Further Mathematics Lesson Note SS1 Second Term
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FURTHER MATHEMATICS SS1
SCHEME OF WORK
WEEK (s) CONTENT
- Review of first term work and introduction to the concept of functions.
- (i) Functions: (one to one, onto, inverse identity, constant, circular, logarithmic, exponential and composite) (ii) Application of functions solution of problems to functions.
- (i) Sequence and series (Arithmetic Progression A.P) (ii) Arithmetic mean series.
- (i) Sequence and series (Geometric progression G.P) (ii) geometric mean G.M.
- Linear inequalities in one variable: (i) Number Solution of x< a, x> a (ii) combined equalities of a<x<b, a
b
- Inequalities in two variables: (a) Drawing of graphs. (b) Definition of region satisfied by simultaneous linear inequalities.
- Review of the first half term and periodic test
- Trigonometrical ratios: (i) revision of sine, cosine and tangent of acute angle. (ii) Derive trigonometrical Ratio of special angle. (iii) Application of trigonometrical ratio of special angles in solving problems without the use of tables.
- Logical Reasoning
- Revision
- Revision
- Examination
- Examination
WEEK 1
REVISION OF FIRST TERM.
Find the value of r if 6r = 51
SOLUTION
x 64 + r x 8 + 7 x 1 = 5 x 81 + 1 x 9 + 1 x 1
+ 8r + 7 = 405 + 9 + 1
391 = 415
8r = 415
391
=
= 3
binary operation
is defined on the set R of real numbers by a
b =
. find the value of
SOLUTION
a = , b =
=
=
Solve = 3
SOLUTION
= 3
+ 15 =
= 64
15 = 49
=
7 or +7 or -7
New further Mathematics project 1 by Tuttuh-Adegun etal .
Revision exercise, chapter 2, question: 4,5,6 and 20.
Chapter 3, question: 2, 4, 5, 21 and 26.
- Onto function: is the process whereby every element of the co-domain is an image of at least one element in the domain. i.e. f(x) =
or f(x) = 2
+ 3x
9
- Constant function: Is when all the elements in the domain is assigned to a single element in the co-domain. i.e.
- INVERSE FUNCTION: A function f has an inverse if it is both one to one and onto. Or i.e. f(x) =
.
- IDENTITY FUNCTION: Given any function set A , if f: A→A which assigns to each element in A itself such that f(x) = x is called identity function.
COMPOSITE FUNCTION: is defined as two function f: X→Y and g: Y →Z such
- that the co-domain of f is the domain
- EXPONENTIAL FUNCTION: is a function that the input variable X occurs as an exponent (power). i.e. f(x) =
. Or f(x) =
.
- LOGARITHMIC FUNCTION: IS A FUNCTION y =
where b is any number such that b>0, b
1 and x> 0.
EXAMPLE
- Given that h:x→
is a mapping defined on the set A = {-1,0,1,2}, find the range of h.
SOLUTION
h(x) = , x = -1,0,1,2
h(-1) = = -4
h(0) = = -3
h(1) = = 0
h(2) = = 5
A h: x
+2x-3 B
2 5
One-to-one function.
(ii) The function f, g and h on the set of real numbers are defined by f(x) = +1, g(x) =2x + 3 and h(x) = 4x +5 respectively. Determine the formulae for the composite functions: (a) f[g(x)] (b) h{g[f(x)]} SOLUTION (a) f[g(x)] = 1(2x + 3
– 1 (by expansion)
= 412x +9 -1
4
+ 12x + 8 (b) h{g[f(x)]} = 4[2(
= 4[2
= 4(2
+5)+5 = 8
+20+5
8
+25. (iii) Let f:x
be a function defined on the set of real numbers excluding 2. Find
(x). SOLUTION
let y = f(x) = , (cross multiply)
yx – 2y = x + 1 yx – x = 2y + 1 (collect like term)
x(y-1) = 2y + 1
x = . Let x =
=
.
ASSESSMENT: work the following questions on function:
New Further Mathematics project 1, page Exercise 7A
questions 2, 3, 5, 7, and 9
page Exercise 7B questions: 3, 5, 7, 10, 13, 15, 16, 21, 23 and 25.
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WEEK 3
SEQUENCE AND SERIES (ARITHMETIC PROGRESSION A.P)
ARITHMETIC PROGRESSION (AP): is the paternally movement of numbers upwardly or downwardly. i.e. positive or negative movement to form a sequence. E.g. 2,5,8,11… 20,16,12,8…
EXAMPLE:
Given 5 x , find the first three terms of the sequence.
SOLUTION
Let n = 1, 2, and 3
5 x . When n =1
5 x = 5 x 1 = 5 (any value raise to power zero = 1)
When n = 2
5 x = 5 x
=
3
When n = 3
5 x = 5 x
=
Therefore the first three sequence is 5, .
TERM OF ARITHMETIC PROGRESSION.
= a + (n – 1) d. where a = first term ,n = the term under consideration, d = the common difference an
= last term of the given sequence.
EXAMPLE
Find the 20th term of the sequence 2, 5,8,…
SOLUTION
= ? , a = 2 , n = 20 and d = second term – first term
= 5 – 2 = 3
= 2 + (20 – 1) 3
= 2 + 19×3 = 2 + 57
= 59. The 20th term of the sequence is 59.
EXAMPLE
The 8th term of an A.P. is 25 and the 3rd term is 15. Find the first term and the 60th term of the sequence.
SOLUTION
= 25, a =?, d = ?, and n = 8.
= a +7d = 25……… (i)
= a + 2d =15….. (ii)
Subtract (ii) from (i) 5d =10
d = 10 = 2
Substitute 2 for d in (i) a + 7(2) = 25
a = 25 – 14 = 11.
To find the 60th term, where a = 11, d = 2 and n = 60.
= 11 + (60-1) 2 = 11+ 59 x 2
= 11 + 118 = 129.
ARITHMETIC SERIES (A.P)
Arithmetic series is the successive addition of subtraction of a given sequence. i.e. 2+5+8+11+…..
FORMULAE FOR A.P. SERIES.
=
{2a + (n-1) d}
=
{a + L}, when given the first and last term.
EXAMPLE
Find the sum of the 20th term of the linear series 10+13+16+…
SOLUTION
n = 20, a = 10, =? and d = 13-10 = 3
=
{2(10) + (20 – 1) 3}
= 10 (20 + 19 x 3) = 10 (77)
= 770
EXAMPLE
The first and last term of an A.P. is given as 0 and 120 respectively. Find the difference and the number of terms if the sum is 50.
SOLUTION
=
{a +L}, a = 0 , L =120, n = ?,
= 50.
50 = { 0 + 120}
50 x 2 = 120n
n = 100
.
To find the difference : =
{2a + (n-1)d}
50 = {2(0) + (
50 = {0+ –
}
50 = –
-5d = 50 x 72 (divide through by 5)
-d = 10 x72 = -720.
The difference of the A.P. is -720
ASSESSMENT: Attempt the following questions:
New further Mathematics project 1 by M.R Tuttuh-Adegun etal page Exercise 5 .
Questions 1 c, d and e.
Questions 3, 6, 8, 9, 10, 11, 12 and 13.
= 54 x –
=
Therefore, the 6th term is .
EXAMPLE
The second term of an exponential sequence is 35 and the fourth term is 875. Find the fifth term.
SOLUTION
= ar = 35 ………….(i)
= a
= 875……..(ii)
= 25
r =
r =
5
Substitute +5 and -5 for r in (i)
35 = a (+5) or 35 = a(-5)
a = 7 or a = -7
The fifth term of the exponential sequence
=
) =7 x 625
= 4375
And where r = -5
= 7 x 625 = 4375.
Therefore , the first term of the sequence is 7, the common ratio is 5 and the fifth term is 4375.
SERIES
Series is a partial sum of a given sequence. i.e. a + ar + +
+…+
.
EXAMPLE
Find the first 6th term of the sequence 2+6+18+…
SOLUTION
=
, r > 1. r = 6
= 3
=
=
= 728
Therefore, the sum of the sixth term of the sequence is 728.
SUM TO INFINITY
INFINITY: is an abstract concept describing something without any bound.
Find the sum to infinity of the sequence:
- 1+
+….
SOLUTION
=
, where a = 1 , r =
.
=
= 1
=
.
(ii) 5,2 ,1
,
=
=
= 1
= 1 x
=
ASSESSMENT: Work the following problems:
1 if 1++
+….are series of a geometric progression. Find the sum to infinity of the series.
2. The nth term of a sequence is given by
= 3 x
.write down the first four terms of the sequence.
3. the 2nd term of an exponential sequence and the 4th term is 81.find the:
(i) common ratio.
(ii) first term of the sequence.
(iii) Sum of the first five terms of the sequence.
4. the sum of the first nth of a sequence is 20 – , where n > 1, find the :
(i) sum of the first five terms.
(ii) fifth term of the sequence
5. find for each of the following geometric progression:
WEEK 5
LINEAR INEQUALITY (IN ONE VARIABLE)
INEQUALITIES: Are symbols or signs used for combining elements in arithmetic operations instead of equality sign. i.e. <,>,.
< = less than
> = greater than
= less than or equal to
= greater than or equal to.
CHARACTERISTICS OF INEQUALITIES
- They are reversed when the order of operation is reversed. i.e. x + y < 7, then 7 > x + y.
- They are affected by negative signs or numbers. They change to the opposite inequalities when combined with negative sign. i.e. –(a+b
c) is –a-b
-c.
- Inequalities are represented by number line and Cartesian graph. (a) NUMBER LINE is a line with an arrow indicating the direction of the numbers that satisfy the given inequality. i.e. arrow with empty circle is illustrated by < and >. While arrow with shaded circle are illustrated with
and
. (b) CARTESIAN GRAPH is a form of representation where shaded area is used to distinguish between the required inequality and the opposite end at the equality region.
- Inequalities can be combined as complete or incomplete solution. i.e. a<x<b (complete solution) and x < a, x > b (incomplete solution).
- Two opposite inequalities are used when a quadratic expression is given. i.e. (x-3)(x-2) > o
x-3 > 0 and x-2 < 0. They are used by combining the lowest factor with the given inequality and the highest factor with the opposite sign.
- Inequalities with variables in the denominator are evaluated by cross multiplication with the combining inequalities having different values according to the nature of the variables. i.e.-
< p
x > py.
RANGE OF INEQUALITIES
Range of inequalities is the value between which the solution of the inequalities lies or cannot lie. i.e. -3 < x <2 means values from -3 to 2.
EXAMPLE
Solve the inequality 5x + 7 6(x +3)
SOLUTION
5x + 7 6(x+3) open the bracket with +6
5x + 7 6x + 18
5x-6x
18 -7
-x 11
x
-11
EXAMPLE
Solve the inequality -9x -10 < 0
SOLUTION
-10x + x -10 < 0 (by factor method)
−10)(x+1)< 0
x-10 < 0 or x+1 > 0
< 10 or x > -1 (by combination)
< x < 10. The required range.
Find the solution set of -x-6 > 0, using number line.
SOLUTION
-x-6 > 0, (by factor method)
(x2)(x – 3)>0
x+2 >0 or x-3<0
X > -2 or x < 3, by combination
-3 -2 -1 0 1 2 3
ASSESSMENT: Solve the following inequalities graphically:
– 3
2+x
+ 4 < x +3 < 4x -5
+
2x + 5
- 0
+ 5x + 3
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WEEK 6
INEQUALITIES IN TWO VARIABLES.
This symbolic sign involves two unknown (variables x and y) and the Cartesian graph is required for the solution to this inequalities.
CARTESIAN GRAPH: is the use of shaded region to distinguish between the required inequality and the opposite end at the equality region (boundary).
The boundary line is dotted or broken lines (<,>) or the full thick line (
EXAMPLE
Show graphically the region represented by x + y 5
SOLUTION
Take intercept of x
y = 0, x + 0 5
x
5
Take y-intercept
X=0, 0 + y 5
y
5.
Therefore, (x,y)= (5,0) and (0,5).
EXAMPLE
Given x 0, y
0 and 3x +4y
12. Show graphically the region satisfied by the inequalities.
SOLUTION
x 0, y
0
X-intercept, y=0
3x+4(0) 12
x
12÷3=4
Y-intercept, x=0
3(0) + 4y 12
y
12
=3
Therefore, (x,y)= (0,0), (4,0) and (0,3).
ASSESSMENT: solve the following inequalities graphically and shade the required region:
- Y >0, y<x, 2x + 3y > 6,and 2x +3y < 12
- X> 0, y > 0, 2x + y < 10 and x +2y
-2
- 4y – 3x + 5
0
- X – 4y +
0
WEEK 7
REVIEW AND PERIODIC TEST.
| c |
| b |
| a |
From the triangle above.
=
=
By Pythagoras theorem, c = =
C =
=
=
or
From the triangle above. considering angle
=
,
and
Angle
a
b = 1
C = 1
From the triangle above =
By Pythagoras theorem a = =
a =
and
APPLICATION OF TRIGONOMETRIC RATIO (SPECIAL ANGLE) TO PROBLEMS.
EXAMPLE
Without tables, solve the following:
The angle of elevation of the top of a tower from a point on the horizontal ground 80m from the foot of the tower is 60°. Find the height of the tower.
SOLUTION
10cm
xcm
80m
=
h = 80m x
h = 80m x = 80
m
Therefore, the height of the tower is 80m.
EXAMPLE
In the diagram below, find the lettered side.
10cm
SOLUTION
=
X = 10CM X
X = 10CM X =
CM.
ASSESSMENT: Evaluate the following trigonometric questions:
1 use tables to evaluate the following:
- Exercise 10 , questions 1 a – f
- Question 2 a – f
Question 6 a – f.
2. Question 20 a- d
3. Question 26, 28 and 30,
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