IGCSE 数学

You must know the words for the different kinds of number. Examiners give marks for using them correctly.

Counting numbers and integers

• natural numbers 自然数 — the counting numbers $1, 2, 3, 4, \dots$
• integers 整数 — whole numbers, positive, negative or zero: $\dots, -2, -1, 0, 1, 2, \dots$

Factors and multiples

• A factor 因数 of a number divides into it exactly, leaving no remainder 余数. The factors of $18$ are $1, 2, 3, 6, 9, 18$.
• A multiple 倍数 of a number is that number times an integer. Multiples of $6$ are $6, 12, 18, 24, \dots$
• A common factor 公因数 of two numbers is a factor of both.
• A common multiple 公倍数 of two numbers is a multiple of both.

Prime, square and cube numbers

• A prime number 质数 has exactly two factors: $1$ and itself. The first primes are $2, 3, 5, 7, 11, 13, \dots$ Note that $1$ is not prime.
• A square number 平方数 is a whole number times itself: $1, 4, 9, 16, 25, \dots$
• A cube number 立方数 uses a whole number three times: $1, 8, 27, 64, \dots$

Rational, irrational and reciprocal

• A rational number 有理数 can be written as a fraction 分数 $\frac{a}{b}$ of two integers. Examples: $\frac{3}{4}$, $5$, $0.7$.
• An irrational number 无理数 cannot be written this way. Examples: $\pi$ and $\sqrt{2}$.
• The reciprocal 倒数 of a number is $1$ divided by that number. The reciprocal of $4$ is $\frac{1}{4}$; the reciprocal of $0.25$ is $4$; the reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$.

Prime factors, HCF and LCM

Every integer above $1$ is prime, or can be written as a product 乘积 of prime numbers. To write a number as a product of its prime factors 质因数, keep dividing by the smallest prime that fits.

Worked example. Write $72$ as a product of its prime factors.

The highest common factor (HCF) 最大公因数 of two numbers is the largest factor they share. The lowest common multiple (LCM) 最小公倍数 is the smallest multiple they share. Prime factors give a quick method.

Worked example. Find the HCF and LCM of $72$ and $120$.

First write each as a product of primes:

• HCF: take the lowest power of each prime that appears in both: $2^{3} \times 3 = 24$.
• LCM: take the highest power of every prime that appears: $2^{3} \times 3^{2} \times 5 = 360$.

A set 集合 is a collection of objects. Each object in the set is an element 元素 of the set. You should know this notation:

A Venn diagram 维恩图 draws each set as a circle inside a rectangle (the universal set). Core uses two sets; Extended may use three.

Worked example. $\mathscr{E} = \{1,2,3,4,5,6,7,8,9,10\}$, $A = \{\text{even numbers}\}$, $B = \{\text{multiples of } 3\}$.

• $A = \{2,4,6,8,10\}$ and $B = \{3,6,9\}$.
• $A \cap B = \{6\}$ — the only number in both.
• $A \cup B = \{2,3,4,6,8,9,10\}$ — the numbers in either set.
• $n(A \cup B) = 7$.

You may also see a set written as a rule, e.g. $C = \{x : 1 \leqslant x \leqslant 5\}$ means "all values $x$ such that $1 \leqslant x \leqslant 5$".

• A power 幂 (also called an index 指数, plural indices) shows how many times to multiply a number by itself: $2^{5} = 2 \times 2 \times 2 \times 2 \times 2 = 32$.
• A square root 平方根 of a number gives that number when squared: $\sqrt{169} = 13$ because $13^{2} = 169$.
• A cube root 立方根 works the same way for cubes: $\sqrt[3]{8} = 2$ because $2^{3} = 8$.

You should be able to recall the squares from $1^2$ to $15^2$ (and their roots), and the cubes of $1, 2, 3, 4, 5$ and $10$.

Worked example. Work out $5^{2} \times \sqrt[3]{8}$.

When you multiply or divide powers of the same base 底数, use these rules:

Some special powers:

(Fractional powers like $a^{\frac{m}{n}}$ are Extended.)

Worked examples.

• $2^{-3} \times 2^{4} = 2^{-3+4} = 2^{1} = 2.$
• $(2^{3})^{2} = 2^{6} = 64.$
• $2^{3} \div 2^{4} = 2^{3-4} = 2^{-1} = \dfrac{1}{2}.$
• $7^{-2} = \dfrac{1}{7^{2}} = \dfrac{1}{49}.$
• $81^{\frac{1}{2}} = \sqrt{81} = 9.$
• $8^{-\frac{2}{3}} = \dfrac{1}{8^{\frac{2}{3}}} = \dfrac{1}{\left(\sqrt[3]{8}\right)^{2}} = \dfrac{1}{2^{2}} = \dfrac{1}{4}.$

Standard form 科学记数法 writes a number as $A \times 10^{n}$, where $1 \leqslant A < 10$ and $n$ is an integer. It is used for very large or very small numbers.

To convert, count how many places the decimal point moves:

• $4\,500\,000 = 4.5 \times 10^{6}$ — the point moves $6$ places left, so the power is positive.
• $0.00072 = 7.2 \times 10^{-4}$ — the point moves $4$ places right, so the power is negative.

Worked example. Work out $(3 \times 10^{5}) \times (2 \times 10^{-2})$.

Multiply the front numbers and add the powers:

A surd 根式 is a root that is irrational, such as $\sqrt{5}$. Leave it in exact form instead of rounding. Two useful rules:

Simplify a surd by taking out the largest square factor.

Worked example. Simplify $\sqrt{20}$ and $\sqrt{200} - \sqrt{32}$.

To rationalise the denominator 分母有理化 means to remove a surd from the bottom of a fraction (the denominator 分母). Multiply the top and bottom by a value that clears the surd.

Worked example. Rationalise $\dfrac{10}{\sqrt{5}}$ and $\dfrac{1}{-1+\sqrt{3}}$.

A fraction has a numerator 分子 (the top) and a denominator (the bottom).

• proper fraction 真分数: numerator smaller than denominator, e.g. $\frac{3}{4}$.
• improper fraction 假分数: numerator the same or larger, e.g. $\frac{7}{4}$.
• mixed number 带分数: a whole number plus a fraction, e.g. $1\frac{3}{4}$.

Change between improper and mixed: $\frac{7}{4} = 1\frac{3}{4}$ because $7 \div 4 = 1$ remainder $3$.

A decimal 小数 uses place value after a point. A percentage 百分比 means "out of $100$", so $37\% = \frac{37}{100} = 0.37$.

Converting between forms

Write a fraction in its simplest form 最简形式 by dividing the top and bottom by their HCF: $\frac{18}{24} = \frac{3}{4}$ (both divided by $6$).

Recurring decimals (Extended)

A recurring decimal 循环小数 repeats the same digits forever. Dots mark the repeating part: $0.1\dot{7} = 0.1777\ldots$ and $0.\dot{1}2\dot{3} = 0.123123\ldots$

To turn a recurring decimal into a fraction, multiply so the repeating parts line up, then subtract.

Worked example. Write $0.1\dot{7}$ as a fraction.

Let $x = 0.1777\ldots$ Only the $7$ repeats, so use $10x$ and $100x$:

Order of operations

Work in this order — the order of operations 运算顺序: brackets first, then indices (powers and roots), then multiply and divide (left to right), then add and subtract (left to right).

Worked example. Work out $-6 \times -3 + 7 \times 2$.

Do the multiplications first: $-6 \times -3 = 18$ and $7 \times 2 = 14$. Then add: $18 + 14 = 32$.

Negative numbers

• Adding a negative: $5 + (-3) = 5 - 3 = 2$.
• Subtracting a negative: $5 - (-3) = 5 + 3 = 8$.
• Multiplying or dividing: same signs give a positive; different signs give a negative. So $-6 \times -3 = 18$ but $-12 \div 4 = -3$.

A change in temperature from $-5\,{}^{\circ}\text{C}$ to $3\,{}^{\circ}\text{C}$ is a rise of $8\,{}^{\circ}\text{C}$.

Calculating with fractions

• Multiply: multiply the tops, multiply the bottoms: $\frac{2}{3} \times \frac{4}{5} = \frac{8}{15}$.
• Divide: multiply by the reciprocal of the second fraction: $\frac{2}{3} \div \frac{4}{5} = \frac{2}{3} \times \frac{5}{4} = \frac{10}{12} = \frac{5}{6}$.
• Add or subtract: use a common denominator (the LCM of the bottoms).

Worked example. Work out $1\frac{7}{15} - \frac{4}{5}$, giving the answer in its simplest form.

Change the mixed number to an improper fraction, then use denominator $15$:

Use these symbols to compare numbers by magnitude 大小 (size):

To put a mixed list in order, change every value to a decimal first.

Worked example. Put $34\%$, $\frac{1}{3}$ and $\frac{3}{10}$ in order, smallest first.

As decimals: $34\% = 0.34$, $\frac{1}{3} = 0.333\ldots$, $\frac{3}{10} = 0.3$. So the order is

Find a percentage of an amount. $15\%$ of $\$80 = 0.15 \times 80 = \$12$.

Write one amount as a percentage of another. A score of $18$ out of $25$ is $\frac{18}{25} \times 100\% = 72\%$.

Percentage increase or decrease uses

Worked example. A price rises from $\$40$ to $\$50$. Find the percentage increase.

The change is $\$10$, so $\frac{10}{40} \times 100\% = 25\%$ increase.

A quick way is a multiplier 乘数. To increase by $15\%$, multiply by $1.15$; to decrease by $15\%$, multiply by $0.85$.

Simple and compound interest

Interest 利息 is money paid for borrowing or for saving. The principal 本金 is the starting amount.

• Simple interest 单利 pays the same amount each year, worked out on the principal only:
  $$I = \frac{P \times r \times t}{100},$$
  where $P$ is the principal, $r$ is the rate per year (as a percentage) and $t$ is the number of years.
• Compound interest 复利 adds the interest on each year, so the next year earns interest on a larger total:
  $$\text{final value} = P\left(1 + \frac{r}{100}\right)^{t}.$$

Worked example. Find the value of $\$500$ saved at $4\%$ compound interest for $3$ years.

Reverse percentages (Extended)

A reverse percentage 逆百分比 problem gives the amount after a change and asks for the original. To solve it, divide by the multiplier — do not just take the percentage off.

Worked example. A coat costs $\$60$ after a $20\%$ increase. Find the original price.

$\$60$ is $120\%$ of the original, so the original price is $60 \div 1.2 = \$50$.

When a quantity changes by the same percentage in each time period, it shows exponential growth 指数增长 (it gets bigger) or exponential decay 指数衰减 (it gets smaller). Use the compound formula. Depreciation 折旧, where something like a car loses value each year, is decay.

Worked example. A car worth $\$20\,000$ loses $15\%$ of its value each year. Find its value after $4$ years.

The multiplier is $0.85$, so

A ratio 比 compares quantities, written like $a:b$. Simplify it like a fraction by dividing by the HCF: $20:30:40 = 2:3:4$.

Dividing in a ratio. Share $\$48$ in the ratio $3:5$.

The total number of parts is $3 + 5 = 8$. One part is $48 \div 8 = \$6$. So the shares are $3 \times 6 = \$18$ and $5 \times 6 = \$30$.

Proportion 比例 means two ratios are equal. Use it for recipes, map scales 比例尺 and finding best value.

Worked example. $3$ pens cost $\$1.80$. Find the cost of $7$ pens.

One pen costs $1.80 \div 3 = \$0.60$. So $7$ pens cost $7 \times 0.60 = \$4.20$.

A rate 比率 compares two quantities measured in different units, such as price per kilogram, or distance per hour.

Average speed 平均速度 uses

Worked example. A cyclist travels $45\text{ km}$ in $3$ hours $45$ minutes. Find the average speed.

First change the time to hours: $3$ h $45$ min $= 3.75$ h. Then

Other rates work the same way. Density 密度 is found from mass 质量 and volume 体积:

Other examples are flow rate, fuel consumption and population density 人口密度. If a rate needs a special formula (such as pressure 压强), the question will give it to you.

Rounding

• decimal places (d.p.) 小数位: digits counted after the point. $3.14159$ to $2$ d.p. is $3.14$.
• significant figures (s.f.) 有效数字: digits counted from the first non-zero digit. $5764$ to $1$ s.f. is $6000$; $0.004067$ to $2$ s.f. is $0.0041$.

Rule: look at the next digit. If it is $5$ or more, round up; if it is less, round down.

Estimation

To estimate 估算 an answer, round every number to $1$ s.f., then calculate.

Worked example. Estimate $\dfrac{41.3}{9.79 \times 0.765}$.

A rounded value could really be anything that rounds to it. The smallest possible value is the lower bound 下界; the largest is the upper bound 上界. For a value rounded to the nearest unit, the bounds lie half a unit on each side.

Worked example. A height $h$ is $635\text{ m}$, correct to the nearest metre. Give the bounds.

So the lower bound is $634.5\text{ m}$ and the upper bound is $635.5\text{ m}$.

Bounds in calculations (Extended)

Combine the bounds to get the bound you want.

Worked example. A rectangle is $8\text{ cm}$ by $5\text{ cm}$, each side to the nearest cm. Find the largest possible area 面积.

Use the upper bounds of both sides: $8.5 \times 5.5 = 46.75\text{ cm}^{2}$. (The smallest area uses the lower bounds: $7.5 \times 4.5 = 33.75\text{ cm}^{2}$.)

For a divided quantity such as $\text{speed} = \dfrac{\text{distance}}{\text{time}}$, the largest speed comes from the largest distance divided by the smallest time.

• $60$ seconds $= 1$ minute, $60$ minutes $= 1$ hour, $24$ hours $= 1$ day, and $1$ year $= 365$ days.
• The 24-hour clock writes a time as four digits: $3.15$ p.m. is $15\,15$.

Worked example. A film starts at $19\,35$ and lasts $70$ minutes. Find the time it finishes.

$70$ min $= 1$ h $10$ min. Adding $1$ hour gives $20\,35$; adding $10$ minutes gives $20\,45$.

For timetables and time zone 时区 problems, add or subtract the time difference between the places.

Work with money as ordinary decimals, but give answers to $2$ d.p. (so $\$4.8$ is written $\$4.80$).

Currency conversion. Use the exchange rate 汇率 as a multiplier.

Worked example. The exchange rate is $\$1 = €0.92$. Convert $\$150$ to euros, and convert $€138$ back to dollars.

In algebra 代数 we use letters to stand for numbers. A letter whose value can change is a variable 变量. To substitute 代入 means to put a number in place of a letter.

Worked example. Find the value of $3x^{2} - 2y$ when $x = 4$ and $y = 5$.

A term 项 is a single part of an expression 表达式, such as $5a$ or $-9b$. Like terms 同类项 have exactly the same letters; you may add or subtract them. The number in front of the letter is the coefficient 系数.

Worked example. Simplify $2a^{2} + 3ab - 1 + 5a^{2} - 9ab + 4$.

Collect like terms: $2a^{2} + 5a^{2} = 7a^{2}$, $3ab - 9ab = -6ab$, $-1 + 4 = 3$. So the answer is

To expand 展开 means to multiply out brackets 括号. Multiply every term inside by the term outside; for two brackets, multiply every term in the first by every term in the second.

Worked examples.

For three brackets (Extended), expand two first, then multiply by the third:

To factorise 因式分解 is the opposite of expanding: write the expression as a product of brackets. Always take out the common factor 公因式 first.

Worked example. $9x^{2} + 15xy = 3x(3x + 5y)$, because $3x$ divides both terms.

The following patterns are Extended.

Grouping (four terms): take a common factor from each pair.

Difference of two squares 平方差: $a^{2} - b^{2} = (a + b)(a - b)$.

Perfect square 完全平方: $a^{2} + 2ab + b^{2} = (a + b)^{2}$.

Quadratic 二次 expressions $ax^{2} + bx + c$: find two numbers that multiply to $a \times c$ and add to $b$, then split the middle term.

Worked example. Factorise $2x^{2} + 7x + 3$.

Here $a \times c = 6$ and $b = 7$. The numbers $1$ and $6$ work. Split and group:

For $ax^{3} + bx^{2} + cx$, take out the common $x$ first: $2x^{3} + 7x^{2} + 3x = x(2x^{2} + 7x + 3) = x(x + 3)(2x + 1)$.

Completing the square 配方法 rewrites $x^{2} + bx + c$ as $(x + p)^{2} + q$. Take half of the $x$-coefficient, square it, then balance.

Worked example. Write $x^{2} + 6x + 1$ in completed-square form.

Half of $6$ is $3$, and $3^{2} = 9$:

When the $x^{2}$ has a coefficient, take it out of the first two terms first:

An algebraic fraction 分式 has algebra on the top or bottom. Add and subtract using a common denominator; multiply and divide as with ordinary fractions.

Worked examples.

To simplify a rational expression 有理式, factorise the top and bottom, then cancel common brackets.

The laws of indices 指数 work with letters too: $a^{m} \times a^{n} = a^{m+n}$, $a^{m} \div a^{n} = a^{m-n}$, and $(a^{m})^{n} = a^{mn}$.

Worked examples.

You can also solve simple index equations by writing both sides with the same base 底数.

Worked example. Solve $2^{x} = 32$. Since $32 = 2^{5}$, you get $x = 5$.

An equation 方程 says two expressions are equal. To solve a linear 一次 equation, do the same operation to both sides until the unknown 未知数 is alone.

Worked example. Solve $5 - 2x = 3(x + 7)$.

Fractional equations (Extended)

A fractional equation 分式方程 has the unknown in a denominator. Multiply both sides by the denominator to clear it.

Worked example. Solve $\dfrac{x}{2x + 1} = 4$.

Simultaneous equations

Simultaneous equations 联立方程 are two equations solved together. For two linear equations, add or subtract to remove one letter.

Worked example. Solve $2x + y = 7$ and $3x - y = 8$.

Adding removes $y$: $5x = 15$, so $x = 3$. Then $y = 7 - 2(3) = 1$.

For one linear and one quadratic equation (Extended), substitute the linear into the curve.

Worked example. Solve $y = x + 2$ and $y = x^{2}$.

so $x = 2$ (giving $y = 4$) or $x = -1$ (giving $y = 1$).

Solving quadratic equations (Extended)

There are three methods.

• By factorising: $x^{2} + 5x + 6 = 0 \Rightarrow (x + 2)(x + 3) = 0 \Rightarrow x = -2$ or $x = -3$.
• By completing the square: $x^{2} + 6x + 1 = 0 \Rightarrow (x + 3)^{2} = 8 \Rightarrow x + 3 = \pm 2\sqrt{2} \Rightarrow x = -3 \pm 2\sqrt{2}$.
• By the quadratic formula 求根公式, which is given in the exam:
  $$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}.$$

Worked example (formula). Solve $2x^{2} + 3x - 1 = 0$. Here $a = 2$, $b = 3$, $c = -1$:

Changing the subject

To change the subject 公式变形 of a formula means to rearrange it so a chosen letter is alone on one side.

Worked example. Make $r$ the subject of $A = \pi r^{2}$ (Extended, because of the power).

When the letter appears twice (Extended), collect those terms and factorise. To make $x$ the subject of $y = \dfrac{x + 1}{x - 1}$:

An inequality 不等式 uses $<$, $>$, $\leqslant$ or $\geqslant$. Solve it like an equation, but reverse the sign if you multiply or divide by a negative number.

Worked example. Solve $-3 \leqslant 3x - 2 < 7$.

Add $2$ to all parts, then divide by $3$:

On a number line 数轴, use an open circle for $<$ or $>$ (value not included) and a closed circle for $\leqslant$ or $\geqslant$ (value included).

Regions (Extended)

An inequality in two letters describes a region 区域 of the graph. Draw the boundary line (broken for $<$ or $>$, solid for $\leqslant$ or $\geqslant$) and shade the unwanted side. You may also be asked to list the inequalities that define a given region.

A sequence 数列 is a list of numbers that follow a rule. The term-to-term rule 递推规则 tells you how to get the next term from the one before.

To find the rule for the term in position $n$ (the $n$th term), look at how the terms change.

Linear sequence (the terms go up by the same amount each time). The difference is the multiple of $n$.

Worked example. Find the $n$th term of $2, 5, 8, 11, \dots$

The terms go up by $3$, so start with $3n$. Since $3 \times 1 = 3$ but the first term is $2$, subtract $1$: the $n$th term is $3n - 1$.

Quadratic sequence (the differences themselves change by the same amount). The second difference equals $2 \times$ the coefficient of $n^{2}$.

Worked example. Find the $n$th term of $2, 5, 10, 17, \dots$

First differences are $3, 5, 7$; the second difference is $2$, so the $n^{2}$ part is $1n^{2}$. Subtracting $n^{2}$ ($1, 4, 9, 16$) from the sequence leaves $1, 1, 1, 1$. So the $n$th term is $n^{2} + 1$.

A cubic 三次 sequence such as $1, 8, 27, 64, \dots$ has $n$th term $n^{3}$. An exponential sequence 指数数列 such as $2, 6, 18, 54, \dots$ multiplies by a fixed number each time; here the $n$th term is $2 \times 3^{\,n-1}$.

Two quantities are in direct proportion 正比例 if one is always a fixed multiple of the other: $y \propto x$ means $y = kx$, where $k$ is a constant 常数. They are in inverse proportion 反比例 if one rises as the other falls: $y \propto \dfrac{1}{x}$ means $y = \dfrac{k}{x}$. The symbol $\propto$ is read "is proportional to". You can also have proportion to a square, square root, cube or cube root.

Worked example. $y$ is in direct proportion to $x$, and $y = 12$ when $x = 3$. Find $y$ when $x = 7$.

First find $k$: $12 = k \times 3$, so $k = 4$ and $y = 4x$. Then $y = 4 \times 7 = 28$.

The gradient 斜率 (steepness) of a graph shows a rate of change 变化率.

• A travel graph 行程图 (distance–time graph) has gradient equal to speed; a flat part means the object is not moving.
• A conversion graph 换算图 is a straight line used to change between two units (for example, miles and kilometres).

Speed–time graphs (Extended)

On a speed–time graph the gradient is the acceleration 加速度 (or deceleration 减速度 if the speed falls), and the area 面积 under the graph is the distance 距离 travelled.

Worked example. A car speeds up from rest to $20\text{ m/s}$ in $8\text{ s}$, then stays at $20\text{ m/s}$ for $12\text{ s}$. Find the acceleration and the total distance.

Acceleration $= \dfrac{20}{8} = 2.5\text{ m/s}^{2}$. The distance is the area: a triangle plus a rectangle,

To draw a graph, make a table of values 数值表: choose values of $x$, work out $y$, then plot the coordinates 坐标 and join them with a smooth curve.

The points where a graph crosses the $x$-axis (the horizontal axis 坐标轴) are the roots 根 — the solutions of $y = 0$.

You can solve an equation by reading a graph. The intersection point 交点 of a line and a curve gives the solution of the two equations together.

For exponential growth 指数增长 and exponential decay 指数衰减, the graph of $y = a\,b^{x} + c$ rises (or falls) faster and faster and flattens towards a horizontal line.

A quick sketch should show the right shape and the key features: where it crosses the axes, any symmetry 对称, and any line the curve gets close to.

For a parabola, completing the square gives the turning point 转折点 (the lowest or highest point). For example $y = (x + 3)^{2} - 8$ has its turning point at $(-3, -8)$.

An asymptote 渐近线 is a line that the curve gets closer and closer to but never touches — for example, the $x$-axis for $y = \dfrac{a}{x}$, or the line $y = b$ for $y = a\,r^{x} + b$.

Differentiation 微分 finds the gradient of a curve at any point. You can estimate it by drawing a tangent 切线 (a line that just touches the curve) and measuring its gradient.

The exact rule: if $y = ax^{n}$, then the derivative 导数 is

Differentiate a sum term by term.

Worked example. If $y = x^{3} + 2x^{2} - 5x$, then $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2} + 4x - 5$.

A stationary point 驻点 (turning point) is where the gradient is zero, so set $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0$.

Worked example. Find the turning point of $y = x^{2} - 6x + 5$.

The turning point is $(3, -4)$. To decide whether a turning point is a maximum 最大值 or a minimum 最小值, check the sign of the gradient on each side, or use the second derivative (positive means a minimum).

A function 函数 turns each input into one output. We write $f(x)$, for example $f(x) = 3x - 5$, so $f(2) = 1$. The set of allowed inputs is the domain 定义域; the set of possible outputs is the range 值域.

The inverse function 反函数 $f^{-1}(x)$ undoes the function. To find it, write $y = f(x)$, swap the roles, and make $x$ the subject.

Worked example. Find the inverse of $f(x) = 3x - 5$.

A composite function 复合函数 applies one function after another: $gf(x)$ means "do $f$ first, then $g$".

Worked example. If $f(x) = 2x$ and $g(x) = x + 3$, then

A point on a graph is described by its coordinates 坐标, sometimes called Cartesian coordinates, written $(x, y)$. The first number is the across value and the second is the up value.

• The two number lines are the axes 坐标轴: the horizontal 水平 $x$-axis and the vertical 竖直 $y$-axis.
• They cross at the origin 原点, the point $(0, 0)$.
• The axes split the grid into four quadrants 象限.

So the point $(3, -2)$ is found by going $3$ to the right and $2$ down.

Most straight lines can be written as

where $m$ is the gradient 斜率 (steepness) and $c$ is the intercept 截距 — the $y$-value where the line crosses the $y$-axis.

• A line like $x = k$ (for example $x = 3$) is vertical.
• A line like $y = k$ (for example $y = 3$) is horizontal.

A line may also be given as $ax + by = c$. Rearrange it into $y = mx + c$ to read off the gradient and intercept.

Worked example. Find the gradient and $y$-intercept of $5x + 4y = 8$.

So the gradient is $-\tfrac{5}{4}$ and the $y$-intercept is $2$.

To draw $y = mx + c$, the quickest way is:

1. Mark the intercept $c$ on the $y$-axis.
2. Use the gradient to step to more points (for $m = 3$, go $1$ right and $3$ up).
3. Join the points with a straight line.

You can also make a small table of values 数值表: choose two or three $x$-values, work out $y$, and plot the points.

A line segment 线段 is the straight piece between two points.

To find the length 长度 of the segment between $(x_1, y_1)$ and $(x_2, y_2)$, use Pythagoras' theorem 勾股定理 on the horizontal and vertical gaps:

To find the midpoint 中点 (the point exactly in the middle), average the coordinates:

Worked example. Find the length and midpoint of the segment from $(1, 2)$ to $(4, 6)$.

Parallel 平行 lines never meet, so they have the same gradient.

Worked example. Find the equation of the line parallel to $y = 4x - 1$ that passes through $(1, -3)$.

The gradient is also $4$. Put the point in:

so the line is $y = 4x - 7$.

Two lines are perpendicular 垂直 if they meet at a right angle 直角. Their gradients multiply to $-1$:

In words: flip the fraction and change the sign.

Worked example. Find the gradient of a line perpendicular to $2y = 3x + 1$.

Rearrange: $y = \tfrac{3}{2}x + \tfrac{1}{2}$, so the gradient is $\tfrac{3}{2}$. The perpendicular gradient is $-\tfrac{2}{3}$.

Perpendicular bisector

The perpendicular bisector 垂直平分线 of a segment cuts it in half at a right angle. To find its equation: get the midpoint, then use the perpendicular gradient through that midpoint.

Worked example. Find the perpendicular bisector of the segment joining $(-3, 8)$ and $(9, -2)$.

• Midpoint: $\left( \frac{-3 + 9}{2}, \frac{8 + (-2)}{2} \right) = (3, 3)$.
• Gradient of the segment: $\frac{-2 - 8}{9 - (-3)} = \frac{-10}{12} = -\tfrac{5}{6}$.
• Perpendicular gradient: $\frac{6}{5}$.

Through $(3, 3)$: $\; 3 = \tfrac{6}{5}(3) + c \Rightarrow c = 3 - \tfrac{18}{5} = -\tfrac{3}{5}$. So the bisector is

A corner where two lines meet is a vertex 顶点. Two lines are parallel 平行 if they never meet, and perpendicular 垂直 if they meet at a right angle. Angles are named by their size:

We name an angle with three letters, e.g. angle $ABC$ is the angle at $B$.

Learn these basic facts. Each one is a valid reason in the exam.

• Angles at a point add up to $360^{\circ}$.
• Angles on a straight line add up to $180^{\circ}$.
• Vertically opposite angles 对顶角 (made by two crossing lines) are equal.

Worked example. Three angles on a straight line are $x$, $50^{\circ}$ and $70^{\circ}$. Find $x$.

When a line crosses two parallel lines:

• Corresponding angles 同位角 (in matching positions, an "F" shape) are equal.
• Alternate angles 内错角 (opposite sides of the crossing line, a "Z" shape) are equal.
• Co-interior angles 同旁内角 (a "C" shape) add up to $180^{\circ}$; we say they are supplementary 互补.

Worked example. A straight line crosses two parallel lines. One angle is $110^{\circ}$. The co-interior angle $y$ satisfies $110 + y = 180$, so $y = 70^{\circ}$.

A triangle 三角形 has three sides and angles that add up to $180^{\circ}$.

Worked example. A triangle has angles $x$, $2x$ and $90^{\circ}$. Find $x$.

A quadrilateral 四边形 has four sides and angles that add up to $360^{\circ}$.

A polygon 多边形 is a shape with straight sides. A regular polygon 正多边形 has all sides and all angles equal.

For a polygon with $n$ sides:

The interior angle 内角 and the exterior angle 外角 at each corner add up to $180^{\circ}$.

Worked example. Find each interior angle of a regular hexagon.

The exterior angle is $\dfrac{360^{\circ}}{6} = 60^{\circ}$, so each interior angle is $180^{\circ} - 60^{\circ} = 120^{\circ}$.

• A shape has line symmetry 轴对称 if a mirror line splits it into two matching halves.
• A shape has rotational symmetry 旋转对称 if it fits onto itself as you turn it. The order is how many times it fits in one full turn.

For solids (Extended), a flat slice that splits the solid into mirror halves is a plane of symmetry 对称面, and a line you can spin it around is an axis of symmetry 对称轴.

Use these to find unknown angles, always giving the reason.

For both levels:

• The angle in a semicircle is $90^{\circ}$.
• The angle between a tangent and a radius is $90^{\circ}$.

Extended (theorems I):

• The angle at the centre is twice the angle at the circumference (standing on the same arc).
• Angles in the same segment are equal.
• Opposite angles of a cyclic quadrilateral 圆内接四边形 add up to $180^{\circ}$.
• The alternate segment theorem 弦切角定理: the angle between a tangent and a chord equals the angle in the other segment.

Extended (theorems II): equal chords are the same distance from the centre; the perpendicular bisector of a chord passes through the centre; two tangents from the same outside point are equal in length.

Worked example. A, B, C are on a circle. The angle at the circumference $ABC$ is $40^{\circ}$. Find the angle $AOC$ at the centre $O$.

The angle at the centre is twice the angle at the circumference: $2 \times 40^{\circ} = 80^{\circ}$.

Two shapes are similar 相似 if one is an enlargement of the other: same angles, and all sides multiplied by the same scale factor 比例因子 $k$. (Shapes that are exactly the same size and shape are congruent 全等.)

For similar shapes and solids:

Worked example. Two similar solids have lengths in the ratio $2 : 3$. The smaller has volume $40\text{ cm}^{3}$. Find the volume of the larger.

The volume ratio is $2^{3} : 3^{3} = 8 : 27$. So the larger volume is $40 \times \dfrac{27}{8} = 135\text{ cm}^{3}$.

A bearing 方位角 gives a direction as a three-figure angle, measured clockwise 顺时针 from north, from $000^{\circ}$ to $360^{\circ}$. So due east is $090^{\circ}$ and due south is $180^{\circ}$.

Worked example. The bearing of $B$ from $A$ is $025^{\circ}$. Find the bearing of $A$ from $B$.

The return (back) bearing differs by $180^{\circ}$: $025^{\circ} + 180^{\circ} = 205^{\circ}$.

• To construct 作图 a triangle from three given sides, draw the base with a ruler, then use a pair of compasses 圆规 to mark each other side as an arc. Leave the construction arcs showing.
• A net 展开图 is a flat shape that folds up into a solid. You can use a net to work out surface areas.
• A scale drawing 比例图 shows a real object smaller (or larger) by a fixed scale, such as $1\text{ cm}$ to $5\text{ m}$.

Common solids 几何体 and their parts:

A flat side of a solid is a face 面, two faces meet at an edge 棱, and the whole outside is its surface 表面.

質量（g, kg）、長さ（mm, cm, m, km）、面積、体積、容量（ml, リットル — 容器の内側の空間）にはメートル法の単位を使う。

単位変換を行う際は、平方と立方に注意すること：

• 長さ：$1\text{ m} = 100\text{ cm}$。
• 面積：$1\text{ m}^{2} = 100^{2} = 10\,000\text{ cm}^{2}$。
• 体積：$1\text{ m}^{3} = 100^{3} = 1\,000\,000\text{ cm}^{3}$ 。
• 容量：$1\text{ litre} = 1000\text{ cm}^{3}$、したがって$1\text{ m}^{3} = 1000$リットル。

** worked example（例題）.** $3\text{ m}^{2}$を$\text{cm}^{2}$に変換します。

周長とは、図形の外周の全距離です。面積とは、その内部にある平面的な空間の広さです。ここで$b$は底辺、$h$は垂直な高さです。

** worked example（例題）.** 台形には平行な辺が$6\text{ cm}$と$10\text{ cm}$、高さが$4\text{ cm}$あります。面積を求めます。