Download 5. Polynomials多項式---------------------------------------------------------

Pentecostal School
五旬節中學
Secondary 4/5 Mathematics
中四及中五級數學
Concise Revision Notes
溫習筆記精讀
Table of Contents 內 容
1.
Indices, Logarithms and Surds 指數、對數及根式 -----------------------------
1
2.
Quadratic Equation 一 元 二 次 方 程 式 -----------------------------------
2-3
3.
Function 函數 -------------------------------------------------------------------------------
4
4.
Theorems of Geometry (Plane & Circle) 幾 何 定 理 ﹝ 平 面 及 圓 形 ﹞
5
5.
Polynomials 多項式 ----------------------------------------------------------
6
6.
Rate, Ratio, Proportion and Variation 比 率 、 比 例 、 相 比 、 變 --------- 7
7.
Trigonometry I: Ratios and Graphs 三 角 幾 何 ： 比 及 圖 像 -------------
8-11
8.
Trigonometry II: Applications 三 角 幾 何 ： 應 用 ------------------------
12
9.
Arithmetic and Geometric Sequence 算 術 及 幾 何 數 列 -----------------
13
10. Linear & Quadratic Inequalities In One Variable 一 元 不 等 式 --------- 14
11. Probabilit y 或 然 率 ---------------------------------------------------------
15
12. Statistics 統 計 學 -----------------------------------------------------------
16-17
13. Coordinate Geometry (Straight Line & Circle) 坐標幾何﹝直線及圓形﹞------
18-19
14. Linear Inequalities In Two Variables & Linear Programming 二元不等式及線性規劃- 20
15. Approximate Solutions of Equations & Bisection 方程式近似值及分半法-------
21
16. Areas of Plane Figures & Volumes of Solids 面 積 及 體 積 -------------
22
17. Percentage 百 分 數 ---------------------------------------------------------- 23
18. Glossary (Question/Answer) 詞彙﹝問題/答案﹞------------------------------- 24
Law of Indices 指數
律
1. a 0  1
2. a1  a
a  a
 a  an
3. a
n
4. a
a  
a
 a  na

n
am  an  amn
am
1
 a m  n (m  n) or n  m (m  n)
n
a
a
5.
6.
a 
m n
7.
 a m n
1
m
8.
a  am
n
m
a a
1
 am
10.
m
a
Example:
2 x  8 y  2 x  23 y  2 x  3 y
(i)
a n  2  a n 1 a n  2 1  a 
(ii)

1 a
an2
an2
m
9.
n
a b 
a b 
3 1 2
(iii)
1 2 4

a 6b 2
1
1
  4 6 8 2  2 6
4 8
a b
a b
ab
(iv) If 9 x  2  36, then 3x 
Solution:
9 x  2  32 
x2
So
 32 x  4  32 x  34
36 36
32 x  4 
3
81
36 6 2
3x 
 
81 9 3
Law of Logarithms
對數律
Definition:
If N  10 x , then log 10 N  x .
1. log 1  0
2. log 10  1
3. log 100  2
4. log 0  undefined
5. log M  log N  log( M  N )
M 
6. log M  log N  log  
N
n
7. log M  n log M
Example:
(i)
If log x  a  2, then x 
Solution:
log  x  a   log 100,
so x  100  a.
log 8  log 4
(ii)
log 16
3 log 2  2 log 2

4 log 2
3  2log 2  5

4 log 2
4
Surds &
Rationalization 無理
根式的有理化
1.
a  b  ab
3.
a a a
4.
1
1
a
a



a
a
a
a
5.
6.
7.

a

a
a
1
a
2.
a
a

 a
a
a
b  a  b  a b


b

1
a b

a b
a b

a b
ab
Example:
(i)
27  12
 9 3 4 3
(ii)
3 32 3  3
1
2 3 2
a

b
a
b


(iii)
2 3 2
2 32 3  2 2
2 3 2

10
b
a

a b
a b
b




a  b 2  a 2  2ab  b2
a  b 2  a 2  2ab  b2
a  ba  b  a 2  b2
2.
3.
4.
5.
 b  b2  4ac
2a

 b  b2  4ac
 b  b2  4ac
,  
2a
2a
such that
2x2 + 4x  3 = 0.
Solve
Identities 恆等式
1.
x
Example 2:
a b  a a b
a a  b b
ba ab


ab a b

2. Quadratic Formulae 一元二次公式
1
1

2 3 2 2 3 2
Solution:
 (4)  (4) 2  4  (2)  (3)
1

 1 
10
2  (2)
2
 (4)  (4) 2  4  (2)  (3)
1

 1 
10
2  (2)
2
3.
(a  b)(a 2  ab  b2 )  a3  b3
(a  b)(a 2  ab  b2 )  a3  b3
Sum and Product of Roots
 
b
a
根的和及積
 
and
c
a
Example 3:
Form a quadratic equation whose roots are
1
Quadratic Equations 一元二次方程
Solving Quadratic
Equation 解方程式
If a2b+c0 where a, b and c are real
numbers, then its root(s) 根 can be solved by
1.
Factorisation 因式分解
1
1
10 and  1 
10 .
2
2
Solution:
    (1 
1
1
10 )  (1 
10)  2
2
2
    (1 
1
1
3
10 )  (1 
10)  
2
2
2
So if a = 2, then c = 3
and
b = +4.
The required equation is 2x2 + 4x  3 = 0.
20
(x)(x-)=0
 roots (x) =  or 
Example 1:
Solve 2x25x3=0.
4. Graphical Method 圖像解
(a)
 y  ax 2  bx  c

 y  0y=ax2+bx+c
Solution:
2x2  3x  2=0
(2x + 1)(x  2)=0
x =1/2
y=0


or 2.
c
Example 4:
y  5x  3
Solve graphically 2x25x3 = 0.
Solution:
Step 1:
Draw the graph of y = 2x25x3.
y  2x 2
y  2 x 2  5x  3
Step 2: Read from the graph the x-intercepts
截線, x=1/2 or 3.
Given that
=b24ac, then either
(1) If >0,  and  are 2 distinct 分別 roots
 y  ax 2

 y  bx  c
(b)
Roots and
Discriminant  根
和判別式
or
(2)
y=ax2
If =0,  and  are equal 相等 roots
or
(3) If <0,  and  are not real (no solution).
(2)
(3)
(1)
y= bxc


Example 5:
 y  2x2
Solve graphically 
.
 y  5x  3
Example 6:
Solution:
Determine the discriminant of the following
quadratic equations and state the nature of the
roots.
Step 1: Draw the graphs of the given
equations.
Step2: Read from the graph the
x-coordinates of the points of intersection of
the two graphs,
x = 1/2
or 3.
(1) 2 x 2  x  1  0
(2) x 2  2 x  1  0
(3) x 2  2 x  2  0
Solution:
(1)  = (1)24(2)(1) = 9.
So there are two distinct roots.
(2)  = (2)24(1)(1) = 0.
x2  x  2  0
=(1)24(1)(2)
=9, so there are 2
( x  1)( x  2)  0
x  1 or 2
points of
intersections.
x 2  x  6  2x  4
So there are two equal roots.
(3)  = (2)24(1)(2) = 4.
So there is no real root.
Check by graphical method:
(1) y  2 x 2  x  1
Relation of Vertex and Coefficients of
Quadratic Equation 二元方程式的角點與係
數關係
y=ax2+bx+c
(3) y  x 2  2 x  2
(2) y  x 2  2 x  1
y=a(x-h)2+k
c
h
k
Number of points intersections and
Discriminant of simultaneous equations: One
quadratic and one Linear 聯立方程式的交接
(h,k)
h
點數目及判別式
b
2a
and
k c
 y  a 2  bx  c
Given that 
where a, b, c, m,
y

mx

n

Example 8:
n are integers, then the number of points of
Solution:
intersections can be determined by the
discriminant of combined equation:
h
ax 2  (b  m) x  (c  n)  0
1
 1
So the vertex is   ,1 
8
 4
(1) >0
(2) =0
Find the vertex of
(1)
1
 ;
2(2)
4
b2
4a
y  2x 2  x 1 .
k  (1) 
(1) 2
1
 1
4(2)
8
(see the graph (1) drawn in the figure on the
left)
(3) <0
Functions 函數
Meaning of Function 函數的意義
A function expresses the relationship between
varying 變動 quantities
Example 7:
y  x  x  6
.

y

2
x

4

2
Solve
Solution:
(1) In the formula A=r2,
the area of a circle A is expressed in terms
of its radius r. The value of A depends 依
賴 on the value of r. A and r are varying
quantities called variables. Since A
depends on r we say that r is the
independent variable 自變數 and A the
dependent variable 倚變數. We say A is a
function of r.
(2)
In the graph y = 2x25x3,
the value of y depends on the value of x.
We also see that corresponding to one
value of x there is only and exactly one
value of y. We say y is a function x.
x 1 0 1 2 3 4
y 4 3 6 5 0 9
Concept of Function 函數的概念
A variable y is a function of the variable x
defined on a set of numbers D if to every value
of x in the set D there corresponds 對應 only
one value of y.
The set of numbers D of x is called domain of
the function 函數域.
Example:
(a) y  x  2 for  1  x  2 .
x
y
1
1
0
2
1
3
2
4
2
(b) y  x
for  1  x  2 .
x
y
1
1
0
0
1
2
4
2
Note:
If y  x  1, then y is not a function
of x because one value of x
corresponds to two values of y.
x
y
3
2
2
Notation of Function 函數的符號
Suppose y is a function of x. We can use f to
denote the function using the notation
y  f (x) .
The symbol f (x) is read as “f of x” and f is to
represent the rule of correspondence.
Examples
(a) Constant Function 常值函數
f ( x)  2
(b) Linear Function 綫性函數
f ( x)  2 x  1
(c) Quadratic Function 二次函數
f ( x)  2 x 2  4 x  1
(d) Trigonometric Function 三角幾何函數
f ( x)  sin x for 0  x  2
(e) Logarithmic Function 對數函數
f ( x)  2 x
Evaluating Function 求出量值
We may think of function as a machine that has
an input and output; if a number x in the
domain is fed into the machine, then the
number f(x) will appear as the output.
Example 1:
Let f ( x)  x3  2 x 2  9 x  18 , find f ( 2) .
Solution:
f (2)  (2)3  2(2) 2  9(2)  18
0
Example 2:
Let f ( x)  x 2  x  3 . If f ( k )  k , find the
value(s) of k.
Solution:
Since
f (k )  (k ) 2  (k )  3
k2  k  3  k

k 2  2k  3  0
k  3k  1  0
k = 3 or 1
Theorems of Geometry 幾 何 定 理
Angles and
Parallels 角與平行
線
1.
adj. s on a line 直線上的鄰角(a+b=180)
2.
s at a point 同頂角(a+b+c+d=360)
3.
vert. opp. s 對頂角(a=d)
4.
corr. s, // lines 同位角(b=f)
5.
alt. s, //lines 錯角(d=e)
6.
int. s, //lines 同旁內角(d+f=180)
7.
corr. s equal 同位角相等
8.
alt. s equal 錯角相等
a
c
e
b
d
f
9.
int. s supp.同旁內角互補
AB AC BC


DE DF EF
Angles of Polygon 多邊形的角
1.
 sum of  (內角和) (a+b+c=180)
2.
exterior  of  (外角) (a+b=d)
3.
 sum of polygon
Pythagoras Theorem 畢氏定理
a
b
sum of exterior  s of polygon
4.
c
1. a2+b2=c2
b
2. Converse of Pythagoras theorem 畢氏定理逆定理
Parallelograms 平行四邊形
e
=(n-2)x180 多邊形內角和
a
c d
f
多邊形外角和 (d+e+f=360)
Congruent
Triangles 全等三角
形
1.
opposite sides, //gram 平行四邊形對邊(AB=DC)
2.
opposites, //gram 平行四邊形對角(A=C)
3.
diagonals, //gram 平行四邊形對角線(AC=BD)
4.
opposite sides equal 兩組對邊相等
5.
2 sides equal and //一組對邊相等且平行
6.
opposite s equal 對角相等
3. A.S.A.(SAA,AAS)
diagonals bisect each other 對角線互相平分
D
C
L1
c
xc e
a
L2
y
d
f
A
B
b
c
L3
Mid-point Theorem, Intercept Theorem and Equal Ratios
4. R.H.S.
1.
mid-point theorem 中點定理 (a=b/2,L2//L3)
5. corr. s of  s 全等的對應角
2.
intercept theorem 截線定理 (if c=d, then e=f)
3.
equal ratios theorem 等比定理(
1.
S.S.S
2.
S.A.S.
6. corr. sides of  s 全等的對應邊
Isosceles Triangle
等腰三角形
1.
base s, isos. 等腰底角(a=b)
2.
sides opp. = s 等角對邊相等
 from centre bisects chord 圓心至弦的垂線平分
弦
2. line joining centre and mid-point of chord  chord
圓心至弦中點的連線垂直弦
3. =chords, equidistant from centre 等弦的圓心等距
(x=y)
1.
a
Angles of Circle 圓內的角
Equilateral Triangle
等邊三角形
properties of equilateral  (3 sides equal & 60)
A
D
1. A.A.A.
6
2
2.
3 sides proportional 三邊成比例 9
3E
F
3.
2 sides proportional and included  equal
三邊成比例且夾角相等
4.
B
6
C
Properties of Similar Triangles 相似三角形性質
a
y
 at centre twice  at circumference
圓心角是圓周角兩倍(b=2a)
b
2.  in semi-circle 半圓上的圓周角(c+d=90)
3. s in the same segment
同弓內的圓周角(a=c)
1.
x
c
d
Arcs, Chords and
Angles 孤，弦和角
全等的性質﹝三邊相等和內角 60﹞
Similar Triangles 相似三角形
c x
 )
d y
Chords of Circle 圓內的弦
b
1.
7.
4
= arcs, = s at centre 同圓心角所對的弧相等
= chords, = s at circumference 同圓周角所對的
弦相等
3. proportional arcs, proportional s at circumference
相比同圓周角所對的弦相比
1.
2.
Cyclic
Quadrilaterals 圓內
接四邊形
1.
2.
2x+3x2+4x+5x2=6x+8x2
3x2y2(2xy)2+xy=xyx2y2
Evaluating 估值
Suppose f(x)=3x22x+1
opposite s, cyclic quad.圓內四邊形對角
(b+c=180)
exterior , cyclic quad.圓內接四邊形外角(a=d)
Concyclic Points 共圓點
opposite s supplementary 對角互補
b
exterior  = interior opposite 
外角等於內對角
3. converse of s in the same segment
同弓形內的圓周角逆定理
f(0)=3(0)22(0)+1=1
a
Remainder Theorem 餘式定理
1.
2.
c
d
When a polynomial f(x) is divided by factor 因子
(x-a), the remainder 餘式 R is equal to f(a).
Example:
Tangent to Circle 圓的切線
tangent  radius 切線與半徑垂直
y
a
converse of tangent  radius
b
切線與半徑垂直的逆定理
x
3. tangent from external point
由外點引切線(x=y, a=b)
d
4.  in alternate segment 交錯弓形的圓周角(c=d)
5. converse of  in alternate segment
交錯弓形的圓周角的逆定理
f(1)=3(1)22(1)+1=2
then
To find the remainder of (3x22x+1)(x1):
1.
2.
f(1) =3(1)22(1)+1
=2
c
Polynomials 多項式
Definition 定義
1. A polynomial in one variable 元, such as , of
degree 次數 n is a algebraic expression of the
form 代數表達形式
Check by long division method 長除法驗算:
3x  1
x  1 3x  2 x  1
2
3x 2  3x
 x 1
 x 1
 2
Factor Theorem 因子定理
If f(x) is a polynomial with f(a)=0, then (x-a) is
a factor of the polynomial.
Example:
Factorise 因子分解 3x22x1.
Since f(1)=3(1)22(1)1=0,
an x n  an 1 x n 1  ....  a2 x 2  a1 x  a0
where n is a non-negative integer 非負數整數
So (x1) is a factor.
Also since
 1
  1
  1
f     3   2   1  0
 3
 3 
 3 
2
and the coefficients 係數 an , an 1 ,...., a1 , a0 are
real numbers 真實數 with an  0 .
Example: 3x22x+1
Degree=2; Variable=x, Number of monomials 單項
=3, Order=Descending 降冪。
Addition and Subtraction 加和減
Coefficients of monomials whose variables and
degree are both equal can be simplified 化簡 by
adding or subtracting.
Example:
2x+3x=5x
2x+2xy+3y+3xy=2x+3y+5xy
So
1  is

x  
3

another factor. It can be further
simplified as (3x+1).
So
3x22x1(x1)(3x1).
Check by long division method 長除法驗算:
3x  1
x  1 3x 2  2 x  1
3x 2  3x
 x 1
 x 1
0
Equality 對等
If
Ax2+Bx+C3x22x+1,
then A=3, B=-2, C=1.
Identities of Sum and Difference of Cube 立方和立
方差恆等式

 x  a x
x 3  a 3  x  a  x 2  ax  a 2
x3  a3
2
 ax  a 2


Example:
x3+1=(x+1)(x2x+1)
27x364=(3x)3(4)3=(3x4)(9x2+12x+16)
Ratio, Proportion and Variation
Rate 比率
Rate compares two quantities of different kinds and has
units.
Example:
The speed of light is 3108 m per second.(m/s). It travels
round the earth 7 and half times in one second.
Highest Common Factor 最大公因數 &
Ratio 比例
Lowest Common Multiple 最小公倍數
Ratio compares two quantities of the same kind and has
Example:
To find the HCF and LCM of 12x2yz and 8xy3.

4 xy 12 x 2 yz , 8 xy3
       3 xz,  2 y
2
no units.
Example:
In a map whose scale is 1:20 000 means that 1 cm on the
map represents 20 000 cm (or 20 km) actual distance.
H.C.F.=4xy, LCM=24x2y3z
Si mplification of Fractio ns 化 簡 分 數
(a)
Common factors of numerator 分子 and
Proportion 相比
A proportion is an equality of two ratios.
Example:
denominator 分母 can cancel 相約 each other.
2:5=4:10=5:12.5
Example:
2 4
5


......
5 10 12.5
x y
2x  y
 2
x  2 y 2 x  3xy  y 2
( x  y)
(2 x  y )


( x  2 y ) (2 x  y )( x  y )
1

x  2y
(b) When adding or subtracting algebraic fractions
So when a:b=2:5, we can say that a=2k and b=5k.
The k-method
2 4
5


k
5 10 12.5
k  0.4
If
Then 2=5k, 4=10k, 5=12.5k
Example 1:
of different denominators, it can be simplified
If a:b=2:3 and b:c=4:5, find a:b:c.
by finding the LCM of the denominator 通分母.
Solution:
Example:
From the first ratio:
x y
x y
 2
x  2 y x  xy  2 y 2
( x  y) ( x  y)
x y



( x  2 y ) ( x  y ) ( x  y )( x  2 y )
( x  y )( x  y )  ( x  y )

( x  y )( x  2 y )
( x  y )(( x  y )  1)

( x  y )( x  2 y )
x  y 1

x  2y
a=2k, b=3k.
nd
From the 2 ration,
b=4k, c=5k.
So when b=12k, a=8k and c=15k.
So
a:b:c
= 8k:12k:15k
= 8:12:15
Example 2:
If a=2b=4c, find a:b:c.
Solution:
If a=2b, then a:b=2:1.
So
If 2b=4c, then b:c=4:2. So
a=2k and b=k.
b=4k and c=2k.
So when b=4k, a=8k and c=2k.
a:b:c = 8k:4k:2k = 4:2:1
(c)
Example 3:
z varies inversely as both x and y.
k
xy
z
If 1 : 1 : 1  2 : 3 : 4, find a:b:c.
a b c
Example:
y varies directly as x2 and inversely as
Solution:
x =2 and z=9, find y when x=1 and z=4.
1
1
1
 2k ,  3k ,  4k .
a
b
c
Let
Solution:
Let
1
1
1
,b 
,c 
.
2k
3k
4k
So
a
So
a:b:c  (
1 1 1
:
: ) x12k  6 : 4 : 3
2k 3k 4k
k
So
Example 1:
1
12
y
1
2
 1  (4 )
12
y
1
6
From the ratios given, express the variables in terms of k.
If a:b=3:2, b:c=4:3, find (a+b):(b+c).
Then
= kx2 z
y
(1) = k (2)2 9
So l v i n g P ro b le ms b y t h e k - me t ho d
Suppose
Partial Variation 部分變
a:b=3:2=k.
(a) z partly varies as a constant and partly varies
a=3k, b=2k,
directly as x.
a=6k, b=4k.
So
z . If y=1 when
z  k1  k2  x
b=4k, c=3k
(a+b):(b+c)=((6k)+(4k)):((4k)+(3k))
(b) z partly varies directly as x and partly varies directly
as y.
=10k:7k
z  k1  x  k2  y
=10:7
(c) z partly varies directly as x and partly varies
Direct Variation 正變
If
y
inversely as y.
xy
z  k1  x  k2 
direct
Then
x
k
y
1
y
Example:
y varies partly as x and partly as x2. When x=2, y=20 and
So
x ky
when x=3, y=39. Express y in terms of x.
Inverse Variation 反變
If
x
inverse
Solution:
Let
1
y
y
=
k1x+k2x2
(20)= k1(2)+k2(2)2
x
20
Then x  y  k
(39)= k1(3)+k2(3)2
So
(a)
(b)
x k
1
y
Joint Variation 聯變
z varies directly as x and y.
=
2k1+4k2
39
=
3k1+9k2
(1)3:
60
=
6k1+12k2
(3)
(2)2:
78
=
6k1+18k2
(4)
(4)(3): 18 =
6k2
z  k x y
k2
=
3
z varies directly as x and inversely as y.
k1
=
4
y
=
4x+3x2
z  k x
1
y
(1)
So
(2)
Given that: 3602 radians
Trigonometry Ratios 三角幾何比的比
So
1
Angles of Rotation 角的轉向
= 2   radian
360 180
o
o
1 radian = 360  180
A positive angle is formed by rotating anti-clockwisely
2
逆時針 from the horizontal x-axis.

A negative angle is formed by rotating clockwisely 順時
Example:
針 direction starting from horizontal x-axis.
Express 144 in radian measure.
Solution:
Degree measures 度
(a)
The degree of one rotation is 0 to 360.
Example:
144 x
90o


180
4 radian
5
Convert  radian into degrees.
12
Solution:

140o
12
180o
360o
sin  

 15 o
opposite  side y

hypotenuse
r
(x,y)
r
cos 
adjacent  side x

hypotenuse
r
tan  
opposite  side y

adjacent  side x
270o
Radian measures 弧度
180 o
Ratios of Sine 正弦, Cosine 餘弦, Tangent 正切 三角比
70o
(b)
x

CAST Rule 規則
Radian = Arc  Length
All “+”
radius
Sine “+”
The radian of one rotation is 0 to 2.

2
A
S
II
If arc length = radius
III
length, then the
angle measure is 1
Tangent “+”

4 rad
1 rad
2
C
Example:
In Quadrant I:
In Quadrant II:
sin30o
sin150
=+0.5
length, then the
In Quadrant III: sin210
=0.5
angle measure is 4
In Quadrant IV: sin330
=0.5
radians.
Ratios of 180 and 360.
If arc length = 4 radius
IV
Cosine “+”
T
radian.
I
3
2
=+0.5
opp=+y
180
Conversion of Degrees and Radians 度與弧度的轉換

adj=+x
hyp=1
opp=+y
adj=x
hyp=1
y
adj=x

2) cos   sin 90
y
y
hyp=1
opp=y
adj=+x
hyp=1
180
360

cos   cos180
tan    tan 180


     cos180
     tan 180



     cos360   
     tan 360   
sin    sin 180o     sin 180o     sin 360o  
o
o
o
o

 
1) sin   cos 90 o  
x
x
opp=y
Identities 恆等式
y
o
o
o
1
tan 90 o  
1
4) tan( 90 o   ) 
tan 
sin 
5) tan  
cos 
2
6) sin   cos 2   1
3) tan  


7) sin 2   1  cos 2 
8) cos 2   1  sin 2 
Example 1:
Example:
sin 30o  cos(900  30o )  cos 60o
sin 10o   sin 170o   sin 190o   sin 350o
cos 40o  sin( 90o  40o )  sin 50o
1
1
tan 60o 

o
o
tan( 90  60 ) tan 30o
cos10o   cos1700   cos190o   cos 350o
tan 10o   tan 170o   tan 190o   tan 350o
Example 2:
Ratios of 90 and 270.


If tan 90o    2 , find
90+
opp=+x
Solution:

adj=y
x
hyp=1
opp=+y
adj=+x
y
y
y
hyp=1
x
x
x
opp=x
adj=+y
270
hyp=1
opp=x
adj=y
270+

cos   sin 90


     sin 270
sin 3   sin  cos 2 
sin 
sin  sin 2   cos 2 

cos
sin 

cos
 tan 
1

tan 90o  
1

2





     sin 270

 
sin    cos 90o     cos 270o     cos 270o  
o
o
o
1
1
1


tan 90o  
tan 270o  
tan 270o  





30o 30o
45 2
3

60
1
o
60
o
45o
1
sin 70o   cos1600   cos 2000   cos 340o
2
2
o
Example
1
1
0
0
30
45
60
Sine
1
2
Cosine
1
3
2
1
2
1
2
3
2
1
2
cos 700   sin 160o   sin 200o   sin 340o
tan 70o  

Special Angles 特殊三角比
hyp=1
tan   
sin 3   sin  cos 2  .
sin 
1
1
1


tan 160o
tan 200o
tan 340o
90
1
0
Tangent
0
1
3
1
3
Undefined
Example 2
Solve (cos   3)(3 sin   2)  0 for 02.
Solution:
Since
Simplify 化簡 Trigonometric Ratios

either
cos3=0
or
An expression with more than one trigonometric ratios
Example 1:
3sin2=0
3
+2
cos =3

sin=2/3
So
can be simplified by using the identities and CAST rule.
/2
or
Since the maximum value of
cos(90 o  A)  cos(  A)
sin( 360 o  A)
(sin A)  ( cos A)

 sin A
  cos A
3


+2
0/2
3/2
cos is 1, so cos=3 is undefined.
And
since sin=2/3 when 0.730 (in Quadrant I)or
0.7302.41(in Quadrant II)
Example 2
1  sin 
cos

cos
1  sin 
1  2 sin   sin 2   cos 2 

cos  (1  sin  )
2  2 sin 

cos  (1  sin  )
k
k
k 2 1
2  (1  sin  )


cos  (1  sin  )
1
1
270
2

cos

Graphs 圖像
A. Sine Curve 正弦

Graph of y=sin
k 2 1
90
180
270
360
Example 3
If cos  1 , find tan(  270 o ) .
k
1. The shape of Sine Curve is waveform 波浪.
Solution:
2. The maximum and minimum values of sine curve are
Refer to the diagram.
tan(  270 o )

1
k 2 1
Solving 解 Trigonometric Equations
Example 1
Solve cos=3sin.
Solution:
Since
So
sin 
1

cos
3
1
tan  
3
From the Table of Special Angle, =30 in Quadrant I.
Using the CAST rule, tan is also “+” in Quadrant III, so
=180+30=210.
So cos=3sin when 30or 210.
respectively +1 and –1.
3. From the graph,
sin=0 when =0o, 180o or 360 o,
sin=+1 when =90,
sin=1 when =270.
Transformations of Sine Curve 正弦圖像的變換
2. The maximum and minimum values of Cosine Curve
sin+1
are respectively +1 and –1.
Up 1 unit
3. From the graph,
cos=0 when =90o or 270o,
sin
cos=+1 when =0 or 360,
cos=1 when =180.
90
180
sin(+90)
270
360
Transformations of Cosine Curve 餘弦圖像的變換
sin
Shift left 左移 90 Magnify 放大
x2
cos
cos
2sin
shrink
1. The graph of sin+1 is equivalent to moving the graph
1
cos 縮小
2
90
sin up 1 unit.
Invert
180
倒置
270
360
2. The graph of 2sin is equivalent to increase the
magnitude of the graph of sin 2 times.
cos
Down
3. The graph of sin(+90) is equivalent to shifting the
下移
cos1
graph of sin 90 left.
4. The graph of sin(2) is equivalent to compress 壓縮
the graph of the curve sin by half. (See diagram
below)
1. The graph of cos is equivalent to invert the graph of
cos.
2. The graph of cos/2 is equivalent to shrink the graph
of cos into half.
3. The graph of cos1 is equivalent to moving the graph
of co down 1 unit
4. The graph of cos(/2) ie equivalent to inflate 膨脹
graph of cos() by twice. (See diagram below)
B. Cosine Curve 餘弦
Graph of y=cos
90
180
270
1. The shape of Cosine Curve is V-shaped.
360
C.
Tangent Curve 正切
3. tan vs tan(+90)
Graph of y=tan
90
Shift left 左移
180
270
tan(+90)
tan
360
90
tan(+90)
tan
180
270
360
1. The shape of Tangent Curve is fragmented 斷開 and
exponential 指數式.
2. There is no maximum value or minimum value.
3. From the graph, tan=0 when =0o, 180o or 360 o.
4. tan vs tan(90)
Transformations of Tangent Curves 正切圖像的變換
Shift right + Invert
1. tan vs tan
右移及倒置
Invert 倒置
tan
tan
tan
tan
tan
tan
90
180
tan
tan(90)
270
tan
90
tan
tan(90)
180
270
360
360
tan
2. tan vs tan+1
Shift up 上移
tan+1
Trigonometry (Applications) 應用三角幾何
tan+1
tan
tan
tan+1
Length of Arc 弧長
90
180
270
360
tan
= 2  r  
360 
where 
is the subtended angle
懸吊角 at the centre in degrees.
OR
= r
if
 is in radians
r
Area of Sector 扇形的面積
Sector
=   r2   .
 扇形
360
OR
Arc (l)
弧
= 1 r l
2
where l is arc length.
Sine Formula 正弦定理
a
b
c


sin A sin B sin C
Area of Triangle 三角形的面積
1
Area   a  b  sin 
b
2
where  is the angle included 夾角
between any 2 sides of a triangle.

Bearings 方位角
a
The compass bearing 羅盤方位角 of B from A is N30E
The compass bearing of A from B is S60W
The true bearing 真方位角 of B from A is 030.
The true bearing of A from B is 240.
N (0/360)
Angle between 2 Planes 平面夾角
The angle between two planes is the angle formed along the
edge 邊緣 at where two perpendicular lines of the respective
planes intersect 交接.
N (0/360)
Example:
C
Cosine Formula 餘弦定理
b
a  b  c  2  b  c  cos A
b 2  a 2  c 2  2  a  c  cos B
c 2  a 2  b 2  2  a  b  cos C
2
2
2
a
A
B
c
Given a cube of side l unit.
E
B
F
W
1
E (90)
G
H
1
60
1
S (180)
W (270)
E (90)
A
B
(1) The angle of inclination of F above plane ABCD is
FAC.
FC
1
tan FAC 

AC
2
FAC35.
(2) The angle between the plane ABCD and plane is BCFG
is 90.
(3) The angle between the plane ABCD and plane ABFE is
FBC or EAD.
FC 1
tan FBC 
 1
BC 1
FBC45.
(4) AF2=AC2+FC2=(12+12)+12=3
Note:
Common mistake about angle between irregular planes 不規
則平面:
Since AD is not perpendicular to CD, so we cannot say the
angle between plane ABCD and plane DEFG is 70. The
angle between the planes should be 20.
A
S (180)
Angle of Elevation(Inclination) 仰角 and Angle of
Depression 俯角
The angle of elevation of B from A is 40.
The angle of depression of B to A is 40.
B
eye level
40
eye of sight
視線
40
A
C
D
30
eye level 眼睛水平線
A
B
70
120
E
D
60
20
F
C
G
14
S (1) 
Arithmetic Sequence
等差數列
(a) General Term 通項
T (n)  a  (n  1)d where
a=first term, T(n)=nth term,
d=common difference.
(b) Sum of Terms 項和
n
S (n)  T1  Tn 
2
Example:
Given the sequence: 1, 5, 9, 13, 17,
then a=1 and d=4.
T (1)  1  (1  1)  4  1
T (3)  1  (3  1)  4  13
and
T (5)  1  (5  1)  4  17
T (101)  1  (101  1)  4  401
1
S (1)  1  1  1
2
3
S (3)  1  13  21
2
5
S (5)  1  17   45
2
101
1  101  5151
S (101) 
2
Geometric Sequence 等比數列
(a) General Term 通項
T (n)  aR n 1
where a=1st term, T(n)=nth term, R=common
ratio.
(b) Sum of Terms 項和
a Rn  1
if R>1
or
S ( n) 
R 1
a 1  Rn
S ( n) 
if R<1
1 R
(c) Sum to Infinity 無限和
a
S ( ) 
where 1R1
1 R
Example 1:
Given the sequence: 1, 5, 25, 125, 625, 3125, then
a=1 and R=5.
T (1)  1  511  1




T (3)  1  53 1  25
T (5)  1  55 1  625






1  51  1
1
5 1
1  53  1
S (3) 
 31
5 1
1  55  1
S (5) 
 781
5 1
1  510  1
S (10) 
 2441406
5 1
Example 2:
Given the sequence 243, 81, 27, 9,
1
then a=243 and R= .
3
Arithmetic & Geometric Sequence


11
1
T (1)  243   
 3
 243
1
T (3)  243   
 3
3 1
1
T (5)  243   
 3
5 1
 27
and
3
10 1
1
T (10)  243   
 3

1
1

4
3
81
1

1 
243  1    

 3  

S (1) 
 243
1
1
3
3

 1  

243  1   

 3  

S (3) 
 351
1
1
3
5

 1  

243  1   

 3  

S (5) 
 363
1
1
3
10


1


243  1    

 3  

S (10) 
 364.4938272
1
1
3
243
S   
 364.5
So
1
1
3
and
T (10)  1  5101  1953125
15
Linear and Quadratic Inequalities
1
x4
2
1
x  (2)  4  (2) .
2
x8
Linear Inequalities 一
元一次不等式
(2)
(a) Meaning and Representation.
(i) Statement 文字: x is greater than 2.
Algebraic expression 代數式: x  2
Graphical representation 圖像表達
Both sides with negative factor.
 2x  4
 2 x  (1)  4  (1)
x  4
Similarly,
 2 x  4
 1
 1
 2 x      4    
 2
 2
x2
Quadratic Inequalities 一元二次不等式
(a) Greater than or equal to
(i) Algebraic Method 代數式解
x2  2x  3  0
0
2
(ii) Statement: x is greater than or equal to 2.
Algebraic expression: x  2
Graphical representation
x  3x  1  0
0
2
(iii) Statement: x is less than 2.
Algebraic expression: x  1
Graphical representation
x  3  0
x  3  0
....or....

x  1  0
x  1  0
x  3or  x  1
(ii) Graphical Method 圖像解
y
y=x22x3
1
0
(iv) Statement: x is less than or equal to 1.
Algebraic expression: x  1
Graphical representation
1
1
0
(b) Properties of Inequalities 不等式的性質
(i) Addition and Subtraction Laws 加減定律
x23
3
x
Or simply,
x  2  (2)  3  (2)
1
0
(b) Less than or equal to
(i) Algebraic Method
x2  2x  3  0
x5
Similarly,
x23
x  2  (2)  3  (2) .
x  3x  1  0
x 1
(ii) Multiplication and Division 乘除定律
(1) Both sides with positive factor.
2x  4
2 x  (2)  4  (2) .
x  3  0
x  3  0
....or....

x  1  0
x  1  1
no. solution..or....  1  x  3
(ii) Graphical Method
y
2
y=x 2x3
x2
Similarly
16
3
1
3
Properties 特性
x
(a) Addition Rule 相加規則
P A  or  B  P A  PB
where either Event A or Event B can happen at
different time.不同時間出現的事件
Example:
In a deck of playing cards, what is the
probability of drawing an Ace?
Solution:
P(E)=P(Ace of Spade)+P(Ace of Heart)+
P(Ace of Diamond)+P(Ace of Club)
1
1
1
1
4
1






52 52 52 52 52 13
Spade 葵扇， Heart 紅心，Diamond 紅磚， Club 梅花
(b) Multiplication Rule 相乘規則
P A  and  B  P A  PB
where both Event A and Event B must happen
at the same time.同時間出現事件
(i) Independent Events 獨立事件
The outcome of Event B does not depend
on the outcome of Event A.
Example:
Solution:
If the probabilities of giving birth to a boy and
a girl are both 50%. In a family of 3 children,
what is probability having 3 girls?
Solution:
P(3 girls)=P(1st child is a girl)
P(2nd child is a girl)P(3rd child is a
girl)
1 1 1 1
   
2 2 2 8
(ii) Dependent Events 依賴事件
The outcome of Event B depends on the
outcome of Event A.
Example:
A bag contains 2 gold coins and 3 copper
coins. A man draws at random a coin from the
bag each time without replacement, i.e., not
putting back the coin drawn. He will stop
drawing once he gets a gold coin. What is the
probability that he will stop in 3 draws?
Solution
P. of getting a gold coin in 3 draws
= P.of getting a gold coin in 3rd draw after
NOT getting any gold coin in the first 2
draws
3 2 2 1
  
=
5 4 3 5
(c) Complimentary Rule 互補規則
P( A)  P( A)  1
Or simply,
1
0
3
Probability 或然率
Definition 定義
Number  of  favourable  outcomes
Total  number  possible  outcomes
Meaning 意義
0  P( E )  1
P(E)=0 means event impossible (0%) to happen.
P(E)=1 means event certain (100%) to happen.
P E  
Counting of Event
事件
(a) Listing 表列
Example:
List the number of possible outcomes
from throwing two dice together at the
same time.
Solution:
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
So the number of possible outcomes 可能結
果 is 36.
(b) Drawing a Tree Diagram 樹形圖
Example:
Draw a tree diagram to show the number of possible
outcomes by tossing a coin 3 times.
Solution:
head (hhh)
tail (hht )
head 
head (hth)
tail 
tail (htt )
coin 
head (thh)
head 
tail (tht )
tail 
head (tth)
tail 
tail (ttt)
So the number of possible outcomes is 8.
head 
17
1 1  2  2  3  3  4  4  1 7  2  8  3  9
5
16
(b) Mean (group) 分組平均值 of Table 2
above.
6 2  55  58

 4.8125
16
where Event A and Event A are opposite to
each other.
Example:
In a test there are two questions. The
probability that Mary answers the first
question correctly is 0.3 and the probability
that Mary answers the second question
correctly is 0.4. What is the probability that
she answers at least one question correctly?
P(E) =P(at least ONE correct)
=1P(Both Incorrect)
 1  0.7  0.6

Median 中位值
The middle datum when the data is arranged in
order is the median.
medain(discrete )  x n 1
if n is odd.
 0.58
2
xn  xn
medain(discrete) 
2
2
1
if n is even.
2
Example:
(1) In a set of odd number data: 1, 3, 5, 6, 7.
x5 1  x3
Since
Statistics 統計學
Frequency Distributions 頻數分佈
The display of data and the related frequency is
called frequency distribution.
2
So
median =5
(2) In a set of even number data: 1,3, 5, 6, 7, 9.
Since x 6  x3 and x 6  x4 .
Type of data
2
(a) Discrete Data 分立數據
Example:
In a set of discrete data:
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 7, 8, 8, 9, 9, 9.
Frequency Distribution Table 1.
Data
1 2 3 4 7 8 9
Frequenc 1 2 3 4 1 2 3
y
(b) Grouped Data 分組數據
Example:
Frequency Distribution Table 2.
Class
13
47
79
Interval
Class
2
5
8
Mark
Class
0.53.5 3.57.5 7.59.5
Boundaries
Frequency
6
5
5
Mean 平均值
The sum of data divided by the number of data is
the mean.
x  x  x3   xn
mean(discrete )  1 2
n
x  f  x  f  x3  f 3   xn  f n
mean( group )  1 1 2 2
sum  of  frequency
Example: In a set of discrete data:
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 7, 8, 8, 9, 9, 9.
(a) Mean (discrete) 分立平均值
So median 
2
1
56
 5.5
2
Mode 眾數值
The value of the most frequent data is the mode.
The class mark of the most frequent class is the
modal class.
Example: In a set of discrete data:
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 7, 8, 8, 9, 9, 9.
Mode (discrete) 分立眾數值=4
Modal Class 分組眾數值 of Table 2=13.
Histogram 直方圖
The presentation of data grouped into intervals as
rectangles is called histogram.
Example: Draw a histogram of frequency table
shown below:
Mark
2
5
8
Frequency
6
5
5
frequency
6
5
class
2
5
8
Cumulative Frequency
18
Measure of dispersion related to the sum of the
difference of each datum from the mean of all data.
(1) Standard deviation (discrete) 分立標準差
Polygon 累積頻數多邊
形

The presentation of the sum of rectangles of a
histogram is called cumulative frequency polygon
(curve).
Example: Draw
a cumulative frequency curve
累積頻數圖 of the cumulative frequency table
累積頻數表 shown below:
Class Less than 3.5 7.5 9.5
Frequency
6
11
16
Cumulative frequency
16
Cumulative frequency
curve
11
x1  x 2  x2  x 2  xn  x 2
where
n
x1 , x2 ,...xn is the set of data and x is the mean.
Example: In a set of data: 1, 2, 6, and 10.
1 2  6  7
4
Mean =
4
Standard Deviation
2
2
2
2
  (1  4)  (2  4)  (6  4)  (7  4)  2.55
4
Standard Deviation (group) 分組標準
(2)
差

x1  x 2 
f1  x2  x   f 2   xn  x   f n
sum of frequency
2
2
Example: In the frequency distribution table:
Class mark 1
2
6
7
Frequency
1
3
5
2
Standard deviation (group)
6

(1  4) 2  1  (2  4) 2  3  (6  4) 2  5  (7  4) 2  2
1 3  5  2
1.92
2
5
8
class
Normal Distribution 正態分佈圖
frequency
Dispersion 離差
(a) Range 分佈域
The interval between the highest class
boundary and the lowest class boundary is the
range.
(b) Inter-quartile range 四分位數間區
34% 34%
13.5%
The interval that contains the middle 50% of the data is
the inter-quartile range.
2.35%
Example: Given the cumulative frequency curve
below:
cumulative frequency
16
12
75%
8
50%
4
2.35%
Class
(a) 68% of the data would be between x  1 .
(b) 95% of the data would be between x  2 .
(c) 99.7% of the data would be between x  3 .
Example: Given that x =4 and  =2, what is the
percentage of students whose marks are above 8?
Since 6 is equivalent to x  2 , so the percentage of
students whose marks are above 8 is:
25%
1
2
3
4
5
2.35% 
6
class
Rang =101=9
Inter-quartile range =83=5
Median=6
(c) Standard Deviation 標準差
7
8
9
13.5%
10
100%  99.7%
 2.5%
2
Weighted Mean 加權平均值
The average of relative importance of certain class
mark is called weighted mean xw .
19
x1  w1  x2  w2  x3  w3    xn  wn
w1  w2  w3    wn
where x1 , x2 ,...xn is the set of data and w1 , w2 ,...wn
are the weighted importance.
Example:
The table below shows the scores and the weights
of a student in Chinese, English and Mathematics.
Chinese English Maths
Score
70
60
50
Weights
2
3
2
70  2  60  3  50  2
Weighted mean 
23 2
=60
Standard Score 標準分數
The score mark relative to mean and standard
deviation is called standard score.
xx

where
Equation of straight line 直線的方程
(a) Slope Intercept Form 斜率軸截方式:
y  mx  c
where m=slope; c=y-intercept y 軸截距
(b) Two-points Form 兩點式:

y  y1 
where (x1,y1) and (x2,y2) are any 2 points of
a line on a coordinate plane.
(c) General Form 通式:
Ax  By  C  0
where A, B and C are real numbers.
Example:
A line segment L1 has endpoints (x1,y1) and (x2,y2).
P(x,y) divides L1 into two parts in the ratio h:k. L1
cuts the y-axis at the value c. L1 is perpendicular
to L2 and parallel to L3.
y
L2
L3
(x1,y1)
L1

x=mark, x =mean; and  =standard deviation.
Example: Given that the mean and standard
deviation of a group of students in a test are 60
marks and 10 marks respectively. What is the
standard score if a student gets 80 marks in that
test?
80  60
Standard Score  =
2
10
k
c
P(x,y)
h
x
(x2,y2)
(a) dis tan ce ( L1 ) 
Coordinate Geometry 坐標幾何
x2  x1 2   y2  y2 2
(b) P( x, y )  ( h  x1  k  x2 , h  y1  k  y2 )
hk
Distance of a line
segment 線段的距離
hk
y2  y1
x2  x1
(d) Equation L1 in Two-points Form:
(c) Slope of L1 (m) 
dis tan ce  x2  x1    y2  y2 
Point of division 線段的截點
2
y2  y1
 x  x1 
x2  x1
2
y  y1 
y2  y1
 x  x1 
x2  x1
h  x1  k  x2 h  y1  k  y2
P ( x, y )  (
,
)
hk
hk
x  x2 y1  y2
P ( x, y )  ( 1
,
) if P is the midpoint.
2
2
(e) Equation L1 in Slope Intercept Form:
y  mx  c
(f) Slope of L2 x slope of L1=1,
(g) Slope of L3= Slope of L1.
Slope of straight line
直線的斜率
Equation of a circle
圓形的方程
Slope (m) 
y2  y1
x2  x1
(a) Centre-radius Form 圓心-半徑方式:
x  a 2  ( y  b)2  r 2
where (a, b) is the centre and r is the radius.
(b) General Form 通式
x 2  y 2  Dx  Ey  F  0
where
(1) The slopes of the lines parallel to each other
are equal.
(2) The product of the slopes of lines
perpendicular to each other is 1.
20
2
Examples:
2
D
E
 D  E
a   ;b   ; r 2        F
2
2
 2 2
(x+2)2+(y-4)2=5
P(x,y)
y
L2
(x,y)
r
(a,b)
x
L1
x 2  y 2  Dx  Ey  F  0
L3
Number of
Intersections of
Circle and Line 圓
形與線的相交點
 Ax  By  C  0
To solve 
2
2
 x  y  Dx  Ey  F  0
(1) To find the distance between the line segment
whose endpoints are (2,4) and (2,1):
dis tan ce   2  2  4  1  5
(2) To find the coordinates of P(x,y) if it is a point
of circle whose radius is (-2,4):
x2
 2 ,
so x  6.
2
y 1
 4,
so y  7 .
2
The coordinates of P is (-6,7).
(3) To find the slope of L1 passing through
(-2,4) and (2,1):
2
,
the possible solution(s) could be:
(a) the line intersects the circle at 2 points, so
there are 2 solutions and the discriminant >0.
(b) the line touches the circle at a point, so there is
only 1 solution and the discriminant =0.
(c) the line does not intersect the circle, so there is
no solution and the discriminant <0.
y
2
4 1
3

22
4
(4) To find the equation of L1 :
3
x  2
4
 4 y  4  3x  6
3x  4 y  10  0
y 1 
L3
L1
(5) To find the equation of L2 if it is a tangent line
to the circle given:
Since slope of L1 x slope of L2 = 1,
4
So
slope of L2 = .
3
L2
The required equation is:
x
4
x  2
3
3y  3  4x  8
y 1 
4x  3y  5  0
(6) To find the equation of given circle in General
Form:
L1 intersects the circle at 2 points.
L2 intersects the circle at 1 point.
L3 does not intersect the circle, i.e., it lies outside
the circle.
21
x  (2)2  ( y  4) 2
x
2
 
 52
Linear Inequalities in Two Variables and Linear
Programming

 4 x  4  y 2  8 y  16  25
Linear Inequalities in
Two Variables 兩元
線性不等式
x 2  y 2  4x  8 y  5  0
(7) To find the coordinates of the points of
intersection of L1 and given circle:
3x  4 y  10  0...(1)
Solve  2
2
 x  y  4 x  8 y  5  0...(2)
 3x  10
...(3)
From (1): y 
4
Sub (3) into (2):
(a) A straight line can divide a coordinate plane into 3
regions:
(1)
the region along the line,
(2)
the region above the line and
(3)
the region below the line.
  3x  10 
  3x  10 
x2  
  4 x  8
 5  0
4
4




2
2
16 x  (9 x  60 x  100)  64 x  (96 x  320)
 80  0
25 x 2  100 x  300  0
2
2
(b) A solid line means the points on the line are
included. A dotted line means the points on the
line are not included.
Example:
x 2  4 x  12  0
( x  2)( x  6)  0
x  2 or x  6
Put x=2 into (3): y=1.
Put x=-6 into (3): y=7.
So the points of intersections are (2,1)
and (6,7).
(8) To find the coordinates of point of contact if L2
is a tangent line to the given circle:
4 x  3 y  5  0
Solve  2
2
x  y  4x  8 y  5  0
2
 4x  5 
 4x  5 
2
So x  
  4 x  8
50
 3 
 3 
Region IV
Region I
Region III
Region II
25 x 2  100 x  100  0
Region I is the common solution of
2 x  y  5  0
.

x  3 y  6  0
x2  4x  4  0
x  2x  2  0
When x=2, y=1.
The point of contact is (2, 1).
(9) To show that L3 does not intersect the given
circle:
 x  y  6
Solve  2
2
x  y  4x  8 y  5  0
2
x 2   x  6  4 x  8 x  6  5  0

Region II is the common solution of
2 x  y  5  0
.

x  3 y  6  0
Region III is the common solution of
2 x  y  5  0

x  3 y  6  0

x 2  x 2  12 x  36  4 x   8 x  48  5  0
2 x  24 x  79  0
2
Region IV is the common solution of
2 x  y  5  0

x  3 y  6  0
  (24)  4279  56
So there is no solution.
2
22
The minimum value is 7.
Linear Programming 線 性 規 劃
Approximate Solution of Equations 公式近似值
Graphical Solution 圖像解法
Given that f(x)=0 which cannot be solved by any
simple algebraic methods, then its approximate
solution must between x1 and x2 where f(x1) >0
and f(x2)<0. The values of x1 and x2 can be
determined by drawing the graph of f(x).
Example 1:
Find the approximate solution of x 2  5 .
Solution:
Step 1: Let f(x) = x 2  5 .
Step 2: Draw the graph of y=f(x)= x 2  5 .
Step 3: From the graph drawn, f(x)=0 when x is
approximately +2.2 or 2.2 because
f(3)>0 and f(1)<0 or f(2)<0 and f(3)>0.
To find the maximum 最大 or minimum 最少
value of a linear function under certain number of
constraints 約束 is called linear programming.
Example:
What is the minimum value under the following
2 x  y  5  0
x  3 y  6  0

constraints: 
if P=2x+y.
x

0

 y  5
Solution:
Step 1:
2 x  y  5  0
x  3 y  6  0

Draw the graphs of 
x  0
 y  5
Step2:
Shade the region under the given constraints.
x=0
P= 2x+0
y=5
P= 2x4
P= 2x7
Example 2:
Solve  x3  2 x 2  5 x  4  0
y   x 3  2 x 2  5x  4
Step 3:
By moving the line y=2x to the lowest vertex of
the region of boundary which is (3,1),
so
P=2(3)+(1)= 7.
The minimum value is 7.
Alternate Method:
Since the vertices bounded by the region under the
constraints are (3,1), (0,2 ) and (0,5),
so
P(3,1)=2(3)+(1)= 7,
P(0, 2)=2(0)+( 2)= 2,
P(0,5)=2(0)+ (5)=5.
From the graph drawn, the roots of the equation
are approximately 3.2, 0.6 and 1.9.
23
= cross-section area 橫切面 x length
Bisection Method 分半方法
It is a numerical method to find the approximate
value of the root(s) of an equation with one
unknown which can be expressed as f(x). An
approximate value of the root is obtained by
taking the
average x1
l
and x2 where
f(x1) >0
2.
and
f(x2)<0.
a
Example 1:
Find
the root
of
 x3  2 x 2  5 x  4  0
where 1<x<2 correct to 1 decimal place.
Solution:
Let f ( x)   x3  2 x 2  5x  4
f(x1)>0
f(x2)<0
x  x2 f(m)
m 1
=
(+ve)
(ve)
2
1
2
1.5
+
1.5
2
1.75
+
1.75
2
1.875

1.75
1.875
1.8125
+
1.8125
1.875
1.84375
+
1.84375
1.875
1.859375

1.84375
1.859375 1.8515625

1.851562 1.859375
5
1.9 (1dp) 1.9 (1dp)
Surface area of cuboid 方體
= 2(bw + bl + wl)
Volume of cuboid = bwl
h
b
3.

Area of parallelogram 平行四邊形 = bh
Area of parallelogram = absin
l
4.
w
Area of
trapezium 梯形
(a  b)h
2
b
=
a
So the root is 1.9 correct to 1 decimal place.
Example 2:
Find the root of  x3  2 x 2  5 x  4  0 where 1
< x < 0 correct to 1 decimal place.
Solution:
Let f ( x)   x3  2 x 2  5x  4
f(x1)>0
f(x2)<0
x1  x2 f(m)
m

=
(+ve)
(ve)
2
0
+
1
0.5
0.5
1
0.75

0.5
0.75
0.625

+
0.5
0.625
0.5625
0.5625
0.625
0.6
0.6
(1dp)
(1dp)
h
5.
Area
b

of
circle 圓形
= r2
Surface area of sphere 球體 = 4 r2
Volume of
sphere =
a
h
b
4
  r3
3
So the root is 0.6 correct to 1 decimal place.
rr
Area and Volume of Figures and Solids
1. Area of right angled triangle = (bh) 2
Area of any triangle = absin  2
Volume of triangular prism 角柱體
24
6.
Curved surface area of cylinder = 2rh
Volume of cylinder 柱體 = r2h
x% 
2.
r
x
100
Conversion 轉換
(a) Decimal to Percentage
0.75 = 75%
(b) Percentage to Decimal
h
75% = 0.75
7.
(c) Percentage to Fraction
Curved
surface area of cone =  rl
75% 
1
Volume of cone 圓柱體=    r 2  h
3
75
100
(d) Fraction to Percentage
3 3
  100%  75%
4 4
3.
r2
h
(a) A value increases by r%:
+ h2 = l2
l
Percentage Change 改變
r 

New value = x1 

 100 
The value of x increases to y:
8.
Ratio of
r
areas of
similar figures
= (a:b)2
Ratio of volume of similar figures = (a:b)3
a
b
% increase =
yx
 100%
x
(b) Percentage Decreases
A value decreases by r%:
r 

New value = x1 

 100 
9.
D
The value of x decreases to y:
C
% decrease =
E
A
F
4.
B
(1) Areas of triangles with equal base and heights
are equal:
area of △ADF = area of △FCB
(2) Areas of triangles with unequal base but with equal
heights are proportional:
area of △ADF : area of △DFC = 1:2
(3) Areas of similar triangles is proportional to the
square of their lengths:
area of △AEF : area △CED=(1:2)2=1:4
Percentages 百分比
1.
Definition 定義
25
x y
 100%
x
Profits and Loss 盈利與虧蝕
Pr ofit
 100%
(a) Profit % =
Cost  price
Loss
 100%
(b) Loss % =
Cost  price
5.
Discount 減價
(a) Discount = Marked price 標價 - Selling
price 售價
Discount
 100%
Marked  price
(b) Discount %=
(c) Discount= Marked price x Discount %
(d) Selling price
= Marked price x (1 - Discount %)
6.
Simple Interest 單利息
(a) Simple interest (I)
PRn
=
100
( P = Principal 本金; R= rate 利率; n =
number of terms 期數)
(b) Amount (A)本利和
=P+I
Rn 

= P 1 

 100 
7.
Compound Interest 複息
(a) Amount (A)
R 

= P 1 

 100 
n
(b) Compound Interest
 A P
n
R 

 P 1 
 P
 100 
8.
Growth 增長
R 

New value 新值 = P1 

 100 
n
P=original value 原值, R=growth rate 增長率,
n=terms 期數
9
Depreciation 折舊
R 

New value 新值 = 1 

 100 
n
P=original value 原值, R=depreciation rate 折
舊率, n=terms 期數
26
Glossary (Question/Answer) 詞彙﹝問題/答案﹞
According 根據
Let 設
Unknown 未知數
Calculate 計算
List 列出
Verify 驗證
Check 驗算
Make x the subject 令 x
為主項
What 問
Compare 比較
Complete 完成
Construct 作
Correct to…準確至
Decimal places 點數後
Deduce 推算
Determine 判斷
Draw 繪出﹝畫出﹞
Estimate 估計
Expand 展開
Explain 解釋
Express 表達
Evaluate 求出量值
Factorize 因式分解
Fill in 填上
Find 求
Given that 已知
Hence 然後
If., then 若…則
In terms of 以…表示
Justification 論據
Mark 標明﹝記號﹞
When 當
Measure 量度
Which of the following
下列何者
Prove 證明
Write down 寫出
Plot 標明位置
Read 讀出
Refer 參看
Represent 代表
Round off 四捨五入
Satisfy 滿足
Shade 塗上陰影
Show that 求證
Significant figures 有效
數字
Simplify 化簡
Solve 解
Solution 解答
State 說出
Suppose 假設
Table 表
Using or otherwise 利用
或其他方法
24
14