Mathematics Lesson Notes SS1 Second Term
General Mathematics SS1 Second Term – Edudelight.com
SCHEME OF WORK
WEEK(S) TOPICS
- Revision of first term work
- Quadratic equation by: (a) factorization (b) Completing the square method .
- General form of quadratic equation leading to formula method
- Solution to quadratic equation by graphical method
Ideal of sets: (a) Universal sets, infinite and finite sets, Empty set, sub –set: (b) ideal and Notation for Union and intersection of sets
- Ideal of sets: (a) universal set, finite and infinite sets empty set and sub-set of set. (b) idea and notation for union and intersection of sets.
- Complements of sets: (a) disjoints and Null. (b) Venn diagram and its use in solving problems involving two and three sets relation to real life situations.
- Review of the first half term´s work and periodic test.
- (a) Introduction of circle and its properties: (b) Calculation of length of arc and perimeter of a sector. © Area of sectors and segments . Area of triangle.
- Trigonometric ratios: (a) Sine, Cosine, Tangent of acute angles. (b) use of tables for trigonometric rations: © Determination of length of chord.
10. ANGLE OF ELEVATION AND DEPRESSION
11. LOGIC: (i) Simple statements (ii) Negation (iii) compound statement (iv) conditional and Bi-conditional statement (v) inverse and Contra positive statement (vi) Conjunction and Dis junction .
12 and 13 REVISION and. EXAMINATION
WEEK I
REVISION OF FIRST TERMS WORK
CHANGE OF SUBJECT OF FORMULA
1. Make x the subject of the relation
SOLUTION
. q(1+ax) = p(1 ax) cross multiply
q + ax q= p a x p a x q + a x p = p q
X( aq + ap) = p
X = or .
2 make s the subject of the formula V = .
SOLUTION. = square both sides
T = , S = T
VARIATION
1. If x y, x = 1.5, y = 3.6 find y when x = 1.9.
SOLUTION
X y, x = k x y, when x = 1.5 and y = 3.6
1.5 = k x 3.6 divide through by 3.6
K = = . Then x = relationship between x and y.
When x = 1.9 1.9 =
Y = = 4.56.
2. If V
SOLUTION
V
K = . The relationship between V and T is V = .
Find T, when V + !2. T =
General Mathematics SS1 Second Term – Edudelight.com
WEEK 2
QUADRATIC EQUATION
Quadratic expression is any algebraic statement that has its highest power of the unknown as 2.
Example x 12, 4
EQUATION is any algebraic statement that is divided into two parts by an equality sign (=). Example, .
METHOD OF SOLVING QUADRATIC EQUATION
- Factorization method
- Completing the square method
FACTORIZATION METHOD
Factorize the following expression:
- 6 18a (ii) + 36 (iii) = 0
SOLUTION
QUADRATIC EXPRESSION
(i) 6 18a = 6a (a 3) i.e. taking out the highest common factors.
(ii) . Product of +36 and sum of 8e x 3e. Sum )
+36 =[
e[e 8] 3[e = [e ][e 8]
QUADRATIC EQUATION
(iii) +2x +1 = 0. i.e. x1= and 1x 1= 1
+ x +x + 1 = 0 ( +x)+(X+1)
X(x +1) + 1(X+ 1) = 0 (X+1)(X+1) =0
Either (X+1) = 0 or (X+1) = 0
X+1=0 X= 1 twice
COMPLETING THE SQUARE METHOD
4Y + 4 = 0
SOLUTION
4Y = 4 (Re-arrange the equation)
(Divide through by coefficient of )
4Y+ ( = . Half Y, square it and add it to both sides.
=
(Y = 0 Y 2 =
Y = 0 +2 Y = +2 or 2
OR using method of difference of two squares
(Y 2)(Y+2) = 0
Y 2 = 0 or Y + 2 = 0
Y = 2 or Y = 2.
ASSESSMENT: Solve the following questions:
Using factor method, solve the following quadratic equation:
- 10p 24 = 0. (b) 3 + 2k 1 =0. (c) 10 = 41m +45.
- Using completing the square method solve:
(a) + 2d 2 = 0. (b) 2 10x +7 = 0
General Mathematics SS1 Second Term – Edudelight.com
WEEK 3
GENERAL FORM OF QUADRATIC EQUATION
ax2 + bx + C = 0, is the general form of quadratic equation. Where a is the coefficient of x2, b is the coefficient of X and C is the constant value of the equation.
Deriving the quadratic formula from the general equation. The following steps must be followed:
Step 1. Re-arrange the equation
a + bX = c
Step 2. divide through by the coefficient of
+
Step 3. half X , square it and add it to both sides. i.e.
= .
+ = i.e.
(X + = or
X +
X +
X =
X =
ASSIGNMENT: NEW CONCEPT MATHEMATICS FOR SENIOR SECONDARY SCHOOLS BOOK1
EXERCISE 8.5 16-20
General Mathematics SS1 Second Term – Edudelight.com
WEEK 4
GRAPHICAL SOLUTION TO QUADRATIC EQUATION
GRAPH is a line or diagrammatical representation of information of a given data or distributions.
- Draw the graph of y = + 2x 3 for values of x from 2 to 4. using the scale of 2cm to 1 unit on x-axis and 1 cm to 1 unit on y-axis.
- Use your graph to solve the equation +2x 3=0.
- Find the least (minimum) value of y and the corresponding value of x.
SOLUTION
ASSESSMENT: Solve the following graph problems.
ASSIGNMENT: NEW CONCEPT MATHEMATICS FOR SENIOR SECONDARY SCHOOLS BOOK EXERCISE 8.7 nos. 6 and 20
General Mathematics SS1 Second Term – Edudelight.com
WEEK 5
IDEA OF SET
SET is defined as the collection of well-defined objects , information, items under consideration. Example : set of spanners, television set , Mathematical set, set of ages of students, etc..
UNIVERSAL SET: This is the mother set of all sets or is the set that contains the entire element under consideration or is the set from which other sets are form.
FINITE SET is a set with definite elements or elements that can be listed i.e. 1 to 10.
INFINITE SET is a set that the elements cannot be listed i.e. all natural numbers.
EMPTY SET is a set that contain no single element or has nothing inside as element. { }.
SUBSET OF SET is a set that has all its elements contained in another set.
SET NOTATION is the symbolic representation of a particular set
INTERSECTION OF SET is the common elements between two or more sets.
UNION OF SET is the collection of all the elements between two or more sets without repetition.
CARDINALITY OF SET is the number of elements in a given set.
EXAMPLE
Given that = { -4,-3, -2, -1, 0, 1, 2, 3, 4} is a universal set with subsets
A = { -2, 0, 3}, B = { -4, -2, 2, 3} , C = { -4 , 0, -2} and D = {1,2,3}
find : (i) A ∩ B (ii) Bᴜ C (iii) n(B) (iv) C∩ D
Union of set = ᴜ Empty set = { }, null or Ø
Cardinality of set A = n(A)
universal set = E or μ
Intersection of set = ∩
SOLUTION
- A ∩B ={-2,0,3} ∩ {-4,-2,2,3} = {-2,3}
- BᴜC = {-4,-2,2,3}ᴜ{-4,0,-2} = {-4,-2,0,2,3}
- n(B) = n{-4,-2,2,3} = 4
- C∩ D = {-4,0, -2} ∩{1,2,3} = { } or
ASSESSMENT: Find the solution set of the following questions. New Concept Mathematics for SS1 Exercise 8.9 D page 48.
Exercise B1: 1. a, b, c, d and e.
Question: 3, 4, 5, 6, 8 and 10.
General Mathematics SS1 Second Term – Edudelight.com
WEEK 6
COMPLEMENT, DISJOINT, NULL AND VENN DIAGRAM OF SET
COMPLEMENT OF SET is the element in the universal set that is not in the given set. It is denoted with or .
DISJOINT SET: given two non empty set A and B, if there exist no common element between the two set A and B, the outcome of both set is called disjoint set.
NULL SET is a set that has no element in it. It is denote with NULL or
VENN DIAGRAM is the use of diagram to represent set information.
COMPLEMENT OF SET
Example
Let = {1,2,3,…8,9}, A = { 1,2,3,4}, B = {2,4,6,8}
C = {3,4,5,6}. Find:
- (ii) (iii) C
SOLUTION
(i) = {5,6,7,8,9}
(ii) A C= {1,2,3,4}} {3,4,5,6} = {3,4}
( C ={1,2,5,6,7,8,9}
(Iii) B’ = {1,3,5,7,9,}
C n B’ = {3,4,5,6}n{1,3,5,7,9} = {3,5}.
DISJOINT SET
If = {1,2,3,4,5,6,7,8} and A ={1,3,5,7}, B = {2,4,6,}. Find A n B?
SOLUTION
AnB = { } or
TWO SET RELATIONSHIP
In a certain Local Government Headquarters, there are 70 labourers, 45 of them are permanent staff. If 36 of them are paid overtime weekly including all the daily staff, how many permanent staffs are paid overtime weekly.
| 1,3,5,7 |
| 2,4,6 |
SOLUTION
N( ) = 70, Let x = those paid weekly. Permanent staff (P.S) = 45 and Daily and weekly Staff = 3
μ = 70
70 = 45-x+x+36-x
70 = 81-x
70 -81 = -x
-11 = -x
i.e. x = 11
Therefore, the number that are paid overtime weekly is 11 labourers.
| x |
THREE SET RELATIONSHIP
In a group of 120 students, 72 play football, 65 play table tennis and 53 play hockey, if 35 of the students play both football and table tennis , 30 play both football and hockey, 21 play both table tennis and hockey and each of the students play at least one of the three games.
- Draw a Venn diagram to illustrate this information.
- how many of them play: (i) all the three games? (ii) exactly two or three games? (iii) exactly one of the three games? (iv) football only.
SOLUTION
n ) =120 , n(F.B) = 72, n(T.T.) = 65 ,n(H.) = 53, n(F.B.& H) = 30, n(F.B. & T.T.) = 35, n(T.T. & H) = 21, n(three games) = x
General Mathematics SS1 Second Term – Edudelight.com
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