AS Further Mathematics Unit 1 – Key Concepts from Jan 2021 Mark Scheme | AS高等数学第一单元——2021年1月评分方案知识点精讲

📚 AS Further Mathematics Unit 1 – Key Concepts from Jan 2021 Mark Scheme | AS高等数学第一单元——2021年1月评分方案知识点精讲

The January 2021 mark scheme for AS Further Mathematics Unit 1 offers a clear window into how examiners assess core pure topics such as complex numbers, matrices, roots of polynomials, series, and proof. This article breaks down the essential knowledge and techniques that candidates must master, drawing directly from the types of questions and marking points seen in that paper.

2021年1月AS高等数学第一单元的评分方案清晰地展示了考官如何评估复数、矩阵、多项式根、级数和证明等核心纯数知识点。本文直接从该试卷的题目和评分要点出发，逐一精讲考生必须掌握的核心知识与解题技巧。

1. Complex Numbers – Basic Operations | 复数基本运算

c rules: (a + bi) ± (c + di) = (a ± c) + (b ± d)i. Multiplication requires the use of i² = –1, giving (a + bi)(c + di) = (ac – bd) + (ad + bc)i.

复数表示为 z = a + bi，其中 a 是实部，b 是虚部。加减法遵循常规代数法则：(a + bi) ± (c + di) = (a ± c) + (b ± d)i。乘法需利用 i² = –1，得到 (a + bi)(c + di) = (ac – bd) + (ad + bc)i。

Division by a complex number is performed by multiplying numerator and denominator by the complex conjugate. The conjugate of c + di is c – di. This eliminates the imaginary part from the denominator, leaving a real denominator.

复数除法通过分子分母同乘以共轭复数实现。c + di 的共轭为 c – di，这样可将分母中的虚部消去，得到实数分母。

The mark scheme often awards method marks for showing the expansion and the use of i² = – al simplification error occurs. Accuracy marks depend on correct real and imaginary parts.

评分方案通常对展示展开过程及正确使用 i² = –1 赋予方法分，即使最终化简出现小错。答案分要求实部和虚部均正确。

2. Modulus and Argument Representation | 模与辐角的表示

The modulus of z = a + bi is |z| = √a² + b² st be taken with the quadrant to give the correct principal argument, usually in the range –π < θ ≤ π.

复数 z = a + bi 的模为 |z| = √a² + b²。辐角 arg(z) 是从正实轴量起的角度 θ，满足 tan θ = b/a。须根据象限确定正确的主辐角，通常范围是 –π < θ ≤ π。

pactly as r cis θ. This form is essential when multiplying or dividing complex numbers, as |z₁z₂| = |z₁||z₂| and arg(z₁z₂) = arg(z₁) + arg(z₂).

模 - 辐角形式写作 z = r(cos θ + i sin θ) 或简记为 r cis θ。在复数乘除中此形式至关重要，因为 |z₁z₂| = |z₁||z₂|，且 arg(z₁z₂) = arg(z₁) + arg(z₂)。

In the Jan 21 scheme, a common task was to express a complex number in modulus-argument form and then use de Moivre's theorem for powers, although de Moivre is more typical of A2, AS questions often focus on simple multiplication/division arguments.

在2021年1月方案中，常见任务是写出复数的模 - 辐角形式，然后可能涉及幂运算，不过棣莫弗定理更多出现在A2阶段，AS题目通常聚焦于简单的乘除辐角计算。

3. Polynomial Equations with Complex Roots | 涉及复数根的多项式方程

For a polynomial with real coefficients, complex roots always occur in conjugate pairs. If a + bi is a root, then a – bi is also a root. This principle allows the construction of quadratic factors and the determination of unknown coefficients.

对于实系数多项式，复数根总是以共轭对形式出现。若 a + bi 是根，则 a – bi 也必定是根。利用这一原理可构造二次因式并确定未知系数。

and a real cubic equation, the mark scheme expects candidates to identify the conjugate 2 – i, form a quadratic factor (x – (2+i))(x – (2–i)) = x² – 4x + 5, and then divide the cubic to find the remaining real root.

已知一个复数根，例如 2 + i，和一个实系数三次方程，评分方案期望考生识别共轭根 2 – i，构造二次因式 (x – (2+i))(x – (2–i)) = x² – 4x + 5，再通过除法求出剩下的实数根。

Sometimes the question gives a quadratic factor directly and asks for the other roots. Mark points are allocated for setting the quadratic factor to zero and solving it using the quadratic formula or completing the square.

有时题目直接给出一个二次因式并要求其他根。分数点在于将二次因式设为零，并使用求根公式或配方法求解。

4. Relationships Between Roots and Coefficients | 根与系数的关系

For a quadratic equation ax² + bx + c = 0 with roots α and β, the sum α + β = – c/a. For a cubic ax³ + bx² + cx + d = 0 with roots α, β, γ, the relationships are α+β+γ = –b/a, αβ+βγ+γα = c/a, and αβγ = –d/a.

对于二次方程 ax² + bx + c = 0，设根为 α 和 β，则有 α + β = –b/a，αβ = c/a。对于三次方程 ax³ + bx² + cx + d = 0，设根为 α, β, γ，则有 α+β+γ = –b/a，αβ+βγ+γα = c/a，αβγ = –d/a。

These symmetric sums allow the evaluation of expressions like α² + β² = (α+β)² – the mark scheme, method marks are given for substituting the sums and products correctly.

这些对称和式使得求 α² + β² = (α+β)² – 2αβ 等表达式成为可能，无需单独求出各根。评分方案中对正确代入和与积给方法分。

e shifted. The strategy is to use substitution y = f(x) and manipulate the symmetric sums.

对于更高次方程，题目可能要求构造新多项式，使其根为原根之倒数或平移值。策略是使用代换 y = f(x) 并处理对称和式。

5. Matrix Arithmetic and Determinant | 矩阵的算术与行列式

matrix results in a 2×2 matrix where each entry is the dot product of the corresponding row and column.

同阶矩阵可通过对应元素加减进行加减运算。两个 2×2 矩阵相乘，结果仍为 2×2 矩阵，每个元素为对应行与列的点积。

The determinant of a 2×2 matrix A = [a b; c d] is det(A) = ad – inant is zero. Determinants are crucial for finding inverses and solving matrix equations.

2×2 矩阵 A = [a b; c d] 的行列式为 det(A) = ad – bc。若行列式为零，则矩阵为奇异矩阵（不可逆）。行列式是求逆矩阵与解矩阵方程的关键。

In the Jan 21 paper, a typical mark scheme step required showing explicitly the calculation ad – bc and setting it equal to a given value to find an unknown element. The method mark was awarded for forming the equation, not just the answer.

在2021年1月试卷中，典型的评分步骤要求明确展示 ad – bc 的计算并将其设为给定值以求出未知元素。方法分赋给列出方程的过程，而不仅仅是最终答案。

6. Inverse Matrices and Linear Systems | 逆矩阵与线性方程组

The inverse of a 2×2 non-singular matrix A is A⁻¹ = 1/(ad – bc) [d –b; – inverse, we can solve the matrix equation AX = B by pre-multiplying both sides: X = A⁻¹B.

非奇异 2×2 矩阵 A 的逆矩阵为 A⁻¹ = 1/(ad – bc) [d –b; –c a]。利用逆矩阵可解矩阵方程 AX = B，只需两边左乘逆矩阵：X = A⁻¹B。

When solving a system of linear equations, it is important to write the coefficient matrix and column vectors correctly. The mark scheme often gives a mark for stating the matrix form and another for using the inverse to find the solution vector.

解线性方程组时，正确写出系数矩阵和列向量至关重要。评分方案通常为 1 分写出矩阵形式，另 1 分利用逆矩阵求出解向量。

If the determinant is zero, the system has either no solution or infinitely many solutions. In AS Further Maths, this is often tested by asking for the value of a parameter that makes the equations inconsistent.

若行列式为零，则方程组无解或有无穷多解。在AS高等数学中，常通过要求找出使方程组无解的参数值来考查此知识点。

7. Geometric Transformations Using Matrices | 矩阵几何变换

bout the origin, and stretches parallel to axes. The columns of the matrix are the images of the unit vectors i and j.

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2×2 矩阵可表示平面上的线性变换。常见变换包括关于坐标轴或直线 y = ±x 的反射、绕原点的旋转、平行于坐标轴的伸缩。矩阵的列是单位向量 i 和 j 的像。

For a rotation by θ anticlockwise, the matrix is [cos θ –sin θ; sin θ cos θ]. A stretch by factor m in the x-direction and n in the y-direction is [m 0; 0 n]. Identifying transformations from a given matrix was a recurring mark scheme requirement.

逆时针旋转 θ 角的矩阵为 [cos θ –sin θ; sin θ cos θ]。x 方向拉伸 m 倍、y 方向拉伸 n 倍的矩阵为 [m 0; 0 n]。根据给定矩阵识别变换是评分方案中反复出现的要求。

Questions also ask to find the matrix that represents a combination of transformations, for example a reflection followed by a stretch. The order of multiplication matters: applying A then B corresponds to the matrix product BA.

题目也经常要求找出表示复合变换的矩阵，例如先反射再拉伸。乘法顺序至关重要：先应用 A 再应用 B 对应的矩阵乘积为 BA。

8. Summation of Series – Standard Results | 级数求和——标准公式

The standard results for sums of integers and powers are vital: ∑r = ½ n(n+1), ∑r² = ⅙ ², where the sum runs from r = 1 to n. These are used to evaluate summations such as ∑(3r² – 2r + 1).

整数及幂次求和的标准结果至关重要：∑r = ½ n(n+1)、∑r² = ⅙ n(n+1)(2n+1)、∑r³ = ¼ n²(n+1)²，其中求和从 r=1 到 n。这些公式用于计算诸如 ∑(3r² – 2r + 1) 的求和。

The mark scheme allocates marks for stating the correct standard formula and substituting it fully into the expression. Simplification to a factorised form, such as kn(n+1)(an+b), is frequently required to gain full marks.

评分方案对写出正确标准公式并完整代入表达式给予分数。常常需要化简成因式分解形式，如 kn(n+1)(an+b)，才能得到满分。

formula for ∑r³ but still obtaining a consistent simplified polynomial may still earn method marks.

系数错误会被扣分，但常见的评分注解允许在方法清晰的情况下进行后续补偿。例如，即使用错了 ∑r³ 的公式，但持续得出一致的简化多项式仍可能获得方法分。

9. Method of Differences | 差分法

eries where terms cancel in a telescoping manner. A typical expression is 1/(r(r+1)) = 1/r – e of differences reveals that only the first and last few terms remain after cancellation.

差分法用于求和那些项能像望远镜一样相消的级数。典型表达式如 1/(r(r+1)) = 1/r – 1/(r+1)。将和式写成差分序列可发现，消去后仅剩首尾几项。

To find ∑_r=1^n 1/(r(r+1)), we write the expanded form: (1/1 – 1/2) + (1/2 – 1/3) + … + (1/n – 1/(n+1)). Cancelling leads to 1 – gives method marks for showing at least the first two and the last term of the expansion.

求 ∑_r=1^n 1/(r(r+1)) 时，将其展开为 (1/1 – 1/2) + (1/2 – 1/3) + … + (1/n – 1/(n+1))。相消后得 1 – 1/(n+1) = n/(n+1)。评分方案对展示至少前两项及最后一项的方法给分。

For rational functions with linear factors, partial fractions are used to create the difference form. The method is then identical. Candidates must be careful with the limits and the non-cancelling terms.

对于带线性因式的有理函数，先用部分分式化为差分形式，然后方法相同。考生需注意求和的上下限及无法消去的项。

10. Proof by Mathematical Induction | 数学归纳法证明

e true for n = k. (iii) Inductive step: prove that if true for k, then true for k+1. (iv) Conclusion: by mathematical induction, the statement holds for all n ∈ ℕ.

归纳法证明遵循严格的结构：(i) 奠基：验证 n=1（或最小值）时命题成立。(ii) 归纳假设：假设 n=k 时命题为真。(iii) 归纳递推：证明若 k 成立则 k+1 成立。(iv) 结论：由数学归纳法，命题对所有自然数 n 成立。

Common AS Further Maths induction questions include proving summation formulas like ∑r(r!) = (n+1)! – 1, divisibility statements such as 3²ⁿ + 7 is divisible by 8, and matrix power results. The mark scheme strictly checks the inductive step algebra.

AS 高等数学常见的归纳法题目包括证明求和公式如 ∑r(r!) = (n+1)! – 1、可除性命题如 3²ⁿ + 7 能被 8 整除，以及矩阵幂结果。评分方案严格检查归纳递推中的代数推导。

A common error is to forget stating the conclusion or to write 'true for n = k+1' pothesis. The mark scheme almost always has a final mark for a clear, complete conclusion.

常见错误是忘记写出结论，或写出“n=k+1 成立”却未与归纳假设相连接。评分方案几乎总有一个明确完整的结论分。

11. Inequalities Involving Absolute Values | 含绝对值的不等式

Solving inequalities like |2x – 3| < 5 requires considering both cases. Graphical or algebraic approaches both work. Algebraically, |f(x)| < a (a > 0) leads to –a < f(x) < a. For |f(x)| > a, we solve f(x) < –a or f(x) > a.

解如 |2x – 3| < 5 的不等式需考虑两种情况。图像法或代数法皆可。代数上，|f(x)| < a（a>0）等价于 –a < f(x) < a。对于 |f(x)| > a，则解 f(x) < –a 或 f(x) > a。

When inequalities involve more than one absolute term, a sketch or critical value method is safest. The mark scheme expects correct critical values and a check of intervals to determine the solution set, expressed in set notation or on a number line.

当不等式含多个绝对值项时，草图或临界值法最稳妥。评分方案期望正确的临界值，并通过对区间的检验得出解集，用集合记号或数轴表示。

Quadratic inequalities such as x² – 5x + 6 ≥ 0 are solved by factorising to (x – 2)(x – Mark points are split between finding critical values and giving the correct final region.

二次不等式如 x² – 5x + 6 ≥ 0 通过因式分解为 (x – 2)(x – 3) ≥ 0，并考察各区间的乘积符号来求解。分数点分配在找出临界值与给出正确最终区域上。

12. Numerical Methods – Newton-Raphson | 牛顿拉夫森数值方法

The Newton-Raphson method uses the iterative formula x_n+1 = x_n – f(x_n)/f'(x_n) to approximate a root. An initial value x₀ is chosen, and successive approximations are calculated until the desired accuracy is achieved.

牛顿拉夫森法使用迭代公式 x_n+1 = x_n – f(x_n)/f'(x_n) 逼近根。选定初始值 x₀，反复计算逐次逼近值直至达到所需精度。

The mark scheme for AS Unit 1 typically asks for the derivative f'(x) to be stated, one iteration performed, or the method applied twice. Marks are given for correct substitution and evaluation, and often for explaining why a particular x₀ leads to convergence or divergence.

AS 第一单元的评分方案通常要求给出导数 f'(x)，执行一次迭代或使用该方法两次。正确代入与计算给分，还常要求解释为何某特定的 x₀ 会导致收敛或发散。

When the iteration fails, it is usually because f'(x) is zero or the iteration oscillates. Candidates may be asked to suggest a better starting value or to sketch the function to illustrate the tangent line crossing the x-axis.

当迭代失败时，通常是因为 f'(x) 为零或迭代振荡。可能会要求考生推荐一个更好的初始值，或画出函数及切线交于 x 轴的草图加以说明。

Another numerical method that sometimes appears is the interval bisection method, though less frequently in pure AS further maths. However, the focus remains on algebraic manipulation and the correct application of the iteration formula.

另一偶尔出现的数值方法是区间对分法，但在AS纯数部分较少见。不过，重点始终是代数操作和正确应用迭代公式。

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