📚 AS Further Maths Unit 2 Mark Scheme Jan19: Key Topic Insights | AS 进阶数学 Unit 2 评分标准精讲（Jan19）
The January 2019 marking scheme for AS Further Mathematics Unit 2 offers a precise insight into examiner expectations. By analysing how marks are awarded for method, accuracy, and specific reasoning steps, students can learn to structure their working efficiently. This article reconstructs the key question types from that paper, explaining exactly what earns each mark and how to avoid common mistakes.
2019 年 1 月的 AS 进阶数学 Unit 2 评分方案为考生明确展示了评分标准。通过分析步骤分、答案分以及特定推理环节的给分方式，学生能够学会高效组织解题过程。本文重构了该试卷中的主要题型，详细说明如何获得每一分，并避开常见失分点。
1. Complex Numbers: Equating Real and Imaginary Parts | 复数：实虚部对应相等
When solving equations involving complex numbers, one of the most frequently tested skills is recognizing that two complex numbers are equal only if their real parts are equal and their imaginary parts are equal. In a typical mark scheme question, you might be asked to find real numbers a and b such that (a + 3i)(2 + i) = b + 11i. The M1 mark is awarded for expanding the left-hand side to obtain (2a – 3) + (a + 6)i. Equating real and imaginary parts then leads to two linear equations.
在解涉及复数的方程时，最常考查的技能之一就是识别两个复数相等当且仅当它们的实部相等且虚部相等。在典型的评分题目中，可能会要求找出实数 a 和 b，使得 (a + 3i)(2 + i) = b + 11i。M1 步骤分授予展开左边得到 (2a – 3) + (a + 6)i 的过程。然后令实部和虚部分别相等，得到两个一次方程。
A further A1 mark is given for correctly solving 2a – 3 = b and a + 6 = 11, yielding a = 5, b = 7. Note that many candidates lose accuracy marks by making sign errors during expansion. Always write the i² term as –1 immediately to avoid confusion.
随后的 A1 答案分用于正确求解 2a – 3 = b 与 a + 6 = 11，得出 a = 5, b = 7。请注意，不少考生在展开时因符号错误而丢失答案分。务必立即将 i² 写成 –1，避免混淆。
2. Modulus-Argument Form and Multiplication | 模-辐角形式与乘法
Questions on modulus-argument form routinely appear in Unit 2. The mark scheme awards B1 for correctly calculating the modulus r = √(x² + y²) and the argument θ = arctan(y/x), with careful attention to the quadrant. For a complex number z = –√3 + i, the modulus is r = √(3 + 1) = 2. The argument lies in the second quadrant: θ = π – arctan(1/√3) = 5π/6.
模-辐角形式的题目在 Unit 2 中经常出现。评分方案中，正确计算模 r = √(x² + y²) 与辐角 θ = arctan(y/x) 且象限判断正确可获得 B1 独立分数。对于复数 z = –√3 + i，其模为 r = √(3 + 1) = 2。辐角在第二象限：θ = π – arctan(1/√3) = 5π/6。
When the question asks for the product z₁z₂ in modulus-argument form, M1 is earned for multiplying the moduli and adding the arguments. If z₁ = 2(cos π/3 + i sin π/3) and z₂ = 3(cos π/4 + i sin π/4), the product modulus is 2 × 3 = 6, and the argument is π/3 + π/4 = 7π/12. The final A1 requires the answer in a simplified exact form, not decimal approximations.
当题目要求用模-辐角形式表示乘积 z₁z₂ 时，M1 分用于模相乘、辐角相加的步骤。若 z₁ = 2(cos π/3 + i sin π/3)、z₂ = 3(cos π/4 + i sin π/4)，乘积的模为 2 × 3 = 6，辐角为 π/3 + π/4 = 7π/12。最终 A1 要求以简洁的精确形式给出答案，不可用小数近似。
3. Quadratic Equations with Complex Solutions | 二次方程的复数解
Another standard mark scheme feature is solving a quadratic equation whose discriminant is negative. For instance, z² – 4z + 13 = 0. The M1 mark is for using the quadratic formula or completing the square. Substituting into the formula gives z = [4 ± √(16 – 52)] / 2 = [4 ± √(–36)] / 2. The next A1 is for correctly simplifying the square root of a negative number: √(–36) = 6i.
评分标准中的另一常见特征就是求解判别式为负的二次方程。例如，z² – 4z + 13 = 0。M1 分用于使用求根公式或配方法。代入公式得 z = [4 ± √(16 – 52)] / 2 = [4 ± √(–36)] / 2。下一个 A1 分用于正确化简负数的平方根：√(–36) = 6i。
The final solutions are z = 2 ± 3i. B1 is often given for expressing the conjugate pair clearly. Many candidates forget to write both roots explicitly, so always present them as z = 2 + 3i and z = 2 – 3i. Using i rather than √(–1) throughout ensures the answer meets the mark scheme’s notational requirements.
最终解为 z = 2 ± 3i。通常会单独给一个 B1 分要求清晰地写出一对共轭复数。很多考生忘记明确写出两个根，因此务必以 z = 2 + 3i 和 z = 2 – 3i 的形式给出。全程使用 i 而非 √(–1)，可以确保答案符合评分方案的符号要求。
4. Matrix Multiplication and Dimension Rules | 矩阵乘法与维度规则
Matrix questions in Unit 2 test not only computation but also the understanding of when multiplication is defined. A typical B1 mark is awarded for stating that a product AB exists only if the number of columns of A equals the number of rows of B. For example, a 2×3 matrix multiplied by a 3×2 matrix results in a 2×2 matrix.
Unit 2 中的矩阵题目不只考查计算，还考查对乘法何时可行的理解。典型的 B1 独立分授予阐明乘积 AB 存在当且仅当 A 的列数等于 B 的行数这一条件。例如，一个 2×3 矩阵乘以一个 3×2 矩阵，结果是一个 2×2 矩阵。
To calculate a specific entry, the mark scheme gives M1 for setting up the correct row-column dot product. For instance, to find entry (1,2) of the product, multiply pairs from row 1 of the first matrix and column 2 of the second, then sum. An A1 mark follows for the correct numerical value. Consistent labels and systematic working reduce arithmetic slips.
计算具体元素时，评分方案对正确设置行列点积的步骤授予 M1 分。例如，求乘积的 (1,2) 元素时，将第一个矩阵的第 1 行与第二个矩阵的第 2 列对应元素相乘再求和。随后的 A1 分授予正确的数值。保持清晰的标注和系统的计算能减少算术失误。
5. Determinant and Inverse of 2×2 Matrices | 2×2 矩阵的行列式与逆矩阵
Finding the determinant of a 2×2 matrix M = [[a, b], [c, d]] is a routine B1 mark: det(M) = ad – bc. The mark scheme then tests the condition for invertibility. M1 is awarded for stating that M⁻¹ exists if and only if det(M) ≠ 0. For a singular matrix, the determinant equals zero, leading to no inverse.
计算 2×2 矩阵 M = [[a, b], [c, d]] 的行列式是常规的 B1 分：det(M) = ad – bc。评分方案随后考查可逆性条件。M1 分授予声明 M⁻¹ 存在当且仅当 det(M) ≠ 0。对于奇异矩阵，行列式为零，没有逆矩阵。
When the inverse is required, the formula M⁻¹ = (1/det(M)) [[d, –b], [–c, a]] must be applied. M1 is for correct substitution into the formula; A1 is for full simplification, including the scalar factor. Always check by multiplying M by M⁻¹ to confirm the identity matrix. Leaving the inverse as a matrix of fractions without the scalar factor outside is a frequent mark-losing error.
需要求逆矩阵时，必须使用公式 M⁻¹ = (1/det(M)) [[d, –b], [–c, a]]。M1 分用于正确代入公式；A1 分用于完全化简，包括提取标量因子。养成用 M 乘以 M⁻¹ 验证是否得到单位矩阵的习惯。将逆矩阵写成分数矩阵而未将公因子提出，是常见的丢分错误。
6. Sum and Product of Polynomial Roots | 多项式根的和与积
Questions on roots of polynomials frequently rely on the relationships Σα = –b/a, Σαβ = c/a, and αβγ = –d/a for a cubic equation ax³ + bx² + cx + d = 0. The mark scheme typically awards M1 for recalling these Viète’s formulas and A1 for substituting numerical coefficients correctly. For the equation 2x³ – 3x² + 4x – 5 = 0, the sum of roots is 3/2, the sum of pairwise products is 2, and the product is 5/2.
多项式根的问题常常依赖如下关系：对于三次方程 ax³ + bx² + cx + d = 0，有 Σα = –b/a，Σαβ = c/a，以及 αβγ = –d/a。评分方案通常对回忆这些韦达定理给 M1 分，对正确代入数值系数给 A1 分。对于方程 2x³ – 3x² + 4x – 5 = 0，根的和为 3/2，两两积之和为 2，积为 5/2。
When the question asks to find a new polynomial whose roots are, say, 2α, 2β, 2γ, an M1 mark is earned by substituting the new sum and product into the general cubic form. Use Σ(2α) = 2Σα, Σ(2α)(2β) = 4Σαβ, and (2α)(2β)(2γ) = 8αβγ. Then construct x³ – (sum)x² + (sum of pairwise products)x – product = 0. Accuracy marks follow for arithmetic simplifications.
当题目要求构造新多项式，例如根为 2α, 2β, 2γ，M1 分可通过代入新的和与积到标准三次形式获得。先计算 Σ(2α) = 2Σα，Σ(2α)(2β) = 4Σαβ，以及 (2α)(2β)(2γ) = 8αβγ。然后构造 x³ – (根的和)x² + (两两积的和)x – 积 = 0。随后的 A 分用于算术化简的正确性。
7. Summation of Series: Standard Formulae | 级数求和：标准公式
The Unit 2 mark scheme expects candidates to apply standard summation results: ∑r = n(n+1)/2, ∑r² = n(n+1)(2n+1)/6, and ∑r³ = n²(n+1)²/4. A B1 mark is often given for quoting these correctly at the start. The method mark M1 is awarded for splitting a summation into manageable parts and factorising the final expression.
Unit 2 评分方案期望考生能应用标准求和公式：∑r = n(n+1)/2, ∑r² = n(n+1)(2n+1)/6, 以及 ∑r³ = n²(n+1)²/4。通常一开始正确引用这些公式即可获得 B1 独立分。M1 方法分授予将求和拆分为可处理部分并对最终表达式进行因式分解的过程。
For example, evaluating ∑(r + 2)(r – 3) involves expanding to ∑(r² – r – 6). Then separate: ∑r² – ∑r – ∑6. Substituting the standard formulas for the first two terms and using ∑6 = 6n gives the required expression. The final A1 demands a fully factorised form such as n(n – 5)(n + 1)/3 or equivalent. Marks are lost if the factorisation is incomplete.
例如，求 ∑(r + 2)(r – 3) 的值需要将其展开为 ∑(r² – r – 6)。然后拆分：∑r² – ∑r – ∑6。前两项代入标准公式，并使用 ∑6 = 6n，即得所需表达式。最终 A1 分要求给出完全分解因式的形式，如 n(n – 5)(n + 1)/3 或等价形式。若因式分解不彻底，将丢失分数。
8. Proof by Induction: Summations | 归纳法证明：求和公式
Proof by induction is a high-mark question on the Jan19 mark scheme. The structure is rigid: B1 for the base case (showing true for n = 1), M1 for assuming true for n = k, M1 for attempting the n = k + 1 step, and A1 for completing the algebraic manipulation to the target form. A concluding statement earns the final B1.
数学归纳法证明是 Jan19 评分方案中的高分题目。其结构十分固定：B1 分用于基础情形（证明 n = 1 时成立），M1 分用于假设 n = k 时成立，M1 分用于尝试 n = k + 1 的步骤，A1 分用于完成代数推导得到目标形式。最后的归纳结论可获得另一个 B1 分。
Consider proving ∑(3r – 1) = n(3n + 1)/2. For n = 1, LHS = 2, RHS = 2, so base case holds. Assume ∑(from r=1 to k) (3r – 1) = k(3k + 1)/2. For n = k + 1, add the (k+1)th term: LHS = k(3k + 1)/2 + [3(k + 1) – 1] = k(3k + 1)/2 + 3k + 2. Combine into a single fraction and factorise to obtain (k + 1)(3(k + 1) + 1)/2. This confirms the formula for n = k + 1. Any omission of the final statement “therefore true for all positive integers n” costs the last mark.
以证明 ∑(3r – 1) = n(3n + 1)/2 为例。n = 1 时，左边 = 2，右边 = 2，基础情形成立。假设 ∑(从 r=1 到 k) (3r – 1) = k(3k + 1)/2。对于 n = k + 1，加上第 (k+1) 项：左边 = k(3k + 1)/2 + [3(k + 1) – 1] = k(3k + 1)/2 + 3k + 2。合并为单一分数并因式分解，得到 (k + 1)(3(k + 1) + 1)/2。这就证明了 n = k + 1 时公式也成立。忘记写最终结论”因此对所有正整数 n 成立”将失去最后一分。
9. Solving Modulus Equations | 绝对值方程求解
Modulus equations such as |2x – 1| = x + 3 require careful consideration of cases. The mark scheme awards M1 for splitting into two linear equations without the modulus: 2x – 1 = x + 3 and 2x – 1 = –(x + 3). Solving each gives x = 4 and x = –2/3. A critical A1 mark is reserved for checking whether the solutions satisfy the original equation’s domain condition, typically x + 3 ≥ 0, because the right side must be nonnegative.
绝对值方程如 |2x – 1| = x + 3 需要仔细分情况讨论。评分方案对去掉绝对值号后拆分成两个一次方程 2x – 1 = x + 3 和 2x – 1 = –(x + 3) 的步骤授予 M1 分。分别求解得 x = 4 和 x = –2/3。关键的 A1 分留给了验证这些解是否满足原方程的定义域条件，通常需要 x + 3 ≥ 0，因为右边必须非负。
For x = –2/3, the right side is 7/3 > 0, so it is valid. A common alternative method involves squaring both sides, but then extraneous roots must still be checked against the original. The mark scheme allows either approach, but explicitly penalises the omission of domain checks. Always state ‘checking in original equation’ to secure full marks.
对于 x = –2/3，右边为 7/3 > 0，因此有效。另一种常见方法是将两边平方，但仍需代回原方程检验增根。评分方案允许任一方法，但对省略定义域检查的做法会明确扣分。务必写出”代回原方程检验”，以确保拿到全部分数。
10. Loci in the Complex Plane | 复平面上的轨迹
The Jan19 mark scheme includes a question on sketching loci defined by |z – a| = r or arg(z – a) = θ. B1 is awarded for correctly identifying the locus as a circle or half-line. For |z – (2 + i)| = 3, the mark scheme expects a circle centred at (2, 1) with radius 3. M1 is given for correctly placing the centre and drawing the circle, with A1 for accuracy in the radius and clear labelling of intersections with axes if required.
Jan19 评分方案中包含一道绘制复平面轨迹的题目，轨迹由 |z – a| = r 或 arg(z – a) = θ 定义。B1 分授予正确识别轨迹为圆或半直线的步骤。对于 |z – (2 + i)| = 3，评分方案期望绘制以 (2, 1) 为圆心、半径为 3 的圆。M1 分用于正确标出圆心并绘制圆，A1 分用于半径准确，并在需要时清晰标注与坐标轴的交点。
For the half-line arg(z – i) = π/4, the starting point is at (0, 1) and the line makes an angle of π/4 with the positive real axis. The direction must be indicated by an arrow, and the starting point should be marked with an open circle because z ≠ i. The final mark often depends on whether the angle is measured correctly and whether any part of the ray is drawn behind the starting point, which is not allowed.
对于半直线 arg(z – i) = π/4，起点为 (0, 1)，直线与正实轴成 π/4 角。必须用箭头标明方向，起点处应画空心圆，因为 z ≠ i。最终分数往往取决于角度测量是否正确，以及是否在半直线起点后方画出了任何不应存在的线段。
A combination question might ask for the intersection of the two loci. M1 is given for solving the simultaneous equations geometrically or algebraically. Use the fact that on the half-line, z – i = r(cos π/4 + i sin π/4), giving z = i + r/√2 + i r/√2. Substitute into the circle equation to find r, then find the Cartesian coordinates. Full marks require exact coordinates in surd form.
组合题型可能要求找出两条轨迹的交点。M1 分用于用几何或代数方法联立求解。利用在半直线上 z – i = r(cos π/4 + i sin π/4)，即 z = i + r/√2 + i r/√2。代入圆的方程求出 r，再求出直角坐标。完整得分需要以带根号的精确形式给出坐标。
Published by TutorHao | AS Further Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导