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과목

GAC 수학

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GAC Mathematics is four modules. GAC004 rebuilds the fundamentals in English, from arithmetic and algebra through coordinate geometry, plane and solid geometry, trigonometry and the exponential and logarithmic functions. GAC010 turns to sequences, financial mathematics, probability and statistics. GAC016 is single-variable calculus. GAC024 is discrete mathematics, from sets and counting systems to algorithms, graphs and Boolean logic.

Language is assessed alongside method. A terminology logbook, kept unit by unit and inspected during the semester, is a real component of GAC004, so being able to say what a gradient or a residual is matters as much as computing one.

GAC016 is the module universities most often credit. Several publish a calculus equivalency for it, which makes the derivative and integral work here worth treating as a first-year course rather than as school revision.

The events are a mixture. Expect a short in-class test on the early units, one or two projects built in a spreadsheet, an examination covering everything at the end, and coursework marks that accumulate quietly all semester.

Worked examples, drills and mock papers here are ours, written to the outcomes and the format of those events.

전체 어휘 연습
  • 1

    GAC004 수학 I: 기초 이론

    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 한국어
    in-class test/ɪn klæs test/ 课堂测验(in-class test)
    projects/ˈprɒdʒekts/ 현재 디렉터리(Current dir)
    examination/eɡˌzæmɪˈneɪʃn/ 시험
    coursework/ˈkɔːsjuːɜːk/ 课程作业(coursework)
    terminology logbook/ˌtɜːmɪˈnɒlədʒi ˈlɒɡbʊk/ 용어 일지
    command word/kəˈmænd wɜːd/ 명령어
    1.1

    Terminology and arithmetic review

    Syllabus

    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.

    출처: Cambridge International syllabus

    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 한국어
    integer/ˈɪntɪdʒə/ 정수
    rational number/ˈræʃənl ˈnʌmbə/ 유리수
    irrational number/ɪˈræʃənl ˈnʌmbə/ 무리수
    prime number/praɪm ˈnʌmbə/ 소수
    order of operations/ˈɔːdə ɒv ˌɒpəˈreɪʃnz/ 연산 순서
    factor/ˈfæktə/ 인수
    multiple/ˈmʌltɪpl/ 배수
    highest common factor (HCF)/ˈhaɪɪst ˈkɒmən ˈfæktə/ 최대공약수 (HCF)
    lowest common multiple (LCM)/ˈləʊɪst ˈkɒmən ˈmʌltɪpl/ 최소공배수 (LCM)
    percentage/pəˈsentɪdʒ/ 백분율
    Significant figures/sɪɡˈnɪfɪkənt ˈfɪɡəz/ 유효숫자
    Standard form/ˈstændəd fɔːm/ 표준형(standard form)
    1.2

    Algebra I: introductory algebra

    Syllabus

    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.

    출처: Cambridge International syllabus

    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 한국어
    expression/ekˈspreʃn/ 표현식
    terms/tɜːmz/ 용어
    coefficient/ˌkəʊɪˈfɪʃənt/ 계수
    constant/ˈkɒnstənt/ 일정함
    equation/ɪˈkweɪʒn/ 방정식(equation)
    identity/aɪˈdentɪti/ 정체성
    inequality/ɪniːˈkwɒlɪti/ 불평등
    expand/ekˈspænd/ 전개
    factorise/ˈfæktəraɪz/ 인수분해
    linear equation/ˈlɪnɪə ɪˈkweɪʒn/ 일차 방정식
    quadratic equation/kwɒˈdrætɪk ɪˈkweɪʒn/ 이차 방정식(quadratic equation)
    Simultaneous equations/ˌsɪməlˈteɪnɪəs ɪˈkweɪʒnz/ 연립방정식
    reverses/rɪˈvɜːsɪz/ 반대가 됩니다
    1.3

    Algebra: graphs and coordinate geometry

    Syllabus

    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.

    출처: Cambridge International syllabus

    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 한국어
    gradient/ˈɡreɪdɪənt/ 기울기
    y-intercept/waɪ ˌɪntəˈsept/ y절편
    parallel/ˈpærəlel/ 평행
    perpendicular/ˌpɜːpənˈdɪkjʊlə/ 수직Unless perpendicular lines/planes.
    midpoint/ˈmɪdpɔɪnt/ 중점
    distance/ˈdɪstəns/ 거리
    parabola/pəˈræbələ/ 포물선Unless parabola.
    vertex/ˈvɜːteks/ 정점
    roots/ruːts/ 근(解)
    graphically/ˈɡræfɪkli/ 그래프적으로
    1.4

    Geometry: plane, solid and Euclidean geometry

    Syllabus

    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.

    출처: Cambridge International syllabus

    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 한국어
    polygon/ˈpɒlɪɡən/ 다각형
    Congruent/ˈkɒŋɡruːənt/ 합동
    similar/ˈsɪmɪlə/ 비슷한
    Pythagoras' theorem/paɪˈθæɡərəs ˈθɪərəm/ 피타고라스 정리
    hypotenuse/haɪˈpɒtənjuːs/ 기하변
    Area/ˈeərɪə/ 면적
    volume/ˈvɒljuːm/ 부피
    1.5

    Trigonometry

    Syllabus

    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.

    출처: Cambridge International syllabus

    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 한국어
    sine rule/saɪn ruːl/ 사인 법칙
    cosine rule/ˈkəʊsaɪn ruːl/ 코사인 법칙
    degrees/dɪˈɡriːz/ 도
    radians/ˈreɪdɪənz/ 라디안(radians)
    unit circle/ˈjuːnɪt ˈsɜːkl/ 단원
    periodic/ˌpɪərɪˈɒdɪk/ 주기함수이기 때문에 무한히 많은 해를 가집니다
    bearing/ˈbeərɪŋ/ 방향
    Angles of elevation and depression/ˈæŋɡlz ɒv ˌelɪˈveɪʃn ænd dɪˈpreʃn/ 상승각과 하강각
    1.6

    Exponential and logarithmic functions

    Syllabus

    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.

    출처: Cambridge International syllabus

    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 한국어
    exponential function/ˌekspəˈnenʃl ˈfʌŋkʃn/ 지수 함수
    exponent/ekˈspəʊnənt/ 지수에 위치함
    logarithm/ˈlɒɡərɪθəm/ 로그
    Natural logarithms/ˈnætʃərəl ˈlɒɡərɪθəmz/ 자연로그
    Compound interest/ˈkɒmpaʊnd ˈɪntrest/ 복리
    Exponential decay/ˌekspəˈnenʃl dɪˈkeɪ/ 지수 붕괴
    Additional notes PDF

    Extra practice: four questions with worked answers

    This pack adds practice to three units of GAC004 Mathematics I. It covers arithmetic review (sheet 1.1), introductory algebra (sheet 1.2) and geometry (sheet 1.4). The questions practise basic arithmetic, algebraic methods and plane geometry. Use this pack beside the GAC004 handout and its practice sheets.

    It is original practice material. It does not change your centre's assessment brief, and it is not an official paper.

    Answer the questions in “Practice questions” first. Write your full method. Then compare your method with each worked answer, not only your final number. A clear, stated method shows your reasoning better than a long report.

    Each answer shows the same structure: what is known, why the rule applies, the rule in symbols, then the numbers. One calculation stage sits on each line. The check at the end is part of the answer, so run it yourself on a fresh copy.

    Question Sheet What it practises
    Q1 1.1 Two price changes in a row
    Q2 1.1 Undo one increase
    Q3 1.2 Divide by a negative number
    Q4 1.4 From lengths to areas

    Practice questions

    Answer these first. Do not look ahead.

    Q1. A price of 200 units rises by 10%, then falls by 10%.

    • Find the final price.
    • Explain why the final price is not 200 units.

    Q2. After a 20% increase, a figure is 84. Find the original figure.

    Q3. Solve $9-2x\le3$. Show your solution on a number line 数轴.

    Q4. Two similar 相似 models have a length ratio 长度比 of $3:5$. The smaller model has surface area 表面积 $36\ \text{cm}^2$. Find the larger surface area.

    Worked answers

    Answer to Q1: each change uses the current price

    Known: $P_0=200$ units, rise $r=10\%$, fall $r=10\%$. Each percentage 百分比 uses the current price. So apply a multiplier 乘数 at each stage.

    The increase equation 方程, in symbols:

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

    The decrease, in symbols:

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

    The final price is $\mathbf{198}$ units.

    Why is it not 200? The two multipliers form one combined multiplier:

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

    So the final price keeps $99\%$ of the starting price. The overall loss is $1\%$, not zero.

    The rise and the fall also started from different amounts:

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

    10% of 220 is larger than 10% of 200. The fall removes more than the rise added.

    Check: the overall loss is $1\%$ of the starting price.

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

    Both give 2 units, so the two checks agree.

    Answer to Q2: undo the increase by dividing

    Known: new figure $N=84$, increase $r=20\%$. The increase multiplied the original figure by $1+r/100$. To reverse a percentage change, divide by this multiplier. Never subtract the same percentage back.

    The increase equation, in symbols:

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

    Rearrange for the original figure:

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

    Substitute:

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

    Work out the multiplier:

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

    Divide:

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

    The original figure is $\mathbf{70}$.

    Reverse check: run the increase forward again.

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

    The original 70 returns the stated 84, so the answer is correct.

    The trap: subtracting 20% of the new figure.

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

    That is wrong. The increase was calculated on 70, not on 84.

    Answer to Q3: solve, then check both sides

    Known: the inequality 不等式 $9-2x\le3$. Solve it like an equation. Dividing by a negative number reverses 反向 the inequality sign.

    Subtract 9 from both sides:

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

    Divide both sides by $-2$, and reverse the sign:

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

    The solution is $x\ge3$.

    Boundary 边界 check: at $x=3$ the two sides are equal.

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

    So 3 is included. On the number line, mark a closed circle 实心圆点 at 3 and shade to the right.

    Inside check: $x=4$.

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

    True.

    Outside check: $x=2$.

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

    False. Both checks agree with $x\ge3$.

    Answer to Q4: from lengths to areas

    Known: length ratio $3:5$; smaller surface area $A_s=36\ \text{cm}^2$. Let $k$ be the scale factor 比例因子 from a smaller length $L_s$ to the matching larger length $L_L$. Each area is a product of two lengths. So areas scale as the square 平方 of $k$.

    The general relation between the two areas:

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

    The length ratio gives the scale factor:

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

    The area ratio:

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

    Rearrange for the larger area:

    $$A_L=A_s k^2$$

    Substitute:

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

    The larger surface area is $\mathbf{100}\ \text{cm}^2$.

    Ratio check:

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

    The area ratio $25:9$ is the square of the length ratio $5:3$. Only areas scale by $\frac{25}{9}$. A length scales by $k=\frac{5}{3}$.

    What to do next

    • Try the matching exercises on practice sheets 1.1, 1.2 and 1.4.
    • Run every check yourself on a fresh copy.
    • Record the step where your method differed, then try a new question.

    A stated method with a check shows your reasoning clearly. A bare number does not.

  • 2

    GAC010 수학 II: 확률, 통계 및 금융

    2.1

    이 모듈의 구성 및 채점 기준

    더 낮은 headline rate가 반드시 더 낮은 총액을 산출하는 것은 아니며, 더 큰 표본 조사가 반드시 더 대표성 있는 결과를 제공하지는 않습니다. 수학 II는 이러한 질문을 명시적 규칙, 계산 및 증거 한계로 조사합니다.

    GAC010은 수열, 금융, 확률 및 통계를 다룹니다. 귀속 기관의 현재 지시서에 따라 평가 과제, 가중치, 도구 및 마감일이 결정됩니다. 시험, 프로젝트, 시험 및 과제는 서로 다른 지시를 사용할 수 있으며, 본 원본 연습지에는 공식적인 패턴이나 대학 학점이 설정되어 있지 않습니다.

    주어진 데이터, 공식이 적용되는 이유, 기호 방정식 및 대입 과정을 제시하십시오. 단위를 명시하고 결과를 해석하십시오. 통계 요약은 정의된 관측 집합이나 명시된 모델을 설명할 뿐이며, 인과관계나 전체집단 전반의 결론을 자동으로 입증하지는 않습니다.

    예시에出现的 모든 금리, 수수료 및 지급 규칙은 교육용으로 제작된 조건입니다. 실제 세금 또는 대출 규칙이라고 가정하지 말고, 명시된 시기와 반올림 가정을 따르십시오.

    2.1

    수열 및 급수

    Syllabus

    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.

    출처: Cambridge International syllabus

    수열은 순서가 정해진 숫자의 나열이며, 급수는 그들의 합입니다.

    등차수열은 각 단계마다 고정된 공차 $d$를 더합니다. A 등비수열은 고정된 공비 $r$를 곱합니다.

    $$u_n = a + (n-1)d \qquad\text{and}\qquad u_n = ar^{\,n-1}$$
    • 합은 $S_n = \tfrac{n}{2}\left(2a + (n-1)d\right)$이고 $S_n = \dfrac{a(1 - r^n)}{1 - r}$는 $r \neq 1$에 대해입니다.
    • $|r| < 1$인 등비급수는 무한합 $S_\infty = \dfrac{a}{1-r}$을 가집니다.
    • 항을 총액과 구분하십시오. 일부 금융 모델은 기하학적으로 복리计算的하지만, 다른 모델은 단순 이자 또는 현금 흐름을 사용합니다. 명시된 규칙에 맞는 모델을 선택하십시오.

    해설 예제. 행의 개수가 일정하게 증가함

    세 행이 각각 4개, 7개, 10개의 점으로 구성되어 상수 차이 3을 보여줍니다.

    알려진 정보: 행의 개수는 4, 7, 10이며 동일한 차이를 이어갑니다. 따라서 $a=4$와 $d=3$입니다. 다섯 번째 행의 경우:

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

    첫 five行的 합을 구하려면 다섯 번째 항이 아닌 등차급수 합을 사용하십시오.

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

    연습지 2.1은 해설이 포함된 점층적으로 난이도가 높은 문제들을 제공합니다.

    English 한국어
    test/test/ 테스트(test)
    projects/ˈprɒdʒekts/ 현재 디렉터리(Current dir)
    examination/eɡˌzæmɪˈneɪʃn/ 시험
    coursework/ˈkɔːsjuːɜːk/ 课程作业(coursework)
    sequence/ˈsiːkwəns/ 순서(시퀀스)
    series/ˈsɪəriːz/ 직렬
    arithmetic sequence/əˈrɪθmətɪk ˈsiːkwəns/ 산술 급수
    common difference/ˈkɒmən ˈdɪfrəns/ common difference(등차)
    geometric sequence/ˌdʒiːəʊˈmetrɪk ˈsiːkwəns/ 기하 급수
    common ratio/ˈkɒmən ˈreɪʃɪəʊ/ common ratio(등비)
    sum to infinity/sʌm tʊ ɪnˈfɪnɪti/ 무한 급수의 합
    Gross pay/ɡrəʊs peɪ/ 총급여
    net pay/net peɪ/ 순급여
    deductions/dɪˈdʌkʃnz/ 공제금
    Simple interest/ˈsɪmpl ˈɪntrest/ 단리
    Depreciation/dɪˌpriːʃɪˈeɪʃn/ 감가상각
    discount/ˈdɪskaʊnt/ 할인
    present value/ˈprezənt ˈvæljuː/ 현재 가치
    annuity/əˈnjuːɪti/ 연금
    loan repayment/ləʊn rɪˈpeɪmənt/ 대출 상환
    2.2

    금융: 돈 버기 및 상품 구매

    Syllabus

    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.

    출처: Cambridge International syllabus

    이 단원은 영어로 표현한 일상생활의 산술입니다.

    • 총급여는 당신이 번 금액이며, 순급여는 세금 및 보험 같은 공제 항목이 차감된 후 도착하는 금액입니다.
    • 단리는 $I = Prt$ — 원금에만 대한 이자입니다.
    • 복리는 $A = P(1+r)^n$ — 이자에 대한 이자로, 이로 인해 기하학적(등비)이 됩니다. 기하학적.
    • 감가상각은 기록된 가치의 감소를 의미합니다. 고정 비율 모델은 $V_n=V_0(1-r)^n$를 사용하며, 다른 감가상각 관례는 다른 규칙을 사용할 수 있습니다.
    • 20% 할인은 0.8을 곱하는 것입니다. 연속하여 20%와 10%의 할인을 applied하면, $0.8 \times 0.9 = 0.72$로 줄어듭니다. 이는 28% 감소를 의미하며, 30% 감소가 아닙니다.

    해설 예제. 정규근로시간과 초과근로시간을 별도로 계산

    해결된 급여 명세서는 8시간의 정규근로와 2시간의 초과근로를 분리하여 respective rates를 보여줍니다.

    已知: 8시간당 20단위의 정시 근로와 2시간의 초과근무(정시 임금의 1.5배)를 알 때, 합산 전에 두类别를 분리하여 급여를 계산하십시오.

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

    만약 고定的 공제 금액이 15 단위라면, 총급여에서 한 번 차감하십시오.

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

    연습지 2.2는 해설이 포함된 점층적으로 난이도가 높은 문제들을 제공합니다.

    2.3

    금융: 투자 및 대출

    Syllabus

    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.

    출처: Cambridge International syllabus

    • 명시된 일정 할인율 및 시기 가정에 따라 미래 금액의 현재가치는 다음과 같습니다: $PV = \dfrac{FV}{(1+r)^n}$.
    • 연금은 정기적인 기간에 지정된 납입금을 가집니다. 대출 상환 일정은 오히려 변하는 금액이나 조정된 최종 납입금을 포함할 수 있으며, 이자가 납입 전인지 후인지 명시해야 합니다.
    • APR이라는 명칭은 연율 측정치이며, 그 계산 및 포함되는 비용은 명시된 관례에 따라 달라집니다. 본 연습지는 명시된 APR 관례가 아닌 명확히 정의된 기간율을 사용합니다.
    • ⚠ 명목 12% 교육용利率 중 월利率 1%인 경우 유효 연성장률은 $(1.01)^{12}-1\approx12.68\%$입니다. 다른 복리 주기나 수수료 체계는 다른 계산을 필요로 합니다.

    해설 예제. 이자 이후 납입, 순서대로

    loan-period flow는 1000의 잔고에서 시작하여 2% 이자를 더한 후 300의 납입금을 뺍니다.

    알려진 정보: 초기 잔고 1000 단위, 기간율 2%, 기간 말 납입금 300 단위. 이 관례에 따라 이자는 납입보다 먼저 발생합니다.

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

    다음 기간의 이자는 원금 720을 기준으로 계산되며, 원래 원금은 아닙니다.

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

    연습지 2.3은 해설이 포함된 점층적으로 난이도가 높은 문제들을 제공합니다.

    English 한국어
    APR/ˌeɪ piː ˈɑː/ 연율(APR)
    2.4

    확률: 핵심 개념

    Syllabus

    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.

    출처: Cambridge International syllabus

    확률은 사건이 발생할 가능성을 0부터 1 사이의 척도로 측정합니다.

    • 동등한 가능성이 있는 결과에 대해, $P(A) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}$입니다.
    • 사건들이 상호 배타적일 때 동시에 일어날 수 없습니다: $P(A \text{ or } B) = P(A) + P(B)$입니다.
    • 사건들이 독립적일 때 한 사건이 다른 사건에 영향을 주지 않습니다: $P(A \text{ and } B) = P(A) \times P(B)$입니다.
    • 보완 규칙, $P(\text{not } A) = 1 - P(A)$,는Often "최소 하나"가 포함된 답에 도달하는 가장 빠른 경로입니다. "최소 하나"를 포함하는 답변.
    • 카운팅 기법은 분모를 제공합니다: 순열은 순서 있는 선택을 세고, 조합은 순서 없는 선택을 세고합니다. 선택지 중 조합은 순서를 고려하지 않음.

    조건부 확률은 $P(A\mid B)=P(A\cap B)/P(B)$입니다 ($P(B)>0$에 대해). 합집합은 겹치는 부분을 뺍니다: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$. 제공된 표에서의_passociation_은 자체적으로 인과관계를 입증하지는 않습니다.

    해설 예제. 두 번째 뽑기가 첫 번째에 의존함

    두 개의 빨간색과 하나의 파란색 마커가 있는 주머니에서 교체 없이 두 번 뽑는 트리는 조건부 두 번째 뽑기 확률을 표시합니다.

    알려진 정보: 두 개의 빨간색과 하나의 파란색 마커, 교체 없이 두 번 무작위로 뽑음. 두 개의 빨간색을 얻으려면 빨강-빨강 경로를 따라 그 조건부 확률을 곱하십시오.

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

    첫 번째가 파란색이면 빨간색 마커만 남으므로, 파랑-파랑 경로의 확률은 0입니다. 뽑기는 독립적이지 않습니다.

    연습지 2.4는 단계별로 난이도가 높아지는 문제와 풀이 해설을 포함합니다.

    English 한국어
    Probability/ˌprɒbəˈbɪlɪti/ 확률
    mutually exclusive/ˈmjuːtʃuːəli eksˈkluːsɪv/ 상호 배타적인
    independent/ˌɪndɪˈpendənt/ 독립적임
    2.5

    통계: 데이터 수집 및 표시

    Syllabus

    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.

    출처: Cambridge International syllabus

    • **집단(population)**은 설명하고자 하는 전체 대상이며, **표본(sample)**은 실제로 측정한 대상을 의미합니다. 측정합니다.
    • 집단에 대한 추론에는 적절한 **대표성(representative)**을 갖춘 표본과 선택 및 미응답에 대한 정직한 보고가 필요합니다. 한 교실의 설문 조사는 해당 교실을 설명할 수 있으나, 학교 전체의 비율을 보장하지는 않습니다.
    • **질적 데이터(qualitative data)**는 범주입니다. **양적 데이터(quantitative data)**는 양을 측정하며,either 이산(discrete)(세어짐) 또는 연속(continuous)(측정됨)일 수 있습니다. 이산(discrete)(세어짐) 또는 연속(continuous)(측정됨).
    • **편향(bias)**은 질문 대상, 응답자, 그리고 질문의 wording에 의해 발생하며, 산술 계산만으로는 누락되거나 체계적으로 선택된 관측치를 수정할 수 없습니다.

    해설 예제. 막대 높이가 같더라도 빈도가 다를 수 있음을 숨길 수 있음

    히스토그램은 간격 0부터 10까지 및 10부터 30까지로 밀도 2와 1를 가지며, 각 면적은 빈도 20를 나타냅니다.

    알고 있음: 0~10 클래스의 빈도는 20, 10~30 클래스의 빈도는 20이다. 폭이 다르므로 빈도 밀도를 계산해야 한다.

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

    두 번째 막대는 높이는 절반이지만 폭은 두 배이다. 두 면적 모두 20개의 관측치를 나타낸다.

    연습지 2.5는 단계별로 난이도가 높아지는 문제와 풀이 해설을 포함합니다.

    폭이 다른 히스토그램 클래스에서는 막대 면적이 빈도를 나타내도록 높이를 '빈도 ÷ 클래스 폭'으로 설정한다. 범주는 분리된 막대에, 수치적 연속 구간은 히스토그램에 배치한다.

    English 한국어
    population/ˌpɒpjʊˈleɪʃn/ 인구
    sample/ˈsæmpl/ 표본
    representative/ˌreprɪˈzentətɪv/ 대변인
    Qualitative data/ˈkwɒlɪteɪtɪv ˈdeɪtə/ 정성 데이터
    quantitative data/ˈkwɒntɪteɪtɪv ˈdeɪtə/ 정량적 데이터
    discrete/dɪˈskriːt/ 불연속적
    continuous/kənˈtɪnjuːəs/ 연속적
    Bias/ˈbaɪəs/ 편향
    2.6

    통계: 중심 위치 지표

    Syllabus

    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.

    출처: Cambridge International syllabus

    • **평균(mean)**은 총합을 개수로 나눈 값이다. 모든 값과 외란치(outlier)를 모두 반영한다.
    • **중간값(median)**은 순서대로 나열했을 때 중앙에 있는 값이다. 극단적인 값을 이동시키되 순서상 중앙이 변하지 않으면 중간값은 그대로일 수 있다.
    • **최빈값(mode)**은 가장 빈번하게 나타나는 값이며, 수치 척도를 강요하지 않고도 최빈 범주를 식별할 수 있다. 분포에는 여러 최빈값이 있거나 유일한 최빈값이 없을 수도 있다.
    • ⚠ 질문에 맞는 요약 지표를 선택하라. 높은 외란치는 평균을 중간값 위로 끌어올릴 수 있다. 평균은 여전히 관측당 총합을 측정하지만, 중간값은 순서상 중앙을 locating한다. 둘 다 '대부분의 관측치가 그 값과 같다'고 말하지는 않는다.

    해설 예제. 그룹 크기가 합산 평균을 결정함

    두 그룹이 서로 다른 총합을 제공함: 10 명의 학습자(평균 60)와 20 명의 학습자(평균 75).

    已知: 그룹 A는 10개의 점수로 평균 60이고, 그룹 B는 20개로 평균 75입니다. 합치기 전에 각 그룹의 총점을 복원하십시오.

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

    60과 75의 무가중 평균은 67.5가 되지만, 이는 불균등한 개수를 무시하는 것이다.

    연습지 2.6는 단계별로 난이도가 높아지는 문제와 풀이 해설을 포함합니다.

    English 한국어
    mean/miːn/ 평균
    median/ˈmiːdiːən/ 중位数Unless median.
    mode/məʊd/ 최빈값
    2.7

    통계: 변동성 지표

    Syllabus

    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.

    출처: Cambridge International syllabus

    • **범위(range)**는 최대값에서 최소값을 뺀 값——전체 스펙트럼이므로, 외란치가 크게 변화시킬 수 있다.
    • **사분위 범위(IQR)**는 명시된 사분위 법칙에 따라 중앙 반분을 포괄하는 차이 $Q_3-Q_1$입니다. 상자 그림은 사분위와 중앙값을 보여주며, 수염(whisker) 규칙을 명시해야 합니다.
    • **표준 편차(standard deviation)**는 제곱偏差를 사용하여 퍼짐을 요약한다. 완전한 집단을 위해 사용: $\sigma=\sqrt{\sum(x-\mu)^2/n}$. 표본 추정치는 명시된 규칙 하에 $s=\sqrt{\sum(x-\bar x)^2/(n-1)}$을 사용한다. 분산(s variance)은 제곱 단위를 가지며, 표준 편차는 원래 단위를 가진다.
    • 표준 편차가 클수록 퍼짐이 넓다. 두 데이터 세트는 같은 평균을 공유하면서도 완전히 다른 상황을 나타낼 수 있다.

    해설 예제. 모든 관측치를 읽기보다 오수 요약(five-number summary)을 읽음

    최소값부터 최대값까지 상자 그래프는 최소 2, 첫 번째 사분위 4, 중간값 6, 세 번째 사분위 8, 최대 12를 표시함.

    알고 있는 오수 요약: 최소 2, $Q_1=4$, 중간값 6, $Q_3=8$, 최대 12. 전체 스펙트럼과 중앙 반 스펙트럼은 다르다.

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

    상자는 모든 원본 값, 표본 크기, 혹은 퍼짐의 원인을 식별하지 못한다.

    연습지 2.7는 단계별로 난이도가 높아지는 문제와 풀이 해설을 포함합니다.

    연습지 2.7의 경우, 사분위는 전체 중간값을 제외하고 순서 나열된 반列表의 중간값이다. 수염은 최소값과 최대값까지 이어진다. 이 규칙을 소프트웨어 보간 사분위나 외란치 울타리 수염과 혼동하지 마라.

    English 한국어
    range/reɪndʒ/ 범위Unless range.
    interquartile range (IQR)/ˌɪntəˈkwɔːtaɪl reɪndʒ/ 사분위 범위(IQR)
    box plot/bɒks plɒt/ 상자 그림
    standard deviation/ˈstændəd ˌdiːvɪˈeɪʃn/ 표준편차
    2.8

    분포: 정규 분포

    Syllabus

    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.

    출처: Cambridge International syllabus

    • **정규 분포(normal distribution)**는 평균과 양의 표준 편차로 지정된 연속 대칭 종 모양 모델이다. 이 두 요약 정보만으로 관측 데이터가 이 모델을 따른다고 보장할 수는 없다.
    • 정규 모델 하에서의 근사 경험칙은 다음과 같다: 값의 약 68%는 평균으로부터 표준 편차 1개 이내, 95% 표준 편차 2개 이내, 99.7%는 3개 이내에 위치한다.
    • **z-점수(z-score)**는 어떤 값이 평균으로부터 표준 편차 몇 개 떨어져 있는지를 알려준다: $z = \dfrac{x - \mu}{\sigma}$, 이것이 서로 다른 척도를 비교 가능하게 만드는 이유이다.

    정규 모델을 위해 누적 면적 $\Phi(z)=P(Z\leq z)$과 대칭 $\Phi(-z)=1-\Phi(z)$을 사용한다. 구간은 두 누적 경계값을 빼서 구한다. 기대 빈도는 표본 크기 × 확률이며, 반드시 정수가 아닐 수 있다; 관측 빈도는 정수이다.

    해설 예제. 임계값은 표준화된 값 이상의 영역임

    표준正規 곡선은 z=1을 표시하고 그 이후의 우측 꼬리 영역을 음영 처리함.

    알고 있음: 임의의 측정 모델에서 $\mu=50$과 $\sigma=10$을 가짐. 60보다 큰 값에 대해 임계값을 표준화하라.

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

    제공된 누적 면적의 보완(complement)을 사용하라.

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

    이는 모델 확률이지, 유한 표본에서 정확히 15.87%가 60을 초과한다는 보증이 아니다.

    연습지 2.8는 단계별로 난이도가 높아지는 문제와 풀이 해설을 포함합니다.

    English 한국어
    complement/ˈkɒmplɪmənt/ 보 event
    Counting techniques/ˈkaʊntɪŋ tekˈniːks/ 카운팅 기법
    permutation/ˌpɜːmjuːˈteɪʃn/ 배열
    combination/ˌkɒmbɪˈneɪʃn/ 조합
    normal distribution/ˈnɔːml ˌdɪstrɪˈbjuːʃn/ 표준정규분포
    z-score/zed skɔː/ z 점수
  • 3

    GAC016 수학 III: 미적분 및 고등 응용

    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 한국어
    differentiation/ˌdɪfəˌrenʃɪˈeɪʃn/ 미분
    integration/ˌɪntɪˈɡreɪʃn/ integration
    3.1

    Differentiation

    Syllabus

    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.

    출처: Cambridge International syllabus

    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 한국어
    derivative/dɪˈrɪvətɪv/ 미분Unless derivative.
    gradient of the tangent/ˈɡreɪdɪənt ɒvðə ˈtændʒənt/ 접선의 기울기
    differentiation from first principles/ˌdɪfəˌrenʃɪˈeɪʃn frɒm fɜːst ˈprɪnsɪplz/ 정의에 의한 미분
    power rule/ˈpaʊə ruːl/ 거듭제곱 법칙
    product rule/ˈprɒdʌkt ruːl/ 곱의 법칙
    quotient rule/ˈkwəʊʃənt ruːl/ 비율의 법칙
    chain rule/tʃeɪn ruːl/ 연쇄 법칙
    Stationary points/ˈsteɪʃənəri pɔɪnts/ 정지점
    second derivative/ˈsekənd dɪˈrɪvətɪv/ 이계 도함수Unless second derivative.
    maximum/ˈmæksɪməm/ 최대
    minimum/ˈmɪnɪməm/ 최소값일 때의 거리입니다.
    3.2

    Integration

    Syllabus

    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.

    출처: Cambridge International syllabus

    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 한국어
    fundamental theorem of calculus/ˌfʌndəˈmentl ˈθɪərəm ɒv ˈkælkjʊləs/ 미적분학의 기본 정리
    indefinite integral/ɪnˈdefɪnət ˈɪntɪɡrəl/ 부정적분
    constant of integration/ˈkɒnstənt ɒv ˌɪntɪˈɡreɪʃn/ 적분 상수
    definite integral/ˈdefɪnət ˈɪntɪɡrəl/ 정적분
    Integration by substitution/ˌɪntɪˈɡreɪʃn baɪ ˌsʌbstɪˈtjuːʃn/ 치환적 적분(Integration by substitution)
    3.3

    Advanced applications

    Syllabus

    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.

    출처: Cambridge International syllabus

    • 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 한국어
    Optimisation/ˌɒptɪmaɪˈzeɪʃn/ 최적화(Optimisation)
    Rates of change/reɪts ɒv tʃeɪndʒ/ 변화율
    Area between two curves/ˈeərɪə bɪˈtwiːn tuː kɜːvz/ 두 곡선 사이의 면적
    Marginal cost and revenue/ˈmɑːdʒɪnl kɒst ænd ˈrevənjuː/ 한계 비용 및 수익
  • 4

    GAC024 이산 수학

    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

    Syllabus

    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.

    출처: Cambridge International syllabus

    • 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 한국어
    set/set/ set
    Union/ˈjuːnɪən/ Union
    intersection/ˌɪntəˈsekʃn/ intersection
    set complement/set ˈkɒmplɪmənt/ set complement
    subset/ˈsʌbset/ subset
    relation/rɪˈleɪʃn/ relation
    function/ˈfʌŋkʃn/ function
    Venn diagram/ven ˈdaɪəɡræm/ Venn diagram
    inclusion-exclusion principle/ɪnˈkluːʒn eksˈkluːʒn ˈprɪnsɪpl/ inclusion-exclusion principle
    4.2

    Counting systems

    Syllabus

    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.

    출처: Cambridge International syllabus

    • 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 한국어
    number base/ˈnʌmbə beɪs/ number base
    Decimal/ˈdesɪml/ Decimal
    binary/ˈbaɪnəri/ binary
    hexadecimal/ˌheksəˈdesɪml/ hexadecimal
    place value/pleɪs ˈvæljuː/ place value
    4.3

    Binary applications

    Syllabus

    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.

    출처: Cambridge International syllabus

    • 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 한국어
    Binary arithmetic/ˈbaɪnəri əˈrɪθmətɪk/ Binary arithmetic
    bit/bɪt/ bit
    byte/baɪt/ byte
    Boolean algebra/ˈbuːlɪən ˈældʒɪbrə/ Boolean algebra
    truth table/truːθ ˈteɪbl/ truth table
    Logic gates/ˈlɒdʒɪk ɡeɪts/ Logic gates
    logic circuit/ˈlɒdʒɪk ˈsɜːkɪt/ logic circuit
    4.4

    Algorithms

    Syllabus

    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.

    출처: Cambridge International syllabus

    • 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 한국어
    algorithm/ˈælɡərɪθəm/ algorithm
    flowchart/ˈfləʊtʃɑːt/ flowchart
    Pseudocode/ˈsuːdəʊkəʊd/ Pseudocode
    Tracing/ˈtreɪsɪŋ/ Tracing
    Efficiency/ɪˈfɪʃənsi/ Efficiency
    4.5

    Graphs and networks

    Syllabus

    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.

    출처: Cambridge International syllabus

    • 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 한국어
    graph/ɡræf/ graph
    vertices/ˈvɜːtɪsiːz/ vertices
    edges/ˈedʒɪz/ edges
    degree/dɪˈɡriː/ degree
    tree/triː/ tree
    shortest path/ˈʃɔːtɪst pæθ/ shortest path

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