Further Mathematics Lesson Note SS1 First Term
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FURTHER MATHEMATICS SS1 SCHEME OF WORK FOR FIRST TERM
WEEK(S) TOPICS
- General revision and basic concept of set
- Operation of set and venn diagram: Union, intersection, compliment, and cardinality of Set.
- BINARY OPERATION AND BASIC LAWS: (a) definition of binary operation (b) solve simple operation of binary operations e.g a x b = – 2ab (b)Identity law of binary operation.
- BINARY OPERATION CONTINUES: (a) solve problems and application of laws of binary operation to given problems (b) identity and inverse element (d) draw addition and multiplication table for binary operation of modulo.
- INDICES: (a) basic laws of indices (b) use of indices laws in solving given problems.
- INDICIAL EQUATION AND GRAPH OF EXPONENTIAL FUNCTIONS: e.g Y = , where x 1.
- Review of first half term and periodic test
- LOGARITHMS; (a) state the laws of logarithms (b) application of law of logarithms to given problems (c) solve problems involving change of base
- SURD: (a) definition of surd (b) state the rules of surds (c) solve basic operations involving addition, subtraction, multiplication, division and rationalization of surd with the use of conjugate surd.
- MEASURE OF LOCATION: (a) mean, median and mode (b) estimating mode from histogram of grouped data.(c)estimating the median from histogram of groped data
- REVISION
- EXAMINATION
- EXAMINATION
REFERENCE MATERIAL
New Further Mathematics For Senior Secondary School 1. By Tuthu- Adigun Etal.
WEEK 1
CONCEPT OF SET
SET: set can be defined as a collection of objects according to a well-defined common element, object, items, or properties. E.g. Mathematical set, drum set, set of spanners, set of screw drivers e.t.c.
ELEMENT: This is each member of a set or properties, items in a given set. x A or A = {x}
METHOD OF DESCRIBING SETS
THE SET BUILDER/PROPERTY SET: this is the set that describes the elements of the set by referring to their common properties. Example, W = , Y = {x: x is even numbers between 0 ≤ x ≤ 10}
THE ROSTER/TABULAR/LISTING METHOD: This is the actual listing of all the members of a given set. Example, Y = {2,4,6,8,10}. W = {SUNDAY,MONDAY,TUESDAY,WEDNESDAY,THURSDAY,FRIDAY, SATURDAY)
TYPES OF SETS
FINITE SET: Is a set that its elements can be listed or has an end point. i.e. {1,2,4,6,8,9,10}
INFINITE SET: Is the set that their element is continuous or impossible to list. i.e. all the natural positive numbers {1,2,3,4,…}
EMPTY SET: This is a set that has or contain no element and its represented as {}, Ø or null.
EQUALITY SET: This is two set that has the same elements. i.e. A = {1,3,5,7,9} and B = {1,1,5,3,9,9,7}
EQUIVALENT SET: Two sets are said to be equivalent if they both have the same numbers of elements. i.e. X ={a,b,c,d,e,y,z} and P = {1,2,3,4,5,6,7}
SUB-SETS: Given two sets A and B such that set A consists of all the elements in set B, then set B is a subset of set A
SUPER-SET: Given two set A and B, if the set A is a SUBSET of set B and there exist at least one element in set B which is not in set A , then the set B is a SUPER-SET of set A B ↄ A or A is a proper subset of B, A ϲ B
ASSESSMENT: work out the following:
- If µ = {all the months in the year} A = {all the months in the year that begins with letter J} B = {all the months in the year ending with letter r}
- List all the members of µ (ii) List all the members in A (iii) List all the members of B
- Given that µ = {all the days in a week}. P = {all the days in the week whose letters begins with S}.
- List all the elements in µ. (ii) List the elements in P (iii) list the members .
- List the members of the following sets:
µ = {all positive integers less than or equal to 30}
X = {all even positive numbers less than or equal to 20}
Y = {all odd numbers less than or equal to 19}
Z = {all integers x: 10 x 30} and hence find,
, .
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WEEK 2
OPERATION OF SET AND VENN DIAGRAM
UNIVERSAL SET: is a set that contains all the elements or items under consideration or is called the MOTHER SET.Its denoted byµ or Ɛ.
COMPLIMENTARY SET: is the set of elements in the universal set that is not in the given set or in the subset. Its denoted by or .
CARDINALITY OF SET: Is the number of elements in a given set. Written as n(A) or n(P)
POWER SET: is the set of all the subsets of a given set. Its denoted as p(A) or . Note that in every set there is always an empty set of Ø in addition to the given elements in the set.
INTERSECTION OF SET: given two non-empty set A and B, if there exit a common element in A and B, the common members of the two set is called intersect ion of set.
UNION OF SET: Given two set X and Y, the collection of the element sin both set without repetition is called UNION of set.
DISJOINT SET: This is two or more set that are not related by any element.
VENN DIAGRAM: This is a graphical means of representing set information. It was first used by JOSEPH JOHN VENN AND LEONARD EULER.
EXAMPLE
Given that µ = {all the letters in the alphabet}, A = {a,e,o,I,u} and B = {e,b,c,d,f,h}. Find:
- AƲB (ii) AȠB (iii) (iv) p(B) (v) n(A) + n(B)
SOLUTION
µ = {a, b, c, d, e, …, x, y, z}, A = {a, e, I, o, u} and B = {e, b, c, d, f, h}
- AƲB = {a,b,c,d,e,f,h,I,o,u}
- AȠB = {e}, called unit or singleton set.
- = {all the consonants in the alphabets}
- P(B) = = 64
- N(A) + n(B) = 5 + 6 =11
VENN DIAGRAM
In an examination, 18 candidates passed General Mathematics,17 candidates passed Physics, 11 passed both subjects and one student failed both subjects, find:
- The number of candidates that passed only General Mathematics,
- The number of candidates that passed only Physics
- The total number of candidates that sat for the examination.
SOLUTIONƐ
| DIAGRAM |
- General Mathematics only = 18 – 11 = 7
- Physics only 17 – 11 = 6
- Total number of candidates is 7 + 11 + 6 + 1 = 25 students.
In an examination for promotion to ss2 students of Elias International Secondary School, 60 offered History, 50 Economics and 48 Literature. 30 offered History and Economics, 16 offered History and Literature, and 22 offered Economics and Literature. If 10 candidates offered all the three subjects.
- Find: (i) the number that offered only History. (ii) the number that offered Economics only (iii) the number that offered only Literature.
- How many candidates sat for the examination, assuming that each candidate sat for at least one subjects.
SOLUTION
Let History = H, Economics = E and Literature = L
| Diagram |
n(History) = 60, n(Economics) = 50,n( Literature) = 48.
n(HȠE) =30, n(HȠL) = 16, n(EȠL) = 22, n(HȠEȠL) = 10
- n(HȠ Ƞ = 60 – (20 + 6 + 10) = 24
- n( ȠEȠ ) = 50 – (20 + 12 + 10) = 8
(iii n( Ƞ Ƞ L) = 48 – (6 + 12 + 10) = 20
- The total candidates that sat for the examination 24 + 20 +10 + 12 + 6 + 8 + 20 = 100.
DISJOINT OR EMPTY SET: Given two set A = {2, 4, 6, 8 ,10}, B {1, 3, 5, 7, 9}. Find AȠB?
SOLUTION
| 1,3,5,7,9 |
| 2,4,6,8,10 |
A Ƞ B = Ø Ԑ
Assessment: New further Mathematics project 1 by Tuthu –Adegun find the solution set of the following questions:
Page 15 and 16, exercise 1c. Questions 1, 2, 4. 5, 8 and 10.
Page159 and 160, revision test. Questions 4, 8, 9, 18, 19, 21, and 22.
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WEEK 3
BINARY OPERATION.
Binary operation is any rule of combination of any two elements of a given non – empty set. It is denoted by asterisk (ӿ). E.g. a * b = a + b – 3. The operations are +, -, x, ÷.
CLOSURE PROPERTY: A non – empty set S is said to be closed under a binary operation * if for all a,bε S. a*b ε S.
EXAMPLE
The binary operation * on the set Q of positive rational numbers is defined by p*q = , p,qεQ. Determine:
- 2*1. (b) -3 *1. Is the operation * closed under Q.
SOLUTION
- P =2 and q = 1, then 2 * 1 = = –
2*1 ɆQ or – is not closed.
- -3 * 1 = =
= Ԑ Q, is closed.
COMMUTATIVE PROPERTY: Given a non – empty set S which is closed under a binary operation * if for all a, b Ԑ S, a*b = b*a, then the binary operation is saidto be COMMUTATIVE.
EXAMPLE
The operation * on the set Q of real numbers is defined by x*y = 25xy for x, y Ԑ Q. find under the operation *, the commutative property?
SOLUTION
If x*y = 25xy, then y*x = 25yx or let x=2 and y =4
2*4 = 25(2×4) = 200
4*2 = 25(4×2) = 200
Therefore 2*4 = 4*2 which is commutative, hence y*x = x*y.
ASSOCIATIVE PROPERTY: Given a non –empty set S closed under a binary operation*, then a,bԐ S a*b ԐS. a*b can also combine with cԐS and it becomes (a*b)*c. that is a*(b*c) = (a*b)*c.
EXAMPLE
Given that the binary operation a*(b*c) is associative under the operation *, determine under which basic operations a*(b*c) is associative?
SOLUTION
If a=6, b=3 and c=2. Then, UNDER ADDITION OPERATION
a + (b +c) = (a + b) + c
6 +(3 + 2) = (6 + 3) + 2
11 = 11 associative
SUBTRACTION
A – (b – c) = (a – b) – c
6 – (3-2) = (6-3) -2
5 ≠ 1. Subtraction is not associative under the operation *, Iff a, b and c are equal
HOME WORK: find out whether multiplication and division associate under *.
DISTRIBUTIVE PROPERTY: Given a non – empty set S, closed under the operation* and Δ if for all a, b,cԐS, a*(bΔc) = (a*b) Δ (a*c). Then the operation * is said to be distributive over the operation Δ.
EXAMPLE
The operations * and Δ are defined on the set N of natural numbers on +, -, x, and ÷. Does they distribute
SOLUTION
a*(bΔc) = (a*b) Δ (a*c)
ax(b+c) = (axb) + (axc) = ab + ac. Multiplication distributes under addition.
ax(b-c) = ab – ac. Multiplication distributes under subtraction
ax(bxc) = abc. Multiplication does not distributes under multiplication.
ax(b÷c) = . Multiplication does not distributes under division
this can only distributes iff a,b,c are equal.
ASSESSMENT: work out the following questions:
- Determine if the following properties distributes under their operations:
- a- (b+c) (b) a ÷(b + c) (c) a + (bxc) (d) a + (b – c) (e) a ÷ (b ÷ c)
- the operation * is defined on the set R, of real numbers, by a *b = – 1 for all a,bԐ R.
- Is R closed under *?
- Is the operation * commutative in R
- Is the operation* associative in R
- The operation * on the set R of real numbers, is defined by: x*y = 3x + 2y -1. X,yԐ R. Determine:
- 2*3 (b) -4*5 (c) * (d) 3 * -1.
- Determine whether or not each of the following sets is closed under the given operations defined by :
- a*b = 4(a + b) a,bԐR of real numbers.
- P Δq = p,qԐR of real numbers
- X Θ y = x + y + , x, y Ԑ Q of rational numbers.
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WEEK 4
BINARY OPERATION 2
MODULO: Is a system of arithmetic for integers, where numbers ̋wrap around ̋ upon reaching a certain value.
EXAMPLE
Consider the table of rules of combination for addition and multiplication in modulo 6
SOLUTION
= {0,1,2,3,4,5}
| 0 | 1 | 2 | 3 | 4 | 5 | |
| 0 | 0 | 1 | 2 | 3 | 4 | 5 |
| 1 | 1 | 2 | 3 | 4 | 5 | 0 |
| 2 | 2 | 3 | 4 | 5 | 0 | 1 |
| 3 | 3 | 4 | 5 | 0 | 1 | 2 |
| 4 | 4 | 5 | 0 | 1 | 2 | 3 |
| 5 | 5 | 0 | 1 | 2 | 3 | 4 |
| 0 | 1 | 2 | 3 | 4 | 5 | |
| 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | 2 | 3 | 4 | 5 |
| 2 | 0 | 2 | 4 | 0 | 2 | 4 |
| 3 | 0 | 3 | 0 | 3 | 0 | 3 |
| 4 | 0 | 4 | 2 | 0 | 4 | 2 |
| 5 | 0 | 5 | 4 | 3 | 2 | 1 |
Ԑ
THE IDENTITY ELEMENT: Given a non-empty set S, which is closed under a binary operation * , if there exist an element eԐS such that a*e = e*a = a for all aԐS, then e is called the IDENTITY OR NEUTRAL ELEMENT in S under the operation *.
THE INVERSE ELEMENT: Given a non -empty set S, which is closed under a binary operation * , if xԐS and we can find an element ԐS such that x* = *x = e. where e is the identity element in S under * , the is the inverse of x in S.
EXAMPLE
The operation Δ on the set Q of rational numbers is defined by x Δ y = 9xy for x,yԐQ. find under the operation Δ: (i) the identity element (ii) the inverse element.
SOLUTION
- XΔy = 9xy, let y = e
Then xΔe = eΔx = x
x = 9xe. Therefore, e = .
- Let y = inverse of x
Recall that X* *x = e
e = 9x , then = 9x
Therefore = .
ASSESSMENT: work out the following;
- Construct a table for addition and multiplication of modulo 9.
- New Further Mathematics project 1 by M.R. Tuttuh-Adegun Etal.page 29, exercise2. Questions: 20, 22, 25, 26, 27 and 28.
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WEEK 5
INDICES
INDICES: is expressing numbers in powers corresponding to the given value.
LAWS OF INDICES
- = 1
- =
- ( = =
- = (
EXAMPLE.
1. LAW OF MULTIPLICATION
Simplify the following expressions:
- ii.
SOLUTION
- =
- = =
2. LAW OF DIVISION
Simplify the following expression:
- iii. 2m
SOLUTION
- 2v = (16 = 8
- = (2÷8) = = .
3. LAW OF ZERO POWER OR INDEX
Simplify the following:
- 3[ ii. 13 ÷ 13 iii.
SOLUTION
- 3[ = 3 =3
- 13 = = 1 =1 1 = 1
- = = = = 1
4. LAW OF NEGATIVE POWER
Simplify the following:
- ii. 60 iii. 27
- = =
- ÷ 12 = (60÷12) = 5 =
- 27 = (27 ) = 3 = 3
LAW OF RAISING A POWER TO ANOTHER POWER
Simplify (a) ( (b) -5( 2 f) (c) 5x
SOLUTION
- ( = = = .
- -5(2 f = -5 = -80
- 5x = 5x = 80
FRACTIONAL INDICES:
Simplify the following:
- (b) (64 (c) (
SOLUTION
- = ( = = ( =
- = =
- ( = = 1 ÷ = 1 = 16.
ASSESSMENT: Simplify the following questions:
- 2. 3. 0.12 4. 5. ( 6. (
- (-2 8. (
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WEEK 6
INDICIAL EQUATION/EXPONENTIAL EQUATION
This is one of the equations which has its variables expressed as powers.
GRAPH: is a straight line use to express statements or ideal.
EXAMPLE
INDICIAL EQUATION
Solve the equations:
- 3 = 6 b. = c. + 4 = 0
SOLUTION
- 3 = 6 =
= → Y = 8
- = → =
2m-1 = -4 → 2m = -4 + 1
M =
- . Let = p
–5( + 4 = 0
– 5P + 4 = 0 (by factorization)
(P – 1) (P –4) = 0
Then p = 1 or p = 4
Recall = p
= , x = 0
When p = 4
( ) = , x = 2
Therefore x = 0 or 2
EXPONENTIAL FUNCTION: Is a function in which the variable is in the power i.e. y = where ≤ X ≤ 1.
EXAMPLE
Draw the graph of y = for values of x with intervals of 0.2 from 0 to 1
| X | 0.0 | 0.2 | 0.4 | 0.6 | 0.8 | 1.0 |
| Y = | 1 | 1.6 | 2.5 | 4.0 | 6.3 | 10 |
DIAGRAM OF THE GRAPH
ASSESSMENT: Solve the following questions:
= 0 (b) 32 (c) = (d) – 4( + 1 = 0
- Copy and complete the table of y = for 0 below
| X | 0 | o.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.7 | 0.8 | 0.9 | 1.0 |
| 1 | 1.3 |
- Solve the following:
- = 8. (b) = 243.
MORAL OBJECTIVES: (PROVERBS 8:11) Wisdom is better than rubies and all the things that may be desired are not to be compared to it.
Week 7 Review and periodic test
WEEK 8
LOGARITHMS:
Logarithms to a given base of a number are the power to which the base must be raised to give or make the number.
RELATIONSHIP BETWEEN INDICES AND LOGARITHMS
NUMBER POWER (INDICES) LOGARITHMS
10 = 1
100 2
1000 3
0.0001 -4
-2
LAWS OF LOGARITHMS
- LAW OF MULTIPLICATION
=
EXAMPLE
Simplify
SOLUTION
+ = = = = 2
- LAW OF DIVISION
– =
Simplify (i) – (ii)
SOLUTION
- =
= or 1 + 2
- = =
= = .
- = n
Simplify
Solution
= = = 5 = 5
- = N M =
=
Solution
= = x =
= ( = = 16
ASSESSMENT: Work out the following questions:
- 2 – 6 = 0
- – = 0
- 2 = 8
- – 2 +
- .
MORAL INSTRUCTION: The Lord expresses his mind to those that are close to him.
WEEK 9
SURD
RATIONAL NUMBERS: These are numbers or values that can easily be simplified or broken down into simpler form. They are numbers that can be express as ratios of whole numbers. i.e. , , etc.
IRRATIONAL NUMBERS: These are numbers that cannot be simplified easily. E.g. , , , etc.
SURD: Surds are irrational numbers which are roots of rational integers. E.g. , , , , etc.
CONJUGATE SURD: This is the simplification of two surds with the same value but different signs. E.g. ( + ) ( ).
RULES OF SURDS
- Surds of the same value can only be added and subtracted.
- Surds of differentvalue can be multiplied and divided.
- When rationalizing surd with an operational sign, the sign changes in operation.
SIMPLIFICATION OF SURD
Simplify the following surd:
(2) (3).
SOLUTION
- = = = 2
- = = = 2
- = = = 6
EXPRESSING VALUE AS A SINGLE SURD
Express (1) 2 (2) 7 as a single surd.
SOLUTION
- 2 = = =
- 7 = = =
BASIC OPERATIONS OF SURD
Simplify the following surds:
- (2) (3) (4) (5)
SOLUTION
- =
2 = (2+1) = 3
- 5 = 5
= 3
- = = = 9
- = =
= = =
- = =
= = =
ASSESSMENT: Evaluate the following questions:
- 3
- (i) (ii) (iii) (iv) (3
- (i) (ii) (iii)
- Express as a single surd:
- 12 (ii) 3 (iii) 5
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WEEK 10
MEASURE OF LOCATION (MEAN, MEDIAN AND MODE)
This is an aspect of Mathematics that deals with the measure of central tendencies.
MEAN: Is the average of any given set of numbers.
MEDIAN: This is the meddle number of a given distribution when arranged in ascending order
MODE: this is the number that occur most in a given distribution
The table below shows the distributions children of age x years in a hospital.
| X | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| F | 3 | 4 | 5 | 6 | 7 | 6 | 5 | 4 |
- Find the range of the distribution.
- Calculate the mean age of the children.
- What is the modal class of the children?
- Calculate the median of the distribution.
SOLUTION
| X | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | |
| F | 3 | 4 | 5 | 6 | 7 | 6 | 5 | 4 | 40 |
| FX | 3 | 8 | 15 | 24 | 35 | 36 | 35 | 32 | 188 |
- Range = Highest value Lowest value = 8 1 = 7
- Mean = = = = 4.7 or .
- Mode = 5
- Median = 5
EXAMPLE
The table below shows the number of work-days lost through illness among 500 factory employees during a one-year period.
| Number of days | 0-4 | 5-9 | 10-14 | 15-19 | 20-24 | 25-29 | 30-34 |
| Number of employees | 250 | 158 | 33 | 29 | 15 | 10 | 5 |
- Calculate the mean number of days lost.
- Draw a histogram for the distribution above.
- (i) From your histogram in (b) : estimate the modal days lost (ii) the median days lots of the distribution.
SOLUTION
| NO. OF DAYS | NO. OF EMPLOYEES (F) | MID-VALUE (X) | F X (FX) | CLASS BOUNDARY |
| 0 – 4 | 250 | 2 | 500 | |
| 5 – 9 | 158 | 7 | 1106 | 4.5- 9.5 |
| 10 – 14 | 33 | 12 | 396 | 9.5 – 14.5 |
| 15 – 19 | 29 | 17 | 493 | 14.5 – 19.5 |
| 20 – 24 | 15 | 22 | 330 | 19.5 -24.5 |
| 25 – 29 | 10 | 27 | 270 | 24.5 – 29.5 |
| 30 – 34 | 5 | 32 | 160 | 29.5 – 34.5 |
| 500 |
(a). Mean = = = 6.51
(b) GRAPH
© Solution from the graph.
ASSIGNMENT: Determine the solutions of the following;
- The masses of 40 students , in kg ,were recorded to the nearest kg in the table below:
| Class Interval | Frequency |
| 57 – 61 | 11 |
| 62 – 66 | 16 |
| 67 – 71 | 9 |
| 72 – 76 | 4 |
Calculate: (a) the mean of the distribution (b) draw a histogram for the distribution ,hence estimate the mode from your histogram (c) find the range of the distribution.
- Students taking a teacher-training course are grouped by age as in the table below:
| Age group | 19 – 20 | 20 – 21 | 21- 22 | 22 – 23 | 23 – 24 | 24 – 25 |
| Number in group | 4 | 5 | 10 | 16 | 12 | 3 |
Calculate the: (a) range (b) mean age of the distribution (c) median of the age group (d) draw a histogram for the distribution and from your histogram , find the modal age of the group.
Week 11 Revision
