A-Level Mathematics · ⁨מתמטיקה A-Level⁩

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.