟏𝟏𝟏𝟏°
𝑛𝑛 = 1 → 𝑥𝑥 = 75 + 180(1)
𝑥𝑥 = 𝟐𝟐𝟐𝟐𝟐𝟐° or 𝟐𝟐𝟐𝟐𝟐𝟐°
© Pocket Tutor 2022
2𝑥𝑥 = 210 + 360𝑛𝑛
𝑥𝑥 = 105 + 180𝑛𝑛
As the value of cos 𝑥𝑥 is negative we need
to find where cos is negative. Cos is
negative in the second and third
quadrants, so we take 30 from 180° and
add it to 180°.
Dividing across by 2
or
𝑥𝑥 = 105 + 180(0)
Letting 𝑛𝑛 = 0
or
𝑥𝑥 = 105 + 180(1)
Letting 𝑛𝑛 = 1
𝑛𝑛 = 2 gives angles greater than 360° so
we stop at 𝑛𝑛 = 1
316
𝒃𝒃)
𝒚𝒚�𝟒𝟒 − 𝒚𝒚𝟐𝟐
𝟐𝟐
cos 𝐴𝐴 =
cos 𝐴𝐴 =
𝑦𝑦
2
adjacent
hypotenuse
(2)2 = 𝑥𝑥 2 + 𝑦𝑦 2
4 = 𝑥𝑥 2 + 𝑦𝑦 2
4 − 𝑦𝑦 2 = 𝑥𝑥 2
�4 − 𝑦𝑦 2 = 𝑥𝑥
sin 𝐴𝐴 =
sin 𝐴𝐴 =
opposite
hypotenuse
𝑥𝑥
�4 − 𝑦𝑦 2
→ sin 𝐴𝐴 =
2
2
sin 2𝐴𝐴 = 2 sin 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴
sin 2𝐴𝐴 = 2 �
�4 − 𝑦𝑦 2 𝑦𝑦
�� �
2
2
�4 − 𝑦𝑦 2 𝑦𝑦
sin 2𝐴𝐴 = 2 �
�� �
2
2
sin 2𝐴𝐴 =
2
𝑦𝑦
If cos 𝐴𝐴 = then the side
𝑥𝑥
2
A
Y
adjacent to the angle A is y and
the hypotenuse is 2. As shown
in the diagram.
We then use Pythagoras’
theorem to get 𝑥𝑥 in terms of y.
We can now write sin 𝐴𝐴 in
terms of 𝑦𝑦
We get this line from page 14
of the Maths Tables Book.
Subbing in our values for sin 𝐴𝐴
and cos 𝐴𝐴
Multiplying out the brackets.
𝒚𝒚�𝟒𝟒 − 𝒚𝒚𝟐𝟐
𝟐𝟐
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317
Question 5
a)
2𝑥𝑥 + 3𝑦𝑦 + 1 = 0
2(−2) + 3(1) + 1 = 0
To verify a point is on a point you sub in the x
and y values and show that it equals 0
−4 + 3 + 1 = 0
0=0
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318
b) 𝑩𝑩 = (𝟔𝟔, 𝟏𝟏𝟏𝟏)
Perpendicular distance from 𝐴𝐴 to the line 𝑛𝑛 is equal to |𝐴𝐴𝐴𝐴|
Distance from 𝐴𝐴 to 𝑛𝑛: 2𝑥𝑥 + 3𝑦𝑦 − 51 = 0
|𝑎𝑎𝑥𝑥1 + 𝑏𝑏𝑦𝑦1 + 𝑐𝑐|
From page 18 of The Maths Tables Book.
√𝑎𝑎2 + 𝑏𝑏 2
𝑎𝑎 = 2, 𝑏𝑏 = 3, 𝑐𝑐 = −51, 𝑥𝑥1 = −2, 𝑦𝑦1 = 1
|(2)(−2) + (3)(1) + (−51)|
�(2)2 + (3)2
Distance from 𝐴𝐴 to 𝐵𝐵 = 4√13
|𝐴𝐴𝐴𝐴| = �(𝑥𝑥2 − 𝑥𝑥1
)2
=
52
= 4√13
√13
From page 18 of The Maths Tables Book
+ (𝑦𝑦2 − 𝑦𝑦1
2
)2
4√13 = ��𝑥𝑥2 − (−2)� + �𝑦𝑦2 − (1)�
2
2
�4√13� = (𝑥𝑥 + 2)2 + (𝑦𝑦 − 1)2
3
From the equation of 𝑛𝑛, 𝑥𝑥 = − 𝑦𝑦 +
2
51
2
2
3
51
�4√13� = ��− 𝑦𝑦 + � + 2� + (𝑦𝑦 − 1)2
2
2
2
55 2
3
208 = �− 𝑦𝑦 + � + 𝑦𝑦 2 − 2𝑦𝑦 + 1
2
2
208 =
9 2 165
3025
𝑦𝑦 −
𝑦𝑦 +
4
2
4
832 = 9𝑦𝑦 2 − 330𝑦𝑦 + 3025 + 4𝑦𝑦 2 − 8𝑦𝑦 + 4
0 = 13𝑦𝑦 2 − 338𝑦𝑦 + 2197
𝑦𝑦 2 − 26𝑦𝑦 + 169 = 0
(𝑦𝑦 − 13)(𝑦𝑦 − 13) = 0
𝑦𝑦 = 13
51
3
𝑥𝑥 = − (13) +
2
2
𝑥𝑥 = 6
𝑩𝑩 = (𝟔𝟔, 𝟏𝟏𝟏𝟏)
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2𝑥𝑥 + 3𝑦𝑦 + 51 = 0
2𝑥𝑥 = −3𝑦𝑦 − 51
51
3
𝑥𝑥 = − 𝑦𝑦 −
2
2
Subbing this in for 𝑥𝑥
Squaring the brackets
Multiplying across by 4
Dividing across by 13
Solving the quadratic
Plugging our y value back into:
3
51
𝑥𝑥 = − 𝑦𝑦 −
2
2
To find 𝑥𝑥
Note: This can also be answered by finding
the equation of the line 𝐴𝐴𝐴𝐴, which is
perpendicular to 𝑚𝑚 and then finding where
𝐴𝐴𝐴𝐴 and 𝑛𝑛 meet using simultaneous equations
319
𝟓𝟓 𝟐𝟐
c) (𝒙𝒙 + 𝟏𝟏)𝟐𝟐 + �𝒚𝒚 − � = �
𝟐𝟐
√𝟏𝟏𝟏𝟏
�
𝟐𝟐
𝟐𝟐
The point where circles 𝑠𝑠 and 𝑡𝑡 meet internally divides the line AB in the ratio 1: 3
𝑏𝑏𝑥𝑥1 + 𝑎𝑎𝑥𝑥2 𝑏𝑏𝑦𝑦1 + 𝑎𝑎𝑦𝑦2
,
𝑏𝑏 + 𝑎𝑎
𝑏𝑏 + 𝑎𝑎
𝑏𝑏 = 3, 𝑎𝑎 = 1, 𝑥𝑥1 = −2, 𝑦𝑦1 = 1, 𝑥𝑥2 = 6, 𝑦𝑦2 = 13
All of the relevant formulae are on pg. 18
of the Maths Tables Book
3(−2) + 1(6) 3(1) + 1(13)
,
3+1
3+1
(0,4)
The centre of 𝑠𝑠 is the midpoint between this point and 𝐴𝐴
�
�
𝑥𝑥1 + 𝑥𝑥2 𝑦𝑦1 + 𝑦𝑦2
�
,
2
2
−2 + 0 1 + 4
�
,
2
2
5
Centre = �−1, �
2
Radius = distance from 𝐴𝐴 to the centre
2
��−2 − (−1)� + �1 − (2.5)�
=
2
√13
2
Equation of a circle pg. 19 of the Maths
Tables Book
(𝑥𝑥 − ℎ)2 + (𝑦𝑦 − 𝑘𝑘)2 = 𝑟𝑟 2
2
5 2
√13
�
�𝑥𝑥 − (−1)� + �𝑦𝑦 − � = �
2
2
2
𝟐𝟐
𝟓𝟓 𝟐𝟐
√𝟏𝟏𝟏𝟏
(𝒙𝒙 + 𝟏𝟏) + �𝒚𝒚 − � = �
�
𝟐𝟐
𝟐𝟐
𝟐𝟐
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320
Question 6
a) Proof
Copy from chapter or link
b) 𝟏𝟏𝟏𝟏𝟏𝟏
|𝑋𝑋𝑋𝑋|2 = (4)2 + (3)2
|𝑋𝑋𝑋𝑋|2
= 16 + 9
First use Pythagoras’ theorem to find
side ZC.
|𝑋𝑋𝑋𝑋| = 5
∴ |𝑍𝑍𝑍𝑍| = 5
|𝑋𝑋𝑋𝑋| = 2 × |𝐴𝐴𝐴𝐴|
→ |𝑋𝑋𝑋𝑋| = 2 × 4 = 8
|𝐴𝐴𝐴𝐴| = 8 + 4 = 12
|𝐴𝐴𝐴𝐴| |𝐵𝐵𝐵𝐵|
=
|𝐴𝐴𝐴𝐴| |𝑋𝑋𝑋𝑋|
12 |𝐵𝐵𝐵𝐵|
=
4
5
15 = |𝐵𝐵𝐵𝐵|
15 − |𝑍𝑍𝑍𝑍| = |𝐵𝐵𝐵𝐵|
15 − 5 = |𝐵𝐵𝐵𝐵|
|𝐵𝐵𝐵𝐵| = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
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As we are told in the question that they
are in the ratio 2: 1
So |BX| must be twice |AX|
As the triangles 𝐴𝐴𝐴𝐴𝐴𝐴 and 𝐴𝐴𝐴𝐴𝐴𝐴 are
similar because all of their angles are
equal, hence we can use the formula.
|𝐴𝐴𝐴𝐴| |𝐵𝐵𝐵𝐵|
=
|𝐴𝐴𝐴𝐴| |𝑋𝑋𝑋𝑋|
Enter in the side lengths that you know
and solve for BC.
We know the length XY = 5cm
XYCZ is a parallelogram so XY = ZC
321
Question 7
a) 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝒎𝒎𝟑𝟑
4
Volume of a Sphere = 𝜋𝜋𝑟𝑟 3
3
From page 10 of the Maths Tables
Book
4
𝜋𝜋(3)3 = 113.1
3
Multiplying the previous volume by
1.75
Volume of B= 113.1 × 1.75 = 197.925
Volume of C= 197.925 × 1.75 = 346.36875
Adding All the volumes
Volume of D= 346.36875 × 1.75 = 606.1453125
Volume of E= 606.1453125 × 1.75 = 1060.754296875
Total Volume
= 2324.293359375 = 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝒎𝒎𝟑𝟑
(Note: This question can also be
answered by constructing a
geometric series for the volumes of
the first 5 spheres.)
b) i) 107.3cm
Surface Area of a sphere = 4𝜋𝜋𝑟𝑟 2
4𝜋𝜋𝑟𝑟 2 = 503
𝑟𝑟 2 =
503
4𝜋𝜋
503
= 6.3267
4𝜋𝜋
𝑟𝑟 = �
Bar E = 120 − 2(6.3267) = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟑𝟑cm
© Pocket Tutor 2022
From page 10 of the Maths Tables
Book
Dividing by 4𝜋𝜋
Square rooting both sides.
Taking the diameter (2 by the radius) of
the sphere from the total height
322
ii)
𝑨𝑨 = 𝟕𝟕𝟕𝟕. 𝟑𝟑𝟑𝟑𝟑𝟑, 𝑩𝑩 = 𝟖𝟖𝟖𝟖. 𝟑𝟑𝒄𝒄𝒄𝒄, 𝑪𝑪 = 𝟖𝟖𝟖𝟖. 𝟑𝟑𝟑𝟑𝟑𝟑, 𝑫𝑫 = 𝟗𝟗𝟗𝟗. 𝟑𝟑𝟑𝟑𝟑𝟑, 𝑬𝑬 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑𝟑𝟑
Volume of a cylinder = 𝜋𝜋𝑟𝑟 2 ℎ
𝜋𝜋(1)2 ℎ = 71.3𝜋𝜋
ℎ = 71.3𝑐𝑐𝑐𝑐
𝑇𝑇𝑛𝑛 = 𝑎𝑎 + (𝑛𝑛 − 1)𝑑𝑑
𝑇𝑇5 = 71.3 + (5 − 1)𝑑𝑑 = 107.3
71.3 + 4𝑑𝑑 = 107.3
4𝑑𝑑 = 36
𝑑𝑑 = 9
𝑇𝑇2 = 71.3 + (2 − 1)(9) = 80.3
From page 10 of the Maths Tables Book
Letting it equal to the volume given for A and
subbing in the radius of A.
Dividing across by 𝜋𝜋
Constructing an expression for the arithmetic
sequence, with 𝑎𝑎 = the height of rod A
Subbing in 5 for 𝑛𝑛 and letting it equal the
height of rod E, in order to find 𝑑𝑑
Using the arithmetic sequence to find the
heights of rods B to D.
𝑇𝑇3 = 71.3 + (3 − 1)(9) = 89.3
𝑇𝑇4 = 71.3 + (4 − 1)(9) = 98.3
𝑨𝑨 = 𝟕𝟕𝟕𝟕. 𝟑𝟑𝟑𝟑𝟑𝟑, 𝑩𝑩 = 𝟖𝟖𝟖𝟖. 𝟑𝟑𝟑𝟑𝟑𝟑, 𝑪𝑪 = 𝟖𝟖𝟖𝟖. 𝟑𝟑𝟑𝟑𝟑𝟑, 𝑫𝑫 = 𝟗𝟗𝟗𝟗. 𝟑𝟑𝟑𝟑𝟑𝟑, 𝑬𝑬 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑𝟑𝟑
c)
Diameter of each bar = 2(1) = 2𝑐𝑐𝑐𝑐
Total Distance between the bars 150 − 20 − 20 − 9(2) = 92𝑐𝑐𝑐𝑐
92
= 𝟏𝟏𝟏𝟏. 𝟓𝟓𝟓𝟓𝟓𝟓
8
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Total distance between the bars =
length between the walls – the two
20cm gaps − the diameters of the 9
rods.
Dividing this by 8 as there are 8 gaps.
323
d) 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
|𝑋𝑋𝑋𝑋| = radius of rod A + radius of rod B + distance between rods
Z
|𝑋𝑋𝑋𝑋| = 1 + 1 + 11.5 = 13.5𝑐𝑐𝑐𝑐
|𝑇𝑇𝑇𝑇| = |𝑋𝑋𝑋𝑋| = 13.5
T
W
X
Y
|𝑊𝑊𝑊𝑊| = (Height of rod B + radius of sphere B) − (Height of Rod A + Radius of Sphere A)
Radius of 𝐵𝐵:
4
Volume = 197.25 = 𝜋𝜋𝑟𝑟 3
197.25
4𝜋𝜋 = 𝑟𝑟 3
3
3
3
√47.08997 = 𝑟𝑟 = 3.611
|𝑊𝑊𝑊𝑊| = (80.3 + 3.611) − (71.3 + 3) = 9.611
|𝑇𝑇𝑇𝑇|2 = |𝑇𝑇𝑇𝑇|2 + |𝑊𝑊𝑊𝑊|2
|𝑇𝑇𝑇𝑇|2 = (13.5)2 + (9.611)2
80.3 = height of rod B (from part b ii))
71.3 = height of rod A
3 = radius of sphere A
Using Pythagoras’ theorem to find |𝑇𝑇𝑇𝑇|
|𝑇𝑇𝑇𝑇|2 = 274.621
|𝑇𝑇𝑇𝑇| = 16.57𝑐𝑐𝑐𝑐 = 16.6𝑐𝑐𝑐𝑐
Rod = |𝑇𝑇𝑇𝑇| − Radius of sphere A − Radius of sphere B
𝑅𝑅𝑅𝑅𝑅𝑅 = 16.6 − 3 − 3.62 = 9.95 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏
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324
Question 8
a) i) 𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕
𝑥𝑥 − 𝜇𝜇
𝜎𝜎
√𝑛𝑛
From page 35 of The Maths Tables Book.
4.6 − 4.64
= −1.054
0.12
√10
→ 0.8531
4.7 − 4.64
= 1.58
0.12
√10
→ 0.9429
(1 − 0.8531) = 0.1469
0.9429 − 0.1469 = 𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕
Getting the z-score from page 36 of The Maths Tables Book.
This is the probability of the weight being less than 4.7g
Getting the z-score from page 36 of The Maths Tables Book.
This is the probability of the weight being more than 4.6g
Taking 0.8531 from 1 gives us the probability of the weight
being more than 4.7g
Taking this number from 09429 gives the probability of the
weight being between 4.6 and 4.7g
ii) 𝟎𝟎. 𝟕𝟕𝟕𝟕 < 𝒑𝒑 < 𝟎𝟎. 𝟖𝟖𝟖𝟖
324
= 0.81
400
Finding the proportion of people
that like the bar.
𝑝𝑝(1 − 𝑝𝑝)
𝑛𝑛
0.81 ± 1.96�
0.81(1 − 0.81)
400
0.81 ± 1.96�
0.81 ± 0.038
𝟎𝟎. 𝟕𝟕𝟕𝟕 < 𝒑𝒑 < 𝟎𝟎. 𝟖𝟖𝟖𝟖
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A 95% confidence interval is the
proportion ±1.96 times the
formula on page 34 of the Maths
Tables Book
Adding and Subtracting 0.0038
from 0.81.
Rounding to 2 decimal places
325
b) i)
Statement
1. When forming confidence
intervals (for fixed n and pˆ), an
increased confidence level implies a
wider interval.
Always
Sometimes
Never
True
True
True

2. As the value of p ̂increases (for
fixed n), the estimated standard
error of the population proportion
increases.
3. As the value of 𝑝𝑝ˆ(1 − 𝑝𝑝ˆ)
increases (for fixed n), the estimated
standard error of the population
proportion increases.


4. As n, the number of people
sampled, increases (for fixed 𝑝𝑝ˆ) the
estimated standard error of the
population proportion increases.
𝒊𝒊𝒊𝒊)

𝟏𝟏
𝟒𝟒
𝑓𝑓(𝑝𝑝) = 𝑝𝑝(1 − 𝑝𝑝) = 𝑝𝑝 − 𝑝𝑝2
Multiplying out the expression to make it
easier to differentiate.
1 − 2𝑝𝑝 = 0
Differentiating the expression and letting
it equal 0 to find the maximum.
𝑓𝑓 ′ (𝑝𝑝) = 1 − 2𝑝𝑝
1 = 2𝑝𝑝
𝑝𝑝 =
1
2
1
1
1
𝑓𝑓 � � = �1 − �
2
2
2
=
𝟏𝟏
𝟒𝟒
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Solving for 𝑝𝑝.
Plugging the value we found for 𝑝𝑝 back
into the expression to find the maximum
value of the expression.
326
iii) 𝟑𝟑. 𝟒𝟒𝟒𝟒%
1.96�
𝑝𝑝(1 − 𝑝𝑝)
𝑛𝑛
(0.25)
= 0.346 = 𝟑𝟑. 𝟒𝟒𝟒𝟒%
800
1.96�
1
Subbing the maximum value � � in for 𝑝𝑝(1 − 𝑝𝑝)
4
c) €𝟒𝟒𝟒𝟒𝟒𝟒, 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟒𝟒𝟒𝟒
The initial sum required is equal to the sum of the present values of the payments:
Present Value =
𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉
(1 + 𝑖𝑖)𝑡𝑡
From page 30 of the Maths Tables Book
Payment 1: 20,000, Payment 2: 20,000(1.01), Payment 3: 20,000(1.01)
Present Value of Payment 2:
20,000(1.01)
(1 + 0.024)1
Present Value of Payment 3:
20,000(1.01)2
(1 + 0.024)2
Sum of the Present Values:
20,000 +
𝑆𝑆𝑛𝑛 =
20,000(1.01)25
20,000(1.01)1 20,000(1.01)2 20,000(1.01)3
+
+
+
⋯
+
(1.024)1
(1.024)2
(1.024)3
(1.024)25
𝑎𝑎(1 − 𝑟𝑟 𝑛𝑛 )
1 − 𝑟𝑟
𝑎𝑎 = 20,000, 𝑟𝑟 =
𝑆𝑆𝑛𝑛 =
1.01
, 𝑛𝑛 = 26
1.024
Making an expression for the sum of the
present values using the formula from
page 22 of the Maths Tables Book
1.01 26
� )
1.024
= €𝟒𝟒𝟒𝟒𝟒𝟒, 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟒𝟒𝟒𝟒
1.01
�
1−�
1.024
20,000(1 − �
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327
Question 9
a) 𝟓𝟓𝟓𝟓°
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆
+
𝑏𝑏
𝑎𝑎
𝑆𝑆𝑆𝑆𝑆𝑆(15) 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆
=
10
30
𝑥𝑥
𝑆𝑆𝑆𝑆𝑆𝑆(15)
30 �
� = sin 𝑥𝑥
10
𝑆𝑆𝑆𝑆𝑆𝑆(15)
� = 𝑥𝑥
10
𝑠𝑠𝑠𝑠𝑛𝑛−1 30 �
𝑥𝑥 = 50.94 = 𝟓𝟓𝟓𝟓°
b) i)
Period: 𝟐𝟐𝟐𝟐,
The period is 2𝜋𝜋 as this is equal to 360° and D moves in a circle.
range: [30,10]
The max length is 30 when D is in the position in Diagram 1 at the start of the question.
When D is on the far side of the circle the length is only 10 as the diameter is 20.
ii)
𝛼𝛼
𝑓𝑓(𝛼𝛼)
0°
90°
180°
270°
360°
30
18.28
10
18.28
30
(cm)
D
The triangle on the right shows 𝛼𝛼 = 90
|𝐶𝐶𝐶𝐶|2 = (30)2 − (10)2
|𝐶𝐶𝐶𝐶|2
30cm
Pythagoras’ theorem
= 800
|𝐶𝐶𝐶𝐶| = √800 = 20√2
C
10cm
90° O
|𝐶𝐶𝐶𝐶| = |𝐶𝐶𝐶𝐶| − The radius of the circle
|𝐶𝐶𝐶𝐶| = 20√2 − 10 = 18.28𝑐𝑐𝑐𝑐
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328
The distance at 270° is the same as at 90° as it is simply the same triangle upside down.
iii)
iv)
Diagram 2
The closer the angle 𝐶𝐶𝐶𝐶𝐶𝐶 is to 90° the more the connecting row will be moved by a change in the
angle
Or
The curve is steeper near 𝛼𝛼 = 70/80 than it is at the angles in Diagrams 1 & 2.
c) 𝟕𝟕𝟕𝟕𝟕𝟕
As this is also a radius
|𝑋𝑋𝑋𝑋| = 𝑟𝑟
→ |𝐴𝐴𝐴𝐴| = 31 + 𝑟𝑟
𝑎𝑎2 = 𝑏𝑏 2 + 𝑐𝑐 2 − 2𝑏𝑏𝑏𝑏 cos 𝐴𝐴
Cosine rule from page 16 of
the Maths Tables Book
|𝐵𝐵𝐵𝐵|2 = |𝐴𝐴𝐴𝐴|2 + |𝐴𝐴𝐴𝐴|2 − 2|𝐴𝐴𝐴𝐴||𝐴𝐴𝐴𝐴| cos 𝐴𝐴
2
2
𝑟𝑟 = (31 + 𝑟𝑟) +
(36)2
− 2(31 + 𝑟𝑟)(36) cos(10)
𝑟𝑟
𝑟𝑟 2 = 961 + 62𝑟𝑟 + 𝑟𝑟 2 + 1296 − (62 + 2𝑟𝑟)(36 cos 10)
𝑟𝑟 2 = 𝑟𝑟 2 + 62𝑟𝑟 + 2257 − (2232 cos 10 + 72 cos(10)𝑟𝑟)
𝑟𝑟 2 − 𝑟𝑟 2 = 62𝑟𝑟 + 58.9090 − 70.906𝑟𝑟
0 = 58.9090 − 8.906𝑟𝑟
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329
8.906𝑟𝑟 = 58.9090
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𝑟𝑟 = 𝟕𝟕𝟕𝟕𝟕𝟕
330
2017 Paper 2
Question 1
a)
256
78125
256
4 1 4 1 4 1 4
× × × × × × =
5 5 5 5 5 5 5 78125
Multiplying the probability of Ciara
answering/ not answering the phone on each
of the 7 days
b)
256
15625
1280
6 1 3 4 3 1
� �� � � � × =
5 78125
3 5
5
=
256
15625
Using Bernoulli trial to find the probability of
Ciara answering 3 times in the first 6 days.
Multiplying this by the probability of Ciara
answering on the 7th day.
c)
4 𝑛𝑛
1−� �
5
The probability of Ciara answering at least once, is = to
1 minus the probability of her not answering at all.
d)
4 𝑛𝑛
1 − � � = 0.99
5
4 𝑛𝑛
− � � = −0.01
5
Letting the expression for Ciara answering
at least once = 0.99
Taking 1 from both sides.
4 𝑛𝑛
� � = 0.01
5
Multiplying across by −1
𝑛𝑛 = 20.6377
𝑎𝑎 𝑥𝑥 = 𝑦𝑦 ↔ log 𝑎𝑎 𝑦𝑦 = 𝑥𝑥
log 4 (0.01) = 𝑛𝑛
5
𝑛𝑛 = 21
© Pocket Tutor 2022
Using the law of logs on page 21 of The
Maths Tables Book:
331
Question 2
a)
= −.0957
Using the calculator
b)
Plotting each point and drawing a line of best fit.
c)
As the speed of the car increases the number of kilometres it can travel on one litre of fuel
decreases.
d) i)
Time =
Mary:
Jane:
Distance
Speed
260
= 2.7083 hours
96
260
= 2.3214 hours
112
2.7083 − 2.3214 = 0.3869 hours
0.3869 × 60 = 23.214 minutes
= 23 minutes
© Pocket Tutor 2022
332
ii)
€6.98
Mary:
Jane:
260
= 23.636
11
260
= 28.889
9
Finding the number of litres each
used by dividing the number of
kilometres they drove by the litres
used per litre of each driver.
28.889 − 23.636 = 5.253 litres
Finding the difference in the litres
used.
= €6.98
Multiplying this by the cost per litre.
5.253 × 132.9 = 698.12cents
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333
Question 3
a)
(1, −1)
G divides the line |𝐴𝐴𝐴𝐴| internally in the ratio 2: 1
�
𝑏𝑏𝑥𝑥1 + 𝑎𝑎𝑥𝑥2 𝑏𝑏𝑦𝑦1 + 𝑎𝑎𝑦𝑦2
�
,
𝑏𝑏 + 𝑎𝑎
𝑏𝑏 + 𝑎𝑎
𝑎𝑎 = 2, 𝑏𝑏 = 1, 𝑥𝑥1 = 0, 𝑦𝑦1 = 6, 𝑥𝑥2 = 𝑥𝑥, 𝑦𝑦2 = 𝑦𝑦
�
�
1(0) + 2(𝑥𝑥) 1(6) = 2(𝑦𝑦)
2 4
,
�=� , �
1+2
1+2
3 3
2 6 + 2𝑦𝑦 4
2𝑥𝑥
�= ,
=
3
3
3
3
2𝑥𝑥 = 2, 6 + 2𝑦𝑦 = 4
2
𝑥𝑥 = , 2𝑦𝑦 = −2
2
Taking the equation for a point dividing
a line in a given ratio from page 18 of
the Maths Tables Book
Subbing in our values
Letting the x-coordinate equal the xcoordinate and the y equal the y/
Solving for 𝑥𝑥 and 𝑦𝑦
𝑥𝑥 = 1, 𝑦𝑦 = −1
(1, −1)
b)
(−2, −4)
Centroid =
𝑥𝑥1 + 𝑥𝑥2 + 𝑥𝑥3 𝑦𝑦1 + 𝑦𝑦2 + 𝑦𝑦3 2 4
,
= ,
3 3
3
3
0 + 4 + 𝑥𝑥 2 6 + 2 + 𝑦𝑦 4
= ,
=
3
3
3
3
4 + 𝑥𝑥 = 2,
8 + 𝑦𝑦 = 4
𝑥𝑥 = −2, 𝑦𝑦 = −4
(−2, −4)
© Pocket Tutor 2022
This is the equation for a centroid.
Letting the equation equal the given
coordinates for the centroid.
Solving for 𝑥𝑥 and 𝑦𝑦, the coordinates of
𝐵𝐵
334
c)
The orthocentre is where the perpendiculars from the vertices (corners) to the opposite sides meet.
Slope of side 𝐴𝐴𝐴𝐴:
6 − (−4) 10
=
=5
2
0 − (−2)
Slope of line perpendicular = −
Equation of line:
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
1
𝑦𝑦 − (2) = − (𝑥𝑥 − (4))
5
5𝑦𝑦 + 10 = −1(𝑥𝑥 − 4)
The orthocentre is found at the intersection of
the perpendicular lines from each vertex to the
opposite side.
1
5
We find the slope of the side opposite the vertex
(4,2)
We then find the perpendicular slope.
We then find the equation of a line through the
vertex with this slope.
Plugging in the coordinates of the opposite
vertex (4,2)
5𝑦𝑦 + 10 = −𝑥𝑥 + 4
𝑥𝑥 + 5𝑦𝑦 − 14 = 0
Slope of side AC
We repeat to find another line.
0 − 2 −2
=
= −1
2
6−4
Slope of line perpendicular = 1
Equation of line:
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
Plugging in the coordinates of the opposite
vertex (−2 − 4)
𝑦𝑦 + 4 = 𝑥𝑥 + 2
This is our equation of a line through a second
vertex
𝑦𝑦 − (−4) = 1�𝑥𝑥 − (−2)�
𝑥𝑥 − 𝑦𝑦 − 2 = 0
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We now use simultaneous equations to see
where they meet.
335
Simultaneous equations :
𝑥𝑥 + 5𝑦𝑦 − 14 = 0
𝑥𝑥 − 𝑦𝑦 − 2 = 0 × −1 = −𝑥𝑥 + 𝑦𝑦 + 2 = 0
Multiplying 𝑥𝑥 − 𝑦𝑦 − 2 = 0 by −1
𝑥𝑥 + 5𝑦𝑦 − 14 = 0
Adding the lines
−𝑥𝑥 + 𝑦𝑦 + 2 = 0
6𝑦𝑦 − 12 = 0
6𝑦𝑦 = 12
𝑦𝑦 = 2
𝑥𝑥 − (2) − 2 = 0
𝑥𝑥 − 4 = 0
Subbing our y value into the equation of a line to
find x
𝑥𝑥 = 4
Orthocentre:
(4,2 )
= 𝐶𝐶
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336
Question 4
a)
𝑥𝑥 2 + 𝑦𝑦 2 − 6.5𝑥𝑥 − 12𝑦𝑦 = 0
𝑥𝑥 2 + 𝑦𝑦 2 + 2𝑔𝑔𝑔𝑔 + 2𝑓𝑓𝑓𝑓 + 𝑐𝑐 = 0
(0,0) = (0)2 + (0)2 + 2𝑔𝑔(0) + 2𝑓𝑓(0) + 𝑐𝑐 = 0
𝑐𝑐 = 0
(6.5,0) = (6.5)2 + (0)2 + 2𝑔𝑔(6.5) + 2𝑓𝑓(0) = 0
42.25 + 13𝑔𝑔 = 0
13𝑔𝑔 = −42.25
−42.25
= −3.25
𝑔𝑔 =
13
(10,7) = (10)2 + (7)2 + 2(−3.25)(10) + 2𝑓𝑓(7) = 0
100 + 49 − 65 + 14𝑓𝑓 = 0
84 + 14𝑓𝑓 = 0
14𝑓𝑓 = −84
𝑓𝑓 = −
84
= −6
14
𝑥𝑥 2 + 𝑦𝑦 2 + 2(−3.25)𝑥𝑥 + 2(−6)𝑦𝑦 = 0
The general equation of a circle from
page 19 of the Maths Tables Book
Subbing in the first coordinates for 𝑥𝑥
and 𝑦𝑦
Subbing in the second coordinates for 𝑥𝑥
and 𝑦𝑦
Solving for 𝑔𝑔
Subbing in the third coordinates for 𝑥𝑥
and 𝑦𝑦
And subbing in −3.25 for 𝑔𝑔
Solving for 𝑓𝑓
Subbing our values for 𝑔𝑔, 𝑓𝑓 and c (0) into
the general equation for a circle.
𝑥𝑥 2 + 𝑦𝑦 2 − 6.5𝑥𝑥 − 12𝑦𝑦 = 0
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337
b)
28.44°
To find an angle in a triangle when given three points we use the cosine rule:
2
2
2
𝑎𝑎 = 𝑏𝑏 + 𝑐𝑐 − 2𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏
As we are looking for the Angle C, we rewrite:
Finding the lengths of each side
using the formula on page 18 of
the Maths Tables Book. To find
𝑎𝑎 we sub in points 𝐵𝐵 and 𝐶𝐶 and
so on.
𝑐𝑐 2 = 𝑎𝑎2 + 𝑏𝑏 2 − 2𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
Now we find the lengths of the sides: A, B and C
𝑎𝑎 = �(10 − 6.5)2 + (7 − 0)2 = √61.25
𝑏𝑏 = �(10 −
𝑐𝑐 = �(6.5 −
0)2
0)2
+ (7 −
+ (0 +
0)2
0)2
= √149
𝐴𝐴(0,0)
2
𝐶𝐶
𝑏𝑏
= 6.5
2
Taking the cosine rule from
page 16 of the Maths Tables
Book.
𝐶𝐶(10,7)
𝑎𝑎
𝐵𝐵(6.5,0)
𝑐𝑐
(6.5)2 = �√61.25� + �√149� − 2�√61.25��√149� cos 𝐶𝐶
Subbing our lengths into the cosine rule.
−2�√61.25��√149�
Subtracting by �√61.25� + �√149�
42.25 − 61.25 − 149
cos −1
= cos 𝐶𝐶
42.25 − 61.25 − 149
−2�√61.25��√149�
© Pocket Tutor 2022
= 28.44°
2
2
And dividing by −2�√61.25��√149�
Finding the cos inverse of this gives us
our angle.
338
Question 5
a)
|< 𝐴𝐴𝐴𝐴𝐴𝐴| = |< 𝐴𝐴𝐴𝐴𝐴𝐴|
|< 𝐹𝐹𝐹𝐹𝐹𝐹| + |< 𝐸𝐸𝐸𝐸𝐸𝐸| = 90
|< 𝐸𝐸𝐸𝐸𝐸𝐸| + |< 𝐴𝐴𝐴𝐴𝐴𝐴| = 90
∴ |< 𝐹𝐹𝐹𝐹𝐹𝐹| = |< 𝐴𝐴𝐴𝐴𝐴𝐴|
Both are 90° as |< 𝐴𝐴𝐴𝐴𝐴𝐴| = 90 and they are
on a line
Add up to a right angle as it’s in a rectangle
The remaining angles in a triangle must add
up to 90 so that the 3 angles add up to 180.
As both of them make 90 when added to
|< 𝐸𝐸𝐸𝐸𝐸𝐸|
∴ The two triangles are equiangular, and therefore similar.
b)
31.2cm
|𝐴𝐴𝐴𝐴|2 = (12)2 + (5)2
|𝐴𝐴𝐴𝐴|2 = 169
|𝐴𝐴𝐴𝐴| = 13
|𝐴𝐴𝐴𝐴| 12
=
5
13
|𝐴𝐴𝐴𝐴| = 13 ×
First, we use Pythagoras’ theorem to find |𝐴𝐴𝐴𝐴|
Then as the triangles are similar, we can find
|𝐴𝐴𝐴𝐴|
The matching sides in similar triangles are in
proportion, so we come up with the equation
|𝐴𝐴𝐴𝐴| 12
=
13
5
12
= 31.2𝑐𝑐𝑐𝑐
5
c)
36cm
27 + 12 |𝐴𝐴𝐴𝐴|
=
12
13
39
× 12 = |𝐴𝐴𝐴𝐴|
13
|𝐴𝐴𝐴𝐴| = 36𝑐𝑐𝑐𝑐
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|𝐴𝐴𝐴𝐴| = 27 + 12
Again, we use the rule that matching
sides in similar triangles are in
proportion
Putting hypotenuse over hypotenuse
and |𝐴𝐴𝐴𝐴| over its corresponding side
339
d)
680.4𝑐𝑐𝑚𝑚2
Area GCDE = Area ABCD − Area ADE − Area ABG
Area ABCD = |𝐴𝐴𝐴𝐴| × |𝐴𝐴𝐴𝐴| = 36 × 31.2 = 1123.2𝑐𝑐𝑚𝑚2
Area of a rectangle = length × width
|𝐷𝐷𝐷𝐷|2 = 829.44
perpendicular height.
Area of triangle ADE =
|𝐷𝐷𝐷𝐷|2
=
(31.2)2
−
(12)2
|𝐷𝐷𝐷𝐷| = 28.8
1
(12)(28.8) = 172.8𝑐𝑐𝑚𝑚2
2
Area of triangle ABG = |𝐵𝐵𝐵𝐵|2 = (39)2 − (36)2
|𝐵𝐵𝐵𝐵|2
= 225
|𝐵𝐵𝐵𝐵| = 15
1
Area of a triangle = base ×
2
First, we need to find |𝐷𝐷𝐷𝐷| using
Pythagoras’ theorem.
Following the same method for
triangle 𝐴𝐴𝐴𝐴𝐴𝐴
1
(15)(36) = 270𝑐𝑐𝑚𝑚2
2
Area GCDE = 1123.2 − 172.8 − 270 = 680.4cm2
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340
Question 6
a)
H
52𝑘𝑘𝑘𝑘
As Jack is looking at a point on the horizon.
Angle 𝐴𝐴𝐴𝐴𝐴𝐴 = 90°
|𝐴𝐴𝐴𝐴| = 6371 + .214 = 6371.214
|𝐽𝐽𝐽𝐽|2 = |𝐴𝐴𝐴𝐴|2 − |𝐴𝐴𝐴𝐴|2
|𝐽𝐽𝐽𝐽|2 = (6371.214)2 − (6371)2
As jack is 214m above the earth’s surface
Pythagoras’ Theorem
6371
6371 + .214
|𝐽𝐽𝐽𝐽|2 = 40592367.83 − 40589641
|𝐽𝐽𝐽𝐽|2 = 2726.834
|𝐽𝐽𝐽𝐽| = 52.21 = 52𝑘𝑘𝑘𝑘
𝐴𝐴
b)
24091𝑘𝑘𝑘𝑘
|𝐴𝐴𝐴𝐴| = The radius =6371
Angle 𝐵𝐵𝐵𝐵𝐵𝐵 = 90 − 53 = 37°
Opposite
sin 𝜃𝜃 =
Hypotenuse
sin(37) =
𝐵𝐵
𝐵𝐵𝐵𝐵
6371
6371 sin(37) = 𝐵𝐵𝐵𝐵
= 3834.1635
= Radius of s1
𝑙𝑙 = 2𝜋𝜋𝜋𝜋
The circumference of a circle, from page 8 of the Maths Tables Book.
𝑙𝑙 = 2𝜋𝜋(3834.1635) = 24091𝑘𝑘𝑘𝑘
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341
Question 7
a)
Volume of a sphere =
4 3
𝜋𝜋𝑟𝑟
3
From page 10 of The Maths Tables Book.
Volume of a cylinder = 𝜋𝜋𝑟𝑟 2 ℎ
1
Volume of a cone = 𝜋𝜋𝑟𝑟 2 ℎ
3
ℎ = 2𝑅𝑅
Empty space in cylinder =
1
𝜋𝜋𝑅𝑅 2 (2𝑅𝑅) − 2( 𝜋𝜋𝑅𝑅2 (𝑅𝑅))
3
2
2𝜋𝜋𝑅𝑅3 − ( 𝜋𝜋 𝑅𝑅3 )
3
=
The empty space in the cylinder is equal to the
volume of the cylinder minus the volume of the 2
cones.
4 3
𝜋𝜋𝑅𝑅
3
b) i)
6√3cm
The centre to 𝐴𝐴 = 12 − 6 = 6
|𝐴𝐴𝐴𝐴|2 = (12)2 − (6)2 = 144 − 36
|𝐴𝐴𝐴𝐴|2 = 108
|𝐴𝐴𝐴𝐴| = 6√3cm
ii)
6cm
6
ℎ1
=
ℎ2 12
𝑟𝑟
6
=
12 12
𝑟𝑟 =
6
× 12
12
𝑟𝑟 = 6cm
The centre of the circle to A to B forms a right angled triangle. The side from the centre to A has
length 12.
We can find |𝐴𝐴𝐴𝐴| using Pythagoras’ theorem.
This is based on the similar triangles rule
that matching sides in two similar
triangles are in proportion to each other.
Hence we can come up with the equation
6
ℎ1
=
ℎ2 12
And then
6
12
=
𝑟𝑟
12
and solve for r
iii)
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342
108𝜋𝜋𝜋𝜋𝑚𝑚2
Surface area of water in sphere = 𝜋𝜋𝑟𝑟 2
2
→ 𝜋𝜋�6√3� = 108𝜋𝜋𝜋𝜋𝑚𝑚2
Surface area of water in cylinder = 𝜋𝜋𝑟𝑟 2 − 𝜋𝜋|𝐶𝐶𝐶𝐶|2
𝜋𝜋(12)2 − 𝜋𝜋(6)2
144𝜋𝜋 − 36𝜋𝜋 = 108𝜋𝜋𝜋𝜋𝑚𝑚2
The surface area of the water in the sphere
is the area of a circle.
The surface area of water in the cylinder is
the area of the circle with radius 12 minus
the area of the circle inside the cone with
radius |𝐶𝐶𝐶𝐶|
c)
360𝜋𝜋 𝑐𝑐𝑚𝑚2
𝑉𝑉 =
1 2
𝜋𝜋𝑟𝑟 ℎ
3
1
1
𝜋𝜋(12)2 (12) − 𝜋𝜋(6)2 (6) = 504𝜋𝜋
3
3
Volume of water in cylinder:
Volume of water in the cylinder equals The
volume of the cylinder up to 6cm minus the
volume of the cone up to 6cm.
The volume of the cone to 6cm = Volume
of the cone – The volume of the smaller
cone from 6cm to the top.
𝜋𝜋𝑟𝑟 2 ℎ
→ 𝜋𝜋(12)2 (6) = 864𝜋𝜋
864𝜋𝜋 − 504𝜋𝜋 = 360𝜋𝜋 𝑐𝑐𝑚𝑚2
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343
Question 8
a)
i)
91.15%
𝑥𝑥 − 𝜇𝜇
𝜎𝜎
50 − 63.5
= −1.35
10
→ 0.9115
= 91.15%
Formula from page 35 of the Maths Tabkes
Book
As we are dealing with a population not a
sample, there is no √𝑛𝑛
ii)
85.2 𝑘𝑘𝑘𝑘
Finding the z-score which corresponds to 1.5%
on page 36 of the Maths Tables Book
1 − 0.015 = 0.985
Subbing it in for 𝑧𝑧 in the equation
1.5% = 0.015
→ 2.17 = 𝑧𝑧
2.17 =
𝑥𝑥 − 63.5
10
2.17(10) + 63.5 = 𝑥𝑥
𝑧𝑧 =
𝑥𝑥 − 𝜇𝜇
𝜎𝜎
Multiplying across by 10 and then adding 63.5
to both sides.
85.2 𝑘𝑘𝑘𝑘 = 𝑥𝑥
iii)
Reject the alternative hypothesis
𝐻𝐻0 The mean weight of children has not changed
𝐻𝐻1 The mean weight of children has changed
𝑧𝑧 =
𝑥𝑥 − 𝜇𝜇
𝜎𝜎
√𝑛𝑛
𝑥𝑥 = 62, 𝜎𝜎 = 10, 𝑏𝑏 = 150
𝑧𝑧 =
62 − 63.5
10
√150
𝑧𝑧 = −1.837
−1.387 > −1.96
∴ We reject the alternative hypothesis.
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Here we are dealing with a sample so we use
√𝑛𝑛
If we get a z-score which is greater than −1.96
and less than 1.96 we reject the alternative
hypothesis.
344
b)
To get the final probability of each line we multiply each of the probabilities in that line.
ii)
1
2
6
17
1
+
+
+
=
12 30 60 96 80
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Adding each of the probabilities of being late.
345
iii)
0.5490
𝑃𝑃(𝑅𝑅|𝐿𝐿) =
𝑃𝑃(𝐿𝐿) =
𝑃𝑃(𝑅𝑅 ∩ 𝐿𝐿)
𝑃𝑃(𝐿𝐿)
17
80
𝑃𝑃(𝑅𝑅 ∩ 𝐿𝐿) =
1
7
1
+
=
12 30 60
𝑃𝑃(𝑅𝑅 ∩ 𝐿𝐿) = The sum of the probabilities of the
times when it rains, and you were late.
7
60 = 28 = 0.5490
17 51
80
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346
Question 9
a)
√3|𝐶𝐶𝐶𝐶|
tan 𝜃𝜃 =
E
opposite
adjacent
tan 60 =
|𝑇𝑇𝑇𝑇|
|𝐶𝐶𝐶𝐶|
√3|𝐶𝐶𝐶𝐶| = |𝑇𝑇𝑇𝑇|
C
60
T
b)
(15)2 + |𝐶𝐶𝐶𝐶|2 = |𝐷𝐷𝐷𝐷|2
Pythagoras’ Theorem.
|𝑇𝑇𝑇𝑇|
|𝐷𝐷𝐷𝐷|
Now, we need to get
|𝐷𝐷𝐷𝐷| in terms of |𝑇𝑇𝑇𝑇|.
�225 + |𝐶𝐶𝐶𝐶|2 = |𝐷𝐷𝐷𝐷|
tan 30 =
1
√3
|𝐷𝐷𝐷𝐷| = |𝑇𝑇𝑇𝑇|
|𝑇𝑇𝑇𝑇|
|𝐷𝐷𝐷𝐷| =
1
√3
�225 + |𝐶𝐶𝐶𝐶|2 = √3|𝑇𝑇𝑇𝑇|
√3
= |𝑇𝑇𝑇𝑇|
225 + |𝐶𝐶𝐶𝐶|2
|𝑇𝑇𝑇𝑇| = �
3
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Finding Tan 30 and
multiplying across by |𝐷𝐷𝐷𝐷|.
15
T
= √3|𝑇𝑇𝑇𝑇|
�225 + |𝐶𝐶𝐶𝐶|2
D
C
Plugging in √3|𝑇𝑇𝑇𝑇| for
|𝐷𝐷𝐷𝐷|
Dividing across by √3
Rewriting as one square
root.
347
c)
5.3m
225 + |𝐶𝐶𝐶𝐶|2
= √3|𝐶𝐶𝐶𝐶|
3
�
225 + |𝐶𝐶𝐶𝐶|2
= 3|𝐶𝐶𝐶𝐶|2
3
225 + |𝐶𝐶𝐶𝐶|2 = 9|𝐶𝐶𝐶𝐶|2
225 = 8|𝐶𝐶𝐶𝐶|
�
2
225
= |𝐶𝐶𝐶𝐶|
8
Letting |𝑇𝑇𝑇𝑇| = |𝑇𝑇𝑇𝑇|
Squaring both sides
Multiplying across by 3
Taking |𝐶𝐶𝐶𝐶|2 from both sides
Dividing by 8 and square rooting both
sides
|𝐶𝐶𝐶𝐶| =5.3m
d)
9.2m
√3|𝐶𝐶𝐶𝐶| = |𝑇𝑇𝑇𝑇|
√3(5.3) = |𝑇𝑇𝑇𝑇|
|𝑇𝑇𝑇𝑇| = 9.2m
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Subbing in our value for |𝐶𝐶𝐶𝐶|
348
e)
54.7°
cos 𝜃𝜃 =
cos 𝜃𝜃 =
|𝐶𝐶𝐶𝐶|
|𝐹𝐹𝐹𝐹|
|𝐶𝐶𝐶𝐶|
|𝑇𝑇𝑇𝑇|
→ cos 𝜃𝜃 =
cos 𝜃𝜃 =
cos −1
1
|𝐶𝐶𝐶𝐶|
√3|𝐶𝐶𝐶𝐶|
|𝐹𝐹𝐹𝐹| = |𝑇𝑇𝑇𝑇| as they are both the
length of the tree.
T
𝜃𝜃
|𝑇𝑇𝑇𝑇| = √3|𝐶𝐶𝐶𝐶| from part a
The |𝐶𝐶𝐶𝐶|′𝑠𝑠 cancel
F
C
√3
1
√3
= 𝜃𝜃
𝜃𝜃 = 54.7°
f)
30.4%
𝑃𝑃 =
54.7 × 2
= 0.30389
360
= 30.4%
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The tree can fall in any one of 360°.
If it falls 54.7° either side of C it is on Conor’s property.
So, we divide 2 × 54.7 by 360
349
2016 Paper 2
Question 1
a)
3𝑥𝑥 − 2𝑦𝑦 + 1 = 0
Slope AC =
6
2
4 − (−2)
=
=−
−9
3
−3 − 6
Perpindicular slope =
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
𝑦𝑦 − 3 =
3
(𝑥𝑥 − 5)
2
2𝑦𝑦 − 6 = 3(𝑥𝑥 − 5)
2𝑦𝑦 − 6 = 3𝑥𝑥 − 15
3
2
We find the slope of AC so we can find the perpendicular
slope which is the slope of the line from B.
If we multiply the perpendicular slope by the slope, we
get −1. To find the perpendicular slope therefore, we
invert the fraction and change the sign.
Equation of a line from pg 18 of The Maths Tables Book
Multiplying across by 2
Subbing the slope and the coordinates of B into the
equation for a line.
3𝑥𝑥 − 2𝑦𝑦 − 9 = 0
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350
b)
(7,6)
Slope AB =
5
3 − (−2)
=
= −5
−1
5−6
1
Perpindicular slope =
5
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
𝑦𝑦 − (4) =
1
(𝑥𝑥 − (−3))
5
5𝑦𝑦 − 20 = 𝑥𝑥 + 3
𝑥𝑥 − 5𝑦𝑦 + 23 = 0
3𝑥𝑥 − 2𝑦𝑦 − 9 = 0
𝑥𝑥 − 5𝑦𝑦 + 23 = 0 × −3 → −3𝑥𝑥 + 15𝑦𝑦 − 69 = 0
3𝑥𝑥 − 2𝑦𝑦 − 9 = 0
−3𝑥𝑥 + 15𝑦𝑦 − 69 = 0
13𝑦𝑦 − 78 = 0
13𝑦𝑦 = 78
𝑦𝑦 = 6
3𝑥𝑥 − 2(6) − 9 = 0
3𝑥𝑥 − 12 − 9 = 0
3𝑥𝑥 = 21
𝑥𝑥 = 7
The orthocentre is where the lines from the
corners of the triangles which are perpendicular
to the opposite side, meet. (These are called
altitudes).
We need to find a second altitude. Finding the
slope perpendicular to AB
Subbing in the coordinates for C into the
equation of a line.
Taking the equation from part a
Multiplying the equation by −3 so the 𝑥𝑥′𝑠𝑠 will
cancel.
Adding the two equations
Solving for 𝑦𝑦
Subbing 6 in for 𝑦𝑦 in the equation of one of the
lines.
Solving for 𝑥𝑥
The coordinates of the orthocentre.
(7,6)
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351
Question 2
a)
𝑥𝑥 − 7𝑦𝑦 + 43 = 0
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
𝑦𝑦 − 6 =
1
�𝑥𝑥 − (−1)�
7
Equation of a line from pg 18 of The Maths
Tables Book
Subbing in the coordinates of X and the given
slope
7𝑦𝑦 − 42 = 𝑥𝑥 + 1
𝑥𝑥 − 7𝑦𝑦 + 43 = 0
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352
b)
(𝒙𝒙 − 𝟔𝟔)𝟐𝟐 + (𝒚𝒚 − 𝟕𝟕)𝟐𝟐 = 𝟐𝟐𝟐𝟐
We know the distance from the centre to
the line 3𝑥𝑥 + 4𝑦𝑦 − 21 is 5.
𝑙𝑙: 3𝑥𝑥 + 4𝑦𝑦 − 21 = 0
𝐷𝐷 =
So we can use the formula for the
perpendicular distance from a point to a
line. We plug in the centre as (−𝑔𝑔, −𝑓𝑓) and
let it equal 5.
|𝑎𝑎𝑥𝑥1 + 𝑏𝑏𝑦𝑦1 + 𝑐𝑐|
√𝑎𝑎2 + 𝑏𝑏 2
𝑎𝑎 = 3, 𝑏𝑏 = 4, 𝑐𝑐 = −21, 𝑥𝑥1 = −𝑔𝑔, 𝑦𝑦1 = −𝑓𝑓
𝐷𝐷 =
|(3)(−𝑔𝑔) + (4)(−𝑓𝑓) + (−21)|
�(3)2 + (4)2
|−3𝑔𝑔 − 4𝑓𝑓 − 21|
√25
Formula found on page 19 of The Maths
Tables Book.
=5
=5
|−3𝑔𝑔 − 4𝑓𝑓 − 21|
=5
5
Multiplying across by 5.
As there are modulus bars the expression
on the left is equal to plus or minus 25.
|−3𝑔𝑔 − 4𝑓𝑓 − 21| = 25
−3𝑔𝑔 − 4𝑓𝑓 − 21 = ±25
−3𝑔𝑔 − 4𝑓𝑓 − 21 = 25
−3𝑔𝑔 − 4𝑓𝑓 = 46
𝑒𝑒𝑒𝑒1
− 3𝑔𝑔 − 4𝑓𝑓 − 21 = −25
− 3𝑔𝑔 − 4𝑓𝑓 = −4
𝑥𝑥 − 7𝑦𝑦 + 43 = 0
(−𝑔𝑔) − 7(−𝑓𝑓) + 43 = 0
−𝑔𝑔 + 7𝑓𝑓 + 43 = 0
𝑒𝑒𝑒𝑒2
This gives us two equations in g and f
We can now plug in (−𝑔𝑔, −𝑓𝑓) for 𝑥𝑥 and 𝑦𝑦
In the equation of the line from part a)
which passes through the centre of the
circle.
7𝑓𝑓 + 43 = 𝑔𝑔
Getting 𝑔𝑔 by itself.
𝑒𝑒𝑒𝑒1: − 3(7𝑓𝑓 + 43) − 4𝑓𝑓 = 46
Solving for 𝑓𝑓
−21𝑓𝑓 − 129 − 4𝑓𝑓 = 46
−25𝑓𝑓 = 175
𝑓𝑓 = −7
𝑔𝑔 = 7(−7) + 43 = −6
Centre= (−𝑔𝑔, −𝑓𝑓) = (6,7)
Equation = (𝑥𝑥 − 6)2 + (𝑦𝑦 − 7)2 = (5)2
(𝑥𝑥 − 6)2 + (𝑦𝑦 − 7)2 = 25
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Subbing in (7𝑓𝑓 + 43) for 𝑔𝑔 in equation 1
Subbing our value for 𝑓𝑓 into our expression
for 𝑔𝑔.
Plugging our centre point into the general
equation for a circle: (𝑥𝑥 − ℎ)2 + (𝑦𝑦 −
𝑘𝑘)2 = 𝑟𝑟 2 .
(ℎ = 𝑔𝑔, 𝑘𝑘 = 𝑓𝑓, 𝑟𝑟 = 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟) pg 19 Maths
Tables Book
Note: If you use equation 2, you get the
equation (𝑥𝑥 + 8)2 + (𝑦𝑦 − 5)2 = 25, this is
353
also correct
Question 3
a)
cos 7𝐴𝐴 + cos 𝐴𝐴
sin 7𝐴𝐴 − sin 𝐴𝐴
7𝐴𝐴 + 𝐴𝐴
7𝐴𝐴 − 𝐴𝐴
cos
2
2
7𝐴𝐴 + 𝐴𝐴
7𝐴𝐴 − 𝐴𝐴
2 cos
sin
2
2
2 cos
6𝐴𝐴
8𝐴𝐴
cos
2
2
8𝐴𝐴
6𝐴𝐴
2 cos
sin
2
2
2 cos
2 cos 4𝐴𝐴 cos 3𝐴𝐴
2 cos 4𝐴𝐴 sin 3𝐴𝐴
cos 3𝐴𝐴
= cot 3𝐴𝐴
sin 3𝐴𝐴
Questions like these get a lot easier with practice LINK
Using the identities on page 13-15 of the Maths Tables
Book:
cos 𝐴𝐴 + cos 𝐵𝐵 = 2 cos
sin 𝐴𝐴 − sin 𝐵𝐵 = 2cos
𝐴𝐴 + 𝐵𝐵
𝐴𝐴 − 𝐵𝐵
cos
, 𝑜𝑜𝑜𝑜 𝑡𝑡ℎ𝑒𝑒 𝑡𝑡𝑡𝑡𝑡𝑡
2
2
𝐴𝐴 + 𝐵𝐵
𝐴𝐴 − 𝐵𝐵
sin
2
2
The 2 cos 4𝐴𝐴′ 𝑠𝑠 cancel
cos 𝐴𝐴
= cot 𝐴𝐴
sin 𝐴𝐴
b)
cos 𝜃𝜃 = ±
√5
3
cos 2𝜃𝜃 = cos 2 𝜃𝜃 − sin2 𝜃𝜃
→ cos 2𝜃𝜃 = cos 2 𝜃𝜃 − (1 − 𝑐𝑐𝑐𝑐𝑠𝑠 2 𝜃𝜃)
1
−1 + 2 cos 2 𝜃𝜃 =
9
−9 + 18 cos 2 𝜃𝜃 = 1
18 cos 2 𝜃𝜃 = 1 + 9
cos 2 𝜃𝜃 =
10
18
5
(cos 𝜃𝜃)2 =
9
5
9
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = �
cos 𝜃𝜃 = ±
√5
3
© Pocket Tutor 2022
Using the identities on page 13-15 of the Maths Tables
Book:
cos 2𝐴𝐴 = cos 2 𝐴𝐴 − sin2 𝐴𝐴
cos 2 𝐴𝐴 + sin2 𝐴𝐴 = 1
→ sin2 𝐴𝐴 = 1 − cos 2 𝐴𝐴
Multiplying across by 9
Dividing by 18
cos 2 𝜃𝜃 = (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐)2
Square rooting both sides.
As 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 is an unknown it is equal to ± the square root.
354
Question 4
a) i)
If two lines, go from the edges of the diameter
of a semicircle and meet on the semicircle (e.g.
at D) they make a 90° angle.
ii)
√𝑥𝑥 = 𝑦𝑦
|𝐴𝐴𝐴𝐴| |𝐵𝐵𝐵𝐵|
=
|𝐵𝐵𝐵𝐵| |𝐵𝐵𝐵𝐵|
𝑥𝑥 𝑦𝑦
=
𝑦𝑦 1
𝑥𝑥 = 𝑦𝑦 2
√𝑥𝑥 = 𝑦𝑦
© Pocket Tutor 2022
The most important thing about similar triangles is
that there sides are in proportion. Because of this if
you put side AB over BD (as a fraction) then this
will be equal to the fraction made with side BD
over BC.
|𝐴𝐴𝐴𝐴| |𝐵𝐵𝐵𝐵|
=
|𝐵𝐵𝐵𝐵| |𝐵𝐵𝐵𝐵|
Enter the values given – x, y and 1 into this
equation and solve to write y on its own.
355
b)
© Pocket Tutor 2022
356
Question 5
a) i)
ii)
13
120
1
1 1 3
Way 1: P(John scoring) × P(David scorng) × P(Mike missing) = � × × � =
40
5 6 4
1
1 5 1
Way 2: � × × � =
24
5 6 4
1
4 1 1
Way 3: � × × � =
30
5 6 4
To solve this question, we find the
probability of each way occurring and
then add them all up.
1
1 1 1
Way 4: � × × � =
120
5 6 4
1
1
1
13
1
+
+
+
=
40 24 30 120 120
Adding the probabilities
b)
𝑥𝑥 = 0.15
𝑃𝑃(𝐴𝐴) = 𝑥𝑥 + 0.1
𝑃𝑃(𝐴𝐴) = The intersection plus A only = 0.1 + 𝑥𝑥
𝑃𝑃(𝐵𝐵) = 0.1 + 0.3 = 0.4
If the events are independent, then
𝑃𝑃(𝐴𝐴 ∩ 𝐵𝐵) = 𝑃𝑃(𝐴𝐴) × 𝑃𝑃(𝐵𝐵)
0.1 = (𝑥𝑥 + 0.1) × 0.4
𝑃𝑃(𝐴𝐴 ∩ 𝐵𝐵) = 𝑃𝑃(𝐴𝐴) × 𝑃𝑃(𝐵𝐵)
Revise independent events here
0.1 = 0.4𝑥𝑥 + 0.04
0.1 − 0.04 = 0.4𝑥𝑥
Dividing across by 0.4
𝑃𝑃(𝐵𝐵) = The intersection plus B only 0.1 + 0.3
0.06
= 𝑥𝑥
0.4
0.15 = 𝑥𝑥
© Pocket Tutor 2022
357
Question 6
a)
1
1
1
1
×
×
=
26 10 10 2600
Probability of chosen letter being M times
the probability of first number being 3 times
probability of second number being 3
b)
loses 29 cent per play
Event
Payout (𝒙𝒙) €
Win Jackpot
1000
Match letter and first
number only
50
Match letter and
second number only
50
Match letter and
neither number
50
Fail to win
0
Probability of letter and one number:
The club loses 29 cent per play
𝒙𝒙. 𝑷𝑷(𝒙𝒙)
9
2600
450
2600
1
2600
9
2600
81
2600
1
26
450 4050
1000 450
+
+
+
= €2.29
2600 2600 2600 2600
2.29 − 2.00 = 29cent
Probability (𝑷𝑷(𝒙𝒙))
×
1
10
×
9
10
=
9
2600
1000
2600
450
2600
4050
2600
0
By multiplying the probability of winning
each way by the payout we get the
expected payout for each way of
winning.
Adding the expected payout of each
outcome gives us the total expected
payout. The total payout is 29 cent
more than the price of playing so the
club loses money.
c)
€3
600
= 0.71
845
2.29 + 0.71 = €3.00
© Pocket Tutor 2022
The club needs to make €600 from 845 people,
this means they need a profit 0f 71 cent per
person.
We add this to the payout of 2.29 giving us €3
358
Question 7
a) i)
1.95m
|𝐶𝐶𝐶𝐶|2 = |𝐶𝐶𝐶𝐶|2 + |𝐷𝐷𝐷𝐷|2
|𝐶𝐶𝐶𝐶|2
=
(2.5)2
|𝐶𝐶𝐶𝐶|2 = 15.25
+
|𝐶𝐶𝐶𝐶| = √15.25 = 3.905
|𝐶𝐶𝐶𝐶| =
|𝐶𝐶𝐶𝐶| =
1
|𝐶𝐶𝐶𝐶|
2
1
(3.905)
2
C
3
Using Pythagoras’
theorem we can find
|CE|.
D
|𝐶𝐶𝐶𝐶| =
2.5
= 1.9525𝑚𝑚 = 1.95𝑚𝑚
1
|𝐶𝐶𝐶𝐶|
2
B
ii)
tan 𝜃𝜃 =
|𝐷𝐷𝐷𝐷| = |𝐹𝐹𝐹𝐹| as they
are opposite sides of a
rectangle.
E
(3)2
Opposite
Adjacent
tan(50) =
|𝐴𝐴𝐴𝐴|
1.95
1.95 tan(50) = |𝐴𝐴𝐴𝐴|
2.3𝑚𝑚 = |𝐴𝐴𝐴𝐴|
C
50
A
1.95
iii)
3m
opposite
sin 𝜃𝜃 =
hypotenuse
𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
|𝐴𝐴𝐴𝐴|
|𝐵𝐵𝐵𝐵|
si𝑛𝑛(50) =
2.3
|𝐵𝐵𝐵𝐵|
|𝐵𝐵𝐵𝐵|𝑠𝑠𝑠𝑠𝑠𝑠(50) = 2.3
|𝐵𝐵𝐵𝐵| =
iv)
2.3
= 3𝑚𝑚
𝑠𝑠𝑠𝑠𝑠𝑠 (50)
Note this can also be solved with
Pythagoras’ theorem and by using cos𝜃𝜃
Using the value for |AB| given in the
previous part of the question
Multiplying across by |BC|
Dividing across by sin(50)
𝟔𝟔𝟔𝟔°
© Pocket Tutor 2022
359
B
𝑎𝑎2 = 𝑏𝑏 2 + 𝑐𝑐 2 − 2𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏
2
3 =
(3)2
+
(2.5)2
|BD| is also equal to 3 as this is an
isosceles triangle. (The triangle BAD
id identical to the triangle BAC
where we found that BC was 3m)
− 2(3)(2.5)𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
9 = 9 + 6.25 − 15𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
3
3
−6.25 = −15𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
As we have three sides and want to
find an angle we use the cosine rule
(page 16 of The Maths Tables Book)
−6.25
= 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
−15
5
= 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
12
5
𝐴𝐴 = 𝑐𝑐𝑐𝑐𝑠𝑠 −1 � � = 65°
12
C
D
2.5
v)
15𝑚𝑚2
1
Area = 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
2
1
(3)(2.5) sin 65 = 3.398654201
2
3.398654201 × 2 = 6.797
9√3
1
(3)(3) sin 60 =
4
2
9√3
× 2 = 7.794
4
6.797 + 7.794 = 14.59
= 15𝑚𝑚2
© Pocket Tutor 2022
The pyramid is made of two isosceles triangles like the
one in part iv) above and two equilateral triangles like
FBC.
First, we will find the area of the two isosceles triangles.
The formula for the area can be found on page 16 of the
Maths Tables Book. Using 65 as the angle as we found in
the previous part.
Now we find the area of the two equilateral triangles.
Equilateral triangles are made up of three 60° angles,
and each side is the same length. So, we can use 3 as
the lengths of 𝑎𝑎 and 𝑏𝑏 and 60 as our angle
Adding these areas
360
b)
√6𝑚𝑚
B
|𝐵𝐵𝐵𝐵|
tan 60 =
|𝐶𝐶𝐶𝐶|
tan 60 =
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 =
3
|𝐶𝐶𝐶𝐶|
3
|𝐶𝐶𝐶𝐶| tan 60 = 3
|𝐶𝐶𝐶𝐶| =
3
tan 60
C
60
opposite
adjacent
A
|𝐶𝐶𝐶𝐶| = √3
F
|𝐶𝐶𝐶𝐶| =
1
|𝐶𝐶𝐶𝐶|
2
1
√3 × 2 = |𝐶𝐶𝐶𝐶|
2
�2√3� = 2𝑥𝑥 2
12 = 2𝑥𝑥 2
6 = 𝑥𝑥 2
√6 = 𝑥𝑥
© Pocket Tutor 2022
|𝐶𝐶𝐶𝐶| = |𝐶𝐶𝐶𝐶| as where the
2
A
2√3 = |𝐶𝐶𝐶𝐶|
|𝐶𝐶𝐶𝐶|2 = 𝑥𝑥 2 + 𝑥𝑥 2
Just looking at the base of the
pyramid.
E
𝑥𝑥
C
𝑥𝑥
D
diagonals intersect is always the
midpoint between the corners.
Using Pythagoras’ theorem, we
can find the length of the sides.
Both sides are equal as it is a
square.
Solving for 𝑥𝑥
361
Question 8
a)
12 hours, [3.1,0.1]
Period: 2π ÷
Range:
π
= 12 hours
6
1.6 + 1.5 = 3.1
The period of a trigonometric function is always 2𝜋𝜋 divided
by the number before the input, (i.e. the 𝑡𝑡)
The range is always ± the number in front of the
trigonometric function, in this case we add and subtract this
number to the constant to find the range, as the constant
pushes the starting point of the graph up to a height of 1.6
1.6 − 1.5 = 0.1
[3.1,0.1]
b)
The maximum height from the range = 3.1𝑚𝑚
c)
𝜋𝜋
ℎ(𝑡𝑡) = 1.6 + 1.5𝑐𝑐𝑐𝑐𝑐𝑐( 𝑡𝑡)
6
𝜋𝜋
𝑑𝑑ℎ
𝜋𝜋
= 1.5 �− sin � 𝑡𝑡�� ×
6
𝑑𝑑𝑑𝑑
6
𝜋𝜋
𝜋𝜋
ℎ′ (2) = 1.5 �− sin � (2)�� ×
6
6
= −0.68ℎ𝑚𝑚/ℎ
To find the rate of change of the height of the
water we differentiate the formula for the height.
To differentiate cos LINK, we change it to −𝑠𝑠𝑠𝑠𝑠𝑠
and then multiply by the derivative of 𝑡𝑡
Plugging 2 into the derivative to find the rate of
change at this point.
This means that the tide is going out at 0.68 kilometers per hour
© Pocket Tutor 2022
362
d) i)
To find this we sub in 0 for midnight and then 3 for 3am, 6 for 6am, 9 for 9am and then
go back to 0 for 12pm. We then repeat this for the times after this.
This is because 𝑡𝑡 is defined as the number of hours since the last high tide.
ii)
The dotted lines
are for part f)
© Pocket Tutor 2022
363
e)
3m
High tide = 3.1
Low tide = 0.1
3.1 − 0.1 = 3𝑚𝑚
f)
9: 30 → 15: 15
= 5 hours and 45 minutes
We can see from the lines in the graph that the
depth is 2m at 9:30 am and stays above 1.5m until
about 3:15 pm. This means that the maximum
amount of time the barge can stay for is the time
between these.
They do leave a margin for error when these things
are being calculated from sketches.
© Pocket Tutor 2022
364
Question 9
a) i)
5.6%
𝑥𝑥 − 𝜇𝜇
𝑧𝑧 =
𝜎𝜎
From page 35 of the Maths Tables
Book. We do not use the √𝑛𝑛 as we are
dealing with a population not a
sample.
→ 0.9441
Finding the proportion which
corresponds to 1.59 by looking at
pages 36&37 of the Maths Tables
Book.
60,000 − 39400
= 1.59
12920
1 − 0.9441 = 0.0559
= 5.59%
5.6%
Multiplying this by 100 to find a
percentage.
ii)
10% → 1 − 0.1 = 0.9
0.9 → 1.28
𝑥𝑥 − 39400
= −1.28
12920
𝑥𝑥 − 39400 = −1.28(12920)
𝑥𝑥 = −16537.6 + 39400
If ten percent get this then the proportion that
doesn’t is 0.9. The closest z-score to 0.9 is 1.28.
As we are looking for the bottom 0.1 we use
−1.28. This means that a proportion of 0.9 will
get more than this amount.
𝑥𝑥 = €22,862
iii)
𝐻𝐻0 : The mean annual income has not changed.
𝐻𝐻1 : The mean annual income has changed.
𝑧𝑧 =
𝑥𝑥 − 𝜇𝜇
𝜎𝜎
√𝑛𝑛
38280 − 39400
= −2.74
12920
√1000
𝐻𝐻0 Is always that the situation has not
changed from the original statistic.
Taking the formula from page 35 of The
Maths Tables Book
Using the √𝑛𝑛 this time as we are dealing
with a sample.
If the z-score we get is < −1.96 or > 1.96
we reject the null hypothesis.
−2.74 < −1.96
∴ We reject the null hypothesis. The mean annual income has changed.
© Pocket Tutor 2022
365
b)
€26472.24 < 𝜇𝜇 < €27475.76
26,974 ± 1.96
𝜎𝜎
√𝑛𝑛
5120
�
26974 ± 1.96 �
√400
26974 ± 501.76
26974 + 501.76 = 27475.76
26974 − 501.76 = 26472.24
The formula to create a confidence interval is on
𝜎𝜎
page 34 of The Maths Tables Book as but as we
√𝑛𝑛
are finding a 95% confidence interval, we multiply
this by 1.96
Subbing in the standard deviation and number of
people in the sample.
Adding and subtracting the result.
Constructing the confidence interval.
€26472.24 < 𝜇𝜇 < €27475.76
c)
The distribution of the sample means will be normally distributed.
d)
1
√𝑛𝑛
Margin for error can be calculated using
= 0.045
1 = √𝑛𝑛(0.045)
1
= √𝑛𝑛
0.045
�
1 2
� = 𝑛𝑛
0.045
people in the sample. 4.5%→ 0.045
1
√𝑛𝑛
, where 𝑛𝑛 is the number of
Multiplying across by the bottom of the fraction.
Dividing across by 0.045
Squaring both sides
𝑛𝑛 = 493.827
© Pocket Tutor 2022
366
2015 Paper 2
Question 1
a)
b) i)
𝟓𝟓
10
=
36 𝟏𝟏𝟏𝟏
Putting the number of W’s over the total number of
outcomes.
ii) 0.3767
𝑃𝑃(𝐿𝐿) = 1 −
5
13
=
18 18
13 13 13
×
×
= 𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑
18 18 18
The probability of a loss is equal to 1 − the probability of
winning.
We multiply the probability of losing each time.
c) 0.0791
2
7
5
13
9 5
� �� � � � ×
= 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
18
18
2 18
© Pocket Tutor 2022
Using Bernoulli trials to find the probability of winning
two of the first nine throws and multiplying this by the
probability of winning the tenth throw.
367
Question 2
a) €𝟖𝟖𝟖𝟖. 𝟑𝟑𝟑𝟑 < 𝝁𝝁 < €𝟗𝟗𝟗𝟗. 𝟓𝟓𝟓𝟓
𝜇𝜇 ± 1.96
𝜎𝜎
√𝑛𝑛
90.45 ± 1.96
90.45 ± 4.06
20.73
√100
90.45 + 4.06 = 94.51
The formula for the error of the mean is on page 34 of The Maths
𝜎𝜎
Tables Book as . We multiply this by 1.96 as we are looking for a
√𝑛𝑛
95% confidence interval.
By adding and subtracting the standard error this gives us, from the
mean we can construct a 95% confidence interval.
90.45 − 4.06 = 86.39
€𝟖𝟖𝟖𝟖. 𝟑𝟑𝟑𝟑 < 𝝁𝝁 < €𝟗𝟗𝟗𝟗. 𝟓𝟓𝟓𝟓
b) We fail to reject the null hypothesis
𝐻𝐻0 : The mean amount spent by shoppers is €94
𝐻𝐻1 : The mean amount spent by shopper is not €94
𝑥𝑥 − 𝜇𝜇
𝜎𝜎
√𝑛𝑛
𝑥𝑥 = 90.45, 𝜇𝜇 = 94, 𝜎𝜎 = 20.73, 𝑛𝑛 = 100
90.45 − 94
= −1.71
20.73
√100
−1.71 > −1.96
∴ We fail to reject the null hypothesis
© Pocket Tutor 2022
The null hypothesis is the statement made by the
shop; the alternative hypothesis is the findings of the
survey. Even though in this question they introduce
the survey first, it is the alternative hypothesis as it is
only a sample reading not a measurement of the
population.
Taking the equation from page 35 of The Maths
Tables Book.
Subbing in our values gives us −1.71
As this is between the range of −1.96 and 1.96 we
fail to reject the null hypothesis.
368
c) 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎
−1.71 → 0.9564
1 − 0.9564 = 0.0436
2 × 0.0436 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
Converting −1.71 on page 37 of The Maths Tables Book. We
take this number from 1 to find the proportion outside of
−1.71 on the left of the curve. We then multiply this by two
to include the proportion on the right - outside 1.71 on the
curve, this is our p-value
There is an 8.72% chance that a sample of this size would have a mean of €90.45 if the population
mean is €94. As this is greater than 5%, we fail to reject the null hypothesis at the 5% level of
significance.
Question 3
a) 𝒕𝒕 = −𝟓𝟓
𝑙𝑙: 3𝑥𝑥 − 4𝑦𝑦 − 12 = 0
3𝑥𝑥 − 12 = 4𝑦𝑦
3
𝑥𝑥 − 3 = 𝑦𝑦
4
𝑚𝑚 =
3
4
⊥ 𝑚𝑚 = −
4
3
𝑚𝑚 𝐴𝐴𝐴𝐴 = −
4
3
4
𝑦𝑦2 − 𝑦𝑦1
=−
3
𝑥𝑥2 − 𝑥𝑥1
4
𝑡𝑡 − (−1)
=−
3
7−4
4
𝑡𝑡 + 1
=−
3
3
𝑡𝑡 + 1 = −4
𝒕𝒕 = −𝟓𝟓
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If 𝑙𝑙 is perpendicular to 𝐴𝐴𝐴𝐴 we can find the slope of
𝐴𝐴𝐴𝐴 by finding the slope of 𝑙𝑙 and then finding the
perpendicular slope.
Rearranging the equation into the form of 𝑦𝑦 =
𝑚𝑚𝑚𝑚 + 𝑐𝑐 to find the slope.
We can find the perpendicular slope by inverting
the slope and changing the sign.
The perpendicular slope is the slope of 𝐴𝐴𝐴𝐴.
Letting this equal the formula for finding the slope
will allow us to solve for 𝑡𝑡
Multiplying across by 3.
Taking 1 from both sides gives us −5
369
𝒃𝒃)
|𝟏𝟏𝟏𝟏 − 𝟒𝟒𝟒𝟒|
𝟓𝟓
|𝑎𝑎𝑥𝑥1 + 𝑏𝑏𝑦𝑦1 + 𝑐𝑐|
The equation for the distance between a point and a line
can be found on page 19 of The Maths Tables Book.
𝑙𝑙: 3𝑥𝑥 − 4𝑦𝑦 − 12 = 0, 𝑃𝑃(10, 𝑘𝑘)
Where the equation of the line is in the form 𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏 =
𝑐𝑐 and the point Is (𝑥𝑥1 , 𝑦𝑦1 )
|3(10) + (−4)(𝑘𝑘) + (−12)|
Subbing in our values
√𝑎𝑎2 + 𝑏𝑏 2
𝑎𝑎 = 3, 𝑏𝑏 = −4, 𝑐𝑐 = −12, 𝑥𝑥1 = 10, 𝑦𝑦1 = 𝑘𝑘
�(3)2 + (−4)2
|30 − 4𝑘𝑘 − 12|
√25
|𝟏𝟏𝟏𝟏 − 𝟒𝟒𝟒𝟒|
𝟓𝟓
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370
𝒄𝒄) 𝒊𝒊) 𝒌𝒌 =
𝟑𝟑
, −𝟒𝟒𝟒𝟒
𝟒𝟒
|𝑎𝑎𝑥𝑥1 + 𝑏𝑏𝑦𝑦1 + 𝑐𝑐|
If the point bisects the angles between them,
it is an equal distance from both lines. So, if
we find the distance from the point to 𝑙𝑙2 in
terms of 𝑘𝑘 and let it equal to the distance we
found in part b) we can find 𝑘𝑘.
√𝑎𝑎2 + 𝑏𝑏 2
𝑙𝑙2 = 5𝑥𝑥 + 12𝑦𝑦 − 20 = 0
|5(10) + 12(𝑘𝑘) − 20|
�(5)2 + (12)2
|50 + 12𝑘𝑘 − 20|
√25 + 144
=
|30 + 12𝑘𝑘|
√169
|18 − 4𝑘𝑘| |30 + 12𝑘𝑘|
=
13
5
=
Letting the expressions for the distance equal
each other.
|30 + 12𝑘𝑘|
13
Due to the modulus the
30 + 12𝑘𝑘
18 − 4𝑘𝑘
=±
13
5
the right-hand side.
234 − 52𝑘𝑘 = 150 + 60𝑘𝑘
234 − 52𝑘𝑘 = −(150 + 60𝑘𝑘)
84 = 112𝑘𝑘
384 = −8𝑘𝑘
234 − 150 = 60𝑘𝑘 + 52𝑘𝑘
234 + 150 = 52𝑘𝑘 − 60𝑘𝑘
84
= 𝑘𝑘
112
5
is equal to ±
Multiplying across by 5 and 13
(18 − 4𝑘𝑘)13 = ±5(30 + 12𝑘𝑘)
234 − 52𝑘𝑘 = ±(150 + 60𝑘𝑘)
18−4𝑘𝑘
Splitting into the plus and the minus
Solving for 𝑘𝑘
48 = −𝑘𝑘
𝟑𝟑
= 𝒌𝒌
𝟒𝟒
− 𝟒𝟒𝟒𝟒 = 𝒌𝒌
ii) 3
𝑘𝑘 > 0 → 𝑘𝑘 ≠ −48 → 𝑘𝑘 =
Distance P to l1 =
3
|18 − 4 � � |
4
→
5
3
4
|18 − 4𝑘𝑘|
5
Taking the positive value for 𝑘𝑘 and subbing
it into our expression for the distance
between P and 𝑙𝑙1 from part b).
|15|
= 𝟑𝟑
5
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Question 4
a) (𝟏𝟏, −𝟔𝟔), 𝟔𝟔√𝟏𝟏𝟏𝟏
(𝑥𝑥 − 1)2 + (𝑥𝑥 + 6)2 = 360
From page 19 of the Maths Tables Book:
Centre: (𝟏𝟏, −𝟔𝟔)
The centre = (ℎ, 𝑘𝑘) where: (𝑥𝑥 − ℎ2 ) + (𝑦𝑦 − 𝑘𝑘)2 = 𝑟𝑟 2 is
the equation of the circle
Radius: 𝑟𝑟 2 = 360
Square rooting both sides
𝑟𝑟 = √360 = 𝟔𝟔√𝟏𝟏𝟏𝟏
b) i) (𝟓𝟓, 𝟔𝟔)
|𝐴𝐴𝐴𝐴| ∶ |𝐾𝐾𝐾𝐾| = 2 ∶ 1
�
𝑏𝑏𝑥𝑥1 + 𝑎𝑎𝑥𝑥2 𝑏𝑏𝑏𝑏1 + 𝑎𝑎𝑦𝑦2
�
,
𝑏𝑏 + 𝑎𝑎
𝑏𝑏 + 𝑎𝑎
𝑎𝑎 = 2, 𝑏𝑏 = 1, 𝑥𝑥1 = 1, 𝑦𝑦1 = −6, 𝑥𝑥2 = 7, 𝑦𝑦2 = 12
�
�
(1)(1) + (2)(7) (1)(−6) + (2)(12)
,
�
1+2
1+2
The point K divides the centre (A) and B in the ratio
2: 1.
So, we can use the expression for an internal divider on
page 18 of The Maths Tables Book to find the point K
Filling in the two points and the ratio.
15 18
, � = (𝟓𝟓, 𝟔𝟔)
3 3
ii) (𝒙𝒙 − 𝟓𝟓)𝟐𝟐 + (𝒚𝒚 − 𝟔𝟔)𝟐𝟐 = 𝟒𝟒𝟒𝟒
Centre (5,6)
Radius = |𝐾𝐾𝐾𝐾|
�(𝑥𝑥2 − 𝑥𝑥1
)2
+ (𝑦𝑦2 − 𝑦𝑦1
)2
�(7 − 5)2 + (12 − 6)2 = 2√10
2
(𝑥𝑥 − 5)2 + (𝑦𝑦 − 6)2 = �2√10 �
(𝒙𝒙 − 𝟓𝟓)𝟐𝟐 + (𝒚𝒚 − 𝟔𝟔)𝟐𝟐 = 𝟒𝟒𝟒𝟒
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The radius is the distance from K to B, using
the formula on page 18 if The Maths Tables
Book.
Subbing the centre and radius into
(𝑥𝑥 − ℎ)2 + (𝑦𝑦 − 𝑘𝑘)2 = 𝑟𝑟 2
372
c) 𝒙𝒙 + 𝟑𝟑𝟑𝟑 − 𝟒𝟒𝟒𝟒 = 𝟎𝟎
Slope AB =
12 − (−6) 18
=
=3
6
7−1
Slope of tangent = −
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
1
𝑦𝑦 − 12 = − (𝑥𝑥 − 7)
3
1
3
3𝑦𝑦 − 36 = −1(𝑥𝑥 − 7)
The slope of the tangent to a circle is perpendicular
to a line from the centre to the edge of the circle.
Finding the perpendicular slope by inverting and
changing the sign of the slope.
Subbing in the slope and the coordinates of the
point B which the tangent passes through.
Multiplying across by 3
3𝑦𝑦 − 36 = −𝑥𝑥 + 7
𝒙𝒙 + 𝟑𝟑𝟑𝟑 − 𝟒𝟒𝟒𝟒 = 𝟎𝟎
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Question 5
a)
Looking at page 14 of the Maths
Tables Book:
𝑆𝑆𝑆𝑆𝑆𝑆(𝐴𝐴 + 𝐵𝐵) = 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝐶𝐶𝐶𝐶𝐶𝐶(𝐴𝐴 + 𝑏𝑏) = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 − 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
Dividing the top and the bottom by
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
sin 𝐴𝐴
= tan 𝐴𝐴 ,
cos 𝐴𝐴
b) 𝒙𝒙 = 𝟐𝟐𝟐𝟐°, 𝟒𝟒𝟒𝟒°, 𝟏𝟏𝟏𝟏𝟏𝟏°, 𝟏𝟏𝟏𝟏𝟏𝟏°, 𝟐𝟐𝟐𝟐𝟐𝟐°, 𝟐𝟐𝟐𝟐𝟐𝟐°
sin 3𝑥𝑥 =
√3
2
3𝑥𝑥 = sin−1
3𝑥𝑥 = 60°
√3
2
cos 𝐴𝐴 cos 𝐵𝐵
cancels to 1
cos 𝐴𝐴 cos 𝐵𝐵
Finding the sin inverse
Or 3𝑥𝑥 = 180 − 60 = 120
3𝑥𝑥 = 60 + 360𝑛𝑛 𝑜𝑜𝑜𝑜 3𝑥𝑥 = 120 + 360𝑛𝑛
𝑥𝑥 = 20 + 120𝑛𝑛
sin 𝐵𝐵
= tan 𝐵𝐵
cos 𝐵𝐵
𝑜𝑜𝑜𝑜 𝑥𝑥 = 40 + 120𝑛𝑛
𝑛𝑛 = 0 → 𝒙𝒙 = 𝟐𝟐𝟐𝟐° 𝑜𝑜𝑜𝑜 𝒙𝒙 = 𝟒𝟒𝟒𝟒°
𝑛𝑛 = 1 → 𝑥𝑥 = 20 + 120(1) 𝑜𝑜𝑜𝑜 𝑥𝑥 = 40 + 120(1)
𝑥𝑥 = 𝟏𝟏𝟏𝟏𝟏𝟏° 𝑜𝑜𝑜𝑜 𝟏𝟏𝟏𝟏𝟏𝟏°
𝑛𝑛 = 2 → 𝑥𝑥 = 20 + 120(2)𝑜𝑜𝑜𝑜 𝑥𝑥 = 40 + 120(2)
Sin is positive in the first quadrant and the second
quadrant. To find the angle in the second
quadrant we take it away from 180°
Writing out the expression for all the possible
values of 𝑥𝑥
Dividing across by the 3
Letting 𝑛𝑛 = 0,1 and 2
𝑛𝑛 = 3 gives us angles greater than 360° which are
outside the range given to us in the question/
𝑥𝑥 = 𝟐𝟐𝟐𝟐𝟐𝟐° 𝑜𝑜𝑜𝑜 𝟐𝟐𝟐𝟐𝟐𝟐°
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Question 6
a)
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b)
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Question 7
a) 𝟐𝟐𝟐𝟐cm
|𝐵𝐵𝐵𝐵| = |𝐻𝐻𝐻𝐻| = 8𝑟𝑟
|𝐴𝐴𝐴𝐴|2 + |𝐵𝐵𝐵𝐵|2 = |𝐴𝐴𝐴𝐴|2
(3𝑟𝑟)2 + (8𝑟𝑟)2 = �20√73�
Having drawn TB.
2
Using Pythagoras’ theorem
9𝑟𝑟 2 + 64𝑟𝑟 2 = 29200
73𝑟𝑟 2 = 29200
𝑟𝑟 2 = 400
𝑟𝑟 = 𝟐𝟐𝟐𝟐cm
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Dividing by 73
Square rooting both sides
377
b) 8000𝐜𝐜𝐦𝐦𝟐𝟐
The total area of the quadrilateral is equal to
the area of the rectangle plus the area of the
triangle.
Area 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 = Area TBKH + Area ABT
𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 = 8𝑟𝑟 × 𝑟𝑟
= 8(20) × 20 = 3200cm2
Subbing in 20 for 𝑟𝑟
1
Area 𝐴𝐴𝐴𝐴𝐴𝐴 = base ×⊥ height
1
(3𝑟𝑟) × 8𝑟𝑟
2
2
Subbing in 20 for 𝑟𝑟
1
3(20) × 8(20) = 4800cm2
2
3200 + 4800 = 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝐜𝐜𝐜𝐜𝟐𝟐
Adding the two areas
c) i) 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟗𝟗°
Angle 𝐻𝐻𝐻𝐻𝐻𝐻 = 2 × Angle 𝐻𝐻𝐻𝐻𝐻𝐻
𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
tan 𝜃𝜃 =
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
160
60
𝜃𝜃 = 69.444°
69.444 × 2 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟗𝟗°
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These angles are equal so
angle 𝐻𝐻𝐻𝐻𝐻𝐻 = 2 × 𝑃𝑃𝑃𝑃𝑃𝑃
160
160
tan 𝜃𝜃 =
60
𝜃𝜃 = tan−1
Angle HAP = Angle HAB +
Angle 𝑃𝑃𝑃𝑃𝑃𝑃.
60
A
𝜃𝜃
B
20√73
378
ii) 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝒎𝒎𝟐𝟐
Area = Sector HAP + Sector KBQ + Area ABKH + PABQ
𝐴𝐴 = 𝜋𝜋𝑟𝑟 2
𝜃𝜃
360
Sector HAP:
𝜃𝜃 = 360 − 138.9 = 221.1°
𝑟𝑟 = 4 × 20 = 80
𝜋𝜋(80)2 �
221.1
� = 12348.55𝑐𝑐𝑚𝑚2
360
Sector KBQ:
𝜃𝜃 = 138.9
𝜋𝜋(20)2
138.9
= 484.85
360
𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 = 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃
Area of a sector from page 9 of The
Maths Tables Book.
The angle is equal to 360° (the full
circle) minus the angle HAP which
we found earlier.
Plugging into the formula.
Angle KBQ = HAP
Plugging into 𝐴𝐴 = 𝜋𝜋𝑟𝑟 2
From part b)
→ 2 × 8000 = 16000
Adding the three areas.
12348.55 + 484.85 + 16000
To the nearest cm2
𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝒎𝒎𝟐𝟐
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𝜃𝜃
360
379
Question 8
a)
0.7 × 0.8 × 0.8 = 𝟎𝟎. 𝟒𝟒𝟒𝟒𝟒𝟒
b) 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎
Probability of missing the first = 1 − 0.7 = 0.3
Probability of missing the second after missing the first = 1 − 0.6 = 0.4
0.3 × 0.4 × 0.6 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎
c) 𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕
Score, Score, Score: 0.448
Score, Miss, Score: 0.7 × 0.2 × 0.6 = 0.084
Miss, Score, Score: 0.3 × 0.6 × 0.8 = 0.144
Miss, Miss, Score: 0.072
0.448 + 0.084 + 0.144 + 0.072 = 𝟎𝟎. 𝟕𝟕𝟕𝟕𝟕𝟕
d) i)
0.6 + 0.2𝑝𝑝𝑛𝑛
𝑃𝑃𝑛𝑛+1 = 𝑃𝑃(𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆, 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆) + 𝑃𝑃(𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀, 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆)
𝑝𝑝𝑛𝑛 × 0.8 + (1 − 𝑝𝑝𝑛𝑛 )(0.6)
0.8𝑝𝑝𝑛𝑛 + 0.6 − 0.6𝑝𝑝𝑛𝑛
0.6 + 0.2𝑝𝑝𝑛𝑛
From part a)
The probability of missing the second is 0.2 as the first was
scored.
The probability of scoring the third is 0.8 as the second was
scored.
From part b)
Adding the probabilities.
𝑝𝑝𝑛𝑛 is the probability that he is successful in his nth
throw, the question is asking us to find an equation
for the probability of success after this nth throw,
which we write as 𝑃𝑃𝑛𝑛+1 .
There’s two throws going on, the probability of
being successful after a successful shot is 0.8. So
this gives us 𝑝𝑝𝑛𝑛 × 0.8 That’s the (score, score)
scenario.
If the p of success is 𝑝𝑝𝑛𝑛 , then the p of failure is
(1 − 𝑝𝑝𝑛𝑛 ). If you miss the first shot then the
probability of success on the next shot is 0.6. This
gives us the (1 − 𝑝𝑝𝑛𝑛 )(0.6)
This is an OR question. You can score twice, or you
can miss your nth shot and score on the next shot.
So add the two probabilities.
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ii)
0.75
𝑝𝑝𝑛𝑛 = 𝑝𝑝𝑛𝑛+1
The question says that we can assume
𝑝𝑝𝑛𝑛 = 𝑝𝑝𝑛𝑛+1
𝑝𝑝𝑛𝑛 = 0.6 + 0.2𝑝𝑝
We found in d(i) of this question that
𝑝𝑝𝑛𝑛+1 = 0.6 + 0.2𝑝𝑝𝑛𝑛
𝑝𝑝𝑛𝑛 − 0.2𝑝𝑝𝑛𝑛 = 0.6
0.8𝑝𝑝𝑛𝑛 = 0.6
So we sub 𝑝𝑝𝑛𝑛 in for 𝑝𝑝𝑛𝑛+1 and then solve
to write 𝑝𝑝𝑛𝑛 on its own.
0.6
𝑝𝑝𝑛𝑛 =
= 0.75
0.8
e) i)
First sub 0.75 in for p
𝑎𝑎𝑛𝑛 = 𝑝𝑝 − 𝑝𝑝𝑛𝑛
𝑎𝑎𝑛𝑛 = 0.75 − 𝑝𝑝𝑛𝑛
𝑎𝑎𝑛𝑛+1 = 𝑝𝑝𝑛𝑛 − 𝑝𝑝𝑛𝑛+1 = 0.75 − (0.6 + 0.2𝑝𝑝𝑛𝑛 )
𝑎𝑎𝑛𝑛+1
:
𝑎𝑎𝑛𝑛
0.75 − (0.6 + 0.2𝑝𝑝𝑛𝑛 )
0.75 − 𝑝𝑝𝑛𝑛
0.15 + 0.2𝑝𝑝𝑛𝑛
0.75 − 𝑝𝑝𝑛𝑛
To find the next term we need to find
𝑎𝑎𝑛𝑛+1 which is 𝑝𝑝𝑛𝑛 − 𝑝𝑝𝑛𝑛+1
Sub in the values we know for these.
To find the ratio, put the next term over
the first term to give
By factorising out 5 from the bottom we
can simplify to find
1
=
5
1
5
In part (i) we’re told that
ii)
𝑎𝑎𝑛𝑛 = 𝑝𝑝 − 𝑝𝑝𝑛𝑛
𝑎𝑎1 = 𝑝𝑝 − 𝑝𝑝1 = 0.75 − 0.7 = 0.05
𝑎𝑎𝑟𝑟 𝑛𝑛−1 :
0.05(0.2)
.2
That’s the first term.
0.15 + 0.2𝑝𝑝𝑛𝑛
0.75 − 𝑝𝑝𝑛𝑛
0.15 + 0.2𝑝𝑝𝑛𝑛
5(0.15 + 0.2𝑝𝑝𝑛𝑛 )
𝑛𝑛−1
𝑎𝑎𝑛𝑛 = 0.75 − 𝑝𝑝𝑛𝑛
𝑛𝑛−1
< 0.00001
0.00001
<
. 05
(𝑛𝑛 − 1)𝑙𝑙𝑙𝑙0.2 < 𝑙𝑙𝑙𝑙
0.00001
. 05
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𝑎𝑎𝑛𝑛 = 𝑝𝑝 − 𝑝𝑝𝑛𝑛 and that
𝑎𝑎𝑛𝑛 is a geometric sequence. These have
the general form of
𝑇𝑇𝑇𝑇 = 𝑎𝑎𝑟𝑟 𝑛𝑛−1 :
𝑟𝑟 =
1
= .2
2
The probability of success on the first
throw is =. 7
𝑎𝑎 = 𝑝𝑝 − 𝑝𝑝1 = 0.75 − 0.7 = 0.05
So we now have a value for 𝑎𝑎 and 𝑟𝑟
We sub these into our formula and use
the laws of logs to solve for n.
381
𝑛𝑛 − 1 <
00001
. 05
ln 0.2
𝑙𝑙𝑙𝑙.
𝑛𝑛 = 6.29
𝑛𝑛 < 7 is the smallest value for n
f) i)
0.75
This is the long term success rate.
ii)
The events are not independent, i.e. the probability of each throw being scored depends on the
outcome of the previous throw.
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382
Question 9
a) 𝜶𝜶 = 𝟏𝟏𝟏𝟏. 𝟓𝟓°
sin 𝜃𝜃 =
15
Opposite
Hypotenuse
1
15
sin 𝛼𝛼 =
2
150
1
15
𝛼𝛼 = sin−1
2
150
1
𝛼𝛼 = 5.739°
2
𝛼𝛼 = 5.739 × 2 = 11.478°
Drawing a line from where Joan
is to the centre of the green
forms this triangle.
150
1
2
𝛼𝛼
𝛼𝛼 = 𝟏𝟏𝟏𝟏. 𝟓𝟓°
As 𝛼𝛼 is the angle for the full
width of the green, the angle in
this triangle is half that as we
are only including half the
green.
Using sin and then multiplying
the angle by two gives us 𝛼𝛼
b) 213m
𝑎𝑎2 = 𝑏𝑏 2 + 𝑐𝑐 2 − 2𝑏𝑏𝑏𝑏𝑏𝑏𝑜𝑜𝑜𝑜𝑜𝑜
|𝐴𝐴𝐴𝐴|2 = (190)2 + (385)2 − 2(190)(385) cos 18
Taking the cosine rule from page 16
of The Maths Tables Book.
|𝐴𝐴𝐴𝐴|2 = 45185.4317
|𝐴𝐴𝐴𝐴| = √45185.4317
|𝐴𝐴𝐴𝐴| = 212.57 = 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐
c) i) 8m
ℎ = −6𝑡𝑡 2 + 22𝑡𝑡 + 8
𝑡𝑡 = 0 → −6(0)2 + 22(0) + 8
ℎ = 𝟖𝟖𝟖𝟖
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𝑡𝑡 = 0 at the point 𝐾𝐾 as Joan hasn’t hit the
ball yet. So, we sub in 0 for 𝑡𝑡 in the equation
and this gives us the height of K
383
ii) 𝟑𝟑°
Distance = Speed × Time
We can find the distance |OB| by finding how long it
takes for the ball to get to B and multiplying this by the
speed of the ball.
Time from 𝐾𝐾 to 𝐵𝐵:
−6𝑡𝑡 2 + 22𝑡𝑡 + 8 = 0
The height of B is zero so if we let the equation for the
height to equal 0 and solve for 𝑡𝑡 we can find the time the
ball takes to get to B
−𝑏𝑏 ± �(𝑏𝑏)2 + 4𝑎𝑎𝑎𝑎
2𝑎𝑎
−(22) ± �(22)2 + 4(−6)(8)
2(−6)
𝑡𝑡 = −`
Using the minus b formula from page 20 of the Maths
Tables Book.
1
𝑜𝑜𝑜𝑜 4
3
Multiplying the time by the speed gives us the length
The ball is at B when 𝑡𝑡 = 4
38 × 4 = 152𝑚𝑚
tan 𝜃𝜃 =
8
152
𝜃𝜃 = tan−1
8
= 𝟑𝟑°
152
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|OB|. We can know use tan 𝜃𝜃 =
angle.
𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
to find the
K
8
O
152
𝜃𝜃
B
384
d) i)
𝒅𝒅 = 𝟐𝟐𝟐𝟐
|𝑪𝑪𝑪𝑪| = 𝟐𝟐𝟐𝟐 − 𝒉𝒉
tan 𝜃𝜃 =
ℎ
𝑑𝑑
𝜃𝜃 = 𝑡𝑡𝑡𝑡𝑛𝑛−1
1
2
1
→ 𝑡𝑡𝑡𝑡𝑡𝑡 𝜃𝜃 =
2
E
G
ℎ 1
=
𝑑𝑑 2
ℎ
𝜃𝜃
D
𝑑𝑑
1
ℎ = 𝑑𝑑
2
5ℎ2 − 50ℎ = 0
5ℎ(ℎ − 10) = 0
5ℎ = 0 ℎ − 10 = 0
ℎ = 0 ℎ = 10
𝒉𝒉 = 𝟏𝟏𝟏𝟏m
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As both are equal to tan 𝜃𝜃
|𝐶𝐶𝐶𝐶| = |𝐶𝐶𝐶𝐶| − |𝐷𝐷𝐷𝐷|
ii) 𝟏𝟏𝟏𝟏𝟏𝟏
4ℎ2 + 625 − 50ℎ + ℎ2 = 625
As we are told in the question
Multiplying across by 2
|𝑪𝑪𝑪𝑪| = 𝟐𝟐𝟐𝟐 − 𝒉𝒉
(2ℎ)2 + (25 − ℎ)2 = (25)2
opposite
adjacent
Multiplying across by d
𝒅𝒅 = 𝟐𝟐𝟐𝟐
𝑑𝑑 2 + (25 − ℎ)2 = (25)2
tan 𝜃𝜃 =
𝑑𝑑
G
25
D
25 − ℎ
C
Using Pythagoras’
theorem.
Subbing in 2ℎ for 𝑑𝑑 which
we found in part i)
Factorising and solving
We know the height above
G can’t be 0 as 𝜃𝜃 does not
equal 0, so we take 10 as
our answer.
385
2014 Paper 2
Question 1
a) i) 𝟕𝟕𝟕𝟕. 𝟏𝟏𝟏𝟏°
𝑎𝑎2 = 𝑏𝑏 2 + 𝑐𝑐 2 − 2𝑏𝑏𝑏𝑏 cos 𝐴𝐴
(150)2
=
(120)2
+
(134)2
− 2(120)(134) cos 𝐵𝐵
22500 − 32356 = −32160 cos 𝐵𝐵
−9856
= cos 𝐵𝐵
−32160
𝐵𝐵 = cos −1
9856
32160
Taking the cosine rule from page16 of the
Maths Tables Book.
Making sure that the length on the left-hand
side is the side opposite the angle we are trying
to find.
Using our calculator to find the cos inverse of
the fraction gives us our answer.
𝐵𝐵 = 𝟕𝟕𝟕𝟕. 𝟏𝟏𝟏𝟏°
ii)
Area =
1
𝑎𝑎𝑎𝑎 sin 𝐶𝐶
2
1
(120)(134)(sin(72.15)) = 7652.97
2
𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝟕𝒎𝒎𝟑𝟑
Taking the equation for the area of a triangle from
page 16 of the Maths Tables Book.
Subbing in the two sides and the angle we just
found. Rounding to the nearest whole number.
b)
If the circumcentre is at D, then: |𝐴𝐴𝐴𝐴| = |𝐵𝐵𝐵𝐵| = |𝐶𝐶𝐶𝐶|
Each of the triangles 𝐸𝐸𝐸𝐸𝐸𝐸. 𝐸𝐸𝐸𝐸𝐸𝐸, 𝐸𝐸𝐸𝐸𝐸𝐸 is right-angled at D and the two sides, the base and the
perpendicular are equal. Hence, by Pythagoras’ theorem, the third side of each, the hypotenuse (the
cables), must be equal.
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386
Question 2
a)
cos 2𝐴𝐴 = cos 2 𝐴𝐴 − sin2 𝐴𝐴
cos(𝐴𝐴 + 𝐵𝐵) = cos 𝐴𝐴 cos 𝐵𝐵 − sin 𝐴𝐴 sin 𝐵𝐵
cos(𝐴𝐴 + 𝐴𝐴) = cos 𝐴𝐴 cos 𝐴𝐴 − sin 𝐴𝐴 sin 𝐴𝐴
cos(2𝐴𝐴) = cos 2 𝐴𝐴 − sin2 𝐴𝐴
Starting with cos 𝐴𝐴 + 𝐵𝐵 from page 14 of the Maths Tables
Book.
Then subbing in 𝐴𝐴 for every 𝐵𝐵.
b) 𝟐𝟐. 𝟓𝟓 𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫
𝑙𝑙 = 𝑟𝑟𝑟𝑟
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾 = |𝑂𝑂𝑂𝑂|𝜃𝜃
𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻 = (|𝑂𝑂𝑂𝑂| + 1.2)𝜃𝜃
𝐾𝐾𝐾𝐾𝐾𝐾𝐾𝐾 + 3𝑚𝑚 = 𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻𝐻
|𝑂𝑂𝑂𝑂|𝜃𝜃 + 3 = (|𝑂𝑂𝑂𝑂| + 1.2)𝜃𝜃
|𝑂𝑂𝑂𝑂|𝜃𝜃 + 3 = |𝑂𝑂𝑂𝑂|𝜃𝜃 + 1.2𝜃𝜃
3 = 1.2𝜃𝜃
𝜃𝜃 =
3
= 𝟐𝟐. 𝟓𝟓 𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫𝐫
1.2
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Taking the equation for the length of an arc from page 9
of the Maths Tables Book.
The arc Helen is running on has a radius of |𝑂𝑂𝑂𝑂| + the
width of Kate’s lane: so, we can write it as |𝑂𝑂𝑂𝑂| + 1.2.
The arc Helen is running on is 3m longer than Kate’s so
we can say that Kate’s arc +3 is equal to Helen’s arc.
Now we can solve for 𝜃𝜃.
Taking |𝑂𝑂𝑂𝑂|𝜃𝜃 from both sides.
387
Question 3
a)
b)
Expected value = Probability of an outcome × the outcome.
1
1
1
2
Game A: (0) + (3) + (5) + (6) = €2.80
5
5
5
5
1
1
1
1
1
1
Game B: (0) + (1) + (2) + (3) + (4) + (5) = €2.50
6
6
6
6
6
6
Adding up the expected values of each of
the outcomes in Game A.
Repeating this for Game B. We can see that
Game B pays out less.
Game B will be more successful in raising funds for charity as it pays out less money.
c)
𝑛𝑛
� � 𝑝𝑝𝑟𝑟 𝑞𝑞 𝑛𝑛−𝑟𝑟
𝑟𝑟
𝑛𝑛 = 6, 𝑟𝑟 = 2, 𝑝𝑝 =
2
6−2
6 1
5
� �� � � �
2 6
6
1
5
, 𝑞𝑞 =
6
6
= 𝟎𝟎. 𝟐𝟐
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Taking the equation for Bernoulli trials from page 33 of the
Maths Tables Book.
𝑛𝑛 = the number of trials, r = the number of successes.
𝑝𝑝 = probability of success, q = probability of failure.
Plugging our values into the equation and into the calculator.
388
Question 4
a) i)
311 sin(100𝜋𝜋𝜋𝜋)
[−𝟑𝟑𝟑𝟑𝟑𝟑, 𝟑𝟑𝟑𝟑𝟑𝟑]
The range can be read from the number before sin.
ii)
1
= 𝟓𝟓𝟓𝟓 𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩𝐩 𝐩𝐩𝐩𝐩𝐩𝐩 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬
0.02
We can see from the graph that it repeats every 0.02 seconds.
Dividing one second by this shows us how many periods there
are in one second.
b) i)
Following the lines on the diagram we can see that the third line is the max value which we
already found to be 311. Then filling in the rest of the numbers following the pattern of a sin
graph.
ii)
𝜎𝜎 = 𝟐𝟐𝟐𝟐𝟐𝟐
You can revise how to use your calculator to find standard
deviation in section 4.2 of Statistics.
c) i)
𝑘𝑘𝑘𝑘 = 𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚
𝑘𝑘(220) = 311
𝑘𝑘 =
311
= 𝟏𝟏. 𝟒𝟒𝟒𝟒𝟒𝟒
220
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Filling the information, we already have into the given
equation, the maximum value is the upper limit of the range we
found in part a) i).
Dividing across by 220 and rounding to 3 decimal places.
389
ii) 𝒃𝒃 = 𝟑𝟑𝟑𝟑𝟑𝟑 𝒂𝒂 = 𝟏𝟏𝟏𝟏𝟏𝟏
𝑏𝑏
= 60
2𝜋𝜋
𝑏𝑏 = 60 × 2𝜋𝜋
The period of a sin graph can be gotten by putting the number in front of
the variable (the 𝑡𝑡) over 2𝜋𝜋.
𝑘𝑘𝑘𝑘 = 𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚
We know that 𝑎𝑎 is the max value of the graph so we plug that in for 𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚
in the equation given to us in the previous part.
𝑏𝑏 = 120𝜋𝜋 = 𝟑𝟑𝟑𝟑𝟑𝟑
𝑘𝑘𝑘𝑘 = 𝑎𝑎
1.414(110) = 𝑎𝑎
𝑎𝑎 = 155.54
𝒂𝒂 = 𝟏𝟏𝟏𝟏𝟏𝟏
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Here we know that the period is 60 so we let that equal
𝑏𝑏
2𝜋𝜋
.
To find a value for 𝑎𝑎 that gives the function a standard deviation of 110,
we fill 110 and the value we just found for 𝑘𝑘 into the equation.
390
Question 5
a) 𝑹𝑹 �−
𝟐𝟐𝟐𝟐
𝟑𝟑
, 𝟎𝟎�
Area ROS =
→
1
base ×⊥ height
2
1
|𝑅𝑅𝑅𝑅| × |𝑂𝑂𝑂𝑂| = Area
2
125
1
|𝑅𝑅𝑅𝑅| × 10 =
3
2
5|𝑅𝑅𝑅𝑅| =
|𝑅𝑅𝑅𝑅| =
25
=
3
125
3
125
÷5
3
The area of a triangle is equal to half the base times the
perpendicular height.
Subbing in 10 for |𝑂𝑂𝑂𝑂| as we can see from the coordinate of 𝑆𝑆
that it is 10 units above 𝑂𝑂.
Multiplying the ten and the half.
Dividing across by 5.
If this is the length of |𝑅𝑅𝑅𝑅| then we know that the coordinate of
R is
25
𝟐𝟐𝟐𝟐
, 𝟎𝟎�
𝑹𝑹 �−
𝟑𝟑
3
25
3
units to the left of 𝑂𝑂.
units left of the coordinates (0,0) gives us (−
25
3
, 0).
b)
slope |𝑅𝑅𝑅𝑅| =
𝑦𝑦2 − 𝑦𝑦1
𝑥𝑥2 − 𝑥𝑥1
6
10 − 0
=
25
0 − (− ) 5
3
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
𝑦𝑦 − 10 =
6
(𝑥𝑥 − 0)
5
5𝑦𝑦 − 50 = 6𝑥𝑥
6𝑥𝑥 − 5𝑦𝑦 + 50
6(−5) − 5(4) + 50 = 0
To show that this point is on the line we need to find
the equation of the line and plug the point into the
equation.
Finding the slope of the line using the equation from
page 18 of the Maths Tables Book.
Taking the equation of a line from page 18 as well and
subbing in the slope and the coordinates of S.
Multiplying across by 5.
Plugging the point (−5,4) into the equation of the line
we just found.
−30 − 20 + 50 = 0
0=0
∴ (−5,4) is on the line.
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391
𝐜𝐜) 𝒎𝒎 =
𝟐𝟐𝟐𝟐
𝟖𝟖
, 𝒄𝒄 =
𝟑𝟑
𝟏𝟏𝟏𝟏
𝑦𝑦 = 𝑚𝑚𝑚𝑚 + 𝑐𝑐
𝐿𝐿𝐿𝐿𝐿𝐿 𝑦𝑦 = 0
𝐿𝐿𝐿𝐿𝐿𝐿 𝑥𝑥 = 0
𝑚𝑚𝑚𝑚 = −𝑐𝑐
𝑦𝑦 = 𝑐𝑐
0 = 𝑚𝑚𝑚𝑚 + 𝑐𝑐
𝑥𝑥 = −
�−
𝑐𝑐
𝑚𝑚
𝑐𝑐
, 0�
𝑚𝑚
Area =
→
𝑦𝑦 = 𝑚𝑚(0) + 𝑐𝑐
1
base ×⊥ height
2
𝑐𝑐 2
125
1
| �− � | =
3
2
𝑚𝑚
𝑐𝑐 2
250
| �− � | =
3
𝑚𝑚
𝑐𝑐 2 =
Then we plug 0 in for 𝑥𝑥 to find the point where the line
crosses the 𝑦𝑦 − axis..
(0, 𝑐𝑐)
1
𝑐𝑐
125
| − × 𝑐𝑐| =
2
𝑚𝑚
3
|−𝑐𝑐 2 | =
First of all, we plug in 0 for 𝑦𝑦 to find the point where the
line crosses the 𝑥𝑥 − axis.
250
𝑚𝑚
3
250
𝑚𝑚
3
𝑦𝑦 = 𝑚𝑚𝑚𝑚 + 𝑐𝑐
(4) = 𝑚𝑚(−5) + 𝑐𝑐
4 = −5𝑚𝑚 + 𝑐𝑐
5𝑚𝑚 + 4 = 𝑐𝑐
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Now we use those points as the lengths of the base and the
height of the triangle as we did in part a).
Subbing them into the formula for the area of a triangle.
The left-hand side is in modulus bars as it is representing
area so it must be positive.
Multiplying across by 2
Multiplying across by 𝑚𝑚
Forcing 𝑐𝑐 2 to be positive.
Plugging the point −5,4 into the equation of a line.
Rearranging to get it in terms of 𝑐𝑐.
392
(5𝑚𝑚 + 4)2 =
250
𝑚𝑚
3
(25𝑚𝑚2 + 40𝑚𝑚 + 16) =
250
𝑚𝑚
3
75𝑚𝑚2 + 120𝑚𝑚 + 48 = 250𝑚𝑚
75𝑚𝑚2 − 130𝑚𝑚 + 48 = 0
(15𝑚𝑚 − 8)(5𝑚𝑚 − 6) = 0
𝑚𝑚 =
8
15
𝑚𝑚 =
𝟖𝟖
𝟏𝟏𝟏𝟏
𝑚𝑚 =
6
5
5𝑚𝑚 + 4 = 𝑐𝑐
5�
8
� + 4 = 𝑐𝑐
15
𝑐𝑐 =
Plugging the expression, we just found for 𝑐𝑐 into
𝑐𝑐 2 =
250
𝑚𝑚
3
Squaring out the bracket and multiplying across by 3.
Factorising and solving the quadratic. If you struggle to
find the factors the minus b formula can also be used
here.
8
6
is the slope of the line RS so we take
15
5
Subbing this value for 𝑚𝑚 into the expression we found
for 𝑐𝑐 earlier.
𝟐𝟐𝟐𝟐
𝟑𝟑
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393
Question 7
a)
Students, Retired people.
b) Median: 1947.9, IQR: 151.5
1773.4, 1784.8, 1803.4, 1813.4, 1828.6, 1867.0, 1903.3, 1954.9, 2049.6, 2060.4
1
Median: (𝑛𝑛 + 1)𝑡𝑡ℎ number.
2
𝑛𝑛 = 10
1
(10 + 1) = 5.5
2
To find the median we find the middle number using
the expression on the left.
As we get 5.5, we add the fifth and the 6th terms
together and divide by 2.
Mean of 5th and 6th terms
1828.6 + 1867.0
= 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏. 𝟖𝟖
2
Upper Quartile:
1
(𝑛𝑛 + 1)𝑡𝑡ℎ number.
2
1
(5 + 1) = 3
2
Find the third term to the right of the median:
1954.9
Lower Quartile:
1
(𝑛𝑛 + 1)𝑡𝑡ℎ number:
2
1
(5 + 1) = 3
2
Find the third term to the left of the median:
1803.4
Inter quartile range: 𝑄𝑄3 − 𝑄𝑄1 :
To find the upper quartile we find the median of the
half above the median.
There are 5 numbers above the median, so we plug 5
in for 𝑛𝑛.
This gives us 3 so the third number above the median
is our upper quartile.
To find the lower quartile we find the median of the
half below the median.
There are 5 numbers below the median, so we plug 5
in for 𝑛𝑛.
This gives us 3, so the third number below the median
is our lower quartile.
Subtracting the lower quartile from the upper quartile
gives us the inter quartile range.
1954.9 − 1803.4 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟓𝟓
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394
c) i)
To find the percentage of people at work in a certain year, we put the number of people at work
that year over the total number of people and multiply by 100.
Repeat this to find the percentage of people who are unemployed and again for people not in
the labour force.
ii)
To find the percentage of people at work or unemployed we repeat the same process as above,
but this time we add the number of people aged under 15 to the total number of people on the
bottom of the fraction.
To find the percentage of people not in the labour force we add the number of people aged
under 15 to the number in the table and then put it over the total number of people plus the
number of people aged under 15.
iii)
Yes, as the percentage of people at work has decreased. This means the government is taking in less
in tax, so it has less to spend. The percentage of people not in the workforce has increased so the
government has increased expenditure on supports.
d) i)
Disagree as there was a greater decline in the number of males employed than there was a decline
in the number of females employed.
ii)
Liam’s graph shows trend over time as well as the numbers whereas Niamh’s graph only shows
percentage in workforce and gives no information about actual numbers.
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395
iii) 𝟓𝟓𝟓𝟓. 𝟐𝟐%
2012:
1773.4
× 100 = 49.4%
3590
2013:
1803.4
× 100 = 50.3%
3586.2
50.3 − 49.4 = 0.9% increase
2014: 50.3 + 0.9 = 𝟓𝟓𝟓𝟓. 𝟐𝟐%
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Finding the percentage of people, aged over 15, at
work in 2012.
Repeating this for 2013.
Finding the increase in percentage from 2012 to
2013.
Using this to predict the percentage of people at
work in 2013.
396
Question 8
a) i)
To find the probability of a someone having the disease and testing positive we multiply the
probability of the person having the disease (0.003) by the probability of the test coming back as
positive given that they have the disease (0.99).
We follow this method for each of the end outcomes.
ii)
𝑃𝑃(Positive test) = P(Has disease & tests positive) + 𝑃𝑃(Doesn′ t have disease & tests positive)
→ 0.00297 + 0.03988 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
iii)
0.00297
Has diseas & tests positive
�=
𝑃𝑃 �
= 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
0.04285
Positive test
iv)
The test is not very useful as a person who tests positive only has the disease 7% of the time.
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397
b) i)
𝐻𝐻0 : The new drug is not more successful than the generic drug.
𝑝𝑝 = 0.51
New drug:
296
= 0.592
500
Margin of error: =
0.592 ± 0.045
0.592 + 0.045,
0.637
1
√500
To test the company’s claim, we can find the
confidence interval for the drug’s success rate at
the 5% level of significance.
= 0.045
0.592 − 0.045
0.547
First finding the drug’s success rate.
The margin of error can be calculated using
1
√𝑛𝑛
.
Adding and subtracting the margin of error from
the drug’s success rate to create a confidence
interval.
0.547 < 𝑝𝑝 < 0.637
0.51 is not within this confidence interval
∴ reject the null hypothesis.
The generic drug’s success rate (0.51) is not within
this interval so we reject the null hypothesis.
ii) 𝟐𝟐𝟐𝟐𝟐𝟐 patients
0.51 + 0.045 = 0.555
𝑥𝑥
< 0.555
500
𝑥𝑥
= 0.555
500
𝑥𝑥 = 500(0.555)
𝑥𝑥 = 277.5
𝟐𝟐𝟐𝟐𝟐𝟐 patients
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To be within the margin of error the success rate must be less
than 0.51 + the margin of error, so 0.555.
𝑥𝑥 is the number of successful cases.
Letting the proportion equal 0.555 to solve for 𝑥𝑥.
Multiplying across by 500.
We round down as the proportion must equal less than 0.555
398
Question 9
𝐚𝐚) 𝐢𝐢) centre: (−𝟐𝟐, −𝟑𝟑), radius: 𝟒𝟒√𝟐𝟐
𝑥𝑥 2 + 𝑦𝑦 2 + 4𝑥𝑥 + 6𝑦𝑦 − 19 = 0
centre: (−𝑔𝑔 − 𝑓𝑓)
centre = (−𝟐𝟐, −𝟑𝟑)
radius: �𝑔𝑔2 + 𝑓𝑓 2 − 𝑐𝑐
→ �(2)2 + (3)2 − (−19) = 𝟒𝟒√𝟐𝟐
We can see from page 19 of the Maths Tables Book,
that when the circle is written in the form:
𝑥𝑥 2 + 𝑦𝑦 2 + 2𝑔𝑔𝑔𝑔 + 2𝑓𝑓𝑓𝑓 + 𝑐𝑐 = 0
The centre is (−𝑔𝑔, −𝑓𝑓).
Also taking the equation for the radius from page 19.
Subbing in 𝑔𝑔, 𝑓𝑓 and 𝑐𝑐.
ii) 𝒓𝒓 = √𝟐𝟐
�(𝑥𝑥2 − 𝑥𝑥1 )2 + (𝑦𝑦2 − 𝑦𝑦1 )2
|𝐷𝐷𝐷𝐷| = �(3 − (−2))2 + (2 − (−3))2
To find the radius of the circle 𝑘𝑘 we need to find the
distance between the centres of the circles and then take
away the radius of the circle ℎ. As the circles are touching
externally this will leave us with the radius of 𝑘𝑘.
𝑟𝑟1 + 𝑟𝑟2 = 5√2
Now letting the sum of the radii equal the distance
between the centres.
𝐷𝐷 = (−2, −3), 𝐸𝐸 = (3,2)
= 5√2
𝑟𝑟1 + 4√2 = 5√2
𝑟𝑟1 = 5√2 − 4√2
𝒓𝒓 = √𝟐𝟐
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Using the distance formula form page 18 of the Maths
Tables Book.
Taking the radius of ℎ away from this distance gives us
the radius of 𝑘𝑘.
399
iii)
Slope of DE:
𝑦𝑦2 − 𝑦𝑦1
𝑥𝑥2 − 𝑥𝑥1
To find the distance from a point to a line we first
need to find the equation of the line.
2 − (−3)
=1
3 − (−2)
Finding the slope between the two centres using the
equation on page 18 of the Maths Tables Book.
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
Subbing the slope and one of the points into the
equation of a line also found on page 18.
𝑦𝑦 − (−3) = (1)�𝑥𝑥 − (−2)�
𝑦𝑦 + 3 = 𝑥𝑥 + 2
Now that we have the equation of the line, we can
use the formula for the distance between a point and
a line from page 19, where a line is written in the
form of 𝑎𝑎𝑎𝑎 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐.
𝑥𝑥 − 𝑦𝑦 − 1 = 0
|𝑎𝑎𝑥𝑥1 + 𝑏𝑏𝑦𝑦1 + 𝑐𝑐|
√𝑎𝑎2 + 𝑏𝑏 2
𝑎𝑎 = 1, 𝑏𝑏 = −1, 𝑐𝑐 = −1, 𝑥𝑥1 = −2, 𝑦𝑦1 = 2
|(1)(−2) + (−1)(2) + (−1)|
5
√2
=
�(1)1
5√2
2
+
(−1)2
5√2 1
= |𝐷𝐷𝐷𝐷|
2
2
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=
|−5|
√2
The point we’re trying to find the distance to is
(−2,2).
Subbing everything in.
The −5 becomes positive due to the modulus bars.
5√2
is half 5√2, the distance we found for |DE|
2
in the previous part.
400
𝟗𝟗 𝟐𝟐
𝟏𝟏 𝟐𝟐
𝒊𝒊𝒊𝒊) �𝒙𝒙 − � + �𝒚𝒚 − � = 𝟐𝟐
𝟐𝟐
𝟐𝟐
Midpoint of |𝐷𝐷𝐸𝐸|:
�
𝑥𝑥1 + 𝑥𝑥2 𝑦𝑦1 + 𝑦𝑦2
�
,
2
2
�
(−2) + 3 (−3) + 2
1 1
� = � ,− �
,
2
2
2 2
𝐷𝐷(−2, −3), 𝐸𝐸(3,2)
1 1
� , − � → (−2,2)
2 2
5
1
→ −2 = −
2
2
−
1
5
→2=+
2
2
Centre of circle 𝑘𝑘: (3,2)
3−
2+
5 1
=
2 2
5 9
=
2 2
Taking the midpoint formula from page 18 of the Maths
Tables Book.
Finding the midpoint of |𝐷𝐷𝐷𝐷|.
To find the translation from the midpoint to 𝐶𝐶, the point
1
(−2,2). We see what we have to add or subtract to to
make it −2. We repeat this for the y coordinates.
2
Now using the translation, we just found to translate the
centre of 𝑘𝑘 to the centre we are trying to find.
1 9
centre of 𝑗𝑗: � , �
2 2
(𝑥𝑥 − ℎ)2 + (𝑦𝑦 − 𝑘𝑘)2 = 𝑟𝑟 2
1 2
9 2
2
�𝑥𝑥 − � + �𝑦𝑦 − � = �√2�
2
2
𝟏𝟏 𝟐𝟐
𝟗𝟗 𝟐𝟐
�𝒙𝒙 − � + �𝒚𝒚 − � = 𝟐𝟐
𝟐𝟐
𝟐𝟐
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Plugging the centre we just found for 𝑗𝑗 into the equation of
a circle from page 19 of the Maths Tables Book and using
the radius we found earlier for the circle 𝑘𝑘, as we are
mapping this circle onto the circle 𝑗𝑗.
401
v) 𝒍𝒍 = 𝟏𝟏𝟏𝟏
k
𝐷𝐷(−2, −3), 𝐹𝐹(3, −3)
|𝐷𝐷𝐷𝐷| = �(𝑥𝑥2 − 𝑥𝑥1 )2 + (𝑦𝑦2 − 𝑦𝑦1 )2
2
��3 − (−2)� + �(−3) − (−3)�
|𝐷𝐷𝐷𝐷| = 5
Length = r1 + |DF| + r2
→ 4√2 + 5 + √2 = 12.07
𝒍𝒍 = 𝟏𝟏𝟏𝟏
2
ℎ
The width of the two cogs is
equal to the radius of the
circle ℎ, plus the length from
𝐷𝐷 (the centre of ℎ) to 𝐹𝐹 in
line with the centre of 𝑘𝑘, plus
the radius of 𝑘𝑘.
We find the point 𝐹𝐹 by taking
the coordinates for the
centre of 𝑘𝑘 and moving it
down until the y value is
equal to the y value of 𝐷𝐷.
Using the distance formula
on page 18 of the Maths
Tables Book to find |𝐷𝐷𝐷𝐷| and
then adding the radii.
b) i)
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402
ii)
|𝐴𝐴𝐴𝐴|2 = |𝐴𝐴𝐴𝐴|2 + |𝐵𝐵𝐵𝐵|2
𝜋𝜋
|𝐴𝐴𝐴𝐴|2
(2)2
= 𝜋𝜋
|𝐴𝐴𝐴𝐴|2
(2)2
+ 𝜋𝜋
Using Pythagoras’ theorem in the right-angled triangle.
|𝐵𝐵𝐵𝐵|2
(2)2
Thus, the area of 𝑢𝑢 = the area of 𝑠𝑠 + area of 𝑡𝑡
The area of a circle = 𝜋𝜋𝑟𝑟 2 . The equation we have is using
the circles is with the diameters, so we divide each one by
(2)2 to convert them to 𝑟𝑟 2 . Multiplying each one by 𝜋𝜋 gives
us an equation for the area of each circle. We can see that,
the area of 𝑢𝑢 = the area of 𝑠𝑠 + area of 𝑡𝑡.
iii)
1
1
area of 𝑢𝑢 = (area of 𝑠𝑠 + area of 𝑡𝑡)
2
2
𝐴𝐴3 + 𝐴𝐴4 + 𝐴𝐴5 = (𝐴𝐴1 + 𝐴𝐴4 ) + (𝐴𝐴5 + 𝐴𝐴2 )
𝐴𝐴3 = 𝐴𝐴1 + 𝐴𝐴2
We proved that the area of 𝑢𝑢 = the area of 𝑠𝑠 + the area
1
1
of 𝑡𝑡 in the last part. It follows that 𝑢𝑢 = (𝑠𝑠 + 𝑡𝑡).
2
1
We can see that 𝐴𝐴4 + 𝐴𝐴3 + 𝐴𝐴5 = 𝑢𝑢
1
2
2
Also 𝐴𝐴4 + 𝐴𝐴1 = 𝑠𝑠 as they cover the area of the circle
2
1
above the diameter. Similarly, 𝐴𝐴5 + 𝐴𝐴2 = 𝑡𝑡.
2
Taking 𝐴𝐴4 and 𝐴𝐴5 from both sides leaves us with
𝐴𝐴3 = 𝐴𝐴1 + 𝐴𝐴2 .
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403
2013 Paper 2
Question 1
a) i)
A sample space is the set of all possible outcomes of an experiment.
ii)
Mutually exclusive events are events which have no outcomes in common.
i.e. 𝑃𝑃(𝐴𝐴 ∪ 𝐵𝐵) = 𝑃𝑃(𝐴𝐴) + 𝑃𝑃(𝐵𝐵)
iii)
Two events are independent if the outcome of one does not depend on the outcome of the other.
i.e. 𝑃𝑃(𝐴𝐴 ∩ 𝐵𝐵) = 𝑃𝑃(𝐴𝐴) × 𝑃𝑃(𝐵𝐵)
b) i)
ii)
𝑃𝑃(𝐸𝐸 ∩ 𝐹𝐹) =
4
30
𝑃𝑃(𝐸𝐸) × 𝑃𝑃(𝐹𝐹) =
4
20 6
×
=
30 30 30
𝑃𝑃(𝐸𝐸 ∩ 𝐹𝐹) = 𝑃𝑃(𝐸𝐸) × 𝑃𝑃(𝐹𝐹) ∴ The events are independent.
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The probability of 𝑃𝑃(𝐸𝐸 ∩ 𝐹𝐹) is the number of
students who study biology and physics over the
total number of students.
Multiplying the probability of a student studying
Physics by the probability of a student studying
Biology. This equals 𝑃𝑃(𝐸𝐸 ∩ 𝐹𝐹), therefore the
events are independent.
404
Question 2
a) i)
𝑥𝑥 − 𝜇𝜇
𝜎𝜎
68 − 60
= 1.6
5
→ 𝟎𝟎. 𝟗𝟗𝟗𝟗𝟗𝟗𝟗𝟗.
ii) 𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
𝑥𝑥 − 𝜇𝜇
𝜎𝜎
52 − 60
= 1.6
5
→ 0.9452.
1 − 0.9452 = 0.0548
0.9452 − 0.0548 = 𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖
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Taking the equation from page 35 of the Maths Tables Book, without the √𝑛𝑛
as we are not dealing with a sample. Plugging in the standard deviation, the
mean and the number given.
To find the probability of the mean being less than 68 we need to convert the
z-score this gives us, using pages 36 & 37 of the Maths Tables Book.
We know the probability of the mean being less than 68 from the previous part.
To find the probability of it being in the given range we just need to find the
probability of the mean being less than 52 and take this from our answer in part
i). This will leave us with the probability of the mean being less than 68 and
greater than 52.
Following the same method as in part i) we find the probability of the mean
being greater than 52.
Taking this probability from 1 gives us the probability of the mean being less
than 52.
Finally taking this number from our answer from part i) gives us our answer.
405
b)
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406
Question 3
a)
Description
Line(s)
A line with a slope of 2
𝑙𝑙
1
𝑙𝑙
A line which intersects the y-axis at (0, −2 )
2
A line which makes equal intercepts in the axes
A line which makes an angle of 150° with the positive sense of the x-axis
Two lines which are perpendicular to each other.
𝑙𝑙: 4𝑥𝑥 − 2𝑦𝑦 − 5 = 0
𝑚𝑚
𝑙𝑙 and 𝑘𝑘
If we rewrite the line 𝑙𝑙 in the form 𝑦𝑦 =
𝑚𝑚𝑚𝑚 + 𝑐𝑐, the number in front of the 𝑥𝑥 is
the slope of the line.
2𝑦𝑦 = 4𝑥𝑥 − 5
𝑦𝑦 = 2𝑥𝑥 −
ℎ
5
→ slope = 2
2
1
5
𝑦𝑦 − intercept → �0, − � = �0, −2 �
2
2
ℎ: 𝑥𝑥 = 3 − 𝑦𝑦
Also, when the line is in this format the
𝑦𝑦 − intercept is (0, 𝑐𝑐).
𝑦𝑦 = −𝑥𝑥 + 3
Rewriting the line ℎ in the form 𝑦𝑦 =
𝑚𝑚𝑚𝑚 + 𝑐𝑐 to find the 𝑦𝑦 − intercept.
𝑥𝑥 − intercept → (0) = −𝑥𝑥 + 3
We can see that the two values are the
same.
𝑦𝑦 − intercept = (0,3)
𝑥𝑥 = 3 → (3,0)
𝑚𝑚: 𝑥𝑥 + √3𝑦𝑦 − 10 = 0
𝑦𝑦 = −
1
𝑥𝑥 +
10
→ 𝑚𝑚 = −
√3
√3
𝑚𝑚1 − 𝑚𝑚2
tan 𝜃𝜃 =
1 + 𝑚𝑚1 𝑚𝑚2
1
√3
1
1
−
−0
√3
tan 𝜃𝜃 =
= √3
1
1
1 + � � (0)
√3
−
𝜃𝜃 = tan−1 �−
1
√3
� = 30°
© Pocket Tutor 2022
Then plugging in 0 for 𝑦𝑦 to find the 𝑥𝑥 −
intercept.
First writing the line 𝑚𝑚 in the form 𝑦𝑦 =
𝑚𝑚𝑚𝑚 + 𝑐𝑐 to get the slope.
Then using the formula from page 19 of
the Maths Tables Book, to find the angle
between this line and the x – axis. (The
slope of the x – axis is 0).
Finding the tan inverse gives us 30° but
−
1
√3
is in the second quadrant, so to find
180 − 30 = 150° the angle we take 30 away from 180.
Rewriting the line 𝑘𝑘 in the form 𝑦𝑦 = 𝑚𝑚𝑚𝑚
407+ 𝑐𝑐 in
order to find the slope.
1
𝑘𝑘: 𝑦𝑦 = − (2𝑥𝑥 − 7)
4
1
7
𝑘𝑘: 𝑦𝑦 = − 𝑥𝑥 +
2
4
𝑚𝑚 = −
1
2
Slope of 𝑘𝑘 × slope of 𝑙𝑙:
1
− × 2 = −1 ∴ perpindicular
2
b) 𝟑𝟑𝟑𝟑°
𝑚𝑚: 𝑥𝑥 + √3𝑦𝑦 − 10 = 0
𝑦𝑦 = −
1
√3
→ 𝑚𝑚 = −
𝑥𝑥 +
1
√3
10
√3
𝑛𝑛: √3𝑥𝑥 + 𝑦𝑦 − 10 = 0
𝑦𝑦 = −√3 + 10
→ 𝑚𝑚 = −√3
tan 𝜃𝜃 = ±
𝑚𝑚1 − 𝑚𝑚2
1 + 𝑚𝑚1 𝑚𝑚2
1
− �−√3�
√3
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = ±
1
1 + �− � �−√3�
√3
−
1
− �−√3�
√3
θ = tan−1 �
�
1
1 + �− � �−√3�
√3
−
𝜃𝜃 = 𝟑𝟑𝟑𝟑°
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Rewriting 𝑚𝑚 in the form 𝑦𝑦 = 𝑚𝑚𝑚𝑚 + 𝑐𝑐 to find its
slope.
Doing the same for 𝑛𝑛.
Taking the equation for the angle between two
lines from page 19 of the Maths Tables Book.
Subbing in our two slopes.
Finding the tan inverse of this expression using
our calculator.
This is the acute angle between the two lines.
408
Question 4
a)
𝑐𝑐1 :
2
2
(𝑥𝑥 − ℎ) + (𝑦𝑦 − 𝑘𝑘) = 𝑟𝑟
(ℎ, 𝑘𝑘) = (−3, −2), 𝑟𝑟 = 2
Taking the equation of a circle from page 19 of the
Maths Tables Book.
2
2
2
�𝑥𝑥 − (−3)� + �𝑦𝑦 − (−2)� = (2)2
(𝒙𝒙 + 𝟑𝟑)𝟐𝟐 + (𝒚𝒚 + 𝟐𝟐)𝟐𝟐 = 𝟒𝟒
𝑐𝑐2 : (𝟏𝟏, 𝟏𝟏), 𝒓𝒓 = 𝟑𝟑
𝑥𝑥 2 + 𝑦𝑦 2 + 2𝑔𝑔𝑔𝑔 + 2𝑓𝑓𝑓𝑓 + 𝑐𝑐
Subbing in our values for ℎ, 𝑘𝑘 and 𝑟𝑟.
Taking the other form of the equation of a circle
from page 19 of the Maths Tables Book.
𝑥𝑥 2 + 𝑦𝑦 2 − 2𝑥𝑥 − 2𝑦𝑦 − 7 = 0
centre = (−g, −f)
2𝑔𝑔 = −2
−𝑔𝑔 = 1
(𝟏𝟏, 𝟏𝟏)
2𝑓𝑓 = −2
− 𝑓𝑓 = 1
Taking the equation for the radius of a circle from
page 19 of the Maths Tables Book.
radius = �𝑔𝑔2 + 𝑓𝑓 2 − 𝑐𝑐
�(−1)2 + (−1)2 − (−7)
√9 = 𝟑𝟑
𝐛𝐛) 𝐢𝐢) �
−𝟕𝟕 −𝟒𝟒
�
,
𝟓𝟓 𝟓𝟓
Ratio of line from 𝑐𝑐1 (−3, −2) to c2 (1,1) = 2: 3
�
𝑏𝑏𝑦𝑦1 + 𝑎𝑎𝑦𝑦2
𝑏𝑏𝑥𝑥1 + 𝑎𝑎𝑥𝑥2
�,�
�
𝑏𝑏 + 𝑎𝑎
𝑏𝑏 + 𝑎𝑎
𝑎𝑎 = 2, 𝑏𝑏 = 3, 𝑥𝑥1 = −3, 𝑦𝑦1 = −2, 𝑥𝑥2 = 1, 𝑦𝑦2 = 1
�
(3)(−3) + (2)(1)
(3)(−2) + 2(1)
�,�
�
3+2
3+2
−𝟕𝟕 −𝟒𝟒
�
�
,
𝟓𝟓 𝟓𝟓
© Pocket Tutor 2022
Given that 𝑐𝑐1 has a radius of 2 and 𝑐𝑐3 has a radius of 3 we
know that there are 5 units between their centres. We
also know that the line between their centres is in the
ratio of 2: 3 due to the lengths of their radii.
Using this information, we can use the equation for finding
a point dividing a line segment in a given ratio, to find the
point of contact. (Page 18 of the Maths Tables Book).
Subbing the ratio and the two centres into this equation
gives us our answer.
409
ii) 𝟒𝟒𝒙𝒙 + 𝟑𝟑𝟑𝟑 + 𝟖𝟖 = 𝟎𝟎
slope of line between the centres:
𝑦𝑦2 − 𝑦𝑦1
𝑥𝑥2 − 𝑥𝑥1
1 − (−2) 3
=
1 − (−3) 4
slope of tangent:
−
4
3
Subbing in the centre coordinates to find the slope
between the centres.
A tangent to a circle is perpendicular to a line from
the centre of the circle. So, the slope of the tangent
and the slope we found multiply together to give
−1.
So, to find the slope of the tangent we invert the
slope we found and change the sign.
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
4
7
4
𝑦𝑦 − �− � = − �𝑥𝑥 − �− ��
3
5
5
𝑦𝑦 +
Taking the equation for the slope of a line from page
18 of the Maths Tables Book.
7
4
4
= − �𝑥𝑥 + �
5
5
3
3𝑦𝑦 +
3𝑦𝑦 +
12
7
= −4 �𝑥𝑥 + �
5
5
12
28
= −4𝑥𝑥 −
5
5
Taking the equation of a line from page 18 of the
Maths Tables Book.
Plugging in the point of contact of the two circles we
found in part b) i), and the slope we just found.
Multiplying across by 3.
Tidying up.
𝟒𝟒𝟒𝟒 + 𝟑𝟑𝟑𝟑 + 𝟖𝟖 = 𝟎𝟎
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410
Question 5
a)
1
1
1
𝑎𝑎𝑎𝑎 sin 𝐶𝐶 = 𝑎𝑎𝑎𝑎 sin 𝐵𝐵 = 𝑏𝑏𝑏𝑏 sin 𝐴𝐴
2
2
2
1
1
1
𝑎𝑎𝑎𝑎 sin 𝐶𝐶
𝑎𝑎𝑎𝑎 sin 𝐵𝐵
𝑏𝑏𝑏𝑏 sin 𝐴𝐴
2
2
=
=2
1
1
1
𝑎𝑎𝑎𝑎𝑎𝑎
𝑎𝑎𝑎𝑎𝑎𝑎
𝑎𝑎𝑎𝑎𝑎𝑎
2
2
2
1
1
1
𝑎𝑎𝑎𝑎 sin 𝐶𝐶
𝑎𝑎𝑎𝑎 sin 𝐵𝐵
𝑏𝑏𝑏𝑏 sin 𝐴𝐴
2
2
=
=2
1
1
1
𝑎𝑎𝑎𝑎𝑐𝑐
𝑎𝑎𝑏𝑏𝑐𝑐
𝑎𝑎𝑏𝑏𝑏𝑏
2
2
2
Taking the formula for the area of a
triangle from page 9 of the Maths Tables
Book.
1
Dividing across by 𝑎𝑎𝑎𝑎𝑎𝑎
2
Cancelling out the relevant terms in each
fraction.
sin 𝐶𝐶 sin 𝐵𝐵 sin 𝐴𝐴
=
=
𝑏𝑏
𝑎𝑎
𝑐𝑐
→
𝑐𝑐
𝑏𝑏
𝑎𝑎
=
=
sin 𝐶𝐶 sin 𝐵𝐵 sin 𝐴𝐴
Inverting the fractions to put it in the form
given in the question.
b) i) 𝟒𝟒𝟒𝟒°, 𝟏𝟏𝟏𝟏𝟏𝟏°
𝑏𝑏
𝑎𝑎
=
sin 𝐴𝐴 sin 𝐵𝐵
Taking the sin rule from page 16 of the Maths Tables
Book.
5
3
=
sin(27) sin 𝑍𝑍
Subbing in the given values, remember that the angle 𝐴𝐴
is the one opposite the side 𝑎𝑎.
sin(27) sin 𝑍𝑍
=
5
3
sin(27)
5�
� = sin 𝑍𝑍
3
0.75665 = sin 𝑍𝑍
𝑍𝑍 = sin−1 (0.75665)
𝑍𝑍 = 𝟒𝟒𝟒𝟒°
and 180 − 49 = 𝟏𝟏𝟏𝟏𝟏𝟏°
© Pocket Tutor 2022
Inverting the fractions.
Multiplying across by 5.
Plugging into the calculator.
Finding the sin inverse to find Z.
Z has two possible values as sin gives us two values
which are less than 180°. The reference angle: 49 and
the angle sin gives in the second quadrant: 180 − 49.
411
ii)
c) 𝟕𝟕𝟕𝟕𝒎𝒎𝟐𝟐
|< 𝑋𝑋𝑋𝑋𝑋𝑋| = 27°
|< 𝑋𝑋𝑋𝑋𝑋𝑋| = 49°
|< 𝑍𝑍𝑍𝑍𝑍𝑍| = 180 − 27 − 49 = 104°
Area =
1
𝑎𝑎𝑎𝑎 sin 𝐶𝐶
2
1
(3)(5) sin 104
2
The value of |< 𝑋𝑋𝑋𝑋𝑋𝑋| which is less than 90° is 49°.
The third angle in the triangle is equal to 180
degrees minus the other two angles.
Taking the equation for the area of a triangle form
page 16 of the Maths Tables Book.
Subbing in the lengths of the sides given in part b)
i), and the angle we just found.
= 7.27 → 𝟕𝟕𝟕𝟕𝒎𝒎𝟐𝟐
© Pocket Tutor 2022
412
Question 6
a)
i)
the perpendicular bisectors of the sides of the triangle
ii)
the bisectors of the angles of the triangle
iii)
the medians of the triangle
b)
In an equilateral triangle the medians are perpendicular to the opposite sides and bisect the angles.
Therefore, the perpendicular bisectors of the sides, the bisectors of the angles and the median are
the same line and intersect at one point.
© Pocket Tutor 2022
413
Question 7
a) i)
The population is divided into different subgroups which have common characteristics. Random
samples are drawn from each subgroup according to their proportion of the population.
ii)
For Example:
Long haul economy class passengers
Long haul business class passengers
Long haul executive class passengers
Short haul passengers
b) i)
231
= 𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐
1000
238
= 𝟎𝟎. 𝟐𝟐𝟐𝟐𝟐𝟐
1000
Putting the number of passengers whose flights were delayed
over the total number of passengers surveyed.
Putting the number of passengers who were not satisfied over
the total number of passengers surveyed.
ii)
No as this would imply that the events are independent. It is unlikely that the events are
independent as it is likely that a passenger whose flight was delayed would not be satisfied with the
service.
c)
Graph ii)
A lot of passengers are likely to have baggage which weighs less than 20kg as this is the baggage
allowance and there is a cost for baggage which weighs more than this.
© Pocket Tutor 2022
414
d) i)
ii)
We can see from the data that the median age is less than the mean age. This means that the graph
is skewed to the right.
e) i)
Null hypothesis: 70% of customers are satisfied with the company’s overall service.
95% confidence interval:
1
664
±
1000 √1000
0.664 ± 0.032
Using
1
√𝑛𝑛
to find the margin of error at a 5% level of
significance.
0.664 − 0.032 = 0.632
Constructing a confidence interval using this margin of
error, by adding it and taking it from the proportion of
people who were satisfied with the service.
0.632 < 𝑥𝑥 < 0.696
The company’s claim does not lie within this
confidence interval therefore we reject the null
hypothesis.
0.664 + 0.032 = 0.696
0.7 is outside this range:
∴ Reject the null hypothesis.
There is evidence to conclude that the company’s claim is not valid in May.
© Pocket Tutor 2022
415
ii)
No
1
1
1
×
≠
2 √1000 √2000
0.0158 ≠ 0.022
f) i)
(Line of best fit is for part iv)
ii)
0.88
You can revise how to use your calculator to do this in
section 7.3 of Statistics.
iii)
Older passengers tend to spend more.
© Pocket Tutor 2022
416
Question 8
NOTE DUE TO AN ERROR IN THIS QUESTION, DIFFERENT VALID APPROACHES GIVE DIFFERENT
ANSWERS.
a) i)
𝑠𝑠𝑠𝑠𝑠𝑠𝜃𝜃 =
oppposite
hypotenuse
sin(36) =
𝑥𝑥
80
80 sin 36 = 𝑥𝑥
𝑥𝑥 = 𝟒𝟒𝟒𝟒. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
80
90°
𝑥𝑥
110
ii)
𝑎𝑎2 = 𝑏𝑏 2 + 𝑐𝑐 2 − 2𝑏𝑏𝑏𝑏 cos 𝐴𝐴
Taking the cosine rule from page 16 of the
Maths Tables Book.
|𝐻𝐻𝐻𝐻|2 = 28341.8
Plugging in the given sides and the angle
opposite the side we are trying to find.
|𝐻𝐻𝐻𝐻|2 = (80)2 + (110)2 − 2(80)(110) cos 124
|𝐻𝐻𝐻𝐻| = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟑𝟑𝟑𝟑km
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Square rooting both sides.
Different methods give different correct
answers to this question.
417
b) 𝟑𝟑𝟑𝟑. 𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓
110
80
80
168.35
𝑎𝑎2 = 𝑏𝑏 2 + 𝑐𝑐 2 − 2𝑏𝑏𝑏𝑏 cos 𝐴𝐴
(80)2 = (110)2 + (168.35)2 − 2(110)(168.35) cos | < 𝑇𝑇𝑇𝑇𝑇𝑇|
6400 = 40441.7225 − 37037 cos|< 𝑇𝑇𝑇𝑇𝑇𝑇|
−34041.7225 = −37037 cos | < 𝑇𝑇𝑇𝑇𝑇𝑇|
−34041.7225
= cos|< 𝑇𝑇𝑇𝑇𝑇𝑇|
−37037
By using the cosine rule we can find the angle
| < 𝑇𝑇𝑇𝑇𝑇𝑇| which will allow us to find the angle
| < 𝑅𝑅𝑅𝑅𝑅𝑅|. Which in turn will allow us to find
|𝑅𝑅𝑅𝑅|.
Filling in the cosine rule with the distances
given and the length of |𝐻𝐻𝐻𝐻| we found in the
previous part.
0.9191 = cos|< 𝑇𝑇𝑇𝑇𝑇𝑇|
|< 𝑇𝑇𝑇𝑇𝑇𝑇| = cos −1 (0.9191)
|< 𝑇𝑇𝑇𝑇𝑇𝑇| = 23.205
|< 𝑅𝑅𝑅𝑅𝑅𝑅| = 36 − |< 𝑇𝑇𝑇𝑇𝑇𝑇| → 36 − 23.205 = 12.795°
|𝑅𝑅𝑅𝑅|2 = (110)2 + (80)2 − 2(110)(80) cos 12.795
|𝑅𝑅𝑇𝑇|2 = 1337.0
|𝑅𝑅𝑅𝑅| = 𝟑𝟑𝟑𝟑. 𝟓𝟓𝟓𝟓𝟓𝟓𝟓𝟓
© Pocket Tutor 2022
Taking the angle | < 𝑇𝑇𝑇𝑇𝑇𝑇| away from the
angle | < 𝑅𝑅𝑅𝑅𝑅𝑅| which was given in part a) of
the question, gives us | < 𝑅𝑅𝑅𝑅𝑅𝑅|.
Now, taking the triangle 𝑅𝑅𝑅𝑅𝑅𝑅, we can use the
cosine rule again to find |𝑅𝑅𝑅𝑅|. |𝑅𝑅𝑅𝑅| = 80
(from part 𝒂𝒂).
Again, there are multiple correct answers for
this part.
418
Question 9
a) 𝒌𝒌 = 𝟒𝟒√𝟐𝟐
|𝑋𝑋𝑋𝑋|2 = 42 + 𝑘𝑘 2
2
𝑘𝑘 =
|𝑋𝑋𝑋𝑋|2
Using Pythagoras theorem to get 𝑘𝑘 in terms
of |𝑋𝑋𝑋𝑋|.
− 16
Using Pythagoras theorem to get 𝑘𝑘 in terms
of |𝑋𝑋𝑋𝑋|.
|𝑋𝑋𝑋𝑋|2 = 82 + 𝑘𝑘 2
𝑘𝑘 2 = |𝑋𝑋𝑋𝑋|2 − 64
|𝑋𝑋𝑍𝑍|2 + |𝑋𝑋𝑋𝑋|2 = (4 + 8)2
|𝑋𝑋𝑍𝑍|2 + |𝑋𝑋𝑋𝑋|2 = 144
𝑘𝑘 2 = |𝑋𝑋𝑋𝑋|2 − 16
2
𝑘𝑘 = |𝑋𝑋𝑋𝑋|2 − 64
2𝑘𝑘 2 = |𝑋𝑋𝑋𝑋|2 + |𝑋𝑋𝑋𝑋|2 − 80
2𝑘𝑘 2 = (144) − 80
𝐾𝐾 2 = 64
2
𝑘𝑘 = 32
Using Pythagoras theorem to get a value for
|𝑋𝑋𝑌𝑌|2 + |𝑋𝑋𝑋𝑋|2 .
Adding our two expressions for 𝑘𝑘.
Subbing 144 in for |𝑋𝑋𝑌𝑌|2 + |𝑋𝑋𝑋𝑋|2 .
Dividing by 2.
Square rooting both sides.
𝑘𝑘 = 𝟒𝟒√𝟐𝟐
b) i)
Length of semi circle =
1
(2𝜋𝜋𝜋𝜋) → 𝜋𝜋𝜋𝜋
2
1
Length of a circle
2
𝑟𝑟2 + 𝑟𝑟3 = 𝑟𝑟1
Equation for the length (circumference)
of a circle from page 8 of the Maths
Tables Book.
From the diagram we can see that 2𝑟𝑟2 +
2𝑟𝑟3 = 2𝑟𝑟1 ∴ 𝑟𝑟2 + 𝑟𝑟3 = 𝑟𝑟1 .
𝜋𝜋𝑟𝑟1 + 𝜋𝜋𝑟𝑟2 + 𝜋𝜋𝑟𝑟3
The perimeter of the arbelos is equal to
the lengths of the three semi circles
added together.
𝜋𝜋𝑟𝑟1 + 𝜋𝜋(𝑟𝑟1 )
Factorising out the 𝜋𝜋
𝜋𝜋𝑟𝑟1 + 𝜋𝜋(𝑟𝑟2 + 𝑟𝑟3 )
= 2𝜋𝜋𝑟𝑟1 which is independent of r2 and r3 .
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Subbing in in 𝑟𝑟1 for (𝑟𝑟2 + 𝑟𝑟3 ).
419
ii)
Area of arbelos =
𝑟𝑟1 = 2 + 4 = 6
1
1
1
𝜋𝜋𝑟𝑟1 2 − ( 𝜋𝜋𝑟𝑟2 2 + 𝜋𝜋𝑟𝑟3 2 )
2
2
2
The area of the arbelos is equal to the area
of the semi-circle with radius 𝑟𝑟1 minus the
areas of the two semi-circles with radii 𝑟𝑟2
and 𝑟𝑟3 .
The equation for the area of a semi-circle is
equal to the area of a circle, 𝜋𝜋𝑟𝑟 2 , divided by
2.
1
1
1
𝜋𝜋(6)2 − ( 𝜋𝜋(2)2 + 𝜋𝜋(4)2 )
2
2
2
18𝜋𝜋 − (10𝜋𝜋) = 8𝜋𝜋
Subbing in the lengths of each radius.
𝑘𝑘 = 4√2
From part 𝑎𝑎 we know that 𝑘𝑘 = 4√2. If this is
𝑘𝑘 4√2
= 2√2
𝑟𝑟 = →
2
2
𝑘𝑘
the diameter the radius is .
2
2
Subbing this into the equation for the area of
a circle from page 8 of the Maths Tables
Book
𝜋𝜋𝑟𝑟 2 → 𝜋𝜋�2√2� = 8𝜋𝜋
8𝜋𝜋 = 8𝜋𝜋
c) i)
𝒓𝒓𝟐𝟐
1
6 − 1 = 𝟓𝟓
1
1
1
𝜋𝜋(6)2 − � 𝜋𝜋(1)2 + 𝜋𝜋(5)2 � = 𝟓𝟓𝟓𝟓𝟓𝟓𝒎𝒎𝟐𝟐
2
2
2
6
2
6 − 2 = 𝟒𝟒
1
1
1
𝜋𝜋(6)2 − � 𝜋𝜋(2)2 + 𝜋𝜋(4)2 � = 𝟖𝟖𝟖𝟖𝟖𝟖𝒎𝒎𝟐𝟐
2
2
2
6
3
6 − 3 = 𝟑𝟑
1
1
1
𝜋𝜋(6)2 − ( 𝜋𝜋(3)2 + 𝜋𝜋(3)2 ) = 𝟗𝟗𝟗𝟗𝟗𝟗𝒎𝒎𝟐𝟐
2
2
2
6
4
6 − 4 = 𝟐𝟐
1
1
1
𝜋𝜋(6)2 − ( 𝜋𝜋(4)2 + 𝜋𝜋(2)2 ) = 𝟖𝟖𝟖𝟖𝟖𝟖𝒎𝒎𝟐𝟐
2
2
2
6
5
6 − 5 = 𝟏𝟏
1
1
1
𝜋𝜋(6)2 − ( 𝜋𝜋(5)2 + 𝜋𝜋(1)2 ) = 𝟓𝟓𝟓𝟓𝟓𝟓𝒎𝒎𝟐𝟐
2
2
2
6
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𝒓𝒓𝟑𝟑
Area of arbelos
𝒓𝒓𝟏𝟏
420
ii) 𝝅𝝅(𝟔𝟔𝟔𝟔 − 𝒙𝒙𝟐𝟐 )
𝑟𝑟1 = 6, 𝑟𝑟2 = 𝑥𝑥, 𝑟𝑟3 = 6 − 𝑥𝑥
1
1
1
𝜋𝜋𝑟𝑟1 2 − ( 𝜋𝜋𝑟𝑟2 2 + 𝜋𝜋𝑟𝑟3 2 )
2
2
2
1
1
1
𝜋𝜋(6)2 − � 𝜋𝜋(𝑥𝑥)2 + 𝜋𝜋(6 − 𝑥𝑥)2 �
2
2
2
1
𝜋𝜋�(6)2 − (𝑥𝑥 2 + (6 − 𝑥𝑥)2 )�
2
1
𝜋𝜋(36 − (𝑥𝑥 2 + 36 + 𝑥𝑥 2 − 12𝑥𝑥))
2
1
𝜋𝜋(12𝑥𝑥 − 2𝑥𝑥 2 )
2
𝟐𝟐
𝝅𝝅(𝟔𝟔𝟔𝟔 − 𝒙𝒙 )
If 𝑟𝑟1 = 6 and 𝑟𝑟2 = 𝑥𝑥 then 𝑟𝑟3 = 6 − 𝑥𝑥
Taking the formula for the area of the arbelos
from the part 𝑏𝑏) 𝑖𝑖𝑖𝑖) and plugging in 6, 𝑥𝑥 and (6 −
𝑥𝑥) for 𝑟𝑟1 , 𝑟𝑟2 and r3 respectively.
1
Factorising out the 𝜋𝜋.
2
Squaring out the bracket.
Tidying up.
Multiplying in by the
1
2
iii) 𝟗𝟗𝟗𝟗𝟗𝟗𝒎𝒎𝟐𝟐
𝜋𝜋(6𝑥𝑥 − 𝑥𝑥 2 )
6𝜋𝜋𝜋𝜋 − 𝜋𝜋𝑥𝑥 2
𝑑𝑑𝑑𝑑
= 6𝜋𝜋 − 2𝜋𝜋𝜋𝜋
𝑑𝑑𝑑𝑑
6𝜋𝜋 − 2𝜋𝜋𝜋𝜋 = 0
6𝜋𝜋 = 2𝜋𝜋𝜋𝜋
6𝜋𝜋
𝑥𝑥 =
2𝜋𝜋
To find the maximum value of a function we
differentiate it then let it equal 0.
Multiplying out the bracket first makes it easier to
differentiate.
Differentiating it and letting it equal 0.
Solving for 𝑥𝑥.
𝑥𝑥 = 3
This is the 𝑥𝑥 value which gives the maximum area.
𝐴𝐴 = 𝜋𝜋(6𝑥𝑥 − 𝑥𝑥 2 )
Subbing this 𝑥𝑥 value into the equation we found
for the area gives us our final answer.
𝜋𝜋(6(3) − (3)2 ) = 𝟗𝟗𝟗𝟗𝟗𝟗𝒎𝒎𝟐𝟐
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421
d)
|< 𝑇𝑇𝑇𝑇𝑇𝑇| = 90°
|< 𝐶𝐶𝑇𝑇𝑇𝑇| = 90°
Hence, |< 𝑆𝑆𝑆𝑆𝑆𝑆| = 90°
|< 𝐹𝐹𝐹𝐹𝐹𝐹| = 90°
Hence, |< CRS| = 90°
If a triangle has base vertices at the edges of a semi-circle
the angle at the circumference of the semi-circle will be
90°.
If |< 𝐶𝐶𝐶𝐶𝐶𝐶| = 90° so must |< 𝑆𝑆𝑆𝑆𝑆𝑆| as they are on a straight
line and must add up to 180°.
Same reason for | < 𝐶𝐶𝐶𝐶𝐶𝐶|
Hence the angles in 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 are right angles and so 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 is a rectangle.
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422
2012 Paper 2
Question 1
a)
Any 3 of the following:
1. Check whether both pairs of opposite sides have the same slope using the slope formula.
2. Check whether both pairs of opposite sides are equal in length using the distance formula.
3. Check whether the midpoints of the diagonals coincide as diagonals in a parallelogram bisect each
other.
4. Check whether the translation from 𝐴𝐴 to 𝐵𝐵 is the same as the translation from 𝐷𝐷 to 𝐶𝐶
5. Check whether a pair of opposite sides have the same slope and are equal in length (slope and
distance formulae).
b)
Taking test 3:
Midpoints of diagonals:
�
𝑥𝑥1 + 𝑥𝑥2 𝑦𝑦1 + 𝑦𝑦2
�
,
2
2
Midpoint between (−4, −2) and (8,7):
�
−4 + 8 −2 + 7
�
,
2
2
5
4 5
� , � → �2, �
2
2 2
Midpoint between (21, −5) and (−17,10):
�
21 − 17 −5 + 10
�
,
2
2
5
4 5
� , � → �2, �
2
2 2
We can show that a quadrilateral is a
parallelogram by showing that the
midpoints between opposite corners are
the same. This is because diagonals in a
parallelogram bisect each other.
Taking the equation for the midpoint
between two points from page 18 of the
Maths Tables Book.
Subbing in the coordinates of the opposing
corners to find the two midpoints.
As the midpoints are equal the
quadrilateral is a parallelogram.
Midpoints are equal ∴ Parallelogram
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423
Question 2
a)
𝒄𝒄𝟏𝟏 : 𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂 = (𝟑𝟑, 𝟓𝟓), 𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑 = √𝟓𝟓
𝒄𝒄𝟐𝟐 : 𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂 = (𝟏𝟏, 𝟏𝟏), 𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑 = 𝟑𝟑√𝟓𝟓
𝑐𝑐1 : 𝑥𝑥 2 + 𝑦𝑦 2 − 6𝑥𝑥 − 10𝑦𝑦 + 29 = 0
Centre = (−𝑔𝑔, −𝑓𝑓)
2𝑔𝑔 = −6
−𝑔𝑔 = 3
Centre = (3,5)
2𝑓𝑓 = −10
− 𝑓𝑓 = 5
Radius = �𝑔𝑔2 + 𝑓𝑓 2 − 𝑐𝑐
�(−3)2 + (−5)2 − 29 = √5
Taking circle one first.
The equation is written in the form of 𝑥𝑥 2 +
𝑦𝑦 2 + 2𝑔𝑔𝑔𝑔 + 2𝑓𝑓𝑓𝑓 + 𝑐𝑐
From page 19 of the Maths Tables Book, we
can see that the centre of the circle equals
(−𝑔𝑔, −𝑓𝑓) and the radius is found using the
expression: �𝑔𝑔2 + 𝑓𝑓 2 − 𝑐𝑐 = 𝑟𝑟.
Plugging the numbers into each of these
finds us the centre and the radius.
𝒄𝒄𝟏𝟏 : 𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂 = (𝟑𝟑, 𝟓𝟓), 𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑 = √𝟓𝟓
𝑐𝑐2 : 𝑥𝑥 2 + 𝑦𝑦 2 − 2𝑥𝑥 − 2𝑦𝑦 − 43 = 0
Centre = (−𝑔𝑔, −𝑓𝑓)
2𝑔𝑔 = −2
−𝑔𝑔 = 1
Centre = (1,1)
We repeat this process for circle two.
2𝑓𝑓 = −2
− 𝑓𝑓 = 1
Radius = �𝑔𝑔2 + 𝑓𝑓 2 − 𝑐𝑐
�(−1)2 + (−1)2 − (−43) = √45 = 3√5
𝒄𝒄𝟐𝟐 : 𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂𝐂 = (𝟏𝟏, 𝟏𝟏), 𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑𝐑 = 𝟑𝟑√𝟓𝟓
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424
b)
Distance between centres:
Centres = (3,5), (1,1)
To prove that the circles are touching we have to
show that the distance between the centres equals
either the sum of the two radii or the difference
between the two radii.
= 2√5
Taking the formula for the distance between two
points from page 18 of the Maths Tables Book and
plugging in the coordinates of the centres.
�(𝑥𝑥2 − 𝑥𝑥1 )2 + (𝑦𝑦2 − 𝑦𝑦1 )2
�(3 − 1)2 + (5 − 1)2 = √20
Radius of circle 2 – Radius of circle 1:
3√5 − √5 = 2√5
Finding the difference between the two radii.
This is the same as the distance between the
centres ∴ the circles are touching.
As the distance between the centres equals the difference between the two radii the circles touch.
c)
𝑐𝑐1 : 𝑥𝑥 2 + 𝑦𝑦 2 − 6𝑥𝑥 − 10𝑦𝑦 + 29 = 0
(4)2 + (7)2 − 6(4) − 10(7) + 29 = 0
16 + 49 − 24 − 70 + 29 = 0
𝑐𝑐2 : 𝑥𝑥 2 + 𝑦𝑦 2 − 2𝑥𝑥 − 2𝑦𝑦 − 43 = 0
(4)2 + (7)2 − 2(4) − 2(7) − 43 = 0
16 + 49 − 8 − 14 − 43 = 0
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To show that this point is on both circles we
plug 4 in for 𝑥𝑥 and 7 in for 𝑦𝑦 in both
equations and show that they equal 0.
If they equal 0 that means the point is on
the circle.
Both equations equal 0 for these
coordinates therefore both circles pass
through this point.
(This question can also be answered by
subtracting the equations of the two circles).
425
d) 𝒙𝒙 + 𝟐𝟐𝟐𝟐 − 𝟏𝟏𝟏𝟏 = 𝟎𝟎
Slope from centre, (3,5), to (4,7)
𝑦𝑦2 − 𝑦𝑦1
𝑥𝑥2 − 𝑥𝑥1
A tangent is perpendicular to a line from the centre of the
circle to the point of tangency.
7−5
=2
4−3
⊥ Slope = −
We can find the equation of the common tangent using the
common point. We just need to find the slope of the tangent.
So, if we find the slope from the centre of one of the circles to
the point of tangency and then find the perpendicular slope,
we have the slope of the tangent.
1
2
Equation of tangent:
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
1
𝑦𝑦 − (7) = − (𝑥𝑥 − 4)
2
2𝑦𝑦 − 14 = −1(𝑥𝑥 − 4)
2𝑦𝑦 − 14 = −𝑥𝑥 + 4
To find the perpendicular slope we invert the given slope and
change the sign.
Taking the equation of a line from page 18 of the Maths
Tables Book.
Plugging in the common point and the slope we found above.
Multiplying across by 2
(This question could also be answered by subtracting the
equations of the two circles).
𝒙𝒙 + 𝟐𝟐𝟐𝟐 − 𝟏𝟏𝟏𝟏 = 𝟎𝟎
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426
Question 3
𝟑𝟑 + 𝟐𝟐√𝟐𝟐
𝑥𝑥 + 𝑦𝑦 = 2
𝑎𝑎
𝑚𝑚 =
𝑟𝑟 − 0
=1
𝑟𝑟 − 0
𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 )
𝑦𝑦 − 0 = 1(𝑥𝑥 − 0)
𝑦𝑦 = 𝑥𝑥
𝑥𝑥 + 𝑦𝑦 = 2
(𝑦𝑦) + 𝑦𝑦 = 2
2𝑦𝑦 = 2
𝑦𝑦 = 1
𝑥𝑥 + 𝑦𝑦 = 2𝑘𝑘
Firstly, note that the circle is tangential to the axis, so its
centre point is the same distance, (the radius, 𝑟𝑟), from both
the x and y axis. This means that we can label the centre
coordinates (𝑟𝑟, 𝑟𝑟).
Now, we find the equation of the line 𝑎𝑎 by using the slope
𝑦𝑦 −𝑦𝑦
formula 𝑚𝑚 = 2 1 and the line formula 𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 ).
𝑥𝑥2 −𝑥𝑥1
Subbing in the origin (0,0) as and the centre (𝑟𝑟, 𝑟𝑟) as our
points. Line 𝑎𝑎 → 𝑦𝑦 = 𝑥𝑥
Now we can find the point where 𝑥𝑥 + 𝑦𝑦 = 2 intersects the
blue line 𝑦𝑦 = 𝑥𝑥 by solving the simultaneous equation.
(Subbing in 𝑦𝑦 for 𝑥𝑥)
These lines intersect at (1,1)
𝑥𝑥 = 1
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427
(𝑥𝑥 − 𝑟𝑟)2 + (𝑦𝑦 − 𝑟𝑟)2 = 𝑟𝑟 2
(1 − 𝑟𝑟)2 + (1 − 𝑟𝑟)2 = 𝑟𝑟 2
2
Next, we need to find the radius of the circle
2
1 − 2𝑟𝑟 + 𝑟𝑟 + 1 − 2𝑟𝑟 + 𝑟𝑟 = 𝑟𝑟
𝑟𝑟 2 − 4𝑟𝑟 + 2 = 0
𝑟𝑟 =
𝑟𝑟 =
𝑟𝑟 =
−𝑏𝑏 ±
√𝑏𝑏 2
2𝑎𝑎
− 4𝑎𝑎𝑎𝑎
2
−(−4) ± �(−4)2 − 4(1)(2)
2(1)
4 ± √16 − 8
2
𝑟𝑟 = 2 ± √2
𝑟𝑟 = 2 + √2
𝑘𝑘 + 1
= 𝑟𝑟
2
Using the circle formula from page 19 of the Maths
Tables Book and plugging in the point (𝑟𝑟, 𝑟𝑟) as its
centre. Subbing in the point (1,1) for 𝑥𝑥 and 𝑦𝑦.
Squaring out the brackets
Then using the minus b formula to find the radius.
Note 𝑟𝑟 = 2 − √2 is too small as this equals 0.59 and
we can tell by looking at the diagram and the point
(1,1) on the circle, and the centre of the circle that this
is not an option.
Next, we must find what k is.
𝑟𝑟, 𝑟𝑟 is the midpoint of the segment from
(1,1) 𝑡𝑡𝑡𝑡 (𝑘𝑘, 𝑘𝑘). Taking the midpoint formula:
𝑥𝑥1 + 𝑥𝑥2
= midpoint 𝑥𝑥 coordinate.
2
𝑘𝑘 + 1 = 2𝑟𝑟
(Page 18 of the Maths Table Book)
𝑘𝑘 = 2�2 + √2� − 1
Subbing in the value we found for 𝑟𝑟
𝑘𝑘 = 2𝑟𝑟 − 1
𝑘𝑘 = 𝟑𝟑 + 𝟐𝟐√𝟐𝟐
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Multiplying across by 2 and getting 𝑘𝑘 by itself.
428
Question 4
a)
Trials are independent of each other.
Probability of success is the same each time.
b) i) 𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑
𝑛𝑛
� � 𝑝𝑝𝑟𝑟 𝑞𝑞 𝑛𝑛−𝑟𝑟
𝑟𝑟
𝑝𝑝 = 0.6,
𝑞𝑞 = 1 − 0.6 = 0.4, 𝑛𝑛 = 6, 𝑟𝑟 = 4
6
� � (0.6)4 (0.4)6−4 = 𝟎𝟎. 𝟑𝟑𝟑𝟑𝟑𝟑
4
ii) 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎
𝑛𝑛
� � 𝑝𝑝𝑟𝑟 𝑞𝑞 𝑛𝑛−𝑟𝑟
𝑟𝑟
𝑝𝑝 = 0.6, 𝑞𝑞 = 0.4, 𝑛𝑛 = 4, 𝑟𝑟 = 1
4
� � (0.6)1 (0.4)4−1 = 0.1536
1
0.1536 × 0.6 = 0.09216
= 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎
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Taking the expression for Bernoulli trials from
page 33 of the Maths Tables Book.
(𝑛𝑛 = the number of trials, 𝑟𝑟 = the number of
successes, 𝑝𝑝 = the probability of success and
𝑞𝑞 = the probability of failure)
Plugging in our values.
To find the probability of her scoring for the
second time on the fifth shot we use Bernoulli
trials to find the probability of her scoring once
in four shots.
We then multiply this result by 0.6 to find the
odds of her scoring her second shot on the fifth
throw.
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Question 5
a) 𝟏𝟏𝟏𝟏𝟏𝟏 batteries
𝑧𝑧 =
𝑥𝑥 − 𝜇𝜇
𝜎𝜎
𝜎𝜎 = 0.1, 𝜇𝜇 = 20 𝑥𝑥 = 20 + 0.25 = 20.25
20.25 − 20
= 2.5
0.1
→ 0.9938
1 − 0.9938 = 0.0062
0.0062 × 2 = 0.0124
0.0124 × 10,000 = 𝟏𝟏𝟏𝟏𝟏𝟏 batteries
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Taking the formula for a z-score from page 35 of the
Maths Tables Book. However, leaving out the √𝑛𝑛 as
these values were not gotten from a sample.
Plugging in the given values.
Finding the value corresponding to 2.5 on pages 36 & 37
of the Maths Tables Book.
Taking this value from 1 to find the probability of the
battery having a diameter larger than the allowed limit.
Multiplying this number by 2 to include the probability
of batteries with a diameter smaller than the limit.
Multiplying this probability by 10,000 to find the
number of batteries which will be rejected.
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b) 𝟗𝟗𝟗𝟗. 𝟑𝟑𝟑𝟑% 𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢
𝑧𝑧 =
𝑥𝑥 − 𝜇𝜇
𝜎𝜎
Repeating the process, we just followed in part a).
𝜎𝜎 = 0.1, 𝜇𝜇 = 20.05 𝑥𝑥 = 20 + 0.25 = 20.25
Firstly, using the formula to find the probability of
the battery’s diameter being over 20.25 mm.
→ 0.9772
To do this find the probability corresponding to 2
on pages 36 & 37 of the Maths Tables Book, then
taking this value from one to find the probability of
the diameter being outside of 20.25.
20.25 − 20.05
=2
0.1
1 − 0.9772 = 0.0228
𝑥𝑥 − 𝜇𝜇
𝑧𝑧 =
𝜎𝜎
Then using the formula again to find the
probability of the battery’s diameter being under
19.75.
𝜎𝜎 = 0.1, 𝜇𝜇 = 20.05 𝑥𝑥 = 20 − 0.25 = 19.75
19.75 − 20.05
= −3
0.1
→ 0.9987
1 − 0.9987 = 0.0013
0.0013 + 0.0228 = 0.0241
0.0241 100
×
= 194.35% → 𝟗𝟗𝟗𝟗. 𝟑𝟑𝟑𝟑% 𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢𝐢
1
0.0124
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We can ignore the minus.
Taking the probability from one to find the
probability of it being outside of 19.75.
Adding the two probabilities.
Putting the result over the probability we found in
part a) and multiplying by 100 to find a percentage.
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Question 6
6a
a)
(i)
(ii)
Bisect 60° to get 30°; bisect again to get 15° (as shown above) OR
Construct a right angle and use it to construct 45° and combine with 60° to get 15°.
b)
NOTE: In 2012 there was a choice between Q6a and Q6b. We have excluded the solution to Q6b as
this material is no longer on the curriculum.
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Question 7
a) i)
The arrears rates have increased a lot, they have gone from mostly between 1 and 5 to between 5
and 15.
ii)
The interest rates have also increased a lot, they have gone from mostly between 2 and 4% to
mostly between 4 and 6%.
iii)
There appears to be a stronger relationship between interest rate and arrears in 2011 (we can see
this as the dots are forming a closer line in 2011 than in 2009).
b)
You would need to know how many mortgage holders are represented by each point on each
diagram.
c)
The direction of causality is a question of whether higher arrears rates cause interest rates to go up
or whether higher interest rates cause higher arrears rates.
d) i)
145,414
= 𝟎𝟎. 𝟑𝟑𝟑𝟑
475,136
ii)
11,644
= 𝟎𝟎. 𝟎𝟎𝟎𝟎
145,414
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iii) 𝟎𝟎. 𝟒𝟒𝟒𝟒
No. in arrears or in negative equity = total – number in neither arrears nor negative equity
→ 475,136 − 317,355 = 157,781
Number in arrears only = number in negative equity or arrears – number in negative equity
157,781 − 145,414 = 12,367
Total number in arrears = Number in arrears only + number in negative equity and arrears
12,367 + 11,644 = 24,011 in arrears
Finding the probability by putting the number in both negative equity and arrears over the total
number in arrears.
11,644
= 𝟎𝟎. 𝟒𝟒𝟒𝟒
24,011
e)
Null hypothesis: The proportion in negative equity is 0.31
Alternative hypothesis: The proportion in negative equity is not 0.31
552
= 0.276
2000
0.276 ±
1
√2000
0.276 ± 0.0224
0.276 + 0.0224 = 0.298
0.276 − 0.0224 = 0.254
0.254 < 𝑝𝑝 < 0.298
∴ We reject the null hypothesis.
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Stating the null and alternative
hypothesis.
We can test the hypotheses by
constructing a 95% confidence interval.
Finding the proportion of houses in
negative equity in the sample.
Finding the margin of error using the
formula
1
√𝑛𝑛
.
Adding and taking away the result to
construct a confidence interval.
0.31 does not fall within this interval so
we can reject the null hypothesis. There
is evidence that the proportion has
changed.
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Question 8
a) 𝟒𝟒𝟒𝟒°
|𝑃𝑃𝑃𝑃|2 = 72 + 242
|𝑃𝑃𝑃𝑃|2 = 625
|𝑃𝑃𝑃𝑃| = 25
𝑎𝑎2 = 𝑏𝑏 2 + 𝑐𝑐 2 − 2𝑏𝑏𝑏𝑏 cos 𝐴𝐴
(25)2 = (20)2 + (12)2 − 2(20)(12)𝐶𝐶𝐶𝐶𝐶𝐶 𝛽𝛽
625 − 400 − 144 = −2(20)(12)𝐶𝐶𝐶𝐶𝐶𝐶 𝛽𝛽
81
= cos 𝛽𝛽
−2(20)(12)
cos 𝛽𝛽 = −0.16875
𝛽𝛽 = cos −1 (−0.16785) = 𝟏𝟏𝟏𝟏𝟏𝟏°
𝑎𝑎2 = 𝑏𝑏 2 + 𝑐𝑐 2 − 2𝑏𝑏𝑏𝑏 cos 𝐴𝐴
(12)2 = (25)2 + (20)2 − 2(25)(20)𝐶𝐶𝐶𝐶𝐶𝐶 (𝛼𝛼 − 𝛾𝛾)
−881 = −1000𝐶𝐶𝐶𝐶𝐶𝐶 (𝛼𝛼 − 𝛾𝛾)
−881
= cos(𝛼𝛼 − 𝛾𝛾)
−1000
We can find |𝑃𝑃𝑃𝑃| using Pythagoras’ theorem.
Finding this allows us to find 𝛽𝛽 by using the
cosine rule which can be found on page 16 of
the Maths Tables Book.
Remember when using the cosine rule that
the side 𝑎𝑎 is the side opposite the angle A
We can also use the cosine rule to find the
angle 𝛼𝛼 minus 𝛾𝛾.
cos(𝛼𝛼 − 𝛾𝛾) = 0.881
𝛼𝛼 − 𝛾𝛾 = cos −1 (0.881) = 28.237°
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 =
tan 𝛾𝛾 =
𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜
𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎
7
24
𝛾𝛾 = tan−1
7
= 16.260
24
𝛼𝛼 = 16.260 + 28.237 = 44.49 = 𝟒𝟒𝟒𝟒°
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We can find the angle 𝛾𝛾 using the tan ratio.
Subbing in the opposite and adjacent and
using tan inverse to find the angle.
Adding 𝛾𝛾 to the angle we found for (𝛼𝛼 − 𝛾𝛾),
gives us 𝛼𝛼.
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b)
A 𝟏𝟏° error in 𝒂𝒂 will cause the greater error in the location of R.
1° error in 𝛼𝛼 causes R to move along an arc of radius length 25.
1° error in 𝛽𝛽 causes R to move along an arc of radius length 12.
As 𝑙𝑙 = 𝑟𝑟𝑟𝑟 (page 9 of the Maths Tables Book) and 𝜃𝜃 is the same for both cases, the point moves
further when there is an error in 𝛼𝛼 as it has a longer radius.
c)
More sensitive to errors in 𝛼𝛼 when |𝑃𝑃𝑃𝑃| > 12
More sensitive to errors in 𝛽𝛽 when |𝑃𝑃𝑃𝑃| < 12
The condition |𝑃𝑃𝑃𝑃| > 12 is true whenever
𝑎𝑎2 = 𝑏𝑏 2 + 𝑐𝑐 2 − 2𝑏𝑏𝑏𝑏 cos 𝐴𝐴
(12)2 = (20)2 + (12)2 − 2(20)(12)𝐶𝐶𝐶𝐶𝐶𝐶 𝛽𝛽
144 − 400 − 144 = −2(20)(12)𝐶𝐶𝐶𝐶𝐶𝐶 𝛽𝛽
−400
= cos 𝛽𝛽
−2(20)(12)
cos 𝛽𝛽 =
5
6
5
𝛽𝛽 = cos −1 � � = 33.56°
6
𝜷𝜷 > 𝟑𝟑𝟑𝟑. 𝟓𝟓𝟓𝟓
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While |𝑃𝑃𝑃𝑃| > 12 the error in 𝛼𝛼 causes R to move along an
arc with a greater radius than, an error in 𝛽𝛽 does.
The radius of the arc due to an error in 𝛽𝛽 is 12. Once |𝑃𝑃𝑃𝑃|
is less than this, the arm is more sensitive to an error in 𝛽𝛽
as the arc it causes R to move along now has the greater
radius.
To find when these conditions are true, we let |𝑃𝑃𝑃𝑃| = 12
and use the cosine rule to calculate 𝛽𝛽.
The cosine rule can be found on page 16 of the Maths
Tables Book.
We know that when 𝛽𝛽 is greater than this angle the length
of |𝑃𝑃𝑃𝑃| is greater than 12 and therefore the arm is more
sensitive to an error in 𝛼𝛼. When 𝛽𝛽 is less than this angle
the opposite is true.
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d)
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