12 5. Algebra 5. Algebra Sequences and series You should be able to: • show that a given sequence is either arithmetic, geometric or neither • find any given term in an arithmetic sequence or a geometric sequence • find the common difference in an arithmetic sequence and the common ratio in a geometric sequence • find the number of terms in an arithmetic or a geometric sequence • find the sum of the first n terms in an arithmetic or a geometric series • apply the concepts of arithmetic and geometric sequences in the solution of problems • determine when it is possible to find the sum of an infinite geometric series • find the sum to infinity of a convergent geometric series • find the rational equivalent of a recurring decimal number. You should know: • ∑ k = m n u k = u m + u m + 1 + u m + 2 + … + u n − 1 + u n ; this sigma notation is very often used as a “shorthand” way of writing series • when a practical situation requires the use of the formula for the general term of a sequence, and when you need to use the formula for the sum of n terms in the sequence. Example An arithmetic sequence and a geometric sequence both have a first term of 1. Their second terms are equal. The sixteenth term of the arithmetic sequence is equal to four times the third term of the geometric sequence. The common difference of the arithmetic sequence is less than 1. Find the twentieth term of each sequence. Since this question gives information about two sequences, it is a good strategy to get an overview what th look like before starting the problem, by writing them in terms the information given. Arithmetic sequence 1, 1 + d, 1 + 2d, 1 + 3d, 1 + 4d, … Geometric sequence 1, r, r 2 , r 3 , r 4 , … 503-05-IB Maths HL-05.indd 12 12-02-2010 09:06:43
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12
5. Algebra
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Sequences and series
You should be able to:
• show that a given sequence is either arithmetic, geometric or neither• fi nd any given term in an arithmetic sequence or a geometric sequence• fi nd the common difference in an arithmetic sequence and the common ratio
in a geometric sequence• fi nd the number of terms in an arithmetic or a geometric sequence• fi nd the sum of the fi rst n terms in an arithmetic or a geometric series• apply the concepts of arithmetic and geometric sequences in the solution of
problems• determine when it is possible to fi nd the sum of an infi nite geometric series• fi nd the sum to infi nity of a convergent geometric series• fi nd the rational equivalent of a recurring decimal number.
You should know:
• ∑ k = m
n
uk = um + um + 1 + um + 2 + … + un − 1 + un; this sigma notation is very
often used as a “shorthand” way of writing series• when a practical situation requires the use of the formula for the general
term of a sequence, and when you need to use the formula for the sum of n terms in the sequence.
Example
An arithmetic sequence and a geometric sequence both have a fi rst term of 1. Their second terms are equal. The sixteenth term of the arithmetic sequence is equal to four times the third term of the geometric sequence. The common difference of the arithmetic sequence is less than 1.
Find the twentieth term of each sequence.
Since this question gives information about two sequences, it is a good strategy to get an overview of what they look like before starting the problem, by writing them in terms of the information given.
The first three sentences in the question translate to the equations
1 + d = r
and
1 + 15d = 4r 2,
which combine to give:
1 + 15d = 4(1 + d) 2
=> d = 3 _
4 or d = 1.
These values give r = 7 _
4 or r = 2.
The solution required is d = 3 _
4 , r =
7 _
4 .
Replacing these values in the formulae for the nth terms:
un = u1 + (n − 1)d
gives
u20 = 1 + (20 − 1) × 0.75 = 15.25,
needing no rounding since it is an exact value.
un = u1rn − 1
gives
u20 = 1 × 1.75 20 − 1 ≈ 41 468.93911…
This answer would be best left as an exact value,
u20 = ( 7 _ 4 )19
.
This solution can be verified with a GDC by drawing a table.
Texas Instruments Casio
The value of X in the first column tells us the position of each term in the sequence. We can see from this screenshot that the sequences satisfy the first two sentences in the original problem.
To find the twentieth terms, scroll to the twentieth row in the table. Very occasionally, GDC output may not confirm every digit of your answer, owing to rounding errors within the calculator.
Be prepared
• Carefully check that you have correctly translated all the information given in the problem into equations.
• Be careful to plan your working out so that you are working with the correct term; for example, the fourth term of an arithmetic sequence is given by u4 = u1 + 3d, not u4 = u1 + 4d.
• Using your GDC can help by quickly generating terms of a sequence in order to check or develop your answer to a problem.
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Exponents and logarithms
You should be able to:
• apply the laws of indices and the laws of logarithms.
You should know:
• the “change of base” rule, an application of the law
log an = n log a,
which allows you to find loga b, for any positive value of a
• the laws of indices and the definition of logarithm allow you to deduce that
loga b = −log1/a b and loga 1 __
b = −loga b
• the laws of logarithms allow you to simplify expressions (like 5 log a + 2 log b − 3 log a2, for instance) by writing them as a single logarithm
• if a f(x ) = ag (x ), then f(x ) = g(x ).
Example
Solve the equation (3 x)(42x + 1) = 6 x + 2, giving your answer in the form
ln a ___
ln b where a, b [ Z.
This is a typical paper 1 problem and looks tricky because the bases of the exponential expressions are different. Using the laws of logarithms, we can simplify the exponential equation to a linear one, which is usually easier to handle.
ln ((3 x)(4 2x + 1)) = ln (6 x + 2)
=> ln (3 x) + ln (4 2x + 1) = ln (6 x + 2)
=> x ln 3 + (2x + 1)ln 4 = (x + 2)ln 6
=> x(ln 3 + 2 ln 4 − ln 6) = 2 ln 6 − ln 4
=> x = 2 ln 6 − ln 4
___________ ln 3 + 2 ln 4 − ln 6
And now using the laws of logarithms to simplify
x = ln 36 − ln 4
___________ ln 3 + ln 16 − ln 6
=> x = ln 9
___ ln 8
Be prepared
• Be careful to ensure that the answer is given in the format required in the question in order to gain full marks.
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Counting principles
You should be able to:
• find the number of ways in which the combination of a succession of independent operations can be carried out
• find the number of possible permutations or combinations of the elements in a set
• solve problems in which permutations and combinations are combined
• solve problems in which further conditions are introduced, such as circular arrangements or restrictions on certain elements
• expand (a + b)n completely (where a and b are real numbers and n is a natural number).
You should know:
• nCr and n r are equivalent ways of denoting the
number n! _______
r!(n − r )!
• the difference between the permutations and the combinations of the elements in a set and when to use each one
• the properties that allow you to simplify expressions with factorial numbers
• the symmetrical properties of the expansion of (a + b)n (both in its coefficients and in the exponents).
Example
A room has nine desks arranged in three rows of three desks. Three students sit in the room. If the students randomly choose a desk, find the probability that exactly two desks in the front row are chosen.
In this type of problem it’s always a good idea to draw a diagram: it helps you get more of a feel for the structure of the problem than you would from only considering formulae. Here,
for example, this would be one of the possible desk choices the students could make: front, left; front, centre; and back, right:
Other 1 Other 2 Other 3
Other 4 Other 5 Other 6
Front 1 Front 2 Front 3
This situation deals with combinations, since the order in which seats are chosen is not important—we are only interested in whether they are occupied.
The total number of ways to select 3 desks from a total of 9 is
9C3 = 9!
___ 6!3!
= 9 × 8 × 7
______ 3 × 2 × 1
= 84
The total number of ways to select 2 desks from 3 front desks is3C2 =
3! ___
2!1! =
3 _
1 = 3
The total number of ways to select 1 desk from the other 6 desks is
6C1 = 6!
___ 5!1!
= 6 _
1 = 6
For each of the 3 selections of the front 2 desks there are 6 selections of other desks, giving a total of 3 × 6 = 18 ways for
the three students to sit down. The answer is 18
__ 84
= 3 __
14 .
Be prepared
• If you find the problem hard, try to break it into smaller pieces and solve them in turn.
• Problems concerning counting methods can often be solved with brief calculations; any challenges come from understanding the problem and planning its solution. So if you find you are taking up a lot of space with calculations, you may be on the wrong path.
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Mathematical induction
You should be able to:
• set out each of the steps involved in a proof by induction clearly and using correct mathematical notation.
You should know:
• the first step in a proof by induction is to verify that it holds true for some first value, k0; this value is usually, but not always, 1
• when verifying that the proposition holds true for n = k0, it is important to show this explicitly, by replacing n = k0 in both sides of the proposition equality
• when proving that if the proposition holds true for n = k then it also holds true for n = k + 1, it is important to show clearly how you use the inductive hypothesis
• it is important to write a “closing statement” to your proof, stating why you can now say that you have proved that the proposition is valid for all natural numbers.
Example
Use mathematical induction to prove that 5n + 9n + 2 is divisible by 4, for n [ Z+.
Let f(n) = 5 n + 9 n+ 2 and let P(n) be the proposition that f(n) is divisible by 4.
The first step is to show that the proposition is true for the smallest value in the domain, which, in this case, is 1.
f(1) = 16,
so P(1) is true since 16 is divisible by 4.
Then set out the inductive hypothesis: assume that P(n) is true for n = k, that is, f(k) = 5 k + 9 k + 2 is divisible by 4. We can write this as
(5 k + 9 k + 2) = 4M, where M [ Z+
We then have to consider the proposition for n = k + 1 and determine how it relates to the inductive hypothesis. We know that we must arrive at a statement that says that
f(k + 1) = 5 k + 1 + 9 k + 1 + 2 is divisible by 4. But the inductive hypothesis is written in terms of the exponent k, so, using 5 k + 1 = 5 k × 5 = 5 k(4 + 1), we write
Both terms are divisible by 4 so f(k + 1) is also divisible by 4.
Therefore P(k) true => P(k + 1) true.
Always finish with the correctly stated, complete conclusion:
Since P(1) is true, and P(k) true => P(k + 1) true, P(n) is proved true by mathematical induction for n [ Z
+.
Be prepared
• The first step of a proof by induction must include complete calculations showing P(n) to be true for the first element of the domain. Remember that you are writing a formal proof; it must read well and be rigorous.
• Always ask yourself “How do I transform the P(k) into P(k + 1)?” The key point is knowing what P(k + 1) looks like and then rearranging it in some way that will bring P(k) into the picture. This means that you need to think about the mathematical relation between P(k) and P(k + 1) and express it well, so that your use of the inductive hypothesis is communicated clearly.
• The final step of a proof by induction must include complete reasoning, showing P(n) to be true for all elements of the domain; the first step and the inductive step must feature in the reasoning.
• All proofs by induction have similar first and final steps, but these steps must be expressed in complete detail to gain full marks.
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Complex numbers
You should be able to:
• identify the real and the imaginary parts of a complex number
• find the modulus and the argument of a complex number
• add, subtract, multiply and divide complex numbers
• represent complex numbers on an Argand diagram
• transform a complex number from its Cartesian form into its modulus–argument form, and vice versa
• use De Moivre’s theorem to find powers and roots of complex numbers
• prove De Moivre’s theorem using mathematical induction.
You should know:
• the different forms in which a complex number can be written: Cartesian (z = a + ib) and polar or modulus–argument form (z = r (cos θ + i sin θ ), z = re iθ or z = r cis θ )
• if two complex numbers are equal, then their real parts are equal and their imaginary parts are equal
• the geometric significance of the modulus and the argument of a complex number
• the relation between the moduli and the arguments of two complex numbers and those of their products, quotients and powers
• the relation between the geometric representation of the nth roots of a complex number
• the relation between the geometric representation of a complex number and that of its conjugate
• the properties of conjugates: (z*)* = z and zz* = |z|2
• if z = a + ib is a root of a polynomial (with real coefficients), then so is its conjugate.
Example
Given that |z| = √___
10 , solve the equation
5z + 10 __ z*
= 6 − 18i, where z* is the conjugate of z.
In approaching this question, it might not be obvious at the beginning exactly how to use |z| = √
__ 10 . However, one of the first
skills we meet in complex numbers is how to divide, and this will help rationalize the denominator, thus simplifying the equation, so this will be the first step:
5z + 10
__ z*
= 6 − 18i => 5z + 10
__ z*
× (z*)*
____ (z*)*
= 6 − 18i
=> 5z + 10
__ z*
× z _
z = 6 − 18i => 5z +
10z ___
z*z = 6 − 18i
Now, since z*z = |z|2 we can write
5z + 10z
___ |z|2 = 6 − 18i.
But |z| = √ __
10 , so |z|2 = 10 and the formula simplifies to
5z + 10z
___ 10
= 6 − 18i
6z = 6 − 18iSo z = 1 − 3i.
Be prepared
• When dealing with an unfamiliar equation, you may find it useful to focus on the skill that can be applied straight away, or that helps simplify the equation considerably. In the example above, division simplified the equation considerably.
• Use your GDC to check your answer by replacing it into the original equation:
Texas Instruments Casio
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N08/5/MATHL/HP2/ENG/TZ0/XX
8808-7202
– 12 –
11. [Total mark: 21]
Part a [Maximum mark: 11]
(a) A box of biscuits is considered to be underweight if it weighs less than 228 grams. It is known that the weights of these boxes of biscuits are normally distributed with a mean of 231 grams and a standard deviation of 1.5 grams. What is the probability that a box is underweight? [2 marks]
(b) The manufacturer decides that the probability of a box being underweight should be reduced to 0.002.
(i) Bill’s suggestion is to increase the mean and leave the standard deviation unchanged. Find the value of the new mean.
(ii) Sarah’s suggestion is to reduce the standard deviation and leave the mean unchanged. Find the value of the new standard deviation. [6 marks]
(c) After the probability of a box being underweight has been reduced to 0.002, a group of customers buys 100 boxes of biscuits. Find the probability that at least two of the boxes are underweight. [3 marks]
Part B [Maximum mark: 10]
There are six boys and five girls in a school tennis club. A team of two boys and two girls will be selected to represent the school in a tennis competition.
(a) In how many different ways can the team be selected? [3 marks]
(b) Tim is the youngest boy in the club and Anna is the youngest girl. In how many different ways can the team be selected if it must include both of them? [2 marks]
(c) What is the probability that the team includes both Tim and Anna? [1 mark]
(d) Fred is the oldest boy in the club. Given that Fred is selected for the team, what is the probability that the team includes Tim or Anna, but not both? [4 marks]
1213
[Taken from paper 2, November 2008]
How do I approach the question?
(a) Since the team comprises girls and boys, this part can be reduced to three smaller problems: how many boys can be selected, how many girls can be selected and how do these two answers combine to fi nd the answer for the number of different team selections. Taking time to choose the right counting method is essential; a diagram can help you in the process of understanding the problem and planning your approach.
(b) Notice that this part still refers to a team selection. Ask yourself how the requirement that Tim and Anna must be in the team would affect the answer to (b). Would it be more than, less than or the same as the answer to (a)? Then a similar approach to that of (a) could be adopted.
(c) The probability can be found from an application of the formula P(A) = n(A)
____ n(U)
.
(d) The word “given” tells you explicitly that Fred has been chosen, and this means that you must take this fact into account in this part by understanding how this information changes the possible teams.
In solving probability problems, the planning stage is vital. It will help to translate “team includes Tim or Anna, but not both” into symbols, for example, by writing the events T: “Tim is included”, A: “Anna is included”. The probability asked for is a compound event, which can be approached by use of a Venn diagram, tree diagram or a probability formula to help plan out the calculation. Then, with your symbols and choice of diagram, try to construct an expression that will fi nd the required probability. Another approach is to carry on the theme of the question—selections—and try to fi nd the number of selections of each type of team mentioned in the question, and then use them to fi nd the required probability.
What are the key areas from the syllabus?
• Counting principles
What do I require from the information booklet?
• Probabilities of combined events
What are the links with other areas of the syllabus?
• The probability of simple and of compound events
503-05-IB Maths HL-05.indd 18 12-02-2010 09:06:45
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This answer achieved 3/10This answer achieved 3/10
Part B
(a) No. of selections = 6C2 × 2 × 5C2 × 2 A0
= 600 A0 A0
(b) No. of selections = 5C1 × 4C1 M1 A1
= 20
(c) P(Tim and Anna) = 20
___ 600
= 1 __
30 = 0.0333 A1 (ft)
(d) P(Tim or Anna | Fred) = P(Tim or Anna ∩ Fred)
________________ P(Fred)
P(Fred) = 6C1 × 5C1 2 _
3 M0
P(Tim or Anna ∩ Fred) =
The working is not explained clearly enough, so no intermediate marks are awarded. The fact that the student multiplies by 2, twice,
would seem to indicate uncertainty as to whether to work with combinations or permutations.
The working is not explained clearly enough, so no intermediate marks are awarded. The fact that the student multiplies by 2, twice,
The student shows clear understanding that the required answer is the product of the number of ways in which the second boy can
be chosen by the number of ways in which the second girl can be chosen.
The student shows clear understanding that the required answer is the product of the number of ways in which the second boy can
The student’s attempt, using conditional probability, shows only partial understanding of what is being asked. Little has
been done apart from writing down a formula (incorrectly, since the condition “but not both” is omitted) and the only probability worked out is also incorrect.
The student’s attempt, using conditional probability, shows only partial understanding of what is being asked. Little has
Although the answer is incorrect, since the required probability is based on the total
number of ways found in (a), a follow-through mark is awarded.
Although the answer is incorrect, since the required probability is based on the total
When an exact answer has already been obtained in fraction form
(such as 1 ___
30 , the probability in part
(c)), it is completely unnecessary to express it as a decimal number, especially when this involves an approximation leading to a more inaccurate answer.
When an exact answer has already been obtained in fraction form
(such as
✓
503-05-IB Maths HL-05.indd 19 12-02-2010 09:06:46
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This answer achieved 6/10
5. A
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This answer achieved 6/10
Although the student achieves full marks here, a more explicit method explaining what each of
the combination numbers results in could have been used.
Although the student achieves full marks here, a more explicit method explaining what each of
✓
Part B
(a) Th e number of ways = 6C2 × 5C2 A1 A1
= 150 ways A1
(b) Number of ways = 5C1 × 4C1 M1 A1
= 20 ways
(c) P(Tim, Anna) = 20
___ 150
A1
= 0.133
(d) P(Tim, Fred) + P(Anna, Fred)
= 5C1 × 4C1 M0
Unlike the previous example, here it is possible for the examiner to infer the reasoning behind the product: the number of ways the boys can be
chosen × the number of ways the girls can be chosen.
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Unlike the previous example, here it is possible for the examiner to infer the reasoning behind the product: the number of ways the boys can be
Correct probability, from two previous correct answers.Correct probability, from two previous correct answers.
The student seems to be intending to follow the same steps as in (a) to (c), and then to work out the sum of two probabilities. This hints at a
misunderstanding of the question, since it gives no indication that the student has recognized the conditional probability implicit in the information given in the question (that Fred is selected for the team).
The student seems to be intending to follow the same steps as in (a) to (c), and then to work out the sum of two probabilities. This hints at a
503-05-IB Maths HL-05.indd 20 12-02-2010 09:06:47
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This answer achieved 8/10This answer achieved 8/10
Part B
6B, 5G, Boy = B
Girl = G
(a) How many ways?
Team → 2B, 2G
= 6C2 × 5C2 A1 A1
= 150 ways A1
(b) T A
= 1 × 5C1 × 4C1 , Tim, Anna, B, G M1 A1
= 20 ways
(c) P(Tim, Anna incl.) = 20
___ 150
= 2 __
15 A1
(d) P(T, A | Fred) = P(A | B)
= P(A ∩ B)
_______ P(B)
F T G G
F A B G
P(Tim) + P(Anna), Fred in the team =
= 1 × 1 × 4C0 × 4C2 + 1 × 1 × 4C1 × 4C1
= 6 + 16 A1
= 22 ways
P(A ∩ B) = 22
___ 150
= 11
__ 75
P(Fred selected) = P(B)
= 1 × 5C1 × 4C2 A0
= 30
P(B) = 30
___ 150
= 1 _
5
P(A | B) = 11
__ 75
× 5 _
1 M1
= 11
__ 15
A0
This student set out the question very clearly, organizing the information given in the question and thus making it easier to
understand where the combination numbers come from.
This student set out the question very clearly, organizing the information given in the question and thus making it easier to
The student’s side annotations and little diagrams add clarity to the working out.
The student’s side annotations and little diagrams add clarity to the working out.
This student obtains the correct probability, given as an exact value, in its simpli� ed form.
This student obtains the correct probability, given as an exact value, in its simpli� ed form.
Here, the formula for conditional probability is used correctly. Again, the auxiliary diagram drawn by the student adds clarity to the
calculation of the number of teams that include Tim but not Anna and then Anna but not Tim. This total is correct, but then there is a mistake in the calculation of the number of teams that include Fred
(the student multiplies by 4 2 instead of
5 2 and gets 30 instead of 50). This leads
to an incorrect � nal answer, although the method applied is correct.
Here, the formula for conditional probability is used correctly. Again, the auxiliary diagram drawn by the student adds clarity to the
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(a) For this part you need to identify that selections are counted by use of combinations.
(b) A common error is to use a formula in a calculation that bears little relevance to the problem—a diagram is a safeguard against this happening.
(c) A common error here is to approach the calculation as a completely new problem and fail to notice the link with parts (a) and (b)—both parts involve a team that includes both Tim and Anna. Before starting on a calculation, it is good practice to keep in mind your previous results.
(d) The interpretation of “given that” to suggest use of the conditional probability formula is correct, but a common dif� culty is to expect that writing down the formula will rapidly � nd the solution. Be clear on the structure of the problem: if we let T be the event “Tim is chosen” and A “Anna is chosen”, then the probability in this part P(T or A but not both) can be written as
P(T ) × P(A′) + P(T ′ ) × P(A),
and this can be used to � nd the answer without quoting the conditional formula.
If a counting method is used, the condition that Fred is chosen reduces the total possible number of teams from 150 to 50. Use counting methods to � nd the total number of teams with “Tim or Anna but not both” (that is, the number of teams with Tim but not Anna plus the number of teams with Anna but not Tim). This number divided by 50 gives the answer, again without need to quote the formula.
Examiner report
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(a) For this part you need to identify that
Examiner report
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M08/5/MATHL/HP1/ENG/TZ2/XX
2208-7213
– 1� –
14. [Maximum mark: 12]
Let w = +cos 25
i 25
π πsin .
(a) Show that w is a root of the equation z5 1 0− = . [3 marks]
(b) Show that ( ) ( )w w w w w w− + + + + = −1 1 1� 3 2 5 and deduce that w w w w� 3 2 1 0+ + + + = . [3 marks]
(c) Hence show that cos 25
�5
π π+ = −cos 12
. [6 marks]
1414
[Taken from paper 1, time zone 2, May 2008]
How do I approach the question?
Although no GDC is permitted, the good news is that De Moivre’s theorem serves as a useful “calculator”; it always gives a quick way to fi nd the power of a complex number expressed in polar form.
(a) This is the fi rst place in the question where De Moivre’s theorem is useful. Be clear about the meaning of “root” and expect to use your knowledge of trigonometric functions to simplify.
(b) Although the question does not say so explicitly, there is a clear link with part (a). Recall that if w is a root, then w 5 − 1 must equal zero. “Show that” and “deduce” are clear indicators that you should communicate clearly your reasoning about the product on the left-hand side and the fact that it is zero too.
(c) “Hence” is an explicit instruction to apply the previous results. Another clue
is that the argument of w, 2π
__ 5 , appears in the expression to be shown, as
does 4π
__ 5 —this hints that De Moivre’s theorem has been applied. Recall that
in paper 1, the unit circle (and all its symmetries) is another important “pen and paper” calculating device.
What are the key areas from the syllabus?
• Complex numbers
• De Moivre’s theorem
• Factor theorem and roots of polynomial equations
• Trigonometric functions
What do I require from the information booklet?
• De Moivre’s theorem
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This answer achieved 3/12
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This answer achieved 3/12
It is not usually helpful, in this type of question, which asks you to “show that…”, to turn the radians
into degrees, so, more often than not, it ends up being a waste of time.
It is not usually helpful, in this type of question, which asks you to “show that…”, to turn the radians
The student has replaced w into the equation and has attempted to apply De Moivre’s theorem to show that z5 − 1 = 0. However, the
working out is not clear and the student does not show the calculation that leads to 0.
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The student has replaced equation and has attempted to apply De Moivre’s theorem to show that
The student correctly expands the right-hand side of the equation, showing that it equals w 5 – 1, but then does not attempt the “deduce”
part of the question.
The student correctly expands the right-hand side of the equation, showing that it equals then does not attempt the “deduce”
The student, having spotted that one argument is double the other, has used a trigonometric identity to
express cos 4π
__ 5 in terms of cos
2π __
5 : not
a useful approach as far as the question is concerned. The student has ignored the word “Hence”, which was clearly saying that the answer to (b) had to be used, in some way, to prove the given relation.
The student, having spotted that one argument is double the other, has used a trigonometric identity to
express cos
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(a) z5 – 1 = 0
cos 2π
__ 5
+ i sin 2π
__ 5
5 – 1 =? 0
by de Moivre’s theorem: (cos 2π + 2 sin 2π) – 1 = 0 M1
1 – 1= 0 A1
∴ cos 2π
__ 5
+ i sin 2π
__ 5
is a root of z5 – 1 = 0 A1
(b) (w – 1) (w4 + w3 + w2 + w + 1)
= w5 + w4 + w3 + w2 + w – w4 – w3 – w2 – w – 1 M1
= w5 – 1 A1
w5 – 1 = 0 and w ≠ 1, ∴ w4 + w3 + w2 + w + 1 must be zero R1
(c) cos 2π
__ 5
+ i sin 2π
__ 5
4
+ cos 2π
__ 5
+ i sin 2π
__ 5
3
+ cos 2π
__ 5
+ i sin 2π
__ 5
2
+
+ cos 2π
__ 5
+ i sin 2π
__ 5
+ 1 = 0 M1
Here the student is successful in replacing w into the equation and applying De Moivre’s theorem, showing clearly that the answer is 0.
Here the student is successful in replacing applying De Moivre’s theorem, showing clearly that the answer is 0.
The expansion of the factorized expression is carried out correctly, to obtain w5 − 1. The � nal reasoning mark is obtained by
showing clear understanding that, since the product is 0 and one of the factors is not, then the other must be 0.
The expansion of the factorized expression is carried out correctly, to obtain reasoning mark is obtained by
The student pays attention to the “Hence” and starts off the question correctly: by substituting
cos 2π
__ 5 + i sin
2π __
5 into the equation
derived in part (b). Then, De Moivre’s theorem is applied correctly, but the student fails to realize that the next step is to identify real and imaginary parts, and to equate the real parts on the right-hand side and the left-hand side.
The student pays attention to the “question correctly: by substituting
cos
503-05-IB Maths HL-05.indd 25 12-02-2010 09:06:50
26
5. A
lgeb
ra
This answer achieved 11/12
5. A
lgeb
ra
This answer achieved 11/12
(a) z5 – 1 = 0 => z5 = 1
=> z = cos 2π
__ 5
n + i sin 2π
__ 5
n M1
n = 0, 1, 2, 3, 4 A1
taking n = 1 z = cos 2π
__ 5
+ i sin 2π
__ 5
= w A1
(b) (w – 1) (w4 + w3 + w2 + w + 1) = w5 + w4 + w3 + w2 + w – w4 – w3 –
– w2 – w – 1 = M1 A1
= w5 – 1
=> w4 + w3 + w2 – w + 1 = 0 R0
(c) 1 + w + w2 + w3 + w4 = 0 (fr om b)
=>1+ cos 2π
__ 5
+ i sin 2π
__ 5
+ cos 2π
__ 5
+ i sin 2π
__ 5
2 + cos
2π __
5 + i sin
2π __
5
3+
+ cos 2π
__ 5 + i sin
2π __
5
4 = 0 M1
1 + cos 2π
__ 5
+ i sin 2π
__ 5
+ cos 4π
__ 5
+ i sin 4π
__ 5
+ cos 6π
__ 5
+ i sin 6π
__ 5
+
+ cos 8π
__ 5
+ i sin 8π
__ 5
= 0 M1
1 + cos 2π
__ 5
+ cos 4π
__ 5
+ cos 6π
__ 5
+ cos 8π
__ 5
+
+ i sin 2π
__ 5
+ sin 4π
__ 5
+ sin 6π
__ 5
+ sin 8π
__ 5
= 0 A1 A1
=> real part = 0
cos 6π
__ 5
= cos 4π
__ 5
M1
cos 8π
__ 5
= cos 2π
__ 5
4 55
2
65
58
=> real part = 1 + 2 cos 4π
__ 5
+ 2 cos 2π
__ 5
= 0 A1
=> cos 4π
__ 5
+ cos 2π
__ 5
= – 1 _
2 AG
(a) A common error is failing to exploit the fact that, for example,
cos (5 × 2π
__ 5 ) = cos (2π) = 1 and this
leads to marks being lost.
(b) A common error is to leave the “deduce” step poorly written. Make sure that the reasoning is written formally, using each premise from parts (a) and (b) to write a complete and coherent argument.
(c) You need to be careful to follow the development of the question, using the command term “Hence”, to establish a link with part (b).
Examiner report
(a) A common error is failing to exploit
Examiner report
The student looks for the � fth roots of 1, writes down a correct expression for the � ve roots, in terms of n, and shows that one of
these values of n (1) produces w as one of the roots of the equation.
5. A
lgeb
ra
The student looks for the � fth roots of 1, writes down a correct expression for the � ve roots, in terms of
The student simply states
“⇒ w4 + w3 + w2 + w + 1 = 0”, without giving any reasoning to support the statement.
The student simply states
“⇒without giving any reasoning to support the statement.
The trigonometric circle diagram shows clearly how the student recognized the two pairs of equivalent cosines in the expression,
which make it possible to obtain a simpli� ed version of the real part. From here, the required equality is then deduced, quite directly.
The trigonometric circle diagram shows clearly how the student recognized the two pairs of equivalent cosines in the expression,
The student rearranges the right-hand side into the form a + bi, making it easy to identify the real part of the complex number, which
is then equated to 0.
The student rearranges the right-hand side into the form making it easy to identify the real part of the complex number, which
