📚 Common Mistakes from AS Maths Unit 1 Jan 2019 | AS数学 Unit 1 2019年1月真题易错点总结
This article analyses the Edexcel AS Mathematics Unit 1 (Pure Mathematics) paper from January 2019. The paper tests core algebraic fluency, coordinate geometry, sequences, differentiation, integration and trigonometry. In our experience, most marks are lost not because students lack knowledge, but because of recurring reasoning slips, sign errors, and incomplete solutions and misread instructions. Below we work through the most common mistakes, explain why they happen, and show the correct method with the exact wording that earns method and accuracy marks.
本文分析2019年1月Edexcel AS数学Unit 1（纯数学）试卷。该卷考查核心代数运算、坐标几何、数列、微分、积分与三角。根据我们的经验，多数失分并非因为知识欠缺，而是由于反复出现的推理失误、符号错误、解答不完整以及误读题意。下面我们逐一梳理最常见的错误，解释其成因，并给出能拿到方法分与准确分的规范写法。
1. Handling Negative and Fractional Indices | 负指数与分数指数处理错误
A typical question required simplification of an expression such as (8x³)⁻²/₃. The most frequent error was to treat the negative sign as making the whole term negative rather than indicating a reciprocal. A second error was to apply the fractional power in the wrong order, squaring first and then taking the cube root, which produces a different and usually messier expression. A third error was to ignore the coefficient 8 and apply the power only to x³.
一道典型题要求化简如 (8x³)⁻²/₃ 的表达式。最常见的错误是把负号当作整个项为负，而不是表示取倒数。第二个错误是分数指数的运算顺序弄反，先平方再开立方，得到不同且通常更复杂的式子。第三个错误是忽略系数 8，只对 x³ 使用该指数。
is to use the reciprocal first: (8x³)⁻²/₃ = 1/(8x³)²/₃. Then apply the fractional power as "root then power" to a root step and a power step, and deal with the negative exponent before anything else.
正确方法是先处理倒数：(8x³)⁻²/₃ = 1/(8x³)²/₃。然后把分数指数分解为"先开根再乘方"：8x³ 的立方根是 2x，平方后得 4x²，因此答案为 1/(4x²)。始终把分数指数拆成开根与乘方两步，并优先处理负指数。
For expressions such as 16⁻¹/₂ + 27²/₃, write each term separately: 16⁻¹/₂ = 1/√16 = 1/4, and 27²/₃ = (∛ 9¼. Do not attempt to combine the bases; the laws of indices apply to the same base only.
对于 16⁻¹/₂ + 27²/₃ 这类式子，应逐项处理：16⁻¹/₂ = 1/√16 = 1/4，27²/₃ = (∛27)² = 3² = 9，所以结果为 1/4 + 9 = 37/4，即 9¼。不要把底数合并；指数法则只适用于同底数。
2. Surds, Rationalising Denominators and Exact Form | 根式、分母有理化与精确形式
Surd questions penalise two habits: giving a decimal approximation when the question says "leave your answer in exact form", and rationalising a denominator incorrectly by multiplying only the numerator. For an expression such as 5/(3 + √2), multiply the numerator and denominator by the conjugate 3 − √2. The denominator becomes 3² − (√2)² = 9 − 2 = 7, giving 5(3 − √2)/7.
根式题的扣分主要来自两个习惯：题目要求"答案保留精确形式"时给出小数近似，以及分母有理化时只乘分子。对于 5/(3 + √2) 这类式子，应把分子分母同乘共轭 3 − √2。分母变为 3² − (√2)² = 9 − 2 = 7，得到 5(3 − √2)/7。
When simplifying √50 + √18, always extract the largest square factor: √50 = √25 × 2 = 5√2 and √18 = √9 × 2 = 3√2, so the sum is 8√2. Students who write √68 or 2√17 have added the radicands, which is invalid because √a + √b ≠ √a + b. The same warning applies to √a × √b = √ab, which is valid, and to √a − √b, which cannot be simplified by subtracting radicands.
化简 √50 + √18 时，要提取最大的平方因子：√50 = √25 × 2 = 5√2，√18 = √9 × 2 = 3√2，所以和为 8√2。写成 √68 或 2√17 的学生是把被开方数相加了，这是不成立的，因为 √a + √b ≠ √a + b。同样要注意 √a × √b = √ab 成立，而 √a − √b 不能通过相减被开方数来化简。
3. Discriminant: Strict Inequality and the Equal-Root Case | 判别式：严格不等式与等根情形
equation has two distinct real roots. The common mistake was to set b² − 4ac ≥ 0 instead of b² − 4ac > 0, thereby including the equal-root case. Two distinct real roots require a strictly positive discriminant.
在判别式题目中，给出含参数 k 的二次方程，要求找出使方程有两个相异实根的 k 的取值范围。常见错误是设 b² − 4ac ≥ 0 而非 b² − 4ac > 0，从而把等根情形也包含了进来。两个相异实根要求判别式严格大于零。
dates who ignore this condition often include k = 0 in the final answer, which would make the equation linear and produce only one root. Write the condition explicitly: k ≠ 0 and b² − 4ac > 0.
同样必须检查 x² 的系数。若方程为 kx² + 3x + 2 = 0，则必须 k ≠ 0，方程才是二次方程。忽略这一条件的学生常把 k = 0 写进最终答案，而 k = 0 时方程退化为一次方程，只会有一个根。应明确写出条件：k ≠ 0 且 b² − 4ac > 0。
e b² − 4ac < 0. Examiners expect the discriminant expression to be seen, then the inequality set up, then the solution of the resulting quadratic inequality. Skipping the intermediate line risks losing method marks even when the final answer is correct.
对于重根或相切情形，用 b² − 4ac = 0。对于无实根情形，用 b² − 4ac < 0。阅卷人希望看到判别式表达式、据此列出的不等式，以及随后解出的二次不等式。即使最终答案正确，跳过中间步骤也可能丢方法分。
4. Quadratic Inequalities and Interval Direction | 二次不等式与区间方向
After solving a quadratic inequality in k, students often write the final interval with the wrong direction. A common example is obtaining k > 5 and k < −1 and then writing −1 < k < 5. This is incorrect because the word "and" means intersection, and no single number can be both greater than 5 and less than −1. The correct answer is k < −1 or k > 5.
解出关于 k 的二次不等式后，学生常写出方向错误的最终区间。常见例子是得到 k > 5 且 k < −1，却写成 −1 < k < 5。这是错误的，因为"且"表示交集，不可能有数同时大于 5 又小于 −1。正确答案是 k < −1 或 k > 5。
The phrase "or" means union, which is why the two separate regions are joined by "or" rather than compressed into one interval. The safest routine is to sketch the parabola: if the coefficient of k² is positive and the inequality is > 0, the solution lies outside the roots. If the inequality is < 0, the solution lies between the roots. Alternatively, test a value from each region in the original inequality.
"或"表示并集，所以两个分离的区域要用"或"连接，而不能压缩成一个区间。最稳妥的做法是画抛物线：若 k² 的系数为正且不等式为 > 0，解在两个根之外；若不等式为 < 0，解在两个根之间。或者从每个区域取一个值代入原不等式检验。
A sign diagram is equally effective and quick to draw. Mark the critical values on a number line, test one value in each interval, and record the sign. Then read off the intervals that satisfy the inequality. This habit also protects against the related error of forgetting to rearrange the inequality to zero before factorising, for example solving x² > 4x by dividing by x instead of writing x² − 4x > 0 and factorising to x(x − 4) > 0.
符号图同样有效且绘制迅速。在数轴上标出临界值，在每个区间取一个值检验并记录符号，然后读出满足不等式的区间。这个习惯还能防止相关错误：忘记先把不等式移项到一边为零再因式分解。例如解 x² > 4x 时不能两边除以 x，而应写成 x² − 4x > 0，因式分解为 x(x − 4) > 0。
5. Domain and Range of Square Root Functions | 平方根函数的定义域与值域
A question involved f(x) = √x² − 9 n, we require x² − 9 ≥ 0. Many students wrote x ≥ 3 only, forgetting x ≤ −3. The condition |x| to x ≤ −3 or x ≥ 3. A few used x > 3, omitting the endpoints, which is also wrong because √0 is defined and equals 0.
有一题涉及 f(x) = √x² − 9。求定义域需 x² − 9 ≥ 0。许多人只写了 x ≥ 3，忘记 x ≤ −3。条件 |x| ≥ 3 给出 x ≤ −3 或 x ≥ 3。少数人写成 x > 3，遗漏端点，这同样是错误的，因为 √0 有定义且等于 0。
The domain must be written as a pair of inequalities joined by "or", or in set notation: {x : x ≤ −3} ∪ rst: x² − 9 = (x − 3)(x + 3), then determine where the product is non-negative. The product is non-negative outside the roots and negative between them.
定义域必须写成用"或"连接的两个不等式，或用集合表示：{x : x ≤ −3} ∪ {x : x ≥ 3}。只写 [3, ∞) 这样的单个区间是不完整的。当根号内是平方减常数时，先因式分解：x² − 9 = (x − 3)(x + 3)，再判断乘积非负的范围。乘积在两根之外非负，在两根之间为负。
st input with the smallest output. Remember that the domain restricts the input x, while the range describes the output f(x). For the related function g(x) = √9 − x² 3 ≤ x ≤ 3 and the range is 0 ≤ y ≤ 3, because the largest value of 9 − x² is 9 when x = 0.
至于值域，由于主平方根非负，值域为 f(x) ≥ 0。学生有时把值域写成 y ≥ 3，把最小输入与最小输出混淆。要记住定义域限制输入 x，值域描述输出 f(x)。对于相关函数 g(x) = √9 − x²，定义域为 −3 ≤ x ≤ 3，值域为 0 ≤ y ≤ 3，因为当 x = 0 时 9 − x² 取最大值 9。
6. Completing the Square, Vertex and Graph Transformations | 配方、顶点与图像变换
write (x + 3)² − 9, forgetting to add back the constant term. Another is to halve the coefficient incorrectly, writing (x + 6)² or (x + 3)² + 8 instead of subtracting.
配方始终是算术失误的高发区。对于 x² + 6x + 1，正确形式是 (x + 3)² − 9 + 1 = (x + 3)² − 8。常见错误是写成 (x + 3)² − 9，忘记把常数项加回来；另一个是系数折半出错，写成 (x + 6)² 或把符号写反为 (x + 3)² + 8。
refore (2, −3), and the minimum value is −3. Students who forget to multiply the subtracted 4 by 2 obtain the wrong vertex.
当 x² 的系数不是 1 时，先把它提出来。对于 2x² − 8x + 5，写成 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3。顶点为 (2, −3)，最小值为 −3。忘记把减去的 4 乘以 2 的学生会得到错误顶点。
ts down. A translation in the x-direction always acts in the opposite sense to the sign inside the bracket. For stretches, y = f(2x) is a horizontal stretch of scale factor ½ parallel to the x-axis, while y = 2f(x) is a vertical stretch of scale factor 2 parallel to the y-axis. Confusing the two, or describing a horizontal stretch as if it were vertical, loses marks even when the graph is drawn correctly.
在图像变换题中，运算顺序很重要。y = f(x + 2) − 3 的图像是 y = f(x) 向左平移 2 个单位、向下平移 3 个单位。x 方向的平移方向总与括号内符号相反。对于伸缩，y = f(2x) 是沿 x 方向、比例因子为 ½ 的水平伸缩，而 y = 2f(x) 是沿 y 方向、比例因子为 2 的竖直伸缩。混淆两者，或把水平伸缩描述成竖直伸缩，即使图像画对了也会丢分。
7. Factor Theorem, Remainder Theorem and Polynomial Division | 因式定理、余数定理与多项式除法
2) to decide whether (x − 2) is a factor instead of f(2). A related slip is to conclude that (x + 3) is a factor after finding f(3) = 0; in fact, f(3) = 0 shows that (x − 3) is a factor, while (x + 3) requires f(−3) = 0.
因式定理指出：(x − a) 是 f(x) 的因式当且仅当 f(a) = 0。常见错误是代入错误的值，例如判断 (x − 2) 是否为因式时去算 f(−2) 而不是 f(2)。相关失误是发现 f(3) = 0 后断定 (x + 3) 是因式；实际上 f(3) = 0 说明 (x − 3) 是因式，而 (x + 3) 需要 f(−3) = 0。
or example, dividing f(x) by (2x − 1) gives remainder f(½). Students who substitute x = 1 instead of x = ½ lose the mark even if their division is otherwise correct.
余数定理指出：f(x) 除以 (x − a) 的余数是 f(a)。当除以 (ax − b) 时，余数是 f(b/a)，而不是 f(b)。例如 f(x) 除以 (2x − 1) 的余数是 f(½)。把 x = 1 而不是 x = ½ 代入的学生，即使除法过程正确也会丢分。
In algebraic long division, the most common presentation error is to omit the zero coefficient for a missing term. When dividing x³ + 2x − 5 by (x − 1), rewrite the dividend as x³ + 0x² + 2x − 5. Without the placeholder, the columns misalign and the quotient is wrong. After division, always verify by expanding: (divisor)(quotient) + remainder should return the original polynomial. In a "show that" question where the quotient is given, full working must still be shown; writing "it works" earns nothing.
在多项式长除法中，最常见的书写错误是遗漏缺失项的零系数。用 (x − 1) 除 x³ + 2x − 5 时，应把被除式写成 x³ + 0x² + 2x − 5。没有占位项会导致列对齐错误，商也就错了。除完后务必展开验证：除式 × 商 + 余数应等于原多项式。在给出商的"证明"题中，仍须展示完整过程；只写"成立"不得分。
8. Coordinate Geometry: Gradients, Perpendicular Lines and Circles | 坐标几何：斜率、垂直关系与圆
ces a sign error. For perpendicular lines, the product of gradients is −1, so the perpendicular gradient is the negative reciprocal: if m = ⅔, the perpendicular gradient is −3/2. A frequent error is to use the reciprocal without the negative sign, giving 3/2.
在直线题中，斜率按 (y₂ − y₁)/(x₂ − x₁) 计算。只在分子或分母中颠倒顺序会造成符号错误。对于垂直直线，斜率之积为 −1，所以垂直斜率为负倒数：若 m = ⅔，垂直斜率为 −3/2。常见错误是只取倒数而忘记负号，写成 3/2。
n the form ax + by + c = 0 with integer coefficients, do not leave fractions in the final line. Multiply through by the denominator and collect all terms on one side.
求过已知点且斜率已知的直线，用 y − y₁ = m(x − x₁)，再整理成要求的形式。若题目要求形如 ax + by + c = 0 且系数为整数，最终式不要留分数。两边乘以分母，并把所有项移到一边。
erms and y terms, complete the square for each, and balance the constant: (x − 3)² − 9 + (y + 4)² − 16 − 11 = 0, so (x − 3)² + (y + 4)² = 36. The centre is (3, −4) and the radius is 6, not 36. Forgetting to take the square root of the right-hand side is one of the most common single-mark losses in the whole paper.
本卷的圆题要求通过配方从 x² + y² − 6x + 8y − 11 = 0 这类方程求圆心与半径。把 x 项与 y 项分组，分别配方并平衡常数：(x − 3)² − 9 + (y + 4)² − 16 − 11 = 0，因此 (x − 3)² + (y + 4)² = 36。圆心为 (3, −4)，半径为 6 而不是 36。忘记对右边开平方是全卷最常见的单分丢失点之一。
For a tangent to a circle, remember that the tangent is perpendicular to the radius at the point of contact. Find the gradient of the radius from the centre to the point, take its negative reciprocal, and then write the tangent line. A sketch is strongly recommended so that the sign of the tangent gradient can be checked against the diagram.
求圆的切线时，要记住切线与接触点处的半径垂直。先求圆心到该点的半径斜率，取其负倒数，再写出切线方程。强烈建议画草图，以便用图形检查切线斜率的正负是否合理。
9. Sequences and Series: nth Term, Sums and Sigma Notation | 数列与级数：通项、求和与Σ记号
sistent error is to use a + nd instead of a + (n − 1)d, which shifts every term by one place. If the first term is 7 and the common difference is 4, the 10th term is 7 + 9 × 4 = 43, not 7 + 10 × 4 = 47.
对于等差数列，通项为 a + (n − 1)d，前 n 项和为 Sₙ = ½n[2a + (n − 1)d] 或 Sₙ = ½n(a + l)，其中 l 为末项。反复出现的错误是用 a + nd 代替 a + (n − 1)d，使每一项都错位一位。若首项为 7、公差为 4，则第 10 项为 7 + 9 × 4 = 43，而不是 7 + 10 × 4 = 47。
) × 3] = ½n(3n + 7). Students who treat the lower limit as the value of the first term, or who substitute n into the expression instead of summing, produce incorrect answers. Always list the first few terms to identify a and d before using a formula.
在Σ记号中，Σ 从 r = 1 到 n 的 (3r + 2) 是首项为 5、公差为 3 的等差数列，其和为 ½n[2 × 5 + (n − 1) × 3] = ½n(3n + 7)。把下限当作首项的值，或把 n 直接代入表达式而不求和的学生会得到错误答案。使用公式前，先列出前几项以确定 a 与 d。
When a question asks for the sum from the 5th term to the 20th term, use S₂₀ − S₄, not S₂₀ − S₅. The sum from the 5th term to the 20th term excludes the first four terms, so subtract the sum of the first four. This off-by-one error appears regularly and is easily avoided by writing the terms in a
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