📚 Common Mistakes in AS Further Maths Unit 1 (Jan 2019) | AS 进阶数学单元1（2019年1月卷）易错点总结
The January 2019 AS Further Mathematics Unit 1 paper (often Further Pure 1) tests foundational skills in complex numbers, series, induction, matrices, and roots of polynomials. Many students lose marks not because they lack understanding, but because they make recurrent, avoidable mistakes. This article collates the most common errors observed in this sitting, alongside precise remedies, to help you avoid them in your own revision and exams.
2019年1月的AS进阶数学单元1试卷（通常为Further Pure 1）考查复数、级数、归纳法、矩阵和多项式方程根等基础技能。许多学生失分并非由于不理解，而是因为重复出现本可避免的错误。本文收集了此次考试中最常见的错误，并给出精确的纠正方法，帮助你在复习和考试中避开这些陷阱。
1. Complex Numbers: Finding Square Roots | 复数：求平方根
A classic question asks for the square roots of a complex number such as 3 + 4i. The most frequent mistake is to forget the ± sign, giving only one root. Students often set (a + bi)² = 3 + 4i, expand correctly to a² – b² + 2abi, but then only solve for the positive case.
求复数平方根（如 3 + 4i）是经典题目。最常见的错误是忘记 ± 号，只给出一个根。学生常设 (a + bi)² = 3 + 4i，正确展开得到 a² – b² + 2abi，但随后只解正数情况。
Always remember that a non-zero complex number has exactly two square roots, negatives of each other. After equating real and imaginary parts to get a² – b² = 3 and 2ab = 4, solve simultaneously. Many make sign errors when substituting b = 2/a into the real equation, leading to a⁴ – 3a² – 4 = 0, but then discard the negative solution for a². Since a is real, a² must be non-negative, so a² = 4 gives a = ±2. The corresponding b values are ±1, yielding roots 2 + i and –2 – i. Do not write 2 + i only.
永远记住非零复数恰有两个平方根，互为相反数。在令实部与虚部分别相等得到 a² – b² = 3 和 2ab = 4 后，联立求解。许多学生在代入 b = 2/a 到实部方程时出现符号错误，导致 a⁴ – 3a² – 4 = 0，随后丢弃 a² 的负解。由于 a 为实数，a² 非负，因此 a² = 4 得 a = ±2。相应 b 值为 ±1，得到根 2 + i 和 –2 – i。切勿只写 2 + i。
2. Argand Diagrams and Loci | 阿甘图与轨迹
Questions involving loci like |z – a| = r or arg(z – a) = θ often confuse students between a circle and a half-line. A common error is to treat arg(z – 2) = π/4 as a full line through the point, or failing to indicate the excluded endpoint.
涉及诸如 |z – a| = r 或 arg(z – a) = θ 的轨迹问题时，学生经常混淆圆与射线。常见错误是将 arg(z – 2) = π/4 当成过该点的完整直线，或未能标明排除的端点。
In a sketch, the locus |z + 3 – 2i| = 5 is a circle centred at (–3, 2) with radius 5. Many mistakenly plot centre (3, –2) due to sign errors. For the half-line arg(z – 2) = π/4, the start point is 2 (on the real axis) and the line makes an angle of 45° with the positive real axis. The point z = 2 itself must be excluded (open circle). Label the angle clearly and draw an arrow to show direction. Never draw a solid line extending both ways.
在草图中，轨迹 |z + 3 – 2i| = 5 是以 (–3, 2) 为圆心、半径为 5 的圆。许多人由于符号错误画出圆心 (3, –2)。对于射线 arg(z – 2) = π/4，起点为点 2（实轴上），直线与正实轴成 45°。点 z = 2 本身必须排除（画空心圆）。清晰标出角度并画箭头表示方向。绝不要画成双向延伸的实线。
3. Summation of Series using Standard Results | 级数求和的标准结果
Standard summation formulae for Σr, Σr², Σr³ are given in the formula booklet, yet many marks are lost by misapplying these to expressions like Σ(3r² + 2r). The typical mistake is to separate into 3Σr² + 2Σr but then forget the factor 3 acts on the whole sum of r², or to plug n = 0 into the formula incorrectly when the lower limit is not 1.
公式表提供了 Σr, Σr², Σr³ 的标准求和公式，但在处理如 Σ(3r² + 2r) 的表达式时仍丢失许多分数。典型错误是拆分为 3Σr² + 2Σr 后忘记因子 3 作用于整个 r² 求和中，或当下限不是 1 时不正确地代入 n = 0。
When the sum starts from r = 5, many students simply substitute n into the standard formula. Instead, use Σ from r=1 to n minus Σ from r=1 to 4. A common error is to write Σ from r=5 to n of r² = n(n+1)(2n+1)/6 – 10, forgetting that the subtraction must be the sum up to 4, which is 4×5×9/6 = 30. Always calculate the subtracted block carefully.
当求和从 r = 5 开始时，许多学生直接将 n 代入标准公式。应使用从 r=1 到 n 的和减去从 r=1 到 4 的和。常见错误是写 Σ(r², r=5..n) = n(n+1)(2n+1)/6 – 10，忽略被减部分必须为截至 4 的和，即 4×5×9/6 = 30。务必仔细计算减去的块。
4. Proof by Induction: Divisibility | 数学归纳法：整除性
Induction proofs for divisibility, such as showing 7ⁿ – 1 is divisible by 6, regularly suffer from weak base cases and algebraic slip‑ups in the inductive step. Some students forget to explicitly state the assumption for n = k, while others try to manipulate f(k+1) without establishing the link to f(k).
整除性归纳证明（如证明 7ⁿ – 1 能被 6 整除）经常因基础步骤薄弱和归纳步骤中的代数失误而失分。部分学生忘记明确写出 n = k 的假设，另一些则试图操控 f(k+1) 而未建立起与 f(k) 的联系。
The correct structure is: prove true for n = 1; assume true for n = k, i.e. 7ᵏ – 1 = 6m; then show true for n = k+1. Write 7ᵏ⁺¹ – 1 = 7·7ᵏ – 1 = 7(7ᵏ – 1) + 6 = 7(6m) + 6 = 6(7m + 1), which is divisible by 6. A common error is to incorrectly expand 7ᵏ⁺¹ as 7·7ᵏ but then write something like 7·7ᵏ – 1 = 7(7ᵏ – 1) – 6, messing up the constant. Always check the algebra with a small value of k.
正确结构为：证明 n = 1 成立；假设 n = k 成立，即 7ᵏ – 1 = 6m；然后证明 n = k+1 成立。写 7ᵏ⁺¹ – 1 = 7·7ᵏ – 1 = 7(7ᵏ – 1) + 6 = 7(6m) + 6 = 6(7m + 1)，可被 6 整除。常见错误是将 7ᵏ⁺¹ 展开为 7·7ᵏ，却写出类似 7·7ᵏ – 1 = 7(7ᵏ – 1) – 6 的错误表达式，弄错常数。始终用较小的 k 值检验代数过程。
5. Matrices: Inverse of a 2×2 Matrix | 矩阵：2×2 逆矩阵
Finding the inverse of a 2×2 matrix A = [[a, b], [c, d]] is a routine skill, but sign mistakes and forgetting the determinant factor top the error list. Many write the inverse as (1/(ad – bc)) [[d, –b], [–c, a]] but then swap a and d incorrectly, keeping –c and –b in the wrong positions, especially when negative numbers are involved.
求 2×2 矩阵 A = [[a, b], [c, d]] 的逆矩阵是常规技能，但符号错误和忘记行列式因子高居错误榜首。许多人写出逆矩阵为 (1/(ad – bc)) [[d, –b], [–c, a]]，却错误地交换 a 和 d 的位置，同时将 –c 和 –b 放错位置，尤其在涉及负数时更是如此。
The correct inverse is 1/(det A) × [[d, –b], [–c, a]]. Do not accidentally write [[–a, b], [c, –d]]. If A = [[5, 2], [3, 1]], det A = 5×1 – 2×3 = –1. So A⁻¹ = –1 × [[1, –2], [–3, 5]] = [[–1, 2], [3, –5]]. A frequent error is to omit the reciprocal or to forget that multiplying by a negative determinant flips all signs.
正确的逆矩阵是 1/(det A) × [[d, –b], [–c, a]]。不要意外写成 [[–a, b], [c, –d]]。若 A = [[5, 2], [3, 1]]，det A = 5×1 – 2×3 = –1。所以 A⁻¹ = –1 × [[1, –2], [–3, 5]] = [[–1, 2], [3, –5]]。常见错误是遗漏倒数，或忘记乘以负行列式会反转所有符号。
6. Roots of Polynomial Equations: Symmetric Functions | 多项式方程的根：对称多项式操作
Given cubic equation x³ – 4x² + x + 6 = 0 with roots α, β, γ, students often struggle to find sums like α² + β² + γ² or 1/α + 1/β + 1/γ. The first error is misquoting Σα, Σαβ, αβγ. The relationships are Σα = 4, Σαβ = 1, αβγ = –6, but signs frequently go wrong.
给定三次方程 x³ – 4x² + x + 6 = 0 的根为 α, β, γ，学生常难以求出 α² + β² + γ² 或 1/α + 1/β + 1/γ 等值。首要错误是错记 Σα, Σαβ, αβγ。关系为 Σα = 4, Σαβ = 1, αβγ = –6，但符号经常出错。
For α² + β² + γ², use (Σα)² – 2Σαβ = 4² – 2×1 = 14. A common slip is to write (Σα)² – Σαβ, dropping the factor 2. For Σ(1/α), rewrite as (αβ + βγ + γα)/(αβγ) = Σαβ/(αβγ) = 1/(–6) = –1/6. Many invert the numerator incorrectly or forget the sign of the product. Always list the root facts clearly before starting manipulation.
求 α² + β² + γ² 时，用 (Σα)² – 2Σαβ = 4² – 2×1 = 14。常见失误是写成 (Σα)² – Σαβ，遗漏因子 2。求 Σ(1/α) 时，改写为 (αβ + βγ + γα)/(αβγ) = Σαβ/(αβγ) = 1/(–6) = –1/6。许多人错误地颠倒分子或忘记乘积的符号。在开始代数操作前，务必清晰列出根的基本关系。
7. Linear Transformations and Invariant Lines | 线性变换与不变直线
Invariant lines under a transformation given by a 2×2 matrix can cause confusion. Students often mistakenly think an invariant line must be a line of invariant points. An invariant line is a line that is mapped onto itself, but points on it may move along the line. The technique is to set y = mx + c and substitute into the transformed coordinates, then equate the image to a scaled version of the original line.
由 2×2 矩阵给出的变换下的不变直线容易引起混淆。学生常误以为不变直线必须是由不动点组成的线。不变直线是指映射到自身的直线，但直线上的点可能沿该线移动。方法是设 y = mx + c 并代入变换后的坐标，然后令像等于原直线的倍数。
When solving, a typical error is to forget to consider the case c ≠ 0 separately, or to assume the same gradient m applies without checking. Some incorrectly write the condition as matrix × (x, mx + c)ᵀ = (x’, mx’ + c)ᵀ without allowing a scaling factor λ. The correct condition is there exists λ such that the transformed point satisfies (X, Y) = (λx, λ(mx + c)), or equivalently Y = mX + c for all x. Substitute and equate coefficients of x and constant terms to find m, c, and λ.
求解时，典型错误是忘记单独考虑 c ≠ 0 的情况，或未经验证就假设相同的斜率 m 适用。有些人错误地将条件写为矩阵 × (x, mx + c)ᵀ = (x’, mx’ + c)ᵀ，而没有引入比例因子 λ。正确条件为存在 λ 使得变换后的点满足 (X, Y) = (λx, λ(mx + c))，或等价于对所有 x 有 Y = mX + c。代入并比较 x 的系数和常数项，求出 m、c 和 λ。
8. Numerical Methods: Interval Bisection Error Bounds | 数值方法：二分法误差界
Questions on interval bisection often ask for the root to a certain accuracy or for the number of iterations required to achieve a given error bound. The most common mistake is to think the error after n iterations is (b – a)/n instead of (b – a)/2ⁿ. The error bound halves each iteration, so huge underestimates of required steps occur.
二分法题目常要求将根精确到某精度，或求达到给定误差界所需的迭代次数。最常见错误是误以为 n 次迭代后的误差为 (b – a)/n，而非 (b – a)/2ⁿ。每次迭代误差界减半，因此常出现对所需步数的严重低估。
Another error is to stop as soon as f(midpoint) changes sign, without evaluating the width of the interval. The accuracy is determined by the interval width, not by the function value being close to zero. Always check the stopping condition based on (b – a)/2ⁿ < tolerance.
另一个错误是，一旦 f(中点) 变号就停止，而不评估区间宽度。精度取决于区间宽度，而非函数值接近零。始终根据 (b – a)/2ⁿ < 容许误差来检验停止条件。
Also, when using linear interpolation or Newton‑Raphson, a frequent slip is to mis‑copy the derivative. For f(x) = x³ – 2x – 5, the derivative is 3x² – 2. Some write 3x² – 5 or 3x – 2. Always differentiate carefully; a wrong derivative corrupts all subsequent iterations.
此外，在使用线性插值或牛顿-拉夫森法时，常见的笔误是导数抄写错误。对于 f(x) = x³ – 2x – 5，导数为 3x² – 2。有人会写成 3x² – 5 或 3x – 2。务必仔细求导；错误的导数会破坏所有后续迭代。
9. Solving Rational Inequalities | 解有理不等式
Rational inequalities such as (x – 1)/(x + 2) ≤ 2 are fertile ground for mistakes. The classic error is to multiply both sides by (x + 2) without considering its sign, thereby ignoring that the inequality direction may reverse. This leads to a completely wrong solution set.
形如 (x – 1)/(x + 2) ≤ 2 的有理不等式是错误高发区。经典错误是两边同乘 (x + 2) 而不考虑其符号，从而忽略不等式方向可能翻转，导致完全错误的解集。
The safe method is to bring all terms to one side: (x – 1)/(x + 2) – 2 ≤ 0 → (x – 1 – 2x – 4)/(x + 2) ≤ 0 → (–x – 5)/(x + 2) ≤ 0 → (x + 5)/(x + 2) ≥ 0. Then find critical values x = –5 and x = –2, and test intervals. The solution is x ≤ –5 or x > –2. Many forget that x = –2 must be excluded from the domain, but write x ≥ –2 erroneously. Always state the domain restrictions explicitly.
安全做法是将所有项移到一边：(x – 1)/(x + 2) – 2 ≤ 0 → (x – 1 – 2x – 4)/(x + 2) ≤ 0 → (–x – 5)/(x + 2) ≤ 0 → (x + 5)/(x + 2) ≥ 0。然后找出临界值 x = –5 和 x = –2，测试区间。解为 x ≤ –5 或 x > –2。许多人忘记 x = –2 必须从定义域中排除，却错误地写成 x ≥ –2。始终显式声明定义域限制。
10. Common Algebraic and Sign Errors | 常见代数与符号错误
Beyond topic‑specific traps, simple algebraic carelessness abounds. Expanding brackets like (2x – 3)² often yields 4x² – 9 rather than 4x² – 12x + 9 because students square each term and ignore the middle term. When simplifying expressions like (x + h)⁻¹ – x⁻¹, signs frequently go wrong.
除具体主题中的陷阱外，简单的代数粗心也随处可见。展开如 (2x – 3)² 的括号时常得到 4x² – 9 而不是 4x² – 12x + 9，因为学生逐项平方并忽略了中间项。化简如 (x + h)⁻¹ – x⁻¹ 的表达式时，符号经常出错。
Another perennial issue: moving terms across an equation sign. When solving 3 – 2x = 7, some write –2x = 7 – 3 = 4, then x = –2. However, there’s a tendency to drop a minus sign or divide incorrectly. In complex number division, such as (1 + i)/(2 – i), many multiply numerator and denominator by (2 + i) but miscalculate the real and imaginary parts during expansion. Always double‑check each step with a scratch‑pad verification using small integer values where possible.
另一个长期存在的问题：移项。在解 3 – 2x = 7 时，有人写 –2x = 7 – 3 = 4，然后 x = –2。但常有丢掉负号或除法错误的情形。在复数除法中，如 (1 + i)/(2 – i)，许多人将分子分母同乘 (2 + i)，却在展开时算错实部和虚部。始终复查每一步，并尽可能用小的整数值在草稿上验证。
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