📚 Edexcel A-Level Maths: Core Revision from PDF Joiner (4)-082 | Edexcel A-Level数学：PDF合并(4)-082核心复习
Welcome to this Edexcel A-Level Mathematics revision guide, drawing on common questions from PDF Joiner (4)-082. The focus is on core pure mathematics topics that frequently appear in exam papers: algebra and indices, quadratics and inequalities, coordinate geometry, sequences and series, calculus, exponentials and logarithms, proof, trigonometry and vectors. Each section pairs the key rule with the typical exam trap, so you can revise quickly and then test yourself on past-paper questions.
欢迎阅读本 Edexcel A-Level 数学复习指南，内容源自 PDF 合并(4)-082 中的常见题型。我们重点回顾考试中频繁出现的核心纯数学主题：代数与指数、二次函数与不等式、坐标几何、数列与级数、微积分、指数与对数、证明、三角与向量。每一节都把核心法则与常见失分点配对讲解，方便你快速复习，再用历年真题自测。
1. Algebraic Manipulation and Indices | 代数运算与指数
torising. Common errors include misapplying the rule for negative indices and forgetting that a⁰ = 1 for any non-zero a.
许多 A-Level 题目的第一步是化简代数表达式。你必须熟练掌握指数法则、去括号和因式分解。常见错误包括错误应用负指数法则，以及忘记任何非零 a 的 a⁰ = 1。
aᵐ × aⁿ = aᵐ⁺ⁿ | aᵐ ÷ aⁿ = aᵐ⁻ⁿ | (aᵐ)ⁿ = aᵐⁿ | a⁻ⁿ = 1/aⁿ | a⁰ = 1 | a1/n = ⁿ√a
When simplifying, always write the answer with positive indices unless otherwise stated. For example, 2x⁻³ should be written as 2/x³. Also rationalise denominators when surds appear, such as 1/√2 = √2 Remember that (2x)³ = 8x³, not 2x³ - the index applies to every factor inside the bracket.
化简时，除非另有说明，答案应使用正指数。例如 2x⁻³ 应写作 2/x³。出现根式时还需有理化分母，如 1/√2 = √2/2。注意 (2x)³ = 8x³ 而不是 2x³——括号内的每一个因子都要乘方。
o squares a² - b² = (a + b)(a - b), and finally try the standard quadratic form. Expanding (x + 3)(x - 5) gives x² - 2x - 15, and this check confirms your factorisation is correct.
因式分解是展开的逆过程，几乎出现在每个主题中。先提取公因式，再看平方差 a² - b² = (a + b)(a - b)，最后尝试标准二次形式。展开 (x + 3)(x - 5) 得 x² - 2x - 15，用这个检查可以确认因式分解是否正确。
- Index laws apply to products inside brackets, not to sums: (a + b)² ≠ a² + b².
- 指数法则作用于括号内的乘积，而不是和：(a + b)² ≠ a² + b²。
- Write fractional powers as roots when the question asks for an exact answer.
- 当题目要求精确答案时，把分数指数写成根式。
2. Quadratics, Discriminant and Inequalities | 二次函数、判别式与不等式
nd by the quadratic formula in general. The discriminant Δ = b² - 4ac tells you whether there are two distinct real roots (Δ > 0), one repeated root (Δ = 0) or no real roots (Δ < 0).
二次方程 ax² + bx + c = 0 的求解方法有：当根为有理数时用因式分解，需要顶点式时用配方法，一般情况下使用求根公式。判别式 Δ = b² - 4ac 可判断根的情况：两个不同实根（Δ > 0）、一个重根（Δ = 0）或无实根（Δ < 0）。
x = (-b ± √b² - 4ac) / 2a | Δ = b² - 4ac | a(x + p)² + q
or x² - 6x + 11, this becomes (x - 3)² + 2, so the vertex is (3, 2) and the minimum value is 2.
配方法把 ax² + bx + c 改写为 a(x + p)² + q，可直接得到最值点 (-p, q) 与对称轴 x = -p。对于 x² - 6x + 11，配方得 (x - 3)² + 2，所以顶点是 (3, 2)，最小值为 2。
nt of x² is positive the solution is x < -2 or x > 3. A common mistake is to write -2 < x < 3, which is the solution of the reversed inequality x² - x - 6 < 0.
二次不等式最好通过画抛物线求解。对于 x² - x - 6 > 0，临界值为 x = -2 与 x = 3，因为 x² 系数为正，解为 x < -2 或 x > 3。常见错误是写成 -2 < x < 3，那是反向不等式 x² - x - 6 < 0 的解。
- Always rearrange an inequality to compare with zero before finding critical values.
- 求临界值前，务必把不等式整理成与 0 比较的形式。
- Never divide an inequality by a variable or an expression whose sign is unknown.
- 切勿用符号未知的变量或表达式除不等式。
3. Simultaneous Equations, Factor Theorem and Polynomial Division | 联立方程、因式定理与多项式除法
Simultaneous equations with one linear and one quadratic are solved by substitution. Rearrange the linear equation to make one variable the subject, substitute into the quadratic, and solve the resulting equation. Both coordinates must then be found, and each pair of values must satisfy both original equations.
一个线性方程与一个二次方程的联立问题用代入法求解。先把线性方程整理成某一变量的表达式，代入二次方程并求解，随后必须求出两套坐标，且每一对数值都要同时满足原方程。
0. The remainder theorem says that when f(x) is divided by (x - k) the remainder is f(k). These two results are the fastest way to find unknown constants in exam questions about polynomials.
因式定理指出：当且仅当 f(k) = 0 时，(x - k) 是 f(x) 的因式。余数定理指出：f(x) 除以 (x - k) 的余数为 f(k)。在涉及未知常数的多项式考题中，这两条结论是最快的解法。
f(k) = 0 ⇔ (x - k) is a factor of f(x) | f(x) = (x - k)Q(x) + f(k)
ainder, so the degree check is easy: the quotient must be one degree lower than the dividend. For example, dividing 2x³ - 5x² + 3x - 4 by (x - 2) gives quotient 2x² - x + 1 with remainder -2.
多项式除法可用长除法或待定系数法。除完后应写成 f(x) = 除数 × 商 + 余数 的形式，便于检验次数：商的次数应比被除数低一次。例如把 2x³ - 5x² + 3x - 4 除以 (x - 2)，得商 2x² - x + 1，余数为 -2。
4. Surds, Rationalising and Exact Forms | 根式、有理化与精确形式
Surds appear whenever an exact answer is required. Simplify by extracting square factors: √50 = 5√2 and √72 = 6√2. Then combine like terms, since 5√2 + 3√2 = 8√2 but 5√2 + 5√3 cannot be simplified.
当题目要求精确答案时就会出现根式。先提出平方因子化简：√50 = 5√2，√72 = 6√2。然后合并同类项，因为 5√2 + 3√2 = 8√2，但 5√2 + 5√3 无法再化简。
To rationalise a denominator containing a surd, multiply top and bottom by the conjugate. For 1/(3 + √2), multiply by (3 - √2)/(3 - √2) to obtain (3 - √2)/7, because (3 + √2)(3 - √2) = 9 - 2 = 7. This technique also works for denominators such as 1/√2 = √2/2.
若分母含根式，用共轭式上下同乘。对于 1/(3 + √2)，乘以 (3 - √2)/(3 - √2) 得 (3 - √2)/7，因为 (3 + √2)(3 - √2) = 9 - 2 = 7。这个方法也适用于 1/√2 = √2/2 这类分母。
- Always give surd answers in simplest form with no fraction in the denominator.
- 根式答案必须化为最简形式，且分母中不含根号。
- When a question says 'show that', keep every line exact until the final statement.
- 遇到 "show that" 类题目，每一步都保持精确值，直到最后结论。
5. Functions, Domain, Range and Transformations | 函数、定义域、值域与图像变换
A function is a rule that maps each input to exactly one output. The domain is the set of allowed inputs, and the range is the set of outputs. The denominator of a fraction cannot be zero, and the expression under a square root cannot be negative, so these restrictions define the domain.
函数是把每个输入映射到唯一输出的规则。定义域是允许输入的范围，值域是输出的范围。分式的分母不能为零，根号内的表达式不能为负，这些限制就定义了定义域。
Reflections are given by y = -f(x) and y = f(-x), and stretches by y = af(x) and y = f(ax).
y = f(x) + a 的图像是 y = f(x) 沿 y 轴正方向平移 a 个单位，而 y = f(x + a) 是沿 x 方向平移 -a 个单位。反射由 y = -f(x) 与 y = f(-x) 给出，伸缩由 y = af(x) 与 y = f(ax) 给出。
| Transformation 变换 | Equation 方程 | Effect on x-coordinates 对横坐标的影响 |
| Translation right 向右平移 | y = f(x - a) | x → x + a |
| Stretch in x 横向伸缩 | y = f(ax) | x → x/a |
| Reflection in y-axis 关于 y 轴反射 | y = f(-x) | x → -x |
omain of f⁻¹ is the range of f, and the graphs of f and f⁻¹ are reflections of each other in the line y = x.
反函数 f⁻¹ 逆转映射，只有当 f 是一一函数时才存在。求法：设 y = f(x)，交换 x 与 y，再解出 y。f⁻¹ 的定义域是 f 的值域，f 与 f⁻¹ 的图像关于直线 y = x 对称。
6. Coordinate Geometry: Lines and Circles | 坐标几何：直线与圆
- x₁). Parallel lines have equal gradients, and perpendicular lines satisfy m₁ × m₂ = -1.
直线方程可写成 y = mx + c 或 y - y₁ = m(x - x₁)。两点间斜率为 m = (y₂ - y₁)/(x₂ - x₁)。平行直线斜率相等，垂直直线满足 m₁ × m₂ = -1。
y to recover the centre and radius. For x² + y² - 4x + 6y - 12 = 0, completing the square gives (x - 2)² + (y + 3)² = 25, so the centre is (2, -3) and the radius is 5.
圆心为 (a, b)、半径为 r 的圆方程为 (x - a)² + (y - b)² = r²。若方程以展开形式给出，对 x 与 y 分别配方即可还原圆心与半径。对于 x² + y² - 4x + 6y - 12 = 0，配方得 (x - 2)² + (y + 3)² = 25，圆心 (2, -3)，半径 5。
To show that a line is a tangent to a circle, substitute the line into the circle equation and show that the resulting quadratic has discriminant zero. To find where a chord meets a circle, solve the two equations simultaneously and give both intersection points.
要证明直线与圆相切，把直线代入圆方程，证明所得二次方程的判别式为零。求弦与圆的交点时，联立两方程求解并写出两个交点坐标。
7. Sequences, Series and Binomial Expansion | 数列、级数与二项展开
e has a common ratio r and nth term uₙ = arⁿ⁻¹, with sum Sₙ = a(1 - rⁿ)/(1 - r) for r ≠ 1.
等差数列有公差 d，第 n 项 uₙ = a + (n - 1)d，前 n 项和 Sₙ = n/2 [2a + (n - 1)d]。等比数列有公比 r，第 n 项 uₙ = arⁿ⁻¹，前 n 项和 Sₙ = a(1 - rⁿ)/(1 - r)（r ≠ 1）。
uₙ = a + (n - 1)d | Sₙ = n/2 (a + l) | uₙ = arⁿ⁻¹ | S∞ = a/(1 - r), |r| < 1
A geometric series converges only when |r| k for the smallest n such that Sₙ exceeds a given value, which requires setting up an inequality and solving it, usually with logarithms.
等比级数仅在 |r| < 1 时收敛，此时无穷和为 S∞ = a/(1 - r)。考题常要求使 Sₙ 超过某值的最小 n，需要建立不等式并求解，通常要用到对数。
The binomial expansion of (a + b)ⁿ uses the formula (a + b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ bʳ tive integer n the expansion is finite with n + 1 terms; the coefficients come from Pascal's triangle or from the nCr button on your calculator. The general term ᵏCᵣ aᵏ⁻ʳ bʳ lets you find a single coefficient without expanding everything.
(a + b)ⁿ 的二项展开用公式 (a + b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ bʳ。当 n 为正整数时，展开式有限，共 n + 1 项；系数来自帕斯卡三角形或计算器上的 nCr 键。通项 ᵏCᵣ aᵏ⁻ʳ bʳ 可让你在不完全展开的情况下求出某一项的系数。
8. Differentiation: Rules and Applications | 微分：法则与应用
is power rule, together with the constant multiple rule and the sum rule, covers every polynomial. Rewrite roots and fractions as powers before differentiating: √x = x1/2 differentiates to ½x-1/2.
微分给出斜率函数。对于 y = xⁿ，dy/dx = nxⁿ⁻¹。这条幂函数法则加上常数倍法则与和法则，可以处理所有多项式。求导前先把根式与分数写成指数形式：√x = x1/2 求导得 ½x-1/2。
d/dx (xⁿ) = nxⁿ⁻¹ | Product: (uv)' = u'v + uv' | Quotient: (u/v)' = (u'v - uv')/v² | Chain: dy/dx = dy/du × du/dx
outside, then multiply by the derivative of the inside. The product rule is needed for x²eˣ or sin x / x. For trigonometric functions, d/dx (sin x) = cos x and d/dx (cos x) = -sin x.
链式法则处理 (3x + 5)⁴ 这类复合函数：先对外层求导，再乘以内层导数。积法则用于 x²eˣ，商法则用于 sin x / x。三角函数方面，d/dx (sin x) = cos x，d/dx (cos x) = -sin x。
tationary point dy/dx = 0; use the second derivative to classify it: if d²y/dx² > 0 it is a minimum, if d²y/dx² < 0 it is a maximum. Optimisation questions require you to form an expression in one variable using the given constraint, differentiate it, solve dy/dx = 0 and justify that the answer is a maximum or minimum.
微分的应用包括切线与法线、增减函数、驻点与最优化。驻点处 dy/dx = 0；用二阶导数判断类型：d²y/dx² > 0 为极小值，d²y/dx² < 0 为极大值。最优化问题需先用给定条件把表达式化为单变量，再求导、解 dy/dx = 0，并说明结果是最大还是最小。
9. Integration: Techniques and Definite Integrals | 积分：方法与定积分
The constant of integration c must always be included for indefinite integrals, and forgetting it is one of the most common mark-losing errors in the whole paper.
积分是微分的逆运算。当 n ≠ -1 时，∫ xⁿ dx = xⁿ⁺¹/(n + 1) + c。不定积分必须写出积分常数 c，忘记它是整张试卷中最常见的丢分点之一。
∫ xⁿ dx = xⁿ⁺¹/(n + 1) + c | ∫ₐᵇ f(x) dx = F(b) - F(a) | ∫ (1/x) dx = ln |x| + c
A definite integral is evaluated as F(b) - F(a) and needs no constant c. The area between a curve and the x-axis is found by integrating, but you must split the interval wherever the curve crosses the axis, because areas below the axis come out negative. The area between two curves is found by integrating (upper - lower).
定积分用 F(b) - F(a) 计算，不需要常数 c。曲线与 x 轴之间的面积用积分求得，但曲线穿过 x 轴处必须分段，因为轴下方的面积为负。两曲线之间的面积用积分 (上曲线 - 下曲线) 求得。
irectly, use numerical methods such as the trapezium rule: ∫ₐᵇ y dx ≈ ½h [y₀ + 2(y₁ + y₂ + … apezium rule overestimates or underestimates, which depends on the concavity of the curve.
当函数无法直接积分时，可用梯形法则等数值方法：∫ₐᵇ y dx ≈ ½h [y₀ + 2(y₁ + y₂ + … + yₙ₋₁) + yₙ]，其中 h = (b - a)/n。题目常要求把估算值与真值比较，并说明梯形法则是高估还是低估，这取决于曲线的凹凸性。
10. Exponentials, Logarithms and Modelling | 指数、对数与建模
An exponential function has the form y = aˣ and its graph always passes through (0, 1). A logarithm is the inverse: logₐ n = x means aˣ = n. The natural exponential eˣ and natural logarithm ln x are the most important pair, since d/dx (eˣ) = eˣ and d/dx (ln x) = 1/x.
指数函数形如 y = aˣ（a > 0），图像必过点 (0, 1)。对数是其逆运算：logₐ n = x 表示 aˣ = n。自然指数 eˣ 与自然对数 ln x 是最重要的一对，因为 d/dx (eˣ) = eˣ，d/dx (ln x) = 1/x。
log(ab) = log a + log b | log(a/b) = log a - log b | log(aⁿ) = n log a | logₐ a = 1 | logₐ 1 = 0
These laws let you solve equations such as 3ˣ = 20 by taking logarithms of both sides to get x = log 20 / log 3. To solve e2x - 5eˣ + 6 = 0, substitute u = eˣ to obtain a quadratic in u, solve for u, then take logarithms to recover x.
这些法则可解 3ˣ = 20 之类的方程：两边取对数得 x = log 20 / log 3。解 e2x - 5eˣ + 6 = 0 时，令 u = eˣ 得到关于 u 的二次方程，解出 u 后再取对数求出 x。
Exponential models describe growth and decay, written as N = N₀ekt ship to linear form, for example plotting log y against x to obtain a straight line of gradient log a, is a standard modelling question.
指数模型描述增长与衰减，写作 N = N₀ekt 或 N = N₀aᵗ。考题通常给出两组数据以求出常数，再要求预测或计算半衰期。把关系化为线性形式（例如以 log y 对 x 作图得到斜率为 log a 的直线）是标准的建模题型。
11. Proof, Trigonometry and Vectors | 证明、三角与向量
Proof questions require clear logical steps, each justified. Methods include proof by deduction, proof by exhaustion and disproof by counter-example. To prove an identity, start from one side and manipulate it until it matches the other; never treat the identity as an equation and move terms across the equals sign.
证明题要求每一步清晰且有依据。方法包括演绎证明、穷举证明和反例反驳。证明恒等式时，从一边出发逐步变形直到等于另一边；切勿把恒等式当作方程来移项。
2sin²θ = 2cos²θ - 1. In radians, arc length is rθ and sector area is ½r²θ, where θ must be in radians.
关键三角恒等式包括 sin²θ + cos²θ = 1、tan θ = sin θ / cos θ，以及二倍角公式 sin 2θ = 2 sin θ cos θ 与 cos 2θ = cos²θ - sin²θ = 1 - 2sin²θ = 2cos²θ - 1。弧度制下，弧长为 rθ，扇形面积为 ½r²θ，其中 θ
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