A-Level 수학

A quadratic 二次式 is an expression of the form $ax^2 + bx + c$, where $a \neq 0$. The letters $a$, $b$, $c$ are the coefficients 系数 (the fixed numbers). Much of this section is about solving the equation $ax^2 + bx + c = 0$.

Completing the square

To complete the square 配方 means to write the quadratic in the form

This form is useful: it shows the vertex 顶点 (turning point) of the curve at $(-p,\ q)$, and it gives a quick way to solve the equation.

Worked example. Write $9x^2 - 36x + 8$ in the form $p(x + q)^2 + r$.

Take the factor $9$ out of the first two terms, then complete the square inside:

So $p = 9$, $q = -2$, $r = -28$.

The discriminant

The discriminant 判别式 of $ax^2 + bx + c$ is

It tells you how many real roots 实根 (real solutions) the equation $ax^2 + bx + c = 0$ has:

Discriminant	Roots
$b^2 - 4ac > 0$	two distinct 相异 real roots
$b^2 - 4ac = 0$	one repeated real root
$b^2 - 4ac < 0$	no real roots

Worked example. Find the values of the constant $k$ for which $3kx^2 + (k + 8)x + 3 = 0$ has two distinct real roots.

Here $a = 3k$, $b = k + 8$, $c = 3$. For two distinct real roots you need $b^2 - 4ac > 0$:

Factorise: $(k - 4)(k - 16) > 0$, so $k < 4$ or $k > 16$. You also need $a \neq 0$, so $k \neq 0$.

Quadratic equations and inequalities

To solve a quadratic equation 二次方程, factorise, complete the square, or use the formula

To solve a quadratic inequality 二次不等式 such as $(k - 4)(k - 16) > 0$, find the two roots, then decide which side of each root makes the statement true. A sketch of the parabola 抛物线 helps: the curve is above the $x$-axis (positive) outside the roots and below it (negative) between them.

Simultaneous equations

To solve a pair of simultaneous equations 联立方程 where one is linear and one is quadratic, use substitution 代入: rearrange the linear equation for one letter, then put that into the quadratic. This gives a single quadratic to solve.

Equations that are quadratic in disguise

Some equations are quadratic in some function of $x$. For example $x^4 - 5x^2 + 4 = 0$ is quadratic in $x^2$: let $u = x^2$, solve $u^2 - 5u + 4 = 0$, then go back to $x$. You will use this idea again in trigonometry.

A function 函数 is a rule that sends each input to exactly one output. Write it as $f(x)$. The composition of functions 复合函数 $fg(x)$ means apply $g$ first, then apply $f$ to the result.

• The domain 定义域 is the set of allowed inputs $x$.
• The range 值域 is the set of outputs the function actually produces.

A function is one-one 一一对应 if different inputs always give different outputs. (No output is repeated.) Only a one-one function has an inverse function 反函数 $f^{-1}$, which reverses the rule.

The composition 复合 of two functions means doing one after the other. $fg(x)$ means "do $g$ first, then $f$": $fg(x) = f(g(x))$. The composite function $fg$ exists only when the range of $g$ lies inside the domain of $f$.

Finding an inverse

To find $f^{-1}$: write $y = f(x)$, make $x$ the subject, then swap letters.

Worked example. The function $f(x) = (x + 3)^2 - 12$ is defined for $x \geqslant 0$. Find $f^{-1}(x)$.

Write $y = (x + 3)^2 - 12$ and solve for $x$:

You take the positive square root because $x \geqslant 0$ means $x + 3 \geqslant 3 > 0$. So

Graphs of inverses and transformations

The graph of $y = f^{-1}(x)$ is the reflection 反射 of $y = f(x)$ in the line $y = x$.

You should know these transformations 变换 of $y = f(x)$:

New equation	Effect on the graph
$y = f(x) + a$	translation 平移 up by $a$
$y = f(x + a)$	translation left by $a$
$y = a\,f(x)$	stretch 伸缩 in the $y$-direction, scale factor $a$
$y = f(ax)$	stretch in the $x$-direction, scale factor $\tfrac{1}{a}$

When two transformations are combined, the order can matter. State each one fully (type, direction, and amount).

Coordinate geometry 坐标几何 studies lines and circles using their equations.

Straight lines

The gradient 斜率 (steepness) of the line joining $(x_1, y_1)$ and $(x_2, y_2)$ is

You can write the equation of a straight line 直线方程 in any of these forms: Two lines are parallel 平行 when their gradients are equal, and perpendicular 垂直 (at right angles) when the product of their gradients is $-1$.

Circles

The circle 圆 with centre 圆心 $(a, b)$ and radius 半径 $r$ has equation

An expanded form like $x^2 + y^2 - 6x + 10y - 27 = 0$ is the same circle: complete the square in $x$ and in $y$ to find the centre and radius.

A tangent 切线 to a circle touches it at one point and is perpendicular to the radius at that point. This right-angle fact solves most circle problems.

Worked example. The points $P(1, 1)$ and $Q(7, 11)$ are the ends of a diameter 直径 of a circle. Find the equation of the circle.

The centre is the midpoint of $PQ$:

The radius is half the length of $PQ$: So the circle is $(x - 4)^2 + (y - 6)^2 = 34$.

Radians

A radian 弧度 is another way to measure angles. One radian is the angle at the centre of a circle that cuts off an arc 弧 equal in length to the radius. The link between radians and degrees 度 is

So to change degrees to radians, multiply by $\dfrac{\pi}{180}$; to change radians to degrees, multiply by $\dfrac{180}{\pi}$.

Arc length and sector area

For a sector 扇形 with radius $r$ and angle $\theta$ in radians:

A chord 弦 cuts the sector into a triangle and a segment 弓形. The segment area is the sector minus the triangle:

Worked example. A sector has centre $O$ and the angle at $O$ is $\tfrac{2}{3}\pi$ radians. Show that the segment cut off by the chord has area about $0.614 r^2$.

Graphs and exact values

You must know the shape of the graphs of the sine 正弦, cosine 余弦 and tangent function 正切 (written $\sin$, $\cos$, $\tan$). The sine and cosine graphs wave between $-1$ and $1$ and repeat every $360^\circ$ ($2\pi$). Learn these exact values:

$\theta$	$30^\circ$	$45^\circ$	$60^\circ$
$\sin\theta$	$\tfrac12$	$\tfrac{1}{\sqrt2}$	$\tfrac{\sqrt3}{2}$
$\cos\theta$	$\tfrac{\sqrt3}{2}$	$\tfrac{1}{\sqrt2}$	$\tfrac12$
$\tan\theta$	$\tfrac{1}{\sqrt3}$	$1$	$\sqrt3$

The notations $\sin^{-1}x$, $\cos^{-1}x$, $\tan^{-1}x$ mean the inverse angle (the principal value 主值).

Identities

An identity 恒等式 is true for every value of the angle. The two you must know are

Use them to rewrite an equation so it contains only one trig function.

Solving trigonometric equations

To solve a trigonometric equation 三角方程, first reduce it to one function, then find every solution in the given interval.

Worked example. Solve $6\sin\theta = 1 + \dfrac{2}{\sin\theta}$ for $-180^\circ < \theta < 180^\circ$.

Multiply through by $\sin\theta$ to clear the fraction. This makes a quadratic in $\sin\theta$:

So $\sin\theta = \tfrac23$ or $\sin\theta = -\tfrac12$.

• $\sin\theta = \tfrac23$: $\theta = 41.8^\circ$ or $\theta = 180^\circ - 41.8^\circ = 138.2^\circ$.
• $\sin\theta = -\tfrac12$: $\theta = -30^\circ$ or $\theta = -150^\circ$.

The four solutions are $\theta = -150^\circ,\ -30^\circ,\ 41.8^\circ,\ 138.2^\circ$.

The binomial expansion

For a positive integer $n$, the binomial expansion 二项展开式 is

where $\binom{n}{r} = \dfrac{n!}{r!\,(n - r)!}$ is a binomial coefficient 二项式系数.

Worked example. Find the first three terms, in ascending powers of $x$, of $(2 - px)^5$.

Arithmetic and geometric progressions

A progression 数列 (sequence) is a list of terms following a rule.

• An arithmetic progression 等差数列 (AP) adds a fixed common difference 公差 $d$ each step. The $n$th term 项 is $u_n = a + (n - 1)d$, and the sum of the first $n$ terms is $S_n = \tfrac{n}{2}\big(2a + (n - 1)d\big)$.
• A geometric progression 等比数列 (GP) multiplies by a fixed common ratio 公比 $r$ each step. The $n$th term is $u_n = ar^{\,n-1}$, and $S_n = \dfrac{a(1 - r^n)}{1 - r}$.

A GP is convergent — it converges 收敛 (settles to a limit) — when $|r| < 1$. Then it has a sum to infinity 无穷和

Worked example. The third term of a GP is $18$ and the sum of the first three terms is $26$. The common ratio is negative. Find the sum to infinity.

From $ar^2 = 18$ you get $a = \dfrac{18}{r^2}$. Put this into $a(1 + r + r^2) = 26$:

The ratio is negative, so $r = -\tfrac34$ and $a = \dfrac{18}{(3/4)^2} = 32$. Then

Differentiation 微分 finds the gradient of a curve 曲线斜率 at each point. The gradient is the limit 极限 of the gradients of shorter and shorter chords, called the derivative 导数.

The rules

Write the derivative as $f'(x)$ or $\dfrac{dy}{dx}$. The basic rule is

Differentiate sums term by term, and use the chain rule 链式法则 for a function inside a function: Differentiating again gives the second derivative 二阶导数 $f''(x)$ or $\dfrac{d^2y}{dx^2}$.

Using the derivative

• Tangent and normal. The gradient of the curve at a point is the gradient of the tangent there. The normal 法线 is perpendicular to the tangent, so its gradient is $-\dfrac{1}{\text{(tangent gradient)}}$.
• Increasing or decreasing. The function is an increasing function 增函数 where $\dfrac{dy}{dx} > 0$, and a decreasing function 减函数 where $\dfrac{dy}{dx} < 0$.
• Rate of change. A derivative is a rate of change 变化率. Linked rates use the chain rule, e.g. $\dfrac{dy}{dt} = \dfrac{dy}{dx}\cdot\dfrac{dx}{dt}$.

Stationary points

A stationary point 驻点 is where $\dfrac{dy}{dx} = 0$. Test its nature with the second derivative: $f''(x) > 0$ gives a minimum point 极小值点, and $f''(x) < 0$ gives a maximum point 极大值点. When $f''(x) = 0$ the test is inconclusive: the point may be a point of inflexion 拐点, where the curve changes concavity (the way it bends) – check the sign of $\dfrac{dy}{dx}$ just before and after to decide.

Worked example. The curve $y = 4x^{1/2} - x$ has a maximum point at $x = a$. Find $a$.

So $a = 4$.

Integration 积分 is the reverse of differentiation. Reversing the power rule gives

The $+\,C$ is the constant of integration 积分常数. If you know one point on the curve, substitute it to find $C$.

Definite integrals and area

A definite integral 定积分 has limits and gives a number:

The area of the region 区域 between a curve and the $x$-axis, from $x = p$ to $x = q$, is $\displaystyle\int_p^q y \, dx$. For the area between two curves, integrate (top curve $-$ bottom curve).

Worked example. The curve $y = 4x^{1/2} - x$ meets the $x$-axis again at $x = 16$. Find the area between the curve and the $x$-axis from $x = 0$ to $x = 4$.

An improper integral 广义积分 has an infinite limit or an endpoint where the integrand is undefined; evaluate it as a limit. For example,

and $\displaystyle\int_0^{1}x^{-1/2}\, dx=\left[2x^{1/2}\right]_0^{1}=2$, which is finite even though the integrand blows up at $x=0$.

Volume of revolution

When a region is turned all the way around an axis it sweeps out a solid. The volume of revolution 旋转体体积 about the $x$-axis is

and about the $y$-axis it is $V = \pi \displaystyle\int x^2 \, dy$. For example, the region under $y = \sqrt{x}$ from $x = 0$ to $x = 4$, turned about the $x$-axis, has volume $\pi\displaystyle\int_0^4 x\, dx = \pi\big[\tfrac{x^2}{2}\big]_0^4 = 8\pi$.

• Show every line of algebra — method marks are lost by jumping steps; use the discriminant $b^2 - 4ac$ to decide the number of real roots.
• Work in radians for circular measure (arc $= r\theta$, sector area $= \frac{1}{2}r^2\theta$) and for calculus of trig functions.
• For differentiation, set $\frac{dy}{dx} = 0$ for stationary points and use the second derivative to classify them.
• For integration, add $+ c$ to an indefinite integral and use limits for area; an area below the axis gives a negative integral.

Two new tools join the algebra from Pure 2.

Partial fractions

A single fraction with a factorised bottom can be split into a sum of simpler fractions. This is called writing it in partial fractions 部分分式, and it makes a rational function 有理函数 (a fraction of polynomials) easy to integrate or expand. Match the form to the bottom:

First check the top is lower degree than the bottom. If it is not ("top-heavy"), divide the polynomial first: dividing gives a quotient 商 plus a remainder 余数 over the original bottom, and you split only that remaining proper fraction.

One fraction with a factorised bottom splits into a sum of simpler fractions.

Worked example. Express $\dfrac{x+4}{(x+1)(x-2)}$ in partial fractions.

Write $\dfrac{x+4}{(x+1)(x-2)} = \dfrac{A}{x+1} + \dfrac{B}{x-2}$, so $x + 4 = A(x-2) + B(x+1)$. Put $x = 2$: $6 = 3B$, so $B = 2$. Put $x = -1$: $3 = -3A$, so $A = -1$. Hence

The binomial expansion for a rational power

The binomial expansion 二项展开式 also works when the power is a fraction or is negative, as long as $|x| < 1$:

For example $(1 + x)^{1/2} = 1 + \tfrac12 x - \tfrac18 x^2 + \cdots$ for $|x| < 1$.

Subtopics 3.2, 3.3 and 3.6 are the same skills you met in Pure 2: the laws of logarithms with $e^x$ and $\ln x$; the identities $\sec^2\theta \equiv 1 + \tan^2\theta$ and $\csc^2\theta \equiv 1 + \cot^2\theta$, the compound- and double-angle formulae, and the $R$-form of $a\sin\theta + b\cos\theta$; and solving an equation numerically by a sign change 变号 and an iterative formula 迭代公式 $x_{n+1} = F(x_n)$. Use them exactly as before.

The three new trig functions are the reciprocals 倒数 of the familiar ones: the secant 正割 $\sec\theta = \dfrac{1}{\cos\theta}$, the cosecant 余割 $\csc\theta = \dfrac{1}{\sin\theta}$, and the cotangent 余切 $\cot\theta = \dfrac{1}{\tan\theta} = \dfrac{\cos\theta}{\sin\theta}$. (Memory aid: match the third letter — sec goes with cosine.) The two Pythagorean identities above come straight from dividing $\sin^2\theta + \cos^2\theta \equiv 1$ by $\cos^2\theta$ or $\sin^2\theta$. For a numerical method, an iteration $x_{n+1}=F(x_n)$ converges 收敛 to the root when successive values get closer together.

The methods are those of Pure 2 (the product, quotient and chain rules, with parametric and implicit curves). One derivative is added — the inverse tangent 反正切:

Worked example. Differentiate $y = \tan^{-1}(3x)$.

Use the chain rule with the result above: $\dfrac{dy}{dx} = \dfrac{1}{1 + (3x)^2} \times 3 = \dfrac{3}{1 + 9x^2}.$

Pure 3 adds several powerful methods.

• A new standard integral: $\displaystyle\int \frac{1}{x^2 + a^2}\,dx = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$.
• Partial fractions: split a rational function first, then integrate each piece as a logarithm.
• The pattern $\dfrac{k\,f'(x)}{f(x)}$: this integrates to $k\ln|f(x)| + C$. For example $\displaystyle\int \frac{2x}{x^2 + 1}\,dx = \ln(x^2 + 1) + C$.
• Integration by parts 分部积分, used for a product: $\displaystyle\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx$.
• Integration by substitution 换元积分: a given change of variable turns a hard integral into an easy one.

Worked example. Find $\displaystyle\int x\cos x\,dx$.

Use integration by parts with $u = x$ and $\dfrac{dv}{dx} = \cos x$, so $\dfrac{du}{dx} = 1$ and $v = \sin x$:

A vector 向量 has both size and direction. Write it as a column, or as $x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$, or as $\overrightarrow{AB}$.

• The magnitude 模长 (length) of $\mathbf{v} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ is $|\mathbf{v}| = \sqrt{x^2 + y^2 + z^2}$.
• A unit vector 单位向量 has magnitude $1$; divide a vector by its magnitude to make one.
• A position vector 位置向量 gives a point's place from the origin; a displacement vector 位移向量 $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ goes from one point to another. Multiplying by a scalar 标量 (a plain number) stretches a vector.

Lines and the scalar product

A straight line through point $\mathbf{a}$ in direction $\mathbf{b}$ has vector equation $\mathbf{r} = \mathbf{a} + t\mathbf{b}$. Two lines may be parallel 平行, may intersect 相交 at a point, or may be skew lines 异面直线 (not parallel and never meeting).

The scalar product 数量积 (dot product) of $\mathbf{a}$ and $\mathbf{b}$ is

where $\theta$ is the angle between them. So $\mathbf{a}\cdot\mathbf{b} = 0$ means the vectors are perpendicular.

Worked example. Find the angle between $\mathbf{a} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$ and $\mathbf{b} = 2\mathbf{i} + 2\mathbf{j} + \mathbf{k}$.

So $\cos\theta = \dfrac{8}{3\times 3} = \dfrac{8}{9}$, giving $\theta = 27.3^\circ$.

A differential equation 微分方程 links a quantity to its rate of change. To solve a first-order equation whose variables are separable 可分离变量, put all the $y$ terms on one side and all the $x$ terms on the other, then integrate both sides. This gives the general solution 通解, which contains a constant. An initial condition 初始条件 (a known value) fixes the constant and gives the particular solution 特解.

Worked example. Solve $\dfrac{dy}{dx} = xy$, given that $y = 1$ when $x = 0$.

Separate the variables and integrate:

Using $y = 1$ at $x = 0$ gives $A = 1$, so $y = e^{x^2/2}$.

A complex number 复数 has the form $z = x + iy$, where $i^2 = -1$. Here $x$ is the real part 实部 and $y$ is the imaginary part 虚部. This $x + iy$ is the Cartesian form 直角坐标形式. Two complex numbers are equal only when their real parts match and their imaginary parts match.

• The conjugate 共轭 of $z = x + iy$ is $z^* = x - iy$. For a polynomial with real coefficients, any non-real roots come in conjugate pairs.
• The modulus 模 is $|z| = \sqrt{x^2 + y^2}$ (its distance from the origin) and the argument 辐角 is the angle the point makes, measured from the positive real axis.
• You can plot $z$ as a point on an Argand diagram 阿干图 (the complex plane).
• The polar form 极坐标形式 is $z = r(\cos\theta + i\sin\theta) = re^{i\theta}$, where $r = |z|$ and $\theta$ is the argument. Multiplying multiplies the moduli and adds the arguments.
• The loci of points satisfying a condition on $z$ are drawn on the Argand diagram — e.g. $|z - a| = r$ is a circle, $|z - a| = |z - b|$ a perpendicular bisector, and $\arg(z - a) = \theta$ a half-line. The square roots of a complex number come from solving $w^2 = z$.

To divide, multiply top and bottom by the conjugate of the bottom.

Worked example. Write $\dfrac{3 + i}{1 - i}$ in the form $x + iy$.

An equation or inequality in $z$ describes a locus 轨迹 (a path or region) on the Argand diagram. For example $|z - a| = r$ is a circle of radius $r$ centred at $a$.

• Split a rational function into partial fractions before integrating or expanding.
• Choose the right integration technique (substitution, by parts, or partial fractions) from the form of the integrand.
• Use the scalar (dot) product for the angle between vectors and to test for perpendicularity; write a line as $\mathbf{r} = \mathbf{a} + t\mathbf{b}$.
• Give complex numbers in the form asked for (Cartesian or modulus-argument) and show them on an Argand diagram.

A force 力 is a push or a pull. It is a vector, so it has size and direction. Because it is a vector, you can split a force into components 分量 (usually horizontal and vertical), and you can add several forces into one resultant 合力.

A particle is in equilibrium 平衡 when the forces are balanced: the vector sum of the forces is zero. In practice this means the components in any direction add to zero.

Friction

When two surfaces touch, the contact force 接触力 between them has two parts: the normal reaction 法向反作用力 $R$, at right angles to the surface, and the friction 摩擦力 $F$, along the surface, which opposes sliding. A "smooth" surface is a model with no friction.

Friction can only grow up to a maximum. At that maximum the body is in limiting equilibrium 极限平衡, about to slip, and the friction is limiting friction 最大静摩擦力. The maximum is set by the coefficient of friction 摩擦系数 $\mu$:

By Newton's third law 牛顿第三定律, the two surfaces push on each other with equal and opposite forces.

Worked example. A block of weight $20\text{ N}$ rests on a rough horizontal table with coefficient of friction $\mu = 0.4$. Find the largest horizontal force that can be applied before the block slides.

The table's normal reaction balances the weight, so $R = 20\text{ N}$. The block is on the point of slipping when friction reaches its maximum $F = \mu R = 0.4 \times 20 = 8\text{ N}$. In equilibrium the applied force equals the friction, so the largest force it can resist is $8\text{ N}$.

Distance 距离 and speed 速率 are scalars (size only). Displacement 位移, velocity 速度 and acceleration 加速度 are vectors (size and direction).

On a velocity-time graph 速度时间图, the area under the graph is the displacement and the gradient is the acceleration (a negative acceleration is a deceleration 减速度). On a displacement-time graph, the gradient is the velocity. More generally, differentiate with respect to time to go from displacement to velocity to acceleration, and integrate to go back.

For motion with constant acceleration 匀加速, use these formulae (the "suvat" equations):

Worked example. A car starts from rest and accelerates at $2.5\ \text{m s}^{-2}$ for $4\ \text{s}$. Find its speed and the distance travelled.

The linear momentum 动量 of a body is $\text{mass} \times \text{velocity}$. It is a vector. In a direct collision of two bodies, the total momentum is unchanged. This is the conservation of linear momentum 动量守恒:

Worked example. A body of mass $2\ \text{kg}$ moving at $3\ \text{m s}^{-1}$ hits a stationary body of mass $1\ \text{kg}$, and they stick together. Find their common speed afterwards.

Newton's laws of motion 牛顿运动定律 connect force and acceleration. The key one is: resultant force $=$ mass 质量 $\times$ acceleration,

The weight 重力 of a body is the force of gravity on it: $W = mg$. Forces in a problem may include weight, friction, tension 张力 in a string, the thrust 推力 (push) in a rod, and air resistance 空气阻力.

For motion on an inclined plane 斜面, split each force into a part along the slope and a part at right angles to it, then apply $F = ma$ along the slope. For connected particles 连接质点 (joined by a string), apply $F = ma$ to each body, or to the whole system.

Worked example. A block of mass $12\ \text{kg}$ is pulled up a rough plane by a rope parallel to the slope. The plane is at $20^\circ$ to the horizontal, the coefficient of friction is $0.4$, and the acceleration is $2\ \text{m s}^{-2}$. Find the tension in the rope.

The normal reaction is $R = mg\cos 20^\circ = 120\cos 20^\circ = 112.8\ \text{N}$, so the friction is $F = \mu R = 0.4 \times 112.8 = 45.1\ \text{N}$. Along the slope, $T - mg\sin 20^\circ - F = ma$:

The work done 功 by a constant force is the scalar product of force and displacement — the force times the distance moved in the direction of the force: $W = Fd\cos\theta$, where $\theta$ is the angle between the force and the motion. Work is measured in joules (J).

Energy comes in forms you can calculate:

• Kinetic energy 动能 (energy of movement): $\text{KE} = \tfrac12 mv^2$.
• Gravitational potential energy 重力势能 (energy of height): $\text{PE} = mgh$.

The work done by the outside forces equals the change in the total energy. When no friction acts, the total energy stays the same — the conservation of energy 能量守恒.

Power 功率 is the rate of doing work. For a force pulling in the direction of motion, $P = Fv$ (power $=$ force $\times$ velocity). Power is measured in watts (W).

Worked example. A car engine works at $12\ \text{kW}$ while the car moves at $20\ \text{m s}^{-1}$ on a level road. Find the driving force.

• Draw a clear force diagram and resolve into perpendicular components; for equilibrium, each direction sums to zero.
• Use SUVAT only for constant acceleration and keep a consistent positive direction.
• Apply $F = ma$ along the direction of motion, including friction ($F = \mu R$) on a rough surface.
• State your assumptions (light inextensible string, smooth pulley, particle) — they are often worth a mark.

Choose a diagram that suits the data. You should be able to draw and read:

• a stem-and-leaf diagram 茎叶图 (keeps the original values and shows the shape); two data sets are compared with a back-to-back 背靠背 version – a shared central stem, one set's leaves increasing to the left and the other's to the right, so you can compare their medians and spreads at a glance;
• a box-and-whisker plot 箱线图 (shows the lowest value, the three quartiles, and the highest value);
• a histogram 直方图 (for grouped data, where the area of each bar shows the frequency);
• a cumulative frequency 累积频数 graph (running totals, used to estimate the median and quartiles).

Averages and spread

A measure of central tendency 集中趋势 is a single "middle" value:

• the mean 平均数 $\bar{x} = \dfrac{\sum x}{n}$ (the average);
• the median 中位数 (the middle value when the data is in order);
• the mode 众数 (the most common value).

A measure of variation 离散程度 shows how spread out the data is:

• the range (of data) 极差 (highest $-$ lowest);
• the interquartile range 四分位距 (upper quartile $-$ lower quartile);
• the standard deviation 标准差 $\sigma = \sqrt{\dfrac{\sum x^2}{n} - \bar{x}^2}$.

You often work from the totals $\sum x$ and $\sum x^2$. The square of the standard deviation is the variance 方差.

Worked example. For $10$ values, $\sum x = 50$ and $\sum x^2 = 300$. Find the mean and standard deviation.

Coding 编码 makes big numbers easier. Replace each value by $t=x-a$ for a convenient assumed mean 假定平均数 $a$. Then $\bar{x}=a+\bar{t}$, while the standard deviation is unchanged (shifting every value does not spread the data). So from the coded totals $\sum(x-a)$ and $\sum(x-a)^2$ you get $\bar x$ and $\sigma$ directly, and two data sets can be compared through their coded totals.

A permutation 排列 is an arrangement where order matters; a combination 组合 is a selection where order does not matter. The numbers are

To arrange objects in a line when some are repeated, divide by the factorial of each repeat count.

Worked example. How many different arrangements are there of the letters of the word NEEDLESS?

There are $8$ letters, with E repeated $3$ times and S repeated $2$ times:

Find a probability by counting equally likely outcomes, or by using permutations and combinations. Combine probabilities with these rules:

• addition for "or": $P(A \cup B) = P(A) + P(B) - P(A \cap B)$;
• multiplication for "and" when events are independent: $P(A \cap B) = P(A)\,P(B)$.

Two events are mutually exclusive events 互斥事件 if they cannot both happen, and independent events 独立事件 if one happening does not change the chance of the other. To test independence, check whether $P(A \cap B) = P(A)\times P(B)$. A conditional probability 条件概率 is the chance of $A$ given that $B$ has happened: $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$.

Worked example. Events have $P(A) = 0.5$, $P(B) = 0.4$ and $P(A \cap B) = 0.2$. Are $A$ and $B$ independent?

Test: $P(A)\times P(B) = 0.5 \times 0.4 = 0.2 = P(A \cap B)$. The values are equal, so $A$ and $B$ are independent.

A discrete random variable 离散型随机变量 $X$ takes separate values, each with a probability. List them in a probability distribution table 概率分布表; the probabilities must add to $1$. Then the expectation 期望 (mean) and variance are

Two special models:

• the binomial distribution 二项分布 $X \sim B(n, p)$, for the number of successes in $n$ independent trials: $P(X = r) = \binom{n}{r}p^r(1-p)^{n-r}$, with $E(X) = np$ and $\mathrm{Var}(X) = np(1-p)$;
• the geometric distribution 几何分布, for the trial on which the first success happens: $P(X = r) = (1-p)^{r-1}p$, with $E(X) = \dfrac{1}{p}$.

Worked example. $X \sim B(10, 0.3)$. Find $P(X = 2)$ and $E(X)$.

The normal distribution 正态分布 models a continuous random variable 连续型随机变量 with a symmetric bell shape. Write $X \sim N(\mu, \sigma^2)$, where $\mu$ is the mean and $\sigma$ is the standard deviation. To use the tables, by standardisation 标准化 convert to the variable $Z \sim N(0, 1)$:

Then $P(X < x) = P\!\left(Z < \dfrac{x - \mu}{\sigma}\right)$, which you read from the normal table $\Phi$.

The normal distribution is also a good approximation 近似 to the binomial when $n$ is large. Because you replace a discrete variable by a continuous one, apply a continuity correction 连续性校正 (adjust by $0.5$).

Worked example. Bags of rice have mass $X \sim N(\mu, 0.14^2)$. Given that $P(X < 1.48) = 0.22$, find $\mu$.

From the table, $P(Z < z) = 0.22$ gives $z = -0.772$. So

• Decide whether order matters: permutations ($^nP_r$) when it does, combinations ($^nC_r$) when it does not.
• For a discrete random variable, check the probabilities sum to $1$ and use $E(X) = \sum x\,P(X=x)$.
• For the normal distribution, standardise with $z = (x - \mu)/\sigma$, sketch and shade, then read the table.
• Apply a continuity correction when approximating a discrete variable by the normal.

The Poisson distribution 泊松分布 $X \sim \mathrm{Po}(\lambda)$ models the number of random events in a fixed interval, when events happen at a steady average rate $\lambda$:

For a Poisson variable the mean and the variance are both equal to $\lambda$. The Poisson distribution is a good approximation 近似 to the binomial distribution when $n$ is large and $p$ is small. The normal distribution (with continuity correction) approximates the Poisson when $\lambda$ is large.

Worked example. $X \sim \mathrm{Po}(3)$. Find $P(X = 2)$.

When you change a variable by a linear rule, the expectation 期望 (mean) and variance 方差 follow these rules:

For two independent variables $X$ and $Y$: Two useful facts: if $X$ has a normal distribution 正态分布 then so does $aX + b$; and the sum of independent Poisson variables is again Poisson.

Worked example. $X$ has mean $5$ and variance $4$. Find $E(3X - 1)$ and $\mathrm{Var}(3X - 1)$.

A continuous random variable 连续型随机变量 can take any value in a range. Its probabilities come from a probability density function 概率密度函数 $f(x)$, with two key properties:

A probability is the area under $f$, and the mean is found by integration:

The cumulative distribution function 累积分布函数 is $F(x) = P(X \leqslant x) = \int_{-\infty}^{x} f(t)\,dt$; the median 中位数 solves $F(m) = 0.5$, and other percentiles 百分位数 solve $F(x) = p$.

Worked example. A continuous variable has $f(x) = \tfrac12 x$ for $0 \leqslant x \leqslant 2$ (and $0$ elsewhere). Find $E(X)$.

The variance uses the same idea, $\mathrm{Var}(X)=\displaystyle\int_{-\infty}^{\infty}x^2 f(x)\,dx-\big(E(X)\big)^2$. For the same $f(x)=\tfrac12 x$ on $[0,2]$: $\displaystyle\int_0^2 x^2\cdot\tfrac12 x\,dx=\left[\tfrac{x^4}{8}\right]_0^2=2$, so $\mathrm{Var}(X)=2-\left(\tfrac43\right)^2=2-\tfrac{16}{9}=\tfrac{2}{9}$.

A sample 样本 is a small group chosen from the whole population 总体. A random sample needs randomness 随机性, so that every member has a fair chance of being chosen.

Some methods are unsatisfactory because they are biased 有偏: sampling only volunteers, or the first 20 people to arrive, over-represents certain kinds of people. A genuinely random sample uses random numbers – number every member of the population, then draw numbers (from a table or a generator) to decide who is in the sample.

The sample mean $\bar{X}$ is itself a random variable, with

By the Central Limit Theorem 中心极限定理, for a large sample $\bar{X}$ is approximately normal, whatever the shape of the population.

From a sample you can find unbiased estimates 无偏估计 of the population mean and variance. A confidence interval 置信区间 gives a range that probably contains the true mean. When the population is normal with known $\sigma$ (or the sample is large), a $95\%$ interval is

You can also find a confidence interval for a population proportion 总体比例 from a large sample.

Worked example. A sample of $n = 64$ has mean $\bar{x} = 50$, from a population with $\sigma = 8$. Find a $95\%$ confidence interval for the population mean.

A hypothesis test 假设检验 uses sample data to judge a claim. You set up two statements: the null hypothesis 原假设 $H_0$ (the claim being tested, usually "no change") and the alternative hypothesis 备择假设 $H_1$ (what you suspect instead). The test is one-tailed 单尾 if $H_1$ points one way (e.g. $\mu > 50$) and two-tailed 双尾 if it allows both ways ($\mu \neq 50$).

You fix a significance level 显著性水平 (often $5\%$), work out a test statistic 检验统计量 from the data, and see whether it lands in the rejection region 拒绝域 (also called the critical region); if it does you reject $H_0$, otherwise the statistic is in the acceptance region.

Two mistakes are possible: a Type I error 第一类错误 is rejecting $H_0$ when it is actually true; a Type II error 第二类错误 is accepting $H_0$ when it is actually false. You can find their probabilities from the rejection region: $P(\text{Type I})=P(\text{statistic in the rejection region}\mid H_0)$ – this equals the significance level – and $P(\text{Type II})=P(\text{statistic in the acceptance region}\mid H_1$ true for a stated value$)$, computed from the binomial, Poisson, or normal distribution.

Worked example. A population is claimed to have mean $50$, with $\sigma = 8$. A sample of $n = 64$ gives $\bar{x} = 52$. Test at the $5\%$ level whether the mean has changed.

$H_0\!: \mu = 50$ and $H_1\!: \mu \neq 50$ (two-tailed). The test statistic is

The critical value at $5\%$ (two-tailed) is $1.96$. Since $2 > 1.96$, you reject $H_0$: there is evidence the mean has changed.

Worked example (a binomial test). A coin is claimed fair but suspected of landing heads too rarely: $H_0\!:p=0.5$, $H_1\!:p<0.5$. In $n=30$ tosses you see $X=9$ heads. Under $H_0$, $X\sim B(30,0.5)$, so the one-tailed tail probability is

Since $0.021<0.05$, reject $H_0$: the coin does seem biased against heads. (For large $n$ the binomial is approximated by a normal; a Poisson test works the same way for rare events. And here, if the rule is "reject when $X\leqslant 9$", then $P(\text{Type I})=P(X\leqslant 9\mid p=0.5)\approx0.021$.)

• Use the Poisson distribution for rare, random, independent events; its mean equals its variance ($= \lambda$).
• When combining independent random variables, variances add (they never subtract).
• For a hypothesis test, state $H_0$ and $H_1$, the significance level, the test statistic, and a conclusion in context.
• For a confidence interval, use the correct $z$ (or $t$) value and interpret it in words.