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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 na n am an amn 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 am 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 an2 an2 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 x2 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 2log 2 5 4 log 2 4 Surds & Rationalization 無理 根式的有理化 1. a b ab 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 ab Example: (i) 27 12 9 3 4 3 (ii) 3 32 3 3 1 2 3 2 a b a b (iii) 2 3 2 2 32 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 ba 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 ba ab ab 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 a2b+c0 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. 20 (x)(x-)=0 roots (x) = or Example 1: Solve 2x25x3=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 2x25x3 = 0. Solution: Step 1: Draw the graph of y = 2x25x3. 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 =b24ac, 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= bxc 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)24(2)(1) = 9. So there are two distinct roots. (2) = (2)24(1)(1) = 0. x2 x 2 0 =(1)24(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)24(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 = 2x25x3, 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 3k 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. opposites, //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=xyx2y2 Evaluating 估值 Suppose f(x)=3x22x+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)22(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)22(1)+1=2 then To find the remainder of (3x22x+1)(x1): 1. 2. f(1) =3(1)22(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 3x22x1. Since f(1)=3(1)22(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 (x1) 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: 3x22x+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 3x22x1(x1)(3x1). 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+C3x22x+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)(x2x+1) 27x364=(3x)3(4)3=(3x4)(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 3108 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 = kx2 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. xy 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 ky 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: 3602 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 cos180 tan tan 180 cos180 tan 180 cos360 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 02. Solution: Since Simplify 化簡 Trigonometric Ratios either cos3=0 or An expression with more than one trigonometric ratios Example 1: 3sin2=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 0.7302.41(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 30or 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 cos1 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 cos1 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 N30E The compass bearing of A from B is S60W 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 FAC35. (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 FBC45. (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 1R1 1 R Example 1: Given the sequence: 1, 5, 25, 125, 625, 3125, then a=1 and R=5. T (1) 1 511 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 11 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 5101 1953125 15 Linear and Quadratic Inequalities 1 x4 2 1 x (2) 4 (2) . 2 x8 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 x2 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 3x 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 3or x 1 (ii) Graphical Method 圖像解 y y=x22x3 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 x23 3 x Or simply, x 2 (2) 3 (2) 1 0 (b) Less than or equal to (i) Algebraic Method x2 2x 3 0 x5 Similarly, x23 x 2 (2) 3 (2) . x 3x 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 2x3 x2 Similarly 16 3 1 3 Properties 特性 x (a) Addition Rule 相加規則 P A or B P A PB 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 PB 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 55 58 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) =1P(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 13 47 79 Interval Class 2 5 8 Mark Class 0.53.5 3.57.5 7.59.5 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 56 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=13. 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 =101=9 Inter-quartile range =83=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 23 2 =60 Standard Score 標準分數 The score mark relative to mean and standard deviation is called standard score. xx 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 ) hk Distance of a line segment 線段的距離 hk 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 ) ( , ) hk hk 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): x2 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 22 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 50 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 2x 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) 4279 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= 2x4 P= 2x7 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(bw + bl + wl) Volume of cuboid = bwl h b 3. Area of parallelogram 平行四邊形 = bh Area of parallelogram = absin 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 = (bh) 2 Area of any triangle = absin 2 Volume of triangular prism 角柱體 24 6. Curved surface area of cylinder = 2rh Volume of cylinder 柱體 = r2h 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 = rl 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 = x1 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 = yx 100% x (b) Percentage Decreases A value decreases by r%: r New value = x1 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 = P1 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