AS AQA Further Maths Support Pack 3 | AS AQA 进阶数学支持包3精讲

📚 AS AQA Further Maths Support Pack 3 | AS AQA 进阶数学支持包3精讲

Welcome to this structured revision guide for the AS AQA International Further Maths (9665) Support Pack 3. This pack consolidates the core techniques you need to master, from complex numbers to proof by induction. Each section pairs a clear English explanation with an equivalent Chinese summary so you can reinforce your understanding in both languages.

欢迎阅读 AS AQA 国际进阶数学（9665）支持包3的系统复习指南。本支持包整合了你需要掌握的核心技巧，从复数到数学归纳法。每一节都提供清晰的英文讲解和对应的中文概要，帮助你用两种语言强化理解。

1. Complex Numbers: Arithmetic and Conjugates | 复数：运算与共轭

Complex numbers extend the real number system by introducing the imaginary unit i, where i² = −1. Write a complex number as z = a + bi, with a as the real part and b as the imaginary part. The conjugate of z is z̄ = a − bi, which is essential for division and for finding real-valued expressions.

复数通过引入虚数单位 i（其中 i² = −1）扩展了实数系。将复数写成 z = a + bi，其中 a 是实部，b 是虚部。z 的共轭为 z̄ = a − bi，它对于除法和求实值表达式至关重要。

To add or subtract, combine real and imaginary parts separately: (2 + 3i) + (4 − i) = 6 + 2i. For multiplication, expand like binomials and replace i² with −1: (2 + 3i)(4 − i) = 8 − 2i + 12i − 3i² = 11 + 10i.

加减法分别合并实部和虚部：(2 + 3i) + (4 − i) = 6 + 2i。乘法像二项式一样展开，并将 i² 替换为 −1：(2 + 3i)(4 − i) = 8 − 2i + 12i − 3i² = 11 + 10i。

For division, multiply numerator and denominator by the conjugate of the denominator:

对于除法，将分子分母同时乘以分母的共轭：

(3 + i) ÷ (2 − i) = [(3 + i)(2 + i)] / [(2 − i)(2 + i)] = (5 + 5i) / 5 = 1 + i

Always simplify the result to the form a + bi. Check your conjugate calculations carefully because signs are easy to lose.

始终将结果化简为 a + bi 的形式。仔细检查共轭计算，因为符号很容易丢失。

2. Solving Quadratic Equations with Complex Roots | 解具有复数根的二次方程

When the discriminant b² − 4ac is negative, a quadratic equation with real coefficients has two complex conjugate roots. Use the quadratic formula with the square root of a negative number expressed using i.

当判别式 b² − 4ac 为负时，实系数二次方程有两个共轭复根。使用二次求根公式，并将负数的平方根用 i 表示。

Consider x² + 4x + 5 = 0. Here a = 1, b = 4, c = 5, so the discriminant is 16 − 20 = −4.

考虑 x² + 4x + 5 = 0。这里 a = 1，b = 4，c = 5，因此判别式为 16 − 20 = −4。

x = [−4 ± √(−4)] / 2 = (−4 ± 2i) / 2 = −2 ± i

The solutions are −2 + i and −2 − i, which are conjugates of each other. This always happens when the coefficients are real and the discriminant is negative.

解为 −2 + i 和 −2 − i，它们互为共轭。当系数为实数且判别式为负时，这种情况总是发生。

You can also verify by substituting back into the original equation. Understanding complex roots helps you factorise quadratics over C.

你也可以通过代回原方程进行验证。理解复数根有助于在复数范围内对二次式进行因式分解。

3. Matrices: Multiplication and Determinants | 矩阵：乘法与行列式

A matrix is a rectangular array of numbers. For matrix multiplication, the number of columns of the first matrix must equal the number of rows of the second. The entry in row r and column c of the product is found by multiplying corresponding elements of row r of the first matrix and column c of the second, then adding.

矩阵是数字的矩形阵列。对于矩阵乘法，第一个矩阵的列数必须等于第二个矩阵的行数。乘积的第 r 行第 c 列元素由第一个矩阵的第 r 行与第二个矩阵的第 c 列对应元素相乘后求和得到。

Let A = [[1, 2], [3, 4]] and B = [[5, 6], [7, 8]]. Then AB = [[1×5 + 2×7, 1×6 + 2×8], [3×5 + 4×7, 3×6 + 4×8]] = [[19, 22], [43, 50]].

设 A = [[1, 2], [3, 4]]，B = [[5, 6], [7, 8]]。则 AB = [[1×5 + 2×7, 1×6 + 2×8], [3×5 + 4×7, 3×6 + 4×8]] = [[19, 22], [43, 50]]。

The determinant of a 2×2 matrix det(A) = ad − bc for A = [[a, b], [c, d]]. It tells whether the matrix is invertible. If det(A) = 0, the matrix has no inverse.

2×2 矩阵 A = [[a, b], [c, d]] 的行列式为 det(A) = ad − bc。它判断矩阵是否可逆。若 det(A) = 0，则矩阵不可逆。

det([[3, 1], [2, 4]]) = 3×4 − 1×2 = 12 − 2 = 10

Remember that matrix multiplication is not commutative: AB ≠ BA in general. Always preserve the order of multiplication.

记住矩阵乘法不满足交换律：一般地 AB ≠ BA。始终保留乘法顺序。

4. Using Matrices for Transformations | 利用矩阵进行变换

A 2×2 matrix can represent a linear transformation in the plane. The point (x, y) maps to the point (x′, y′) via the multiplication: [[x′], [y′]] = M[[x], [y]].

一个 2×2 矩阵可以表示平面上的线性变换。点 (x, y) 通过乘法映射到点 (x′, y′)：[[x′], [y′]] = M[[x], [y]]。

For example, the matrix M = [[0, −1], [1, 0]] represents a rotation of 90° anticlockwise. Applying it to (1, 0) gives (0, 1), which matches the rotation.

例如，矩阵 M = [[0, −1], [1, 0]] 表示逆时针旋转 90°。将它作用于 (1, 0) 得到 (0, 1)，这与旋转一致。

Reflection in the x-axis is represented by [[1, 0], [0, −1]]. Enlargement with scale factor k is [[k, 0], [0, k]].

关于 x 轴的反射由 [[1, 0], [0, −1]] 表示。缩放因子为 k 的伸缩是 [[k, 0], [0, k]]。

To find the image of a shape, apply the matrix to each vertex. To find the matrix of a combined transformation, multiply the matrices in the correct order: if T₁ is followed by T₂, the combined matrix is T₂T₁.

要求一个图形的像，对每个顶点应用矩阵。要求复合变换的矩阵，按正确顺序相乘：若先进行 T₁ 再进行 T₂，则复合矩阵为 T₂T₁。

Check the determinant of a transformation matrix: its absolute value gives the area scale factor of the transformation.

检查变换矩阵的行列式：其绝对值给出变换的面积比例因子。

5. Inequalities with Modulus | 含绝对值的不等式

The modulus of a number x, written |x|, represents its distance from zero on the number line. For any a ≥ 0, |x| < a means −a < x < a, and |x| > a means x < −a or x > a.

数 x 的绝对值，记作 |x|，表示它在数轴上到零的距离。对于任意 a ≥ 0，|x| < a 意味着 −a < x < a，而 |x| > a 意味着 x < −a 或 x > a。

Solve the inequality |2x − 3| ≤ 5. By the definition, this is equivalent to −5 ≤ 2x − 3 ≤ 5.

解不等式 |2x − 3| ≤ 5。根据定义，它等价于 −5 ≤ 2x − 3 ≤ 5。

−5 ≤ 2x − 3 ≤ 5 → −2 ≤ 2x ≤ 8 → −1 ≤ x ≤ 4

For more complex cases, square both sides if both sides are non-negative: |x + 1| > |x − 2| becomes (x + 1)² > (x − 2)². This removes the modulus signs and often solves neatly.

对于更复杂的题目，若两边都非负，可两边平方：|x + 1| > |x − 2| 变为 (x + 1)² > (x − 2)²。这样可以去掉绝对值符号，通常能简洁求解。

Always test a value from your solution interval in the original inequality to avoid sign errors.

始终从解区间取一个值代入原不等式进行检验，以避免符号错误。

6. Sigma Notation and Summations | 西格玛记号与求和

Sigma notation Σ compactly represents the sum of a sequence. The expression Σᵣ₌₁ⁿ aᵣ means the sum of aᵣ as r takes integer values from 1 to n. You need to know standard results for sums of integers, squares and cubes.

西格玛记号 Σ 简洁地表示序列的和。表达式 Σᵣ₌₁ⁿ aᵣ 表示当 r 从 1 到 n 取整数值时 aᵣ 的和。你需要知道整数、平方数和立方数求和的标准结果。

The three key formulas are:

三个关键公式是：

Σ r = n(n + 1)/2

Σ r = n(n + 1)/2
Σ r² = n(n + 1)(2n + 1)/6

Σ r² = n(n + 1)(2n + 1)/6
Σ r³ = [n(n + 1)/2]²

Σ r³ = [n(n + 1)/2]²

For example, Σ₍ᵣ₌₁₎⁵ (2r + 3) = 2Σr + Σ3 = 2(5×6/2) + 5×3 = 30 + 15 = 45. Split the sum, use the standard results, and do not forget to multiply constants.

例如，Σ₍ᵣ₌₁₎⁵ (2r + 3) = 2Σr + Σ3 = 2(5×6/2) + 5×3 = 30 + 15 = 45。拆分求和，使用标准结果，不要忘记乘以常数。

When the lower limit is not 1, write the sum as the difference of two sums: Σᵣ₌₃ⁿ r = Σᵣ₌₁ⁿ r − (1 + 2).

当下限不是 1 时，将和写成两个和的差：Σᵣ₌₃ⁿ r = Σᵣ₌₁ⁿ r − (1 + 2)。

7. Differentiation from First Principles | 从第一原理求导

Differentiation measures the instantaneous rate of change. From first principles, the derivative of f(x) at x = a is defined as the limit as h approaches 0 of [f(a + h) − f(a)] / h.

微分度量瞬时变化率。从第一原理出发，f(x) 在 x = a 处的导数定义为当 h 趋近于 0 时 [f(a + h) − f(a)] / h 的极限。

Find the derivative of f(x) = x² from first principles.

从第一原理求 f(x) = x² 的导数。

f′(x) = lim (h→0) [(x + h)² − x²] / h = lim (h→0) (2xh + h²) / h = lim (h→0) (2x + h) = 2x

When writing this in an exam, show the step before taking the limit. Factorise the numerator so that the h cancels, then substitute h = 0.

在考试中书写时，要展示取极限之前的步骤。对分子因式分解使 h 约去，然后代入 h = 0。

The same method works for polynomial functions. For f(x) = x³, you expand (x + h)³ first, then simplify.

同样的方法适用于多项式函数。对于 f(x) = x³，先展开 (x + h)³，然后化简。

Be careful with notation: f′(x), dy/dx and d/dx(f(x)) all mean the derivative.

注意记号：f′(x)、dy/dx 和 d/dx(f(x)) 都表示导数。

8. Integration of Rational Functions | 有理函数的积分

Integration reverses differentiation. For power functions, the basic rule is ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C, provided n ≠ −1. Rational functions often require rewriting the integrand before integrating.

积分是微分的逆运算。对于幂函数，基本规则是 ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C，条件为 n ≠ −1。有理函数通常需要先改写被积函数再积分。

Integrate ∫ (x² + 1) / x² dx. Divide each term by x² first:

求积分 ∫ (x² + 1) / x² dx。先将每一项除以 x²：

∫ (1 + x⁻²) dx = x − x⁻¹ + C = x − 1/x + C

For fractions like 1/(2x + 1), use the natural logarithm: ∫ 1/(ax + b) dx = (1/a) ln|ax + b| + C. This follows from differentiating ln|ax + b|.

对于像 1/(2x + 1) 这样的分式，使用自然对数：∫ 1/(ax + b) dx = (1/a) ln|ax + b| + C。这是由求导 ln|ax + b| 得到的。

If the denominator is a quadratic, try to express the numerator as a derivative of the denominator plus a constant, or use partial fractions when the denominator factorises.

若分母是二次式，尝试将分子写成分母的导数加常数，或者当分母可因式分解时使用部分分式。

Always include the constant of integration C for indefinite integrals; it represents an arbitrary constant.

不定积分一定要加上积分常数 C；它表示任意常数。

9. Parametric Equations and Gradients | 参数方程与梯度

A parametric curve is defined by two equations x = f(t) and y = g(t), where t is a parameter. This is often useful for describing motion or curves that are not functions of x.

参数曲线由两个方程 x = f(t) 和 y = g(t) 定义，其中 t 是参数。这对于描述运动或不能表示为 x 的函数的曲线通常很有用。

To find the gradient dy/dx from parametric equations, use the chain rule:

要求参数方程的梯度 dy/dx，使用链式法则：

dy/dx = (dy/dt) ÷ (dx/dt), provided dx/dt ≠ 0

If x = t² and y = t³, then dx/dt = 2t, dy/dt = 3t², so dy/dx = 3t² / (2t) = 3t/2 for t ≠ 0. The gradient is expressed in terms of t.

若 x = t²，y = t³，则 dx/dt = 2t，dy/dt = 3t²，所以 dy/dx = 3t² / (2t) = 3t/2（t ≠ 0）。梯度用 t 表示。

To find the equation of a tangent, substitute the value of t into x, y and dy/dx, then use y − y₁ = m(x − x₁).

要求切线方程，将 t 值代入 x、y 和 dy/dx，然后使用 y − y₁ = m(x − x₁)。

Remember that dx/dt must not be zero at the point where you calculate the gradient; otherwise the tangent is vertical.

记住在计算梯度的点处 dx/dt 不能为零；否则切线是竖直的。

10. Series and Recurrence Relations | 级数与递推关系

A recurrence relation defines each term of a sequence using previous terms. For example, uₙ₊₁ = 2uₙ + 1 with u₁ = 0 creates the sequence 0, 1, 3, 7, …

递推关系用前一项或前几项定义序列的每一项。例如，uₙ₊₁ = 2uₙ + 1，u₁ = 0 产生序列 0, 1, 3, 7, …

You may be asked to find a particular term by iteration, or to prove a closed form using induction. Always write out enough terms to spot a pattern, but avoid assuming a pattern without proof.

你可能需要通过迭代求某一项，或者使用数学归纳法证明通项公式。务必写出足够多的项以观察规律，但不要在没有证明的情况下假设规律成立。

Arithmetic series have a constant difference d, with sum Sₙ = n/2[2a + (n − 1)d]. Geometric series have a constant ratio r, with sum Sₙ = a(1 − rⁿ)/(1 − r).

等差数列有恒定公差 d，其和为 Sₙ = n/2[2a + (n − 1)d]。等比数列有恒定公比 r，其和为 Sₙ = a(1 − rⁿ)/(1 − r)。

For a recurring sequence, the behaviour as n increases can be explored numerically: does it converge, diverge, or oscillate? A calculator or spreadsheet helps, but the exam requires analytical reasoning.

对于循环序列，可以通过数值观察 n 增大时的行为：它是收敛、发散还是振荡？计算器或电子表格有帮助，但考试要求分析推理。

11. Proof by Induction | 数学归纳法证明

Proof by induction is a powerful method for statements involving integers. It has three stages: base case, inductive hypothesis, and inductive step.

数学归纳法是一种处理涉及整数命题的强大方法。它包含三个阶段：基础情况、归纳假设和归纳步骤。

First prove the statement is true for the initial value, usually n = 1. Then assume it is true for n = k. Using this assumption, prove it is true for n = k + 1. If both steps hold, the statement is true for all positive integers.

首先证明命题对初始值（通常是 n = 1）成立。然后假设它对 n = k 成立。利用这一假设证明它对 n = k + 1 成立。若两步都成立，则命题对一切正整数成立。

Prove that Σᵣ₌₁ⁿ r = n(n + 1)/2 by induction. Base case: for n = 1, LHS = 1, RHS = 1×2/2 = 1, true.

用归纳法证明 Σᵣ₌₁ⁿ r = n(n + 1)/2。基础情况：n = 1 时，左边 = 1，右边 = 1×2/2 = 1，成立。

Assume true for n = k. Then for n = k + 1:

假设对 n = k 成立。则对于 n = k + 1：

Σᵣ₌₁ᵏ⁺¹ r = Σᵣ₌₁ᵏ r + (k + 1) = k(k + 1)/2 + (k + 1) = (k + 1)(k + 2)/2

This is the formula with n = k + 1, so the induction is complete. Always write the conclusion sentence clearly: “Therefore, by the principle of mathematical induction, the statement holds for all n ≥ 1.”

这正是 n = k + 1 时的公式，因此归纳完成。务必清晰写出结论句：“因此，由数学归纳法原理，命题对所有 n ≥ 1 成立。”

12. Exam Strategy Tips | 考试策略建议

Start each question by reading carefully and identifying the required technique. Show all working steps, especially for method marks. In further maths, partial credit is generous when your reasoning is clear.

开始每道题时仔细阅读并识别所需技巧。展示所有解题步骤，尤其是方法分。在进阶数学中，思路清晰时过程分很慷慨。

Practise using the exact notation from the specification: write complex numbers in the form a + bi, matrices in square brackets, and use the correct summation symbols. Avoid using non-standard abbreviations.

练习使用考试大纲中的标准记号：复数写成 a + bi 形式，矩阵用方括号，使用正确的求和符号。避免使用非标准缩写。

Check your answers by substituting back into original equations or applying inverse operations. For example, if you solve a quadratic, plug the roots back in; if you integrate, differentiate your result.

通过代回原方程或应用逆运算检查答案。例如，解完二次方程后将根代回；完成积分后对结果求导。

Finally, manage your time. If a question feels too hard, move on and return to it later. The support pack exercises are designed to build fluency; repeat them until the methods become automatic.

最后，管理好时间。如果某道题感觉太难，先跳过，稍后再回来。支持包中的练习旨在培养熟练度；重复练习直到这些方法变得自然而然。

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AS AQA Further Maths Support Pack 3 | AS AQA 进阶数学支持包3精讲