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GAC 数学

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GAC Mathematics(数学)は4つのモジュールで構成されます。GAC004では、算数、代数、座標幾何学、平面および立体幾何学、三角関数、指数関数および対数関数を通じて基礎を英語で再構築します。GAC010では、数列、金融数学、確率、統計学に取り組みます。GAC016は単変数微積分です。GAC024は離散数学で、集合や计数体系からアルゴリズム、グラフ理論、ブール論理に至ります。

言語能力も手法 alongside で評価されます。ユニットごとに作成され、学期中に検証される用語ログブックは、GAC004の重要な構成要素であるため、勾配や残差が何を意味するかを説明できることは、それを計算できることと同じくらい重要です。

GAC016は大学が最も頻繁に認定するモジュールです。いくつかの大学でこの科目に対する微積分同等の換算資格を公表しており、これにより、ここで学ぶ微分・積分の知識は高校の復習としてではなく、大学1年生レベルの履修として扱う価値があることを示唆しています。

試験形式は混合です。初期のユニットに関する短い授業内テスト、スプレッドシートで作成する1〜2件のプロジェクト、全範囲を網羅する期末試験、そして学期を通じて静かに蓄積していく coursework(課題)の採点が想定されます。

本サイトにある worked examples(例題)、drills(ドリル)、mock papers(模試)はすべて当サイトオリジナルであり、これらのイベントの到達目標および形式に合わせて作成されています。

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  • 1

    GAC004 数学I:基礎

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    1.1

    What this module is, and how it is marked

    A price can fall by 20% and then rise by 20% without returning to its starting value. Mathematics I helps you explain such results with a method, units and a clear interpretation. The six units connect arithmetic, algebra, graphs, geometry, trigonometry and exponential models.

    Your centre's current assessment brief is the authority for tasks, weights, permitted tools and deadlines. An in-class test 课堂测验 is completed under stated classroom conditions. Projects 项目 can involve collecting data, calculating and reporting; an examination 考试 and coursework 平时作业 may have different instructions. Do not infer an official assessment pattern from these practice sheets.

    A terminology logbook 术语记录本 can record the English term, a meaning in your own words and a small example. Follow your centre's instructions if it is submitted or assessed.

    Read the command word 指令词: solve asks for values satisfying a condition, simplify for an equivalent expression, and justify for a reason. Show a valid method so a reader can check how the result follows. The mark scheme for each task determines its marks.

    The six accompanying practice sheets are original GAC004-aligned material, not official papers. Their solutions award marks for stated mathematical steps rather than for the length of a written report.

    English 日本語
    in-class test/ɪn klæs test/ 授業内テスト
    projects/ˈprɒdʒekts/ プロジェクト
    examination/eɡˌzæmɪˈneɪʃn/ 試験
    coursework/ˈkɔːsjuːɜːk/ 課題
    terminology logbook/ˌtɜːmɪˈnɒlədʒi ˈlɒɡbʊk/ 用語帳
    command word/kəˈmænd wɜːd/ 指示語
    1.1

    Terminology and arithmetic review

    シラバス

    Unit 1 of 6 in GAC004 Mathematics I: Fundamentals (Level I). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    Module purpose: On completion of this module, students should be able to demonstrate an understanding of the basic concepts of mathematics, and the language used, in preparation for tertiary study within an English-speaking environment.

    The module outcomes this unit works towards:

    Learning Objective GAC004.1: Solve elementary math problems using basic arithmetic operations.

    出典: Cambridge International シラバス

    The vocabulary of number is assumed by every later unit.

    An integer 整数 is a whole number, including negative whole numbers and zero. A rational number 有理数 can be written as a fraction of two integers with a nonzero denominator; an irrational number 无理数 cannot, and $\pi$ and $\sqrt{2}$ are the standard examples. A prime number 质数 is a positive integer greater than 1 with exactly two positive factors, itself and 1.

    The order of operations 运算顺序 fixes what a written expression means: brackets, then indices, then multiplication and division, then addition and subtraction, working left to right within a level.

    $$3 + 4 \times 2^2 = 3 + 4 \times 4 = 3 + 16 = 19$$
    • A factor 因数 divides a number exactly; a multiple 倍数 is what you get by multiplying it. 6 is a factor of 24, and 24 is a multiple of 6.
    • The highest common factor (HCF) 最大公因数 and lowest common multiple (LCM) 最小公倍数 come from the prime factors: $24 = 2^3 \times 3$ and $36 = 2^2 \times 3^2$, so the HCF is $2^2 \times 3 = 12$ and the LCM is $2^3 \times 3^2 = 72$.
    • A percentage 百分比 is a fraction with denominator 100. An increase of 15% multiplies by $1.15$; a decrease of 15% multiplies by $0.85$. Reversing a percentage change means dividing, never subtracting the same percentage back.
    • Significant figures 有效数字 are the digits retained to express a value at a stated precision. They do not alone establish the accuracy of a measurement. $0.004\,072$ to three significant figures is $0.004\,07$.
    • Standard form 科学记数法 writes a number as $a \times 10^n$ with $1 \le |a| < 10$ for a nonzero number, with integer n. It is how a calculator shows a very large or very small answer.

    Worked example. a decrease and increase use different bases

    The worked price moves from 80 to 64 to 76.8 because each percentage uses the price at that stage.

    Known: an invented price of 80 units falls by 20%, then rises by 20%. Why use successive multipliers? Each change is calculated from the current price.

    $$P_{new}=P_{old}(1-r/100)$$
    $$P_1=80(1-20/100)=64$$
    $$P_{new}=P_{old}(1+r/100)$$
    $$P_2=64(1+20/100)=76.8$$

    The price is not back at 80. The decrease was 16, while the later increase was 12.8, because the bases differ.

    Continue with practice sheet 1.1. Solve before opening the solutions, and check both the method and the final units.

    English 日本語
    integer/ˈɪntɪdʒə/ 整数
    rational number/ˈræʃənl ˈnʌmbə/ 有理数
    irrational number/ɪˈræʃənl ˈnʌmbə/ 無理数
    prime number/praɪm ˈnʌmbə/ 素数
    order of operations/ˈɔːdə ɒv ˌɒpəˈreɪʃnz/ 計算の順序
    factor/ˈfæktə/ 因数
    multiple/ˈmʌltɪpl/ 倍数
    highest common factor (HCF)/ˈhaɪɪst ˈkɒmən ˈfæktə/ 最大公因数 (HCF)
    lowest common multiple (LCM)/ˈləʊɪst ˈkɒmən ˈmʌltɪpl/ 最小公倍数 (LCM)
    percentage/pəˈsentɪdʒ/ 百分率
    Significant figures/sɪɡˈnɪfɪkənt ˈfɪɡəz/ 有効数字
    Standard form/ˈstændəd fɔːm/ 標準形
    1.2

    Algebra I: introductory algebra

    シラバス

    Unit 2 of 6 in GAC004 Mathematics I: Fundamentals (Level I). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    The module outcomes this unit works towards:

    Learning Objective GAC004.2: Perform basic algebraic operations and solve equations and inequations using algebraic methods.

    出典: Cambridge International シラバス

    Algebra is arithmetic with letters standing for numbers, and it has its own vocabulary.

    In $5x^2 - 3x + 7$, the whole thing is an expression 表达式; $5x^2$, $-3x$ and $7$ are its terms 项; 5 is the coefficient 系数 of $x^2$; and 7 is a constant 常数.

    An equation 方程 says two expressions are equal and is solved for a value. An identity 恒等式 is true for every value in its domain. An inequality 不等式 compares two expressions with $<$, $\le$, $>$ or $\ge$.

    To expand 展开 is to remove brackets; to factorise 因式分解 is to put them back.

    $$(x + 3)(x - 5) = x^2 - 5x + 3x - 15 = x^2 - 2x - 15$$
    • Solving a linear equation 一元一次方程 means doing the same operation to both sides until the letter stands alone.
    • A quadratic equation 一元二次方程 has the form $ax^2+bx+c=0$ with $a\ne0$. Factorise when suitable, or use $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
    • Simultaneous equations 联立方程 are two equations in two unknowns, solved by substitution or by elimination. For two linear equations, a unique solution is their intersection. Parallel distinct lines have no solution; identical lines have infinitely many.
    • ⚠ An inequality reverses 反向 when you multiply or divide both sides by a negative number: from $-2x > 6$ it follows that $x < -3$.

    Worked example. reverse the inequality sign when dividing by a negative

    A number line shows x less than 2, with an open circle at 2.

    Known: solve $7-3x>1$. Subtract 7 from both sides, then divide by negative 3. The last operation reverses the inequality.

    $$7-3x>1\quad\Longrightarrow\quad -3x>-6$$
    $$x<\frac{-6}{-3}=2$$

    The open circle excludes 2. Check $x=1$: $7-3(1)=4>1$; the boundary $x=2$ gives equality and is excluded.

    Continue with practice sheet 1.2. Solve before opening the solutions, and check both the method and the final units.

    English 日本語
    expression/ekˈspreʃn/ 式
    terms/tɜːmz/ 項
    coefficient/ˌkəʊɪˈfɪʃənt/ 係数
    constant/ˈkɒnstənt/ 定数
    equation/ɪˈkweɪʒn/ 方程式
    identity/aɪˈdentɪti/ 恒等式
    inequality/ɪniːˈkwɒlɪti/ 不等式
    expand/ekˈspænd/ 展開
    factorise/ˈfæktəraɪz/ 因数分解
    linear equation/ˈlɪnɪə ɪˈkweɪʒn/ 一次方程式
    quadratic equation/kwɒˈdrætɪk ɪˈkweɪʒn/ 一元二次方程式
    Simultaneous equations/ˌsɪməlˈteɪnɪəs ɪˈkweɪʒnz/ 連立一次方程式
    reverses/rɪˈvɜːsɪz/ 逆
    1.3

    Algebra: graphs and coordinate geometry

    シラバス

    Unit 3 of 6 in GAC004 Mathematics I: Fundamentals (Level I). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    The module outcomes this unit works towards:

    Learning Objective GAC004.3: Graph algebraic relations and use coordinate geometry to solve problems.

    出典: Cambridge International シラバス

    A graph turns an equation into a picture, and coordinate geometry measures that picture.

    A nonvertical straight line can be written $y = mx + c$, where $m$ is the gradient 斜率 and $c$ the y-intercept y 轴截距. The gradient is the rise divided by the run.

    $$m = \frac{y_2 - y_1}{x_2 - x_1}$$
    • Distinct nonvertical lines are parallel 平行 when their gradients are equal. Two nonvertical lines are perpendicular 垂直 when their gradients multiply to $-1$. Vertical lines have undefined gradient; a vertical and a horizontal line are perpendicular.
    • The midpoint 中点 of a segment is the average of the endpoints, and the distance 距离 between two points comes from Pythagoras: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
    • A quadratic graphs as a parabola 抛物线. Its vertex 顶点 is the turning point, and the roots 根 are the x-values where it meets the $x$-axis, the solutions of $y=0$. A repeated root touches the axis without crossing; some quadratics have no real roots.
    • To solve two equations graphically 用图像求解, draw both and read the coordinates of the intersection. The answer is only as accurate as the drawing, which is why an algebraic check matters.

    Worked example. a rising line through two given points

    The line passes through A at (1,2) and B at (4,8), with dashed coordinate-change guides.

    Known: $A=(1,2)$ and $B=(4,8)$. Use gradient because the line is nonvertical.

    $$m=\frac{y_B-y_A}{x_B-x_A}=\frac{8-2}{4-1}=2$$

    The line equation is $y=mx+c$. Substitute A to find the intercept.

    $$c=y_A-mx_A=2-2(1)=0$$

    Thus $y=2x$. The diagram's horizontal change is 3 and vertical change is 6.

    Continue with practice sheet 1.3. Solve before opening the solutions, and check both the method and the final units.

    English 日本語
    gradient/ˈɡreɪdɪənt/ 傾き
    y-intercept/waɪ ˌɪntəˈsept/ y切片
    parallel/ˈpærəlel/ 平行
    perpendicular/ˌpɜːpənˈdɪkjʊlə/ 垂直
    midpoint/ˈmɪdpɔɪnt/ 中点
    distance/ˈdɪstəns/ 距離
    parabola/pəˈræbələ/ 放物線
    vertex/ˈvɜːteks/ 頂点
    roots/ruːts/ 解
    graphically/ˈɡræfɪkli/ グラフによる求解
    1.4

    Geometry: plane, solid and Euclidean geometry

    シラバス

    Unit 4 of 6 in GAC004 Mathematics I: Fundamentals (Level I). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    The module outcomes this unit works towards:

    Learning Objective GAC004.4: Analyze simple problems involving planar shapes and solids and solve problems in Euclidean geometry.

    出典: Cambridge International シラバス

    Plane geometry is about flat shapes, solid geometry about three-dimensional ones, and Euclidean geometry is the reasoning that connects them.

    Angles on a straight line add to $180°$, angles around a point to $360°$, and the interior angles of a plane Euclidean triangle to $180°$. In a simple polygon 多边形 of $n$ sides the interior angles add to $(n - 2) \times 180°$.

    • Congruent 全等 shapes are identical in size and shape; similar 相似 shapes have the same shape with all lengths in one ratio.
    • Pythagoras' theorem 勾股定理 holds in a right-angled triangle: $a^2 + b^2 = c^2$, where $c$ is the hypotenuse 斜边.
    • Area 面积 is measured in square units and volume 体积 in cubic units. A cylinder has volume $\pi r^2 h$; a sphere has volume $\tfrac{4}{3}\pi r^3$ and surface area $4\pi r^2$.
    • ⚠ Scaling is not linear. If every length of a solid is doubled, its area is multiplied by $2^2 = 4$ and its volume by $2^3 = 8$.

    Worked example. a triangular prism has a constant cross-section

    A triangular prism has a right-triangle cross-section with perpendicular sides 3 and 4 cm and length 10 cm; perspective is not to scale.

    Known: perpendicular triangle sides are 3 and 4 cm; prism length is 10 cm. Find the cross-sectional area, then multiply by the prism length.

    $$A=\frac12 bh=\frac12(3)(4)=6\ \text{cm}^2$$
    $$V=AL=6(10)=60\ \text{cm}^3$$

    The sloping triangle side is not its perpendicular height. Volume is not the area of the triangular end.

    Continue with practice sheet 1.4. Solve before opening the solutions, and check both the method and the final units.

    English 日本語
    polygon/ˈpɒlɪɡən/ 多角形
    Congruent/ˈkɒŋɡruːənt/ 合同
    similar/ˈsɪmɪlə/ 相似
    Pythagoras' theorem/paɪˈθæɡərəs ˈθɪərəm/ 三平方の定理
    hypotenuse/haɪˈpɒtənjuːs/ 斜辺
    Area/ˈeərɪə/ 面積
    volume/ˈvɒljuːm/ 体積
    1.5

    Trigonometry

    シラバス

    Unit 5 of 6 in GAC004 Mathematics I: Fundamentals (Level I). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    The module outcomes this unit works towards:

    Learning Objective GAC004.5: Calculate solutions to various problems using trigonometric methods.

    出典: Cambridge International シラバス

    Trigonometry connects the angles of a triangle to its sides.

    In a right-angled triangle, with $\theta$ one of the acute angles, the three ratios are $\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$, $\cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$ and $\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}$.

    For a nondegenerate plane Euclidean triangle, the sine rule 正弦定理 and the cosine rule 余弦定理 apply:

    $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}, \qquad a^2 = b^2 + c^2 - 2bc\cos A$$
    • An angle can be measured in degrees 度 or in radians 弧度, where $\pi$ radians is $180°$. Check which mode your calculator is in before every question.
    • The unit circle 单位圆 extends the ratios beyond $90°$ and explains why $\sin$ and $\cos$ repeat every $360°$: they are periodic 周期的.
    • A bearing 方位角 is measured clockwise from north and always written with three figures, such as $075°$.
    • Angles of elevation and depression 仰角与俯角 are measured from the horizontal, upwards and downwards respectively. Add the observer or instrument height if the answer needs height above the ground. With side-side-angle data, the sine rule may allow two triangles: check both supplementary angles against the remaining angle sum.

    Worked example. height from a horizontal distance and angle

    A right triangle has a 12 m horizontal base and an angle of elevation of 30 degrees; the unknown height is opposite the angle.

    Known: horizontal distance $d=12$ m and elevation $\theta=30^\circ$. Height h is opposite and d is adjacent, so use tangent in degree mode.

    $$\tan\theta=\frac{h}{d}$$
    $$h=d\tan\theta=12\tan30^\circ\approx6.93\ \text{m}$$

    This is the height above the observer's horizontal sight level, not automatically the height above the ground.

    Continue with practice sheet 1.5. Solve before opening the solutions, and check both the method and the final units.

    English 日本語
    sine rule/saɪn ruːl/ 正弦定理
    cosine rule/ˈkəʊsaɪn ruːl/ 余弦定理
    degrees/dɪˈɡriːz/ 度
    radians/ˈreɪdɪənz/ ラジアン
    unit circle/ˈjuːnɪt ˈsɜːkl/ 単位円
    periodic/ˌpɪərɪˈɒdɪk/ 周期的
    bearing/ˈbeərɪŋ/ 方位角
    Angles of elevation and depression/ˈæŋɡlz ɒv ˌelɪˈveɪʃn ænd dɪˈpreʃn/ 仰角と俯角
    1.6

    Exponential and logarithmic functions

    シラバス

    Unit 6 of 6 in GAC004 Mathematics I: Fundamentals (Level I). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    The module outcomes this unit works towards:

    Learning Objective GAC004.6: Solve and graph exponential and logarithmic functions and equations.

    出典: Cambridge International シラバス

    These two functions describe growth and decay, and they undo each other.

    An exponential function 指数函数 has the form $y=a^x$ with $a>0$ and $a\ne1$: the variable is in the exponent 指数. Each unit step multiplies its value by a. It grows when $a>1$ and decays when $0. Its y-intercept is 1 and its values stay positive; the x-axis is a horizontal asymptote.

    A logarithm 对数 answers the reverse question. $\log_a y = x$ means exactly $a^x = y$, so $\log_{10}1000=3$. For real logarithms the argument must be positive, and the base must be positive and different from 1. The graph $y=\log_a x$ passes through $(1,0)$ and has vertical asymptote $x=0$.

    $$\log(mn) = \log m + \log n, \qquad \log\!\left(\frac{m}{n}\right) = \log m - \log n, \qquad \log(m^k) = k \log m$$
    • For positive arguments and the same valid base, the laws turn multiplication into addition, which is what makes a logarithm useful for solving an equation with the unknown in the exponent.
    • Natural logarithms 自然对数 use the base $e \approx 2.718$ and are written $\ln$.
    • Compound interest 复利 is exponential: $A = P(1 + r)^n$ after $n$ periods.
    • Exponential decay 指数衰减 has a base between 0 and 1, and can describe a retained proportion each period under stated assumptions. It does not guarantee that a real process follows that model indefinitely.

    Worked example. repeated reductions multiply rather than subtract a fixed amount

    A model starting at 100 is multiplied by 0.8 each period, giving 80 then 64.

    Known: an invented quantity starts at 100 units and decreases by 20% per whole period. The retained proportion is 0.8, so use a multiplicative model.

    $$Q(n)=Q_0(1-r)^n$$
    $$Q(2)=100(1-0.20)^2=64$$

    The first decrease is 20 units and the second is 16. A fixed subtraction of 20 each period would be a different model.

    Continue with practice sheet 1.6. Solve before opening the solutions, and check both the method and the final units.

    English 日本語
    exponential function/ˌekspəˈnenʃl ˈfʌŋkʃn/ 指数関数
    exponent/ekˈspəʊnənt/ 指数
    logarithm/ˈlɒɡərɪθəm/ 対数
    Natural logarithms/ˈnætʃərəl ˈlɒɡərɪθəmz/ 自然対数
    Compound interest/ˈkɒmpaʊnd ˈɪntrest/ 複利
    Exponential decay/ˌekspəˈnenʃl dɪˈkeɪ/ 指数関数的減少
    Additional notes PDF

    追加練習:解説付き解答の付いた4問

    このパックは、GAC004 Math I の3単元の練習を追加する。算数の復習(シート 1.1)、初歩的な代数(シート 1.2)、幾何学(シート 1.4)を網羅する。問題は基本計算、代数的手法、平面幾何学の練習である。GAC004 配布物およびその練習シートと併用して使用すること。

    これはオリジナルの練習材料である。センターの評価指示を変更するものではなく、公式試験問題でもない。

    「Practice questions」の質問にまず答えよ。解法整个过程を完全に記せ。その後、最終的な数値だけでなく、各解説付き解答との間で自分の解法過程を比較せよ。明確に記述された解法過程は、長いレポートよりも推論をよりよく示す。

    各解答は同じ構造を持つ:既知の情報、なぜその法則が適用されるか、法則を記号で表したもの、そして数値。各行に1つの計算段階がある。最後に検証を行うが、これは解答の一部であるため、新しい用紙で実際に実行してみること。

    問題 シート 練習内容
    Q1 1.1 連続する2回の価格変動
    Q2 1.1 増分の元に戻す
    Q3 1.2 負の数で割る
    Q4 1.4 長さから面積へ

    練習問題

    これらにまず答えよ。先を見ないこと。

    Q1. 200 ユニットの価格が 10% 上昇し、その後 10% 低下する。

    • 最終価格を求める。
    • なぜ最終価格が 200 ユニットでないかを説明せよ。

    Q2。 20%の増後、ある数値は84となる。元の数値を求めよ。

    Q3。 $9-2x\le3$を解け。解答を数直線上に示せ。

    Q4。 2つの相似形モデルの長さ比は$3:5$である。小さい方のモデルの表面積は$36\ \text{cm}^2$である。大きい方の表面積を求めよ。

    worked answers(解説付き解答)

    Q1の解答:各変化は現在の価格を用いる

    既知:$P_0=200$個、上昇$r=10\%$、下降$r=10\%$。各百分率は現在の価格に基づく。したがって、各段階で乗数を適用する。

    上昇の式を記号で表すと:

    $$P_{new}=P_{old}(1+r/100)$$
    $$P_1=P_0(1+r/100)=200(1+10/100)=220$$

    下降を記号で表すと:

    $$P_{new}=P_{old}(1-r/100)$$
    $$P_2=P_1(1-r/100)=220(1-10/100)=198$$

    最終価格は$\mathbf{198}$個である。

    なぜ200ではないのか? 2つの乗数はCombined multiplier(統合乗数)を形成する:

    $$\frac{P_2}{P_0}=(1+r/100)(1-r/100)$$
    $$\frac{P_2}{P_0}=(1+r/100)(1-r/100)=(1+10/100)(1-10/100)=0.99$$

    したがって、最終価格は初期価格の$99\%$分を維持している。全体での損失は$1\%$であり、ゼロではない。

    上昇と下降も異なる金額から始まっている:

    $$\Delta_{rise}=P_1-P_0=220-200=20$$
    $$\Delta_{fall}=P_1-P_2=220-198=22$$

    10% of 220 は 10% of 200 より大きい。下降が取り除く量は、上昇が加えた量より多い。

    確認:全体での損失は初期価格の$1\%$である。

    $$1-0.99=0.01$$
    $$P_0\times0.01=200\times0.01=2$$
    $$P_0-P_2=200-198=2$$

    どちらも2個となるため、2つの確認は一致する。

    Q2の解答:増分を割り算で逆算する

    既知:新しい数値$N=84$、増分$r=20\%$。増分は元の数値に$1+r/100$を掛けた結果である。百分率の変化を逆にするには、この乗数で割る。同じ百分率を引いてはいけない。

    増分の式を記号で表すと:

    $$N=P_0(1+r/100)$$

    元の数値について整理すると:

    $$P_0=\frac{N}{1+r/100}$$

    代入:

    $$P_0=\frac{N}{1+r/100}=\frac{84}{1+20/100}$$

    倍率を求めよ:

    $$1+\frac{r}{100}=1+\frac{20}{100}=1.2$$

    割る:

    $$P_0=\frac{N}{1+r/100}=\frac{84}{1.2}=70$$

    元の数値は $\mathbf{70}$ である。

    逆算の検証:増加分を再度計算して確認する。

    $$N=P_0(1+r/100)=70(1+20/100)=84$$

    元の 70 が指定された 84 を生み出すため、答えは正しい。

    罠:新しい数値の 20% を引き算すること。

    $$84-84\times0.20=67.2$$

    それは誤りである。増加量は 70 に基づいて計算され、84 に基づいて計算されたわけではない。

    Q3 の解答:解き、両側を確認する

    既知条件:不等式 $9-2x\le3$ 。方程式のように解く。負の数で割ると 不等号の向きが反転 する。

    両辺から 9 を引く:

    $$9-2x\le3$$
    $$-2x\le3-9$$
    $$-2x\le-6$$

    両辺を $-2$ で割り、符号を反転させる:

    $$x\ge\frac{-6}{-2}=3$$

    解答は $x\ge3$ である。

    境界点 の検証: $x=3$ において両辺は等しい。

    $$9-2x=9-2(3)=9-6=3$$

    したがって 3 は含まれる。数直線では 3 に 閉じた円 を描き、右側に塗る。

    内部の検証: $x=4$ 。

    $$9-2x=9-2(4)=9-8=1\le3$$

    真(True)。

    外部の検証: $x=2$ 。

    $$9-2x=9-2(2)=9-4=5>3$$

    偽(False)。両方の検証とも $x\ge3$ と一致する。

    Q4 の解答:長さから面積へ

    既知:長さ比$3:5$;より小さい表面積$A_s=36\ \text{cm}^2$。$k$を小さな長さ$L_s$から対応する大きな長さ$L_L$へのスケール因子とする。各面積は2つの長さの积である。したがって、面積は$k$の二乗でスケールされる。

    2つの領域の間の一般的な関係:

    $$\frac{A_L}{A_s}=k^2$$

    長さの比から縮尺係数を求める:

    $$k=\frac{L_L}{L_s}=\frac{5}{3}$$

    面積の比:

    $$\frac{A_L}{A_s}=k^2=\left(\frac{5}{3}\right)^2=\frac{25}{9}$$

    大きい方の面積について式を整理する:

    $$A_L=A_s k^2$$

    代入:

    $$A_L=A_s k^2=36\times\frac{25}{9}=100$$

    大きい方の表面積は $\mathbf{100}\ \text{cm}^2$ です。

    比の確認:

    $$\frac{A_L}{A_s}=\frac{100}{36}=\frac{25}{9}$$
    $$k^2=\left(\frac{5}{3}\right)^2=\frac{25}{9}$$

    面積の比 $25:9$ は、長さの比 $5:3$ の二乗です。面積だけが $\frac{25}{9}$ で拡大されます。長さは $k=\frac{5}{3}$ で拡大されます。

    今後のアクション

    • 練習用紙 1.1、1.2、1.4 にあるマッチング演習を試してください。
    • 新しい写しで、すべてのチェックを自分自身で行ってください。
    • 自身の解法が異なったステップを記録し、その後別の問題に挑戦してください。

    チェック付きの明示的な解法は論理を明確に示しますが、単なる数字だけではそれができません。

  • 2

    GAC010 数学II:確率、統計および金融

    レッスンを視聴
    2.1

    What this module is, and how it is marked

    A cheaper headline rate need not produce the cheaper total, and a larger survey need not produce a more representative result. Mathematics II uses explicit rules, calculations and evidence limits to examine such questions.

    GAC010 covers sequences, finance, probability and statistics. Your centre's current brief determines assessment tasks, weights, tools and deadlines. A test 测验, projects 项目, an examination 考试 and coursework 平时作业 may use different instructions; these original practice sheets do not establish their official pattern or university credit.

    Show the known data, why the formula applies, the symbolic equation and the substitution. State the unit and interpret the result. A statistical summary describes a defined set of observations or a stated model; it does not automatically establish a cause or a whole-population finding.

    All financial amounts, rates, fees and pay rules in the examples are invented teaching conditions. Use their stated timing and rounding assumptions rather than assuming an actual tax or lending rule.

    English 日本語
    test/test/ テスト
    projects/ˈprɒdʒekts/ プロジェクト
    examination/eɡˌzæmɪˈneɪʃn/ 試験
    coursework/ˈkɔːsjuːɜːk/ 課題
    2.1

    Sequences and series

    シラバス

    Unit 1 of 8 in GAC010 Mathematics II: Probability, Statistics and Finance (Level II). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    Module purpose: On completion of this module, students should be able to apply a basic knowledge of the principles of probability and statistics to solving and analysing common financial problems.

    The module outcomes this unit works towards:

    Learning Objective GAC010.1: Solve problems using arithmetic and geometric sequences and series.

    出典: Cambridge International シラバス

    A sequence 数列 is an ordered list of numbers; a series 级数 is their sum.

    An arithmetic sequence 等差数列 adds a fixed common difference 公差 $d$ each step. A geometric sequence 等比数列 multiplies by a fixed common ratio 公比 $r$.

    $$u_n = a + (n-1)d \qquad\text{and}\qquad u_n = ar^{\,n-1}$$
    • The sums are $S_n = \tfrac{n}{2}\left(2a + (n-1)d\right)$ and $S_n = \dfrac{a(1 - r^n)}{1 - r}$ for $r \neq 1$.
    • A geometric series with $|r| < 1$ has a sum to infinity 无穷和 $S_\infty = \dfrac{a}{1-r}$.
    • Keep a term separate from a total. Some financial models compound geometrically, while others use simple interest or cash flows. Match the model to the specified rule.

    Worked example. row counts differ by a constant amount

    Three rows contain 4, 7 and 10 dots, illustrating a constant difference of 3.

    Known: row counts are 4, 7 and 10, continuing with the same difference. Thus $a=4$ and $d=3$. For the fifth row:

    $$u_n=a+(n-1)d$$
    $$u_5=4+(5-1)(3)=16$$

    For the first five rows together, use the arithmetic sum, not the fifth term.

    $$S_n=\frac n2[2a+(n-1)d]$$
    $$S_5=\frac52[2(4)+(5-1)(3)]=50$$

    Practice sheet 2.1 gives a progressively harder set with worked solutions.

    English 日本語
    sequence/ˈsiːkwəns/ 数列
    series/ˈsɪəriːz/ 級数
    arithmetic sequence/əˈrɪθmətɪk ˈsiːkwəns/ 等差数列
    common difference/ˈkɒmən ˈdɪfrəns/ 公差
    geometric sequence/ˌdʒiːəʊˈmetrɪk ˈsiːkwəns/ 等比数列
    common ratio/ˈkɒmən ˈreɪʃɪəʊ/ 公比
    sum to infinity/sʌm tʊ ɪnˈfɪnɪti/ 無限和
    2.2

    Finance: earning money and purchasing goods

    シラバス

    Unit 2 of 8 in GAC010 Mathematics II: Probability, Statistics and Finance (Level II). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    The module outcomes this unit works towards:

    Learning Objective GAC010.2: Apply algebraic methods to solve financial problems and analyse information.

    出典: Cambridge International シラバス

    This unit is the arithmetic of an ordinary life, in English.

    • Gross pay 税前工资 is what you earn; net pay 税后工资 is what arrives after deductions 扣除项 such as tax and insurance.
    • Simple interest 单利 is $I = Prt$ — interest on the original amount only.
    • Compound interest is $A = P(1+r)^n$ — interest on the interest, which is why it is geometric.
    • Depreciation 折旧 is a reduction in recorded value. A fixed-percentage model uses $V_n=V_0(1-r)^n$; other depreciation conventions can use different rules.
    • A discount 折扣 of 20% multiplies by 0.8. Two successive discounts of 20% and 10% multiply by $0.8 \times 0.9 = 0.72$, which is a 28% reduction, not 30%.

    Worked example. calculate regular and overtime hours separately

    A worked pay record separates 8 regular hours from 2 overtime hours, with their respective rates.

    Known: 8 hours at 20 units per hour and 2 overtime hours at 1.5 times that rate. Separate the two categories before adding pay.

    $$G=h_rp+h_okp$$
    $$G=8(20)+2(1.5)(20)=220\ \text{units}$$

    If the invented fixed deduction is 15 units, take it from gross pay once.

    $$N=G-D=220-15=205\ \text{units}$$

    Practice sheet 2.2 gives a progressively harder set with worked solutions.

    English 日本語
    Gross pay/ɡrəʊs peɪ/ 総給与
    net pay/net peɪ/ 手取り
    deductions/dɪˈdʌkʃnz/ 控除額
    Simple interest/ˈsɪmpl ˈɪntrest/ 単利
    Depreciation/dɪˌpriːʃɪˈeɪʃn/ 減価償却
    discount/ˈdɪskaʊnt/ 割引
    2.3

    Finance: investing money and loans

    シラバス

    Unit 3 of 8 in GAC010 Mathematics II: Probability, Statistics and Finance (Level II). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    The module outcomes this unit works towards:

    Learning Objective GAC010.2: Apply algebraic methods to solve financial problems and analyse information.

    出典: Cambridge International シラバス

    • Under a stated constant discount rate and timing assumption, the present value 现值 of a future amount is: $PV = \dfrac{FV}{(1+r)^n}$.
    • An annuity 年金 has specified payments at regular intervals. A loan repayment 贷款还款 schedule may instead have changing amounts or an adjusted final payment; specify whether interest comes before or after payment.
    • The label APR 年利率 is an annual rate measure whose calculation and included charges depend on the stated convention. This sheet uses explicitly defined period rates, not an unstated APR convention.
    • ⚠ A nominal 12% teaching rate with monthly rate 1% has effective annual growth $(1.01)^{12}-1\approx12.68\%$. A different compounding frequency or fee schedule needs a different calculation.

    Worked example. interest then payment, in that order

    A loan-period flow takes an opening balance of 1000, adds 2% interest, then subtracts a payment of 300.

    Known: opening balance 1000 units, period rate 2%, end-of-period payment 300 units. Interest comes before payment under this stated convention.

    $$B_1=B_0(1+r)-M$$
    $$B_1=1000(1.02)-300=720$$

    The next period's interest uses 720, not the original principal.

    $$B_2=B_1(1+r)-M=720(1.02)-300=434.40$$

    Practice sheet 2.3 gives a progressively harder set with worked solutions.

    English 日本語
    present value/ˈprezənt ˈvæljuː/ 現在価値
    annuity/əˈnjuːɪti/ 年金
    loan repayment/ləʊn rɪˈpeɪmənt/ 返済額
    APR/ˌeɪ piː ˈɑː/ 年実質利率
    2.4

    Probability: key concepts

    シラバス

    Unit 4 of 8 in GAC010 Mathematics II: Probability, Statistics and Finance (Level II). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    The module outcomes this unit works towards:

    Learning Objective GAC010.3: Solve problems using probability and counting techniques.

    出典: Cambridge International シラバス

    Probability 概率 measures how likely an event is, on a scale from 0 to 1.

    • For equally likely outcomes, $P(A) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}$.
    • Events are mutually exclusive 互斥 when they cannot both happen: $P(A \text{ or } B) = P(A) + P(B)$.
    • Events are independent 独立 when one does not affect the other: $P(A \text{ and } B) = P(A) \times P(B)$.
    • The complement 对立事件 rule, $P(\text{not } A) = 1 - P(A)$, is often the fastest route to an answer containing "at least one".
    • Counting techniques 计数方法 supply the denominator: a permutation 排列 counts ordered selections, a combination 组合 counts unordered ones.

    Conditional probability is $P(A\mid B)=P(A\cap B)/P(B)$ for $P(B)>0$. A union subtracts its overlap: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. Association in a supplied table does not by itself establish a cause.

    Worked example. a second draw depends on the first

    A two-draw tree for a bag with two red and one blue counter, without replacement, labels conditional second-draw probabilities.

    Known: two red and one blue counter, two random draws without replacement. To get two reds, follow the red-red path and multiply its conditional probabilities.

    $$P(RR)=P(R_1)P(R_2\mid R_1)$$
    $$P(RR)=\frac23\times\frac12=\frac13$$

    A first blue leaves only red counters; the blue-blue path has probability zero. The draws are not independent.

    Practice sheet 2.4 gives a progressively harder set with worked solutions.

    English 日本語
    Probability/ˌprɒbəˈbɪlɪti/ 確率
    mutually exclusive/ˈmjuːtʃuːəli eksˈkluːsɪv/ 排反
    independent/ˌɪndɪˈpendənt/ 独立
    complement/ˈkɒmplɪmənt/ 対立事象
    Counting techniques/ˈkaʊntɪŋ tekˈniːks/ 計数法
    permutation/ˌpɜːmjuːˈteɪʃn/ 順列
    combination/ˌkɒmbɪˈneɪʃn/ 組合せ
    2.5

    Statistics: collecting and displaying data

    シラバス

    Unit 5 of 8 in GAC010 Mathematics II: Probability, Statistics and Finance (Level II). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    The module outcomes this unit works towards:

    Learning Objective GAC010.4: Analyse collected data using statistical methods.

    出典: Cambridge International シラバス

    • A population 总体 is everyone you want to describe; a sample 样本 is who you actually measure.
    • Generalising to a population needs a suitably representative 有代表性的 sample and an honest account of selection and nonresponse. A class survey can describe that class without establishing a school-wide percentage.
    • Qualitative data 定性数据 are categories; quantitative data 定量数据 measure amounts, either discrete 离散 (counted) or continuous 连续 (measured).
    • Bias 偏差 enters through who you ask, who answers, and how the question is worded, and arithmetic alone does not repair missing or systematically selected observations.

    Worked example. equal bar heights can conceal unequal frequencies

    A histogram has intervals 0 to 10 and 10 to 30 with densities 2 and 1; their areas each represent frequency 20.

    Known: class 0 to 10 has frequency 20; class 10 to 30 has frequency 20. Widths differ, so calculate frequency density.

    $$D=\frac f w$$
    $$D_1=\frac{20}{10}=2$$
    $$D_2=\frac{20}{20}=1$$

    The second bar is half as high but twice as wide. Both areas represent 20 observations.

    Practice sheet 2.5 gives a progressively harder set with worked solutions.

    For unequal-width histogram classes, height is frequency divided by class width so that bar area represents frequency. Keep categories in separated bars and numeric continuous intervals in a histogram.

    English 日本語
    population/ˌpɒpjʊˈleɪʃn/ 母集団
    sample/ˈsæmpl/ 標本
    representative/ˌreprɪˈzentətɪv/ 代表的
    Qualitative data/ˈkwɒlɪteɪtɪv ˈdeɪtə/ 定性情データ
    quantitative data/ˈkwɒntɪteɪtɪv ˈdeɪtə/ 量的データ
    discrete/dɪˈskriːt/ 離散
    continuous/kənˈtɪnjuːəs/ 連続
    Bias/ˈbaɪəs/ バイアス
    2.6

    Statistics: measures of central location

    シラバス

    Unit 6 of 8 in GAC010 Mathematics II: Probability, Statistics and Finance (Level II). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    The module outcomes this unit works towards:

    Learning Objective GAC010.4: Analyse collected data using statistical methods.

    出典: Cambridge International シラバス

    • The mean 平均数 is the total divided by the count. It uses every value, and every outlier.
    • The median 中位数 is the middle value in order. Moving an extreme value without changing the ordered middle may leave it unchanged.
    • The mode 众数 is the most frequent value, and it can identify the most frequent category without imposing a numerical scale. A distribution may have more than one mode or no unique mode.
    • ⚠ Choose the summary for the question. A high extreme can pull the mean above the median. The mean still measures total per observation; the median locates the ordered middle. Neither says that most observations equal it.

    Worked example. group size determines a combined mean

    Two groups contribute different totals: 10 learners with mean 60, and 20 learners with mean 75.

    Known: group A has 10 scores with mean 60; group B has 20 with mean 75. Recover each group's total before combining.

    $$\bar x=\frac{n_A\bar x_A+n_B\bar x_B}{n_A+n_B}$$
    $$\bar x=\frac{10(60)+20(75)}{10+20}=70$$

    The unweighted mean of 60 and 75 would be 67.5, which ignores the unequal counts.

    Practice sheet 2.6 gives a progressively harder set with worked solutions.

    English 日本語
    mean/miːn/ 平均値
    median/ˈmiːdiːən/ 中央値
    mode/məʊd/ 最頻値
    2.7

    Statistics: measures of variability

    シラバス

    Unit 7 of 8 in GAC010 Mathematics II: Probability, Statistics and Finance (Level II). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    The module outcomes this unit works towards:

    Learning Objective GAC010.4: Analyse collected data using statistical methods.

    出典: Cambridge International シラバス

    • The range 全距 is the largest minus the smallest — the full span, so an extreme can change it substantially.
    • The interquartile range (IQR) 四分位距 is the difference $Q_3-Q_1$ spanning the central half by the stated quartile convention. A box plot 箱线图 shows quartiles and a median, with whisker rules that must be specified.
    • The standard deviation 标准差 summarises spread using squared deviations. For a complete population, use: $\sigma=\sqrt{\sum(x-\mu)^2/n}$. A sample estimate uses $s=\sqrt{\sum(x-\bar x)^2/(n-1)}$ under that stated convention. Variance has squared units; standard deviation has the original units.
    • A larger standard deviation means a wider spread. Two data sets can share a mean and describe completely different situations.

    Worked example. read a five-number summary rather than every observation

    A minimum-to-maximum box plot marks minimum 2, first quartile 4, median 6, third quartile 8 and maximum 12.

    Known five-number summary: minimum 2, $Q_1=4$, median 6, $Q_3=8$, maximum 12. The overall span and middle-half span are different.

    $$R=x_{max}-x_{min}=12-2=10$$
    $$IQR=Q_3-Q_1=8-4=4$$

    The box does not identify every original value, the sample size or why the spread arose.

    Practice sheet 2.7 gives a progressively harder set with worked solutions.

    For sheet 2.7, quartiles are medians of ordered half-lists, excluding the overall median for an odd count. Whiskers run to the minimum and maximum. Do not mix that convention with software interpolated quartiles or outlier-fence whiskers.

    English 日本語
    range/reɪndʒ/ 全範囲
    interquartile range (IQR)/ˌɪntəˈkwɔːtaɪl reɪndʒ/ 四分位範囲 (IQR)
    box plot/bɒks plɒt/ 箱ひげ図
    standard deviation/ˈstændəd ˌdiːvɪˈeɪʃn/ 標準偏差
    2.8

    Distributions: the normal distribution

    シラバス

    Unit 8 of 8 in GAC010 Mathematics II: Probability, Statistics and Finance (Level II). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    The module outcomes this unit works towards:

    Learning Objective GAC010.5: Use the normal distribution to analyse data.

    出典: Cambridge International シラバス

    • The normal distribution 正态分布 is a continuous symmetric bell-shaped model specified by its mean and positive standard deviation. Those two summaries alone do not establish that observed data follow this model.
    • Under a normal model, the approximate empirical rule is: about 68% of values lie within one standard deviation of the mean, 95% within two, and 99.7% within three.
    • A z-score 标准分 says how many standard deviations a value sits from the mean: $z = \dfrac{x - \mu}{\sigma}$, which is what makes two different scales comparable.

    For a normal model, use cumulative area $\Phi(z)=P(Z\leq z)$ and symmetry $\Phi(-z)=1-\Phi(z)$. An interval subtracts its two cumulative bounds. Expected count is sample size times probability and need not be an integer; an observed count is an integer.

    Worked example. a threshold is an area beyond a standardised value

    A standard-normal curve marks z equal to 1 and shades the right-tail region beyond it.

    Known: an invented measurement model has $\mu=50$ and $\sigma=10$. For values above 60, standardise the threshold.

    $$z=\frac{x-\mu}{\sigma}=\frac{60-50}{10}=1$$

    Use the complement of the supplied cumulative area.

    $$P(X>60)=1-\Phi(1)=1-0.8413=0.1587$$

    This is a model probability, not a guarantee that exactly 15.87% of a finite sample exceeds 60.

    Practice sheet 2.8 gives a progressively harder set with worked solutions.

    English 日本語
    normal distribution/ˈnɔːml ˌdɪstrɪˈbjuːʃn/ 正規分布
    z-score/zed skɔː/ z点
  • 3

    GAC016 数学III:微積分と高度な応用

    レッスンを視聴
    3.1

    What this module is, and how it is marked

    An average speed does not tell you every instantaneous speed, and a signed displacement does not always equal total distance. Calculus makes those distinctions precise.

    GAC016 develops differentiation 微分, integration 积分 and their applications. Your centre's current brief determines assessment tasks and grading requirements; these original practice sheets do not establish a university credit decision or an official examination pattern.

    State the independent variable, domain and units. The notation $dy/dx$ is a derivative with respect to x; a dot over y normally denotes a derivative with respect to time. Those notations describe the same operation only when their independent variables agree.

    For applications, describe what a rate or accumulated value means in the stated model. A negative volume-change rate of 3 cubic centimetres per second means volume is falling at that rate, rather than that the volume itself is negative.

    English 日本語
    differentiation/ˌdɪfəˌrenʃɪˈeɪʃn/ 微分
    integration/ˌɪntɪˈɡreɪʃn/ 積分
    3.1

    Differentiation

    シラバス

    Unit 1 of 3 in GAC016 Mathematics III: Calculus & Advanced Applications (Level III). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    Module purpose: On completion of this module, students should be able to demonstrate a basic understanding of the principles of calculus and how they can be applied to the quantitative analysis of practical and financial situations.

    The module outcomes this unit works towards:

    Learning Objective GAC016.1: Determine the derivative (if it exists) of most mathematical functions and use the derivative to analyse functional behaviour.

    出典: Cambridge International シラバス

    A derivative 导数 is a rate of change: how fast $y$ changes as $x$ changes. Geometrically it is the gradient of the tangent 切线斜率 at a point.

    $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

    That limit is the definition. Differentiation from first principles 从定义求导 uses the difference quotient for nonzero h and then takes its limit. Unequal one-sided limits, as for $|x|$ at zero, prevent a derivative even when the function is continuous.

    • The power rule 幂法则, on a domain where the power is differentiable: if $y = x^n$ then $\dfrac{dy}{dx} = nx^{\,n-1}$.
    • The product rule 乘积法则: $(uv)' = u'v + uv'$.
    • The quotient rule 商法则, where $v\ne0$: $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$.
    • The chain rule 链式法则: if $y = f(g(x))$ then $\dfrac{dy}{dx} = f'(g(x)) \cdot g'(x)$.
    • Stationary points 驻点 occur where $f'(x) = 0$. The second derivative 二阶导数 then classifies a stationary point when it is nonzero: negative gives a local maximum 极大值 and positive a local minimum 极小值. A zero second derivative is inconclusive; use derivative signs or other evidence. A stationary inflection can be neither an extremum.

    Worked example. a tangent slope differs from a secant slope

    The curve y equals x squared has a tangent at (1,1) and a dashed secant joining (1,1) to (2,4).

    Known: $f(x)=x^2$. The secant between x equal to 1 and 2 has slope 3, but the derivative at x equal to 1 is 2.

    $$m_{sec}=\frac{f(b)-f(a)}{b-a}=\frac{4-1}{2-1}=3$$
    $$f'(x)=\lim_{h\to0}\frac{(x+h)^2-x^2}{h}=\lim_{h\to0}(2x+h)=2x$$
    $$m_{tan}=f'(1)=2(1)=2$$

    The average slope over a finite interval need not equal the instantaneous slope at either endpoint.

    Use original practice sheet 3.1 to test the method, domain and interpretation against its solutions.

    English 日本語
    derivative/dɪˈrɪvətɪv/ 導関数
    gradient of the tangent/ˈɡreɪdɪənt ɒvðə ˈtændʒənt/ 接線の傾き
    differentiation from first principles/ˌdɪfəˌrenʃɪˈeɪʃn frɒm fɜːst ˈprɪnsɪplz/ 定義による微分
    power rule/ˈpaʊə ruːl/ 累乗則
    product rule/ˈprɒdʌkt ruːl/ 積の微分法則
    quotient rule/ˈkwəʊʃənt ruːl/ 商の微分法則
    chain rule/tʃeɪn ruːl/ 合成関数の微分法則
    Stationary points/ˈsteɪʃənəri pɔɪnts/ 驻点
    second derivative/ˈsekənd dɪˈrɪvətɪv/ 二階導関数
    maximum/ˈmæksɪməm/ 最大値
    minimum/ˈmɪnɪməm/ 最小値
    3.2

    Integration

    シラバス

    Unit 2 of 3 in GAC016 Mathematics III: Calculus & Advanced Applications (Level III). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    The module outcomes this unit works towards:

    Learning Objective GAC016.2: Use techniques of integration to find indefinite and definite integrals.

    出典: Cambridge International シラバス

    Indefinite integration finds antiderivatives. Definite integration measures signed accumulation, which can differ from geometric area. For a continuous integrand, the fundamental theorem of calculus 微积分基本定理 connects a definite integral to the difference of antiderivative values at its limits.

    • An indefinite integral 不定积分 has no limits and needs the constant of integration 积分常数: $\displaystyle\int x^n\,dx = \frac{x^{\,n+1}}{n+1} + c$ for $n \neq -1$.
    • A definite integral 定积分 has limits and gives a number: $\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)$.
    • Integration by substitution 换元积分法 reverses the chain rule.
    • ⚠ Forgetting $+c$ on an indefinite integral is the single most frequent lost mark in the module, and it is lost on questions you have otherwise answered correctly.

    Worked example. negative and positive areas can cancel

    The line y equals x crosses the axis at zero, with shaded regions on both sides over negative 1 to 1.

    Known: $f(x)=x$ from negative 1 to 1. Its definite integral is zero because the negative and positive contributions cancel.

    $$I=\int_{-1}^{1}x\,dx=\left[\frac{x^2}{2}\right]_{-1}^{1}=\frac12-\frac12=0$$

    Geometric area instead adds the magnitudes of the two triangular regions.

    $$A=-\int_{-1}^{0}x\,dx+\int_0^1x\,dx=\frac12+\frac12=1$$

    Use original practice sheet 3.2 to test the method, domain and interpretation against its solutions.

    English 日本語
    fundamental theorem of calculus/ˌfʌndəˈmentl ˈθɪərəm ɒv ˈkælkjʊləs/ 微積分の基本定理
    indefinite integral/ɪnˈdefɪnət ˈɪntɪɡrəl/ 不定積分
    constant of integration/ˈkɒnstənt ɒv ˌɪntɪˈɡreɪʃn/ 積分定数
    definite integral/ˈdefɪnət ˈɪntɪɡrəl/ 定積分
    Integration by substitution/ˌɪntɪˈɡreɪʃn baɪ ˌsʌbstɪˈtjuːʃn/ 置換積分法
    3.3

    Advanced applications

    シラバス

    Unit 3 of 3 in GAC016 Mathematics III: Calculus & Advanced Applications (Level III). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    The module outcomes this unit works towards:

    Learning Objective GAC016.3: Apply differentiation and integration techniques in a variety of practical problems.

    出典: Cambridge International シラバス

    • Optimisation 最优化 finds the largest or smallest value of a quantity: write the quantity as a function of one variable on a stated feasible domain, differentiate, find candidates and justify the required optimum. Check included boundaries and integer constraints where relevant.
    • Rates of change 变化率 chain together: $\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$.
    • Area between two curves 两曲线间面积 is $\displaystyle\int_a^b (f(x) - g(x))\,dx$ where $f$ is the upper curve on that whole interval. Split at crossings if their order changes.
    • Marginal cost and revenue 边际成本与边际收益 are derivatives of stated continuous cost and revenue models. They are instantaneous rates, not automatically exact discrete-unit changes.

    Worked example. a fixed perimeter leaves one area variable

    A rectangle has width x and length 10 minus x under a fixed perimeter of 20 units.

    Known: a rectangle has perimeter 20, so length plus width is 10. Let width be x; length is $10-x$ and $0.

    $$A(x)=x(10-x)$$
    $$A'(x)=10-2x$$

    The stationary candidate is $x=5$. Since $A''(x)=-2<0$ and the area approaches zero at both domain ends, this candidate is the global maximum. It is a 5 by 5 square with area 25.

    Use original practice sheet 3.3 to test the method, domain and interpretation against its solutions.

    English 日本語
    Optimisation/ˌɒptɪmaɪˈzeɪʃn/ 最適化
    Rates of change/reɪts ɒv tʃeɪndʒ/ 変化率
    Area between two curves/ˈeərɪə bɪˈtwiːn tuː kɜːvz/ 2つの曲線間の面積
    Marginal cost and revenue/ˈmɑːdʒɪnl kɒst ænd ˈrevənjuː/ 限界費用と限界収益
  • 4

    GAC024 離散数学

    レッスンを視聴
    4.1

    What this module is, and how it is marked

    A repeated set member is counted once, a binary carry may exceed a fixed width, and the fewest-edge route may not have the smallest weight. Discrete mathematics makes those rules explicit.

    GAC024 covers sets, counting systems, binary logic, algorithms and networks. Your centre's current brief determines assessment tasks, tools, weights and deadlines. These original practice sheets do not establish official marking rules or a university credit decision.

    State the universe, representation width, allowed inputs or graph assumptions before solving. Show enough working for another reader to reproduce the result and distinguish a mathematical model from its real implementation.

    4.1

    Sets, relations and functions

    シラバス

    Unit 1 of 5 in GAC024 Discrete Mathematics (Level III). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    Module purpose: On completion of this module, students should be able to demonstrate an understanding of the basic principles of discrete mathematics, particularly the utilisation of mathematical logic. They should also be able to demonstrate the application of these skills to practical situations.

    The module outcomes this unit works towards:

    Learning Objective GAC024.1: Demonstrate understanding of the introductory concepts and properties of sets, relations and functions.

    出典: Cambridge International シラバス

    • A set 集合 is a collection of distinct objects. Order and repetition do not matter.
    • Union 并集 $A \cup B$ is everything in either; intersection 交集 $A \cap B$ is what is in both; the set complement 补集 is everything in the stated universe but outside the set.
    • A subset 子集 has all its elements inside another set.
    • A relation 关系 pairs elements of two sets. A function 函数 is a relation where each input in its stated domain has exactly one output. Different inputs may share an output; an inverse relation is a function only when outputs uniquely identify their inputs.
    • A Venn diagram 韦恩图 turns a set problem into a picture, and can show the disjoint regions and their counts. Check that those regions add to the supplied universe total.

    The inclusion-exclusion principle 容斥原理 subtracts the twice-counted overlap once: $|A\cup B|=|A|+|B|-|A\cap B|$.

    Worked example. subtract an overlap only once

    A class universe of 30 is divided into French-only 11, both 7, German-only 8 and neither 4.

    Known: 30 learners, 18 study French, 15 German, and 7 both. The overlap is included in both subject totals.

    $$|F\cup G|=|F|+|G|-|F\cap G|=18+15-7=26$$
    $$N_{neither}=|U|-|F\cup G|=30-26=4$$

    French-only is $18-7=11$ and German-only $15-7=8$. The four disjoint regions sum to 30.

    Practice sheet 4.1 includes progressively harder problems and independently checked solutions.

    English 日本語
    set/set/ 集合
    Union/ˈjuːnɪən/ 並集合
    intersection/ˌɪntəˈsekʃn/ 共通部分
    set complement/set ˈkɒmplɪmənt/ 補集合
    subset/ˈsʌbset/ 部分集合
    relation/rɪˈleɪʃn/ 関係
    function/ˈfʌŋkʃn/ 関数
    Venn diagram/ven ˈdaɪəɡræm/ ベン図
    inclusion-exclusion principle/ɪnˈkluːʒn eksˈkluːʒn ˈprɪnsɪpl/ 重複除外の原理
    4.2

    Counting systems

    シラバス

    Unit 2 of 5 in GAC024 Discrete Mathematics (Level III). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    The module outcomes this unit works towards:

    Learning Objective GAC024.2: Understand the relationships between different counting systems and be able to perform simple binary arithmetic operations.

    出典: Cambridge International シラバス

    • A positional number base 进制 b uses digits from zero to b minus one and place weights $b^i$. Decimal 十进制 uses ten, binary 二进制 two, hexadecimal 十六进制 sixteen.
    • Every digit's value is its place value 位值: in binary the places are 1, 2, 4, 8, 16 and so on.
    • Hexadecimal is shorthand for binary: one hex digit is exactly four bits, so conversion can group a stated-width binary pattern into four-bit blocks. Leading zeros preserve width while leaving the unsigned value unchanged.

    For n unsigned bits, values run from zero to $2^n-1$. Distinguish an unrestricted sum from a stored fixed-width result; a wraparound rule, if explicitly given, keeps the low n bits.

    Worked example. place weights determine the decimal value

    The binary digits 1101 are aligned with place weights 8, 4, 2 and 1.

    Known numeral $1101_2$. Use weights from right to left: 1, 2, 4 and 8.

    $$V=\sum d_i2^i$$
    $$V=1(8)+1(4)+0(2)+1(1)=13$$

    The same value is D in hexadecimal. Leading zeros would not change this nonnegative value but can record an intended width.

    Practice sheet 4.2 includes progressively harder problems and independently checked solutions.

    English 日本語
    number base/ˈnʌmbə beɪs/ 進数
    Decimal/ˈdesɪml/ 十進法
    binary/ˈbaɪnəri/ 2進法
    hexadecimal/ˌheksəˈdesɪml/ 16進法
    place value/pleɪs ˈvæljuː/ 位値
    4.3

    Binary applications

    シラバス

    Unit 3 of 5 in GAC024 Discrete Mathematics (Level III). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    The module outcomes this unit works towards:

    Learning Objective GAC024.2: Understand the relationships between different counting systems and be able to perform simple binary arithmetic operations.

    Learning Objective GAC024.5: Use the basic identities of Boolean algebra to analyse logic circuits and understand the basic principles of propositional logic.

    出典: Cambridge International シラバス

    • Binary arithmetic 二进制运算 adds like decimal, carrying at 2 instead of at 10.
    • A bit 位 is one binary digit; a byte 字节 is eight.
    • Boolean algebra 布尔代数 works on true and false with AND, OR and NOT.
    • A truth table 真值表 lists every Boolean input combination and output. Matching every row proves equivalence for the same finite Boolean inputs; it does not prove physical circuit timing or real-system security.
    • Logic gates 逻辑门 implement stated operations, and a logic circuit 逻辑电路 connects them. Trace the abstract logic according to its connections and input conventions.

    Use inclusive OR and explicit brackets. De Morgan gives $\neg(A\land B)=(\neg A)\lor(\neg B)$. Bitwise NOT inverts only the stated width, not an unspecified infinite representation.

    Worked example. an OR output is inverted by NOT

    Inputs A and B enter an OR-labelled block, whose output enters a NOT-labelled block to give Y.

    Known: $Y=\neg(A\lor B)$. Inclusive OR is false only when both inputs are false; NOT reverses that result. In row order $(A,B)=(0,0),(0,1),(1,0),(1,1)$, the output column is 1, 0, 0, 0. De Morgan gives equivalent expression $(\neg A)\land(\neg B)$.

    Practice sheet 4.3 includes progressively harder problems and independently checked solutions.

    English 日本語
    Binary arithmetic/ˈbaɪnəri əˈrɪθmətɪk/ 二進計算
    bit/bɪt/ ビット
    byte/baɪt/ バイト
    Boolean algebra/ˈbuːlɪən ˈældʒɪbrə/ ブール代数
    truth table/truːθ ˈteɪbl/ 真値表
    Logic gates/ˈlɒdʒɪk ɡeɪts/ 論理ゲート
    logic circuit/ˈlɒdʒɪk ˈsɜːkɪt/ 論理回路
    4.4

    Algorithms

    シラバス

    Unit 4 of 5 in GAC024 Discrete Mathematics (Level III). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    The module outcomes this unit works towards:

    Learning Objective GAC024.3: Construct and analyse algorithms and flowcharts for simple mathematical and general procedures.

    出典: Cambridge International シラバス

    • An algorithm 算法 describes unambiguous steps for a task. A procedure solving the stated finite task must terminate and give the required result for its allowed inputs.
    • A flowchart 流程图 draws it: a decision is a diamond, a process a rectangle.
    • Pseudocode 伪代码 represents its steps without requiring a particular implementation language. State assignment, loop bounds and index conventions before tracing.
    • Tracing 追踪 an algorithm — a table with one column per variable and one row per step — records its actual updates. A trace checks the chosen input; a claim for all allowed inputs also needs a correctness argument.
    • Efficiency 效率 matters: a linear search can stop early but may inspect all n items. Binary search repeatedly discards half of an ordered search range; its logarithmic comparison count requires the sorted-data and bound conventions.

    Worked example. repeat a remainder step until the second number is zero

    A Euclidean-algorithm flowchart tests b equal to zero, otherwise computes a remainder and updates the pair before returning to the test.

    Known: start with positive integers a equal to 10 and b equal to 6. While b is nonzero, compute r as a MOD b, then set a to b and b to r. Pairs after complete iterations are (6,4), (4,2), (2,0), giving output 2. Temporary r preserves the remainder before a and b change. Each nonzero remainder is smaller than the previous positive b, supporting termination.

    Practice sheet 4.4 includes progressively harder problems and independently checked solutions.

    English 日本語
    algorithm/ˈælɡərɪθəm/ アルゴリズム
    flowchart/ˈfləʊtʃɑːt/ フローチャート
    Pseudocode/ˈsuːdəʊkəʊd/ 擬似コード
    Tracing/ˈtreɪsɪŋ/ トレース
    Efficiency/ɪˈfɪʃənsi/ 効率性
    4.5

    Graphs and networks

    シラバス

    Unit 5 of 5 in GAC024 Discrete Mathematics (Level III). The module is taught over about 40 class hours plus 20 hours of independent study, and is assessed at the teaching centre and moderated by ACT — there is no external exam.

    The module outcomes this unit works towards:

    Learning Objective GAC024.4: Identify the basic types, properties and applications of graphs and trees.

    出典: Cambridge International シラバス

    • A graph 图 is a set of vertices 顶点 joined by edges 边. It models anything with connections: roads, friendships, dependencies.
    • For a simple undirected graph with no loops or repeated edges, the degree 度 counts incident edges. Every edge contributes two to the total degree sum.
    • A tree 树 is a connected graph with no cycles, and a finite tree with n vertices has n minus 1 edges. Some hierarchical models use trees, but actual systems can also contain cross-links or cycles.
    • A shortest path 最短路径 problem asks for the cheapest route between two vertices, by total weight under the stated constraints, rather than by the number of edges alone. A minimum spanning tree instead connects every vertex without cycles and minimises total included edge weight.

    Worked example. compare total route weight, not the number of edges

    An undirected network joins A to B with weight 2, B to C with 3, A to C with 8 and C to D with 1.

    Known edge weights are AB = 2, BC = 3, AC = 8 and CD = 1. The path A-C-D has weight 9, while A-B-C-D has weight 6. Therefore the three-edge path is shorter by weight despite having more edges. The minimum spanning tree for this small network uses AB, BC and CD with total 6; the agreement of totals here does not make the tasks identical.

    Practice sheet 4.5 includes progressively harder problems and independently checked solutions.

    English 日本語
    graph/ɡræf/ グラフ
    vertices/ˈvɜːtɪsiːz/ 頂点
    edges/ˈedʒɪz/ 辺
    degree/dɪˈɡriː/ 次数
    tree/triː/ 木
    shortest path/ˈʃɔːtɪst pæθ/ 最短路

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