SUBJECT: FURTHER MATHEMATICS CLASS: SS1
SCHEME OF WORK
| WEEK | TOPIC |
| 1 | Indices: Basic Laws & Application of indices |
| 2 | Indicial and Exponential Equations |
| 3 | Logarithms – Laws and application |
| 4 | General review of basic concept of set theory |
| 5 | Operation of sets and Venn diagrams |
| 6 | Review of First Half Terms Lesson & Periodic Test |
| 7 | Binary operations and basic laws of binary operations (i) Definition (ii) Solution of simple problems on binary operations (iii) Closure, commutative, associative and distributive laws |
| 8 | Binary operations continues: (i) Solution to problems on laws of binary operations (ii) Identity and inverse elements of a given binary operations (iii) Addition and multiplication tables for binary operations |
| 9 | Surds: (i) Definition of surds (ii) Rules and manipulation of surds (iii) Rationalization of surds at the denominator and equality of surds. |
| 10 | Measures of central tendency: (i) Mean, Median and Mode of grouped and ungrouped data (ii) Estimation of mode from the histogram of a grouped data. |
| 11 | Revision |
| 12 | Examination |
REFERENCE(S)
- Further Mathematics project 1 by Tuttuh Adegun et al
- New General Mathematics for SSS1, SSS 2 and SSS 3 by M. F. Macrae et al
WEEK ONE
TOPIC: INDICES
CONTENT
- Basic Concept of Laws of Indices
- Application of Laws of Indices
Basic Concept of Laws of Indices
A number of the form am where a is a real number, a is multiplied by itself m times,
The number a is called the base and the super script m is called the index (plural indices) or exponent.
1. a m x a n = am + n ——————–Multiplication law
Example: p3 x p2 = ( p x p x p) x (p x p) = p 5
Or p3 x p2 = p 3 + 2 = p5
2. am ÷ an = am – n ———————Division law
Example: p6 ÷ p4 = p 6 – 4 = p2
3. (a m )n = amn —————-Power law
Example: (p3)2 = p3 x p3 = p 3 + 3 = p6
Or p3 x 2 = p6
4. am ÷ am = am – m = a0 = 1
am ÷am = am/am = ao = 1
a0 = 1 ………………………Zero Index
Note : Any number raised to power of zero is 1
Example: 3o = 1, co = 1, yo = 1
5. (ab)m = ambm ————-Product power law
e.g. (2xy)2 = 4x2y2
6. a – m = 1/am ————- Negative Index
Example: 2 -1 = ½, and 3 -2 = 1/3 2 = 1/9
7. a1/n = n√a ————– Root power law
Example : 9 ½ = √9 = 3
27 1/3 =3√27 = 3 ie (3)3 = 3
8. a m/n = (a 1/n) m = (n√a)m ———–Fraction Index
or a m/n = (am) 1/n = (n√a)m
Example: 272/3 = 3√27 = 32 = 9.
Evaluation
1. 275/3 2. 10000000000 3. 2x-1 x 22x+2
Application of Laws of Indices
Examples
Solve the following
(i) 32 3/5 (ii) 343 2/3 (iii) 64 2/3 (iv) 0.001 (v) 14 0
Solution:
i) 32 3/5 = (32 1/5) 3 = (5√32) 3
= 2 3 = 8
ii) 343 2/3 = (343 1/3 )2 = (3√343)2
= (7 3)1/3)2
= 72 = 49
iii) 64 2/3 = (64 1/3)2 = (4 3)1/3)2 = 4 2
iv) (0.001)3 = (1/100)3 = (1/10)3)3 = (10 -3)3
= 10 -9 = 1/10 9
v) 14 0 = 1
General Evaluation
Simplify the following (a) 216 4/3 (b) 25 1.5 (c) (0.00001)2 (d) 32 2/5 (e) 81 ¾ (f) 6253/8 x 25
Reading Assignment : Further Mathematics project book 1(New third edition).Chapter 2 pg.4 – 6
Weekend Assignment
1) Evaluate 3 x = 1/81 (a) 4 (b) -4 (c) -2 (d) 2
2) Simplify 2r5 X 9r3 (a) p2 (b) 2p4 (c) P3 (d) 18r8
3) Solve 3-y = 243 (a) -5 (b) 5 (c) 3 (d) -3
4) Solve 25-5n = 625 (a) 1/5 (b) 2/5 (c) 1 1/5 (d) – 2/5
5) Simplify (0.0001)2 (a) 10-5 (b) 10 -3 (c) 10-8 (d) 10-10
Theory
1. 163/2 x 82/3 2. 3x2 x 4x3
321/5 6x7