📚 Higher Mathematics Prelims Papers I-III: Homework 9 Revision Guide | 高等数学预备考试卷 I-III 第九次作业复习指南
This revision guide covers the core skills tested in Higher Mathematics Prelims Papers I-III, Homework 9. Each section pairs an English explanation with a Chinese translation so you can check your understanding while working through typical exam-style questions. The focus is deliberately practical: every rule is followed by the exact type of example that appears on Paper I (non-calculator) and Paper II (calculator), so that you can move from knowing a formula to using it fluently under time pressure.
本复习指南涵盖高等数学预备考试卷 I-III 第九次作业中的核心技能。每节都提供英文讲解与中文翻译配对，帮助你在练习典型考题时检验理解。内容强调实用性：每一条法则之后都配有卷 I（不可使用计算器）与卷 II（可使用计算器）中出现的典型例题，帮助你从"记住公式"过渡到"在限时压力下熟练运用"。
Homework 9 sits at the point in the course where algebra, calculus and geometry start to be examined together. Polynomial division feeds into curve sketching; differentiation feeds into optimisation; integration feeds into area problems. Working through the twelve sections below in order gives you a complete revision cycle rather than a set of disconnected drills.
第九次作业处在课程的一个关键节点：代数、微积分与几何开始在同一道题中综合考查。多项式除法服务于函数图像绘制；微分服务于最优化问题；积分服务于面积计算。按顺序完成下面十二个小节，你得到的是一次完整的复习循环，而不是一组彼此孤立的练习。
1. How Prelims Papers I-III Are Structured | 预备考试卷 I-III 的结构与考点分布
Papers I-III together sample the whole Higher Mathematics syllabus. Paper I is normally non-calculator and rewards exact algebraic manipulation, surds and fractions. Paper II allows a calculator and therefore concentrates on modelling, numerical work, trigonometry and applications of calculus. Paper III, where it appears, tends to mix longer multi-step problems that combine two or more topics.
卷 I-III 合起来覆盖整个高等数学考纲。卷 I 通常不允许使用计算器，重点考查精确的代数变形、根式与分数运算；卷 II 允许使用计算器，因而侧重建模、数值计算、三角以及微积分的应用；若设卷 III，则多把两个以上主题组合成较长的多步综合题。
Marks are allocated roughly in proportion to the number of steps required, and method marks are awarded even when the final answer is wrong. This is why the working must always be written down clearly, with the rule you are applying identified at each stage.
分值的分配大致与所需步骤数成正比，而且即使最终答案错误，过程分（方法分）依然会给。这就是为什么必须把过程清晰地写出来，并在每一步标明所用的法则。
| Paper | Calculator | Typical emphasis | 典型重点 |
| Paper I | No | Exact algebra, surds, polynomial division, standard derivatives | 精确代数、根式、多项式除法、标准导数 |
| Paper II | Yes | Applications, trigonometry, optimisation, areas | 应用、三角、最优化、面积 |
| Paper III | Mixed | Multi-topic problems linking two or more units | 跨主题综合题，联结两个或以上单元 |
2. Polynomials and the Remainder Theorem | 多项式与余数定理
+ x - 8, then the remainder when dividing by (x - 3) is f(3) = 2(27) - 5(9) + 3 - 8 = -2.
当多项式 f(x) 除以 (x - h) 时，余数为 f(h)。这就是余数定理。例如，若 f(x) = 2x³ - 5x² + x - 8，则除以 (x - 3) 的余数是 f(3) = 2(27) - 5(9) + 3 - 8 = -2。
x = h removes the first term entirely and leaves r = f(h). Notice that the divisor must be linear, and that the sign matters: dividing by (x + 4) means evaluating f(-4).
该定理成立的原因在于：除以 (x - h) 总可写成 f(x) = (x - h)q(x) + r，其中 q(x) 是商，r 是常数。令 x = h 代入即可消去第一项，得到 r = f(h)。注意除式必须是线性的，而且符号很关键：除以 (x + 4) 意味着计算 f(-4)。
. In the second type, set up the equation f(h) = known remainder and solve for the letter.
典型考题只要求余数，或者给出余数反求未知系数。后一种题型只需列出方程 f(h) = 已知余数，再解出字母即可。
Remainder = f(h)
3. The Factor Theorem and Solving Polynomial Equations | 因式定理与多项式方程求解
f f(h) = 0. This is used to factorise cubic or quartic polynomials once one root is known.
因式定理直接由余数定理推出：当且仅当 f(h) = 0 时，(x - h) 是 f(x) 的因式。已知一个根后，可用它来分解三次或四次多项式。
tors of the constant term first: for f(x) = x³ - 4x² + x + 6, the candidates are ±1, ±2, ±3, ±6. Since f(-1) = -1 - 4 - 1 + 6 = 0, (x + 1) is a factor. Synthetic division then gives x² - 5x + 6, which factorises as (x - 2)(x - 3), so the roots are -1, 2 and 3.
解多项式方程时，先用综合除法或长除法降低次数，再解剩余的一元二次方程。先试常数项的因数：对 f(x) = x³ - 4x² + x + 6，候选值为 ±1、±2、±3、±6。由于 f(-1) = -1 - 4 - 1 + 6 = 0，可知 (x + 1) 是因式。综合除法得到 x² - 5x + 6，它分解为 (x - 2)(x - 3)，因此根为 -1、2、3。
If the leading coefficient is not 1, also test fractions formed from factors of the constant term over factors of the leading coefficient. Always state the factorised form before quoting the roots, since the factorisation itself often carries marks.
若首项系数不为 1，还要检验"常数项因数 ÷ 首项系数因数"所形成的分数。始终先写出分解式再给出根，因为分解式本身常常占分。
4. Differentiation Rules: Power and Chain | 微分法则：幂法则与链式法则
xⁿ⁻¹. You must also apply the chain rule: if y = [g(x)]ⁿ, then dy/dx = n [g(x)]ⁿ⁻¹ g′(x).
幂法则为 d/dx (xⁿ) = n xⁿ⁻¹。还必须运用链式法则：若 y = [g(x)]ⁿ，则 dy/dx = n [g(x)]ⁿ⁻¹ g′(x)。
differentiate y = (3x² + 2)⁵. By the chain rule, dy/dx = 5(3x² + 2)⁴ × 6x = 30x(3x² + 2)⁴.
例题：求 y = (3x² + 2)⁵ 的导数。由链式法则，dy/dx = 5(3x² + 2)⁴ × 6x = 30x(3x² + 2)⁴。
Before differentiating, rewrite roots and reciprocals as powers: √x = x1/2, 1/x³ = x⁻³, and 1/√x = x−1/2. This single step converts most "hard" s multiplying a function survive unchanged, while constants added to a function disappear.
求导之前先把根式与倒数改写为幂：√x = x1/2，1/x³ = x⁻³，1/√x = x−1/2。这一步能把卷 I 中大多数"看似困难"的题化为常规题。常数乘在函数前保持不变，而加在函数后的常数求导为零。
d/dx (xⁿ) = n xⁿ⁻¹ d/dx [g(x)]ⁿ = n [g(x)]ⁿ⁻¹ g′(x)
5. Product and Quotient Rules | 乘积法则与商法则
For products, use y = u v ⇒ dy/dx = u v′ + v u′. For quotients, use y = u/v ⇒ dy/dx = (v u′ - u v′)/v².
对于乘积，使用 y = u v ⇒ dy/dx = u v′ + v u′。对于商，使用 y = u/v ⇒ dy/dx = (v u′ - u v′)/v²。
The order in the quotient rule matters: the numerator is "denominator times derivative of numerator minus numerator times derivative of denominator", and the whole expression is divided by the square of the denominator. A common error is to reverse the subtraction, which changes the sign of the answer.
商法则中各项顺序很关键：分子是"分母 × 分子的导数 − 分子 × 分母的导数"，整个表达式再除以分母的平方。常见错误是把减法顺序颠倒，这会使答案符号相反。
so dy/dx = [(x - 3)(2) - (2x + 1)(1)]/(x - 3)² = (2x - 6 - 2x - 1)/(x - 3)² = -7/(x - 3)².
例题：求 y = (2x + 1)/(x - 3) 的导数。取 u = 2x + 1，v = x - 3，则 u′ = 2，v′ = 1，所以 dy/dx = [(x - 3)(2) - (2x + 1)(1)]/(x - 3)² = (2x - 6 - 2x - 1)/(x - 3)² = -7/(x - 3)²。
A frequent exam variant asks for the gradient of a curve at a given point after a product or quotient rule. Differentiate first, then substitute; never substitute before differentiating.
常见变式是在用乘积或商法则求导之后，再求曲线在给定点处的斜率。务必先求导、后代入，绝不可先代入再求导。
6. Trigonometric, Exponential and Logarithmic Derivatives | 三角、指数与对数函数的导数
x) = -sin x, and d/dx (tan x) = sec²x. The exponential and logarithmic results are d/dx (eˣ) = eˣ and d/dx (ln x) = 1/x for x > 0.
微积分中的角必须以弧度为单位。标准结果有 d/dx (sin x) = cos x，d/dx (cos x) = -sin x，d/dx (tan x) = sec²x。指数与对数的结果为 d/dx (eˣ) = eˣ，且当 x > 0 时 d/dx (ln x) = 1/x。
= g′(x)/g(x). For example, d/dx [sin(3x)] = 3 cos(3x) and d/dx [ln(2x + 5)] = 2/(2x + 5).
当自变量是函数而非 x 时，再次使用链式法则：d/dx [sin(ax + b)] = a cos(ax + b)，d/dx [ln(g(x))] = g′(x)/g(x)。例如 d/dx [sin(3x)] = 3 cos(3x)，d/dx [ln(2x + 5)] = 2/(2x + 5)。
| Function | Derivative | 函数 |
| sin x | cos x | 正弦 |
| cos x | -sin x | 余弦 |
| tan x | sec²x | 正切 |
| eˣ | eˣ | 指数 |
| ln x | 1/x | 自然对数 |
7. Applications of Differentiation: Tangents, Stationary Points and Optimisation | 微分应用：切线、驻点与最优化
the normal has gradient -1/f′(a). Both formulas require you to evaluate the function and the derivative at the same point.
导数给出切线的斜率。在 x = a 处的切线方程为 y - f(a) = f′(a)(x - a)，法线的斜率为 -1/f′(a)。使用这两个公式时，都需要在同一点处分别计算函数值与导数值。
f″(x) < 0 it is a maximum, and if f″(x) = 0 the test is inconclusive so a sign table must be used.
驻点出现在 f′(x) = 0 处。用符号表或二阶导数判断其性质：若 f″(x) > 0 为极小值点，若 f″(x) < 0 为极大值点；若 f″(x) = 0 则此判别法失效，必须改用符号表。
le, differentiate, solve f′(x) = 0, justify that the answer is a maximum or minimum, and finally answer the question in context with units.
最优化问题是分值最高的应用题。标准流程是：把待优化的量写成单变量函数，求导，解 f′(x) = 0，说明该点确为最大值或最小值，最后结合题意与单位给出答案。
Tangent: y - f(a) = f′(a)(x - a) Normal gradient = -1 / f′(a)
8. Integration: Standard Results and the Reverse Chain Rule | 积分：标准结果与反向链式法则
entiation. The standard results are ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C for n ≠ -1, ∫ (1/x) dx = ln|x| + C, ∫ eˣ dx = eˣ + C, ∫ cos x dx = sin x + C, and ∫ sin x dx = -cos x + C.
积分是微分的逆运算。标准结果为：当 n ≠ -1 时 ∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C；∫ (1/x) dx = ln|x| + C；∫ eˣ dx = eˣ + C；∫ cos x dx = sin x + C；∫ sin x dx = -cos x + C。
le, ∫ (2x + 1)⁴ dx = (2x + 1)⁵/10 + C, because dividing by the derivative of the bracket, 2, and by the new power, 5, gives 1/10.
当被积函数形如 [g(x)]ⁿ g′(x) 的复合函数时，反向链式法则给出 ∫ [g(x)]ⁿ g′(x) dx = [g(x)]ⁿ⁺¹/(n + 1) + C。例如 ∫ (2x + 1)⁴ dx = (2x + 1)⁵/10 + C，因为除以括号内函数的导数 2、再除以新指数 5，得到系数 1/10。
Never forget the constant of integration for indefinite integrals, and always check by differentiating your answer back to the original integrand. This check takes ten seconds and catches most arithmetic slips.
不定积分绝不能漏掉积分常数，并且务必把结果求导回原被积函数进行验算。这个检查只需十秒，却能发现大多数算术失误。
9. Definite Integrals and Areas | 定积分与面积
F(a), where F is any antiderivative of f. The constant of integration is not needed because it cancels.
定积分的计算方法是先求原函数，再代入上下限：∫ 从 a 到 b 的 f(x) dx = F(b) - F(a)，其中 F 是 f 的任意一个原函数。由于常数会相互抵消，因此无需写出积分常数。
positive parts minus the integral of the negative parts. If the curve crosses the axis, split the integral at the root and take the absolute value of each piece; simply integrating across the root gives a difference of areas rather than a total area.
曲线与 x 轴之间从 x = a 到 x = b 的面积，等于正值部分的积分减去负值部分（取其绝对值）的积分。若曲线穿过坐标轴，需在交点处拆分积分并对每段取绝对值；若直接跨过交点积分，得到的只是面积之差而非总面积。
The area between two curves is found by integrating (upper curve − lower curve) between the intersection points. Find the intersections first by solving the two equations simultaneously, then decide which curve lies above on that interval.
两曲线之间的面积，由在交点之间对（上方曲线 − 下方曲线）积分求得。先用联立方程求交点，再判断在该区间内哪条曲线在上方。
∫ₐᵇ f(x) dx = F(b) - F(a)
10. Sequences, Series and the Binomial Expansion | 数列、级数与二项式展开
sequence has a constant ratio r, with uₙ = a rⁿ⁻¹ and, for r ≠ 1, Sₙ = a(1 - rⁿ)/(1 - r).
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 = a rⁿ⁻¹，且当 r ≠ 1 时 Sₙ = a(1 - rⁿ)/(1 - r)。
An infinite geometric series converges only when |r| and its sum is S∞ = a/(1 - r). A common exam trap is a series that looks geometric but has |r| ≥ 1, in which case no finite sum exists and you must say so explicitly.
无限等比级数仅当 |r| < 1 时收敛，其和为 S∞ = a/(1 - r)。常见陷阱是级数看似等比但 |r| ≥ 1，此时不存在有限和，必须明确指出这一点。
L = kL + c, giving L = c/(1 - k), and remember that convergence requires -1 < k < 1.
形如 uₙ₊₁ = k uₙ + c 的递推关系也在考查范围内。若极限 L 存在，可通过解 L = kL + c 得到 L = c/(1 - k)，并记住收敛条件是 -1 < k < 1。
11. Straight Lines, Circles and Vectors | 直线、圆与向量
cular lines satisfy m₁ m₂ = -1. The equation of a line can be written as y - b = m(x - a).
过点 (x₁, y₁) 与 (x₂, y₂) 的直线斜率为 m = (y₂ - y₁)/(x₂ - x₁)。平行直线斜率相同，垂直直线满足 m₁ m₂ = -1。直线方程可写成 y - b = m(x - a)。
In expanded form x² + y² + 2gx + 2fy + c = 0, the centre is (-g, -f) and the radius is r = √g² + f² - c. The radius must be positive for the equation to represent a real circle, which is a standard "show that" question.
圆心为 (a, b)、半径为 r 的圆方程为 (x - a)² + (y - b)² = r²。展开形式 x² + y² + 2gx + 2fy + c = 0 中，圆心为 (-g, -f)，半径 r = √g² + f² - c。要使方程表示一个真实的圆，半径必须为正，这是常见的"证明"型问题。
For vectors, the magnitude of a vector with components (p, q) is √p² + q², and the scalar product is a·b = |a||b| angle between two vectors and to proving perpendicularity, since a·b = 0 implies θ = 90°.
对于向量，分量为 (p, q) 的向量模长为 √p² + q²，数量积为 a·b = |a||b| cos θ，它也等于对应分量乘积之和。数量积是求两向量夹角和证明垂直的最快途径，因为 a·b = 0 意味着 θ = 90°。
12. Exam Technique and Common Errors | 应试技巧与常见错误
Read the whole paper first and start with the questions you can finish quickly. Show every line of working: in a stepped question, later parts usually depend on earlier answers, and a wrong early answer with correct method can still earn most of the marks for the later parts.
先通读整份试卷，从能迅速完成的问题入手。每一步都要写出过程：在分步设问的题目中，后面的部分通常依赖前面的答案；即使前面答案有误，只要方法正确，后面的部分仍能拿到大部分分数。
The most frequent lost marks come from a small number of habits: dropping the negative sign in the quotient rule, forgetting + C, using degrees instead of radians, substituting into the derivative before differentiating, and rounding too early in a multi-step calculation. Keep all figures to full calculator accuracy and round only at the final step.
最常见的失分源于几个习惯
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