AS Further Mathematics: Focused Calculation Practice | AS 进阶数学：计算题专项训练

📚 AS Further Mathematics: Focused Calculation Practice | AS 进阶数学：计算题专项训练

Mastering AS Further Mathematics requires consistent practice of core calculation techniques. This article presents a structured drill covering key topics from complex numbers to differential equations, ensuring you build speed and accuracy for exam success.

掌握 AS 进阶数学需要对核心计算技巧进行持续训练。本文提供结构化的专项训练，涵盖从复数到微分方程的关键主题，帮助你提升解题速度与准确性，在考试中取得成功。

1. Complex Numbers: Arithmetic and Polar Form | 复数：运算与极坐标形式

Addition, subtraction, multiplication, and division of complex numbers must become second nature. Always express answers in the form a + bi, and convert between Cartesian and polar forms using modulus r = √(a² + b²) and argument θ = arctan(b/a) with correct quadrant.

复数的加、减、乘、除运算必须熟练掌握。始终将答案写成 a + bi 的形式，并利用模 r = √(a² + b²) 和辐角 θ = arctan(b/a)（注意象限）在直角坐标形式和极坐标形式之间进行转换。

Example: Given z₁ = 3 + 4i and z₂ = 1 − 2i, find z₁z₂, z₁/z₂, and express z₁ in polar form.

例题：给定 z₁ = 3 + 4i 和 z₂ = 1 − 2i，求 z₁z₂、z₁/z₂，并将 z₁ 表示为极坐标形式。

Product: (3+4i)(1−2i) = 3 − 6i + 4i − 8i² = 3 − 2i + 8 = 11 − 2i. Quotient: multiply numerator and denominator by conjugate: (3+4i)/(1−2i) × (1+2i)/(1+2i) = (3+6i+4i+8i²)/(1+4) = (−5+10i)/5 = −1+2i. Polar form of z₁: r = √(3²+4²)=5, θ = arctan(4/3) ≈ 53.1°, so z₁ = 5(cos 53.1° + i sin 53.1°).

乘积：(3+4i)(1−2i) = 3 − 6i + 4i − 8i² = 3 − 2i + 8 = 11 − 2i。商：分子分母乘共轭：(3+4i)/(1−2i) × (1+2i)/(1+2i) = (3+6i+4i+8i²)/(1+4) = (−5+10i)/5 = −1+2i。z₁ 的极坐标形式：r = √(3²+4²)=5，θ = arctan(4/3) ≈ 53.1°，因此 z₁ = 5(cos 53.1° + i sin 53.1°)。

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2. Roots of Polynomial Equations | 多项式方程的根

For real-coefficient polynomials, complex roots occur in conjugate pairs. Use this property to find all roots when one is known. Sum and product of roots relations can also speed up calculations.

对于实系数多项式，复数根共轭成对出现。利用这一性质可以在已知一根时求出所有根。根与系数的关系也能加速计算。

Example: Given that 1 + 2i is a root of f(z) = z³ − z² + 3z + 5, find all roots.

例题：已知 1 + 2i 是 f(z) = z³ − z² + 3z + 5 的一个根，求所有根。

Since coefficients are real, 1 − 2i is also a root. The corresponding quadratic factor is (z − (1+2i))(z − (1−2i)) = z² − 2z + 5. Divide f(z) by this quadratic: f(z) = (z² − 2z + 5)(z + 1). The remaining root is z = −1. Hence the roots are 1+2i, 1−2i, and −1.

由于系数为实数，1 − 2i 也是根。对应的二次因式为 (z − (1+2i))(z − (1−2i)) = z² − 2z + 5。用该二次式除 f(z)，得 f(z) = (z² − 2z + 5)(z + 1)。剩下的根是 z = −1。因此全部根为 1+2i, 1−2i 和 −1。

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3. Matrices: Determinants and Inverses | 矩阵：行列式与逆矩阵

For a 2×2 matrix A = [[a, b], [c, d]], the inverse is A⁻¹ = (1/(ad−bc)) [[d, −b], [−c, a]], provided det(A) ≠ 0. For 3×3 matrices, use the adjugate method or row reduction.

对于 2×2 矩阵 A = [[a, b], [c, d]]，其逆矩阵为 A⁻¹ = (1/(ad−bc)) [[d, −b], [−c, a]]，要求 det(A) ≠ 0。对于 3×3 矩阵，可使用伴随矩阵法或行简化。

Example: Find the inverse of M = [[3, 1], [5, 2]].

例题：求 M = [[3, 1], [5, 2]] 的逆矩阵。

det(M) = 3×2 − 1×5 = 1. So M⁻¹ = (1/1) [[2, −1], [−5, 3]] = [[2, −1], [−5, 3]]. Check: M M⁻¹ = [[3×2+1×(−5), 3×(−1)+1×3], [5×2+2×(−5), 5×(−1)+2×3]] = [[1,0],[0,1]].

det(M) = 3×2 − 1×5 = 1。因此 M⁻¹ = (1/1) [[2, −1], [−5, 3]] = [[2, −1], [−5, 3]]。验证：M M⁻¹ = [[3×2+1×(−5), 3×(−1)+1×3], [5×2+2×(−5), 5×(−1)+2×3]] = [[1,0],[0,1]]。

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4. Solving Systems of Linear Equations | 线性方程组的求解

Linear systems can be solved using matrix inverses or Gaussian elimination. Express the system as Ax = b, then x = A⁻¹b if A is invertible.

线性方程组可通过矩阵求逆或高斯消元法求解。将方程组表示为 Ax = b，若 A 可逆，则 x = A⁻¹b。

Example: Solve 2x + y = 4 and 3x + 2y = 7 using matrices.

例题：用矩阵方法解方程组 2x + y = 4, 3x + 2y = 7。

Let A = [[2,1],[3,2]], x = [[x],[y]], b = [[4],[7]]. det(A)=4−3=1. A⁻¹ = [[2,−1],[−3,2]]. Then x = A⁻¹b = [[2×4+(−1)×7], [−3×4+2×7]] = [[1],[2]]. So x = 1, y = 2.

令 A = [[2,1],[3,2]], x = [[x],[y]], b = [[4],[7]]。det(A)=1，A⁻¹ = [[2,−1],[−3,2]]。于是 x = A⁻¹b = [[2×4+(−1)×7], [−3×4+2×7]] = [[1],[2]]。解得 x = 1, y = 2。

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5. Vector Geometry: Shortest Distance from a Point to a Line | 向量几何：点到直线的最短距离

To find the distance from a point P to a line r = a + td, use the formula d = |(P − a) × d| / |d| in 3D, or use perpendicular vectors in 2D.

求点 P 到直线 r = a + td 的距离，在三维中使用公式 d = |(P − a) × d| / |d|，在二维中使用垂直向量。

Example: Find the shortest distance from P(1,2,3) to the line r = (1,0,2) + t(2,1,−1).

例题：求点 P(1,2,3) 到直线 r = (1,0,2) + t(2,1,−1) 的最短距离。

Let a = (1,0,2), d = (2,1,−1). P − a = (0,2,1). Cross product: (0,2,1)×(2,1,−1) = i(2×(−1)−1×1) − j(0×(−1)−1×2) + k(0×1−2×2) = (−3, 2, −4). |(P−a)×d| = √(9+4+16) = √29. |d| = √(4+1+1) = √6. Distance = √29/√6 = √(29/6).

令 a = (1,0,2), d = (2,1,−1)。P − a = (0,2,1)。叉乘：(0,2,1)×(2,1,−1) = i(2×(−1)−1×1) − j(0×(−1)−1×2) + k(0×1−2×2) = (−3, 2, −4)。|(P−a)×d| = √(9+4+16) = √29。|d| = √(4+1+1) = √6。距离 = √29/√6 = √(29/6)。

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6. Further Calculus: Integration by Parts | 进阶微积分：分部积分法

Integration by parts follows ∫ u dv = uv − ∫ v du. Choose u using the LIATE rule (Log, Inverse trig, Algebraic, Trig, Exponential) to simplify the integral.

分部积分公式为 ∫ u dv = uv − ∫ v du。根据 LIATE 法则（对数、反三角、代数、三角、指数）选择 u，以简化积分。

Example: Evaluate ∫ x e²ˣ dx.

例题：求 ∫ x e²ˣ dx。

Let u = x, dv = e²ˣ dx ⇒ du = dx, v = ½ e²ˣ

令 u = x, dv = e²ˣ dx ⇒ du = dx, v = ½ e²ˣ

∫ x e²ˣ dx = x·½ e²ˣ − ∫ ½ e²ˣ dx = ½ x e²ˣ − ¼ e²ˣ + C = ¼ e²ˣ (2x − 1) + C.

∫ x e²ˣ dx = x·½ e²ˣ − ∫ ½ e²ˣ dx = ½ x e²ˣ − ¼ e²ˣ + C = ¼ e²ˣ (2x − 1) + C。

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7. Hyperbolic Functions: Definitions and Derivatives | 双曲函数：定义与导数

Hyperbolic functions are defined as: sinh x = (eˣ − e⁻ˣ)/2, cosh x = (eˣ + e⁻ˣ)/2. Their derivatives mimic circular functions: d/dx sinh x = cosh x, d/dx cosh x = sinh x.

双曲函数定义为：sinh x = (eˣ − e⁻ˣ)/2，cosh x = (eˣ + e⁻ˣ)/2。它们的导数与圆函数相似：d/dx sinh x = cosh x，d/dx cosh x = sinh x。

Example: Differentiate f(x) = ln(cosh x) and evaluate ∫ sinh 2x dx.

例题：求 f(x) = ln(cosh x) 的导数，并计算 ∫ sinh 2x dx。

f'(x) = (1/cosh x) · sinh x = tanh x. For the integral, ∫ sinh 2x dx = (1/2) cosh 2x + C.

f'(x) = (1/cosh x) · sinh x = tanh x。对于积分，∫ sinh 2x dx = (1/2) cosh 2x + C。

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8. First-Order Linear Differential Equations | 一阶线性微分方程

The standard form is dy/dx + P(x) y = Q(x). Multiply by the integrating factor e^∫ P dx, then integrate both sides.

标准形式为 dy/dx + P(x) y = Q(x)。乘以积分因子 e^∫ P dx，然后两边积分。

Example: Solve dy/dx + 2y = e⁻ˣ, with y(0)=1.

例题：解微分方程 dy/dx + 2y = e⁻ˣ，且 y(0)=1。

Integrating factor: μ(x) = e^∫ 2 dx = e²ˣ. Multiply: e²ˣ dy/dx + 2e²ˣ y = eˣ. The left side is d/dx (y e²ˣ) = eˣ. Integrate: y e²ˣ = ∫ eˣ dx = eˣ + C. Thus y = e⁻ˣ (eˣ + C) = e⁻ˣ + C e⁻²ˣ. Apply y(0)=1: 1 = 1 + C ⇒ C=0. Solution: y = e⁻ˣ.

积分因子：μ(x) = e^∫ 2 dx = e²ˣ。乘以：e²ˣ dy/dx + 2e²ˣ y = eˣ。左边是 d/dx (y e²ˣ) = eˣ。积分：y e²ˣ = ∫ eˣ dx = eˣ + C。因此 y = e⁻ˣ (eˣ + C) = e⁻ˣ + C e⁻²ˣ。代入 y(0)=1：1 = 1 + C ⇒ C=0。解为 y = e⁻ˣ。

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9. Summation of Series | 级数求和

Standard results for Σ r, Σ r², and Σ r³ are essential. Use algebraic manipulation to break complex sums into standard forms.

Σ r、Σ r² 和 Σ r³ 的标准公式至关重要。通过代数变形，将复杂求和分解为标准形式。

Example: Find ∑_{r=1}^{n} (3r² + 2r − 1).

例题：求 ∑_{r=1}^{n} (3r² + 2r − 1)。

Use Σ r² = n(n+1)(2n+1)/6, Σ r = n(n+1)/2, Σ 1 = n. Then sum = 3·n(n+1)(2n+1)/6 + 2·n(n+1)/2 − n = ½ n(n+1)(2n+1) + n(n+1) − n. Simplify: = n(n+1)(n+0.5) + n(n+1) − n = n(n+1)(n+1.5) − n = n[ (n+1)(n+1.5) − 1 ]. Further expansion gives (n/2)(2n²+5n+1).

利用 Σ r² = n(n+1)(2n+1)/6，Σ r = n(n+1)/2，Σ 1 = n。和为 3·n(n+1)(2n+1)/6 + 2·n(n+1)/2 − n = ½ n(n+1)(2n+1) + n(n+1) − n。化简：= n(n+1)(n+0.5) + n(n+1) − n = n(n+1)(n+1.5) − n = n[ (n+1)(n+1.5) − 1 ]。进一步展开得 (n/2)(2n²+5n+1)。

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10. Proof by Induction: Summation Formulas | 归纳证明：求和公式

Mathematical induction requires a base case, an induction hypothesis, and the inductive step. It is frequently used to verify summation identities.

数学归纳法需要基本步、归纳假设和递推步，常用于验证求和恒等式。

Example: Prove by induction that Σ_{r=1}^{n} r² = n(n+1)(2n+1)/6.

例题：用归纳法证明 Σ_{r=1}^{n} r² = n(n+1)(2n+1)/6。

Base case n=1: LHS=1²=1, RHS=1·2·3/6=1. True. Inductive hypothesis: assume true for n=k. Then for n=k+1: LHS = Σ_{r=1}^{k+1} r² = k(k+1)(2k+1)/6 + (k+1)² = (k+1)[ k(2k+1)/6 + (k+1) ] = (k+1)[ (2k²+k+6k+6)/6 ] = (k+1)(2k²+7k+6)/6 = (k+1)(k+2)(2k+3)/6, which matches RHS. Hence proved.

基本步 n=1：左=1²=1，右=1·2·3/6=1，成立。归纳假设：设 n=k 时成立。则 n=k+1：左 = Σ_{r=1}^{k+1} r² = k(k+1)(2k+1)/6 + (k+1)² = (k+1)[ k(2k+1)/6 + (k+1) ] = (k+1)[ (2k²+k+6k+6)/6 ] = (k+1)(2k²+7k+6)/6 = (k+1)(k+2)(2k+3)/6，等于右端。得证。

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11. Numerical Methods: Iteration for Equations | 数值方法：方程迭代求解

When an equation f(x)=0 cannot be solved algebraically, use an iterative formula x_{n+1} = g(x_n). Check convergence by ensuring |g'(x)| < 1 near the root.

当方程 f(x)=0 无法代数求解时，使用迭代公式 x_{n+1} = g(x_n)。通过确保在根附近 |g'(x)| < 1 来检查收敛性。

Example: The equation x³ − 2x − 1 = 0 has a root near 1.5. Use the iteration x_{n+1} = ³√(2x_n + 1) to find this root to

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