📚 Exercise 1A Mastery: Algebra & Functions | A-Level AQA 数学 1A 练习精讲
Exercise 1A is the foundational gateway to A-Level Mathematics with AQA. This chapter consolidates the core algebraic techniques you mastered at GCSE and elevates them to the standard required for advanced study. In this article, we will dissect every question type you may encounter, provide step-by-step strategies, and explore common pitfalls.
Exercise 1A 是 AQA A-Level 数学的入门基石。它将在 GCSE 阶段掌握的代数技巧提升到高级研究所需要的严谨水准。本文将剖析你可能会遇到的每一类题型，提供逐步解题策略，并探讨常见错误陷阱。
We will cover index laws, surds, expanding and factorising, and solving quadratic equations. Each section is paired with bilingual explanations so you can follow along in English and Chinese. Let’s begin your journey to full marks.
我们将涵盖指数法则、根式、展开与因式分解，以及一元二次方程的求解。每一节都有中英双语对照讲解，帮助你轻松跟上思路。让我们一起向满分出发。
1. Refreshing Index Laws | 回顾指数法则
The laws of indices are the arithmetic rules for manipulating powers. They apply to all real numbers and must be memorised precisely. For any non-zero base a and integers m and n, we have:
指数法则是处理幂运算的算术规则，适用于所有实数，需要准确牢记。对于任何非零底数 a 及整数 m 和 n，有：
aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ, (aᵐ)ⁿ = aᵐⁿ
Additionally, the zero exponent and negative exponent rules are essential. A negative index indicates a reciprocal: a⁻ⁿ = 1/aⁿ. Similarly, a⁰ = 1 for any non-zero a. In Exam 1A, you will frequently simplify expressions that combine these rules in a single fraction.
此外，零指数和负指数法则也是必不可少的。负指数表示倒数：a⁻ⁿ = 1/aⁿ。同样地，对于任何非零 a，a⁰ = 1。在 1A 练习中，你常常需要在一个分式中综合运用这些法则来化简表达式。
Worked Example: Simplify (2x²y³)² ÷ (4x⁴y). First, expand the numerator: (2x²y³)² = 4x⁴y⁶. Then divide: (4x⁴y⁶) ÷ (4x⁴y) = y⁵. Observe that the x terms cancel entirely and the number coefficients cancel. This is the elegance of index laws.
示例：化简 (2x²y³)² ÷ (4x⁴y)。首先展开分子：(2x²y³)² = 4x⁴y⁶。然后相除：(4x⁴y⁶) ÷ (4x⁴y) = y⁵。观察可知，x 项完全相消，数字系数也相消。这就是指数法则的优雅之处。
When dealing with fractional indices, remember that a^(1/n) = ⁿ√a and a^(m/n) = (ⁿ√a)ᵐ. These appear frequently in Exercise 1A questions that ask you to evaluate exactly, i.e. without a calculator.
在处理分数指数时，记住 a^(1/n) = ⁿ√a，a^(m/n) = (ⁿ√a)ᵐ。这些在 1A 练习中频繁出现，考题常常要求你不用计算器进行精确计算。
2. Simplifying Surds | 化简根式
A surd is an irrational number expressed as a root, such as √2 or √3. Simplifying surds means finding the largest perfect square factor to extract outside the root. For example, √18 = √(9 × 2) = 3√2. This operation is essential for solving problems elegantly without decimal approximations.
根式是一个无理数，以根号形式表示，例如 √2 或 √3。化简根式就是找出最大的完全平方数因子并提到根号外。例如，√18 = √(9 × 2) = 3√2。这一操作对于以简便方式解题、避免十进制近似值至关重要。
You will also need to rationalise denominators. If you have 1/√a, multiply the numerator and denominator by √a to obtain √a/a. For binomial denominators like 1/(√a + √b), multiply by the conjugate (√a − √b). This removes the root from the denominator, which is often the required final form.
你还需要学会有理化分母。如果遇到 1/√a，把分子分母同时乘以 √a 得到 √a/a。对于二项式分母如 1/(√a + √b)，乘以共轭式 (√a − √b)。这样分母中就不再含根号，这通常也是标准答案的形式要求。
Worked Example: Rationalise the denominator of 3/(√5 − √2). Multiply top and bottom by (√5 + √2): numerator becomes 3(√5 + √2), denominator becomes (√5)² − (√2)² = 5 − 2 = 3. Hence the result simplifies to √5 + √2.
示例：将分母 3/(√5 − √2) 有理化。将分子分母同乘 (√5 + √2)：分子变为 3(√5 + √2)，分母为 (√5)² − (√2)² = 5 − 2 = 3。因此结果简化为 √5 + √2。
When adding or subtracting surds, treat them like collecting like terms. Only surds with the same radicand can be combined, e.g. 2√3 + 5√3 = 7√3, but 2√3 + 3√2 cannot be simplified further.
根式的加减运算类似于合并同类项。只有根号内数字相同的根式才能合并，例如 2√3 + 5√3 = 7√3，但 2√3 + 3√2 无法继续化简。
3. Expanding and Factorising Quadratics | 展开与因式分解二次式
Expanding involves removing brackets by multiplying every term inside one bracket by every term inside the other. For two binomials (x + a)(x + b), the result is x² + (a + b)x + ab. The acronym FOIL — First, Outer, Inner, Last — is a handy mnemonic.
展开式是通过将一个括号内的每一项乘以另一个括号内的每一项来去除括号。对于两个二项式 (x + a)(x + b)，结果为 x² + (a + b)x + ab。首字母缩略词 FOIL——First, Outer, Inner, Last——是一个实用的记忆方法。
Factorising is the reverse process: given a quadratic like x² + 5x + 6, find two numbers that multiply to 6 and sum to 5. Those numbers are 2 and 3, so x² + 5x + 6 = (x + 2)(x + 3). Mastery of this process is vital for solving equations and sketching graphs.
因式分解是反向过程：给定一个二次式如 x² + 5x + 6，找到两个数相乘得 6 且相加得 5。这两个数是 2 和 3，所以 x² + 5x + 6 = (x + 2)(x + 3)。熟练掌握这一过程对于求解方程和绘制函数图像极为重要。
When the coefficient of x² is not 1, you must use the “ac method”. For 2x² + 7x + 3, multiply a and c: 2 × 3 = 6. Find two numbers that multiply to 6 and add to 7 — these are 6 and 1. Rewrite 7x as 6x + 1x, then factor by grouping.
当 x² 的系数不为 1 时，需要使用 “ac 法”。对于 2x² + 7x + 3，先算 a × c = 2 × 3 = 6。找到两个数相乘得 6、相加得 7——即 6 和 1。将 7x 改写为 6x + 1x，然后分组因式分解。
Worked Example: Factorise 2x² + 7x + 3. After splitting the middle term: 2x² + 6x + x + 3. Group: 2x(x + 3) + 1(x + 3). The common factor is (x + 3), giving the final answer (2x + 1)(x + 3).
示例：对 2x² + 7x + 3 分解因式。将一次项拆分后：2x² + 6x + x + 3。分组：2x(x + 3) + 1(x + 3)。公因式为 (x + 3)，最终答案为 (2x + 1)(x + 3)。
4. Solving Quadratic Equations by Factorisation | 用因式分解法解一元二次方程
A quadratic equation in the form ax² + bx + c = 0 can often be solved by factoring into two linear factors. The key principle is the zero product property: if A × B = 0, then A = 0 or B = 0. Once factorised, set each factor equal to zero and solve for x.
形式为 ax² + bx + c = 0 的一元二次方程通常可以通过因式分解为两个一次因式来求解。关键原理是零积性质：若 A × B = 0，则 A = 0 或 B = 0。一旦因式分解完成，令每个因式等于零，即可解出 x。
Always ensure the equation is rearranged so that one side equals zero before factorising. A common mistake is trying to factorise an expression that is not set to zero, which leads to missing solutions or invalid conclusions.
在因式分解之前，务必先将方程重新整理，使一边等于零。一个常见错误是对没有化为零的表达式直接进行因式分解，这会导致漏解或得出错误结论。
Worked Example: Solve x² − 5x = 6. First, rearrange: x² − 5x − 6 = 0. Then factorise: (x − 6)(x + 1) = 0. Thus x = 6 or x = −1. These are the two roots of the equation, which are also the points where the graph of y = x² − 5x − 6 crosses the x-axis.
示例：解方程 x² − 5x = 6。首先整理：x² − 5x − 6 = 0。再因式分解：(x − 6)(x + 1) = 0。因此 x = 6 或 x = −1。这是方程的两个根，也对应 y = x² − 5x − 6 的图像与 x 轴的交点。
5. Completing the Square | 配方法
Completing the square is a powerful algebraic technique that rewrites a quadratic in the form a(x + p)² + q. For x² + bx, add and subtract (b/2)² to form a perfect square. For example, x² + 6x or x² + 6x + 10 = (x + 3)² + 1.
配方法是一种强大的代数技巧，它将二次式改写为 a(x + p)² + q 的形式。对于 x² + bx，加上并减去 (b/2)² 来构造成完全平方。例如，x² + 6x，或 x² + 6x + 10 = (x + 3)² + 1。
This form is particularly useful for finding the vertex of a parabola. In the form y = (x − h)² + k, the turning point is at (h, k), and the line of symmetry is x = h. It is also the basis for deriving the quadratic formula.
这种形式在寻找抛物线顶点时尤其有用。在 y = (x − h)² + k 的形式中，转折点为 (h, k)，对称轴为 x = h。它也是推导二次求根公式的基础。
When the coefficient of x² is not 1, factor it out first before completing the square. For 2x² + 8x + 5, first factor 2: 2[x² + 4x] + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 3. This technique is essential for a wide range of questions in Exercise 1A.
当 x² 的系数不为 1 时，先提因式，再进行配方。对于 2x² + 8x + 5，先提取 2：2[x² + 4x] + 5 = 2[(x + 2)² − 4] + 5 = 2(x + 2)² − 3。这项技巧对于 1A 练习中的大量题目都是不可或缺的。
6. The Quadratic Formula | 二次求根公式
When a quadratic equation cannot be factorised easily, the quadratic formula provides a reliable method. For ax² + bx + c = 0, the solutions are given by:
当二次方程不容易因式分解时，二次求根公式提供了一种可靠的方法。对于 ax² + bx + c = 0，解由下式给出：
x = (−b ± √(b² − 4ac)) ÷ (2a)
The expression b² − 4ac is called the discriminant, denoted Δ. The discriminant determines the number and type of roots. If Δ > 0, there are two distinct real roots. If Δ = 0, there is exactly one repeated real root. If Δ < 0, there are no real roots (only complex ones).
表达式 b² − 4ac 被称为判别式，记作 Δ。判别式决定根的个数和类型。若 Δ > 0，有两个不等实根；若 Δ = 0，有一个重根；若 Δ < 0，则没有实根（只有复数根）。
In AQA exam questions, you may be asked to find the discriminant and state the nature of the roots without solving. This tests conceptual understanding rather than mechanical computation. Always show the substitution clearly to secure method marks.
在 AQA 考试题中，你可能会被要求计算判别式并说明根的性质，而无需完整求解。这考察的是概念理解而非机械计算。务必清晰地展示代入过程，以便获得方法分。
Worked Example: Find the discriminant of 3x² + 2x − 5 = 0. Here a = 3, b = 2, c = −5. Thus Δ = 2² − 4 × 3 × (−5) = 4 + 60 = 64. Since Δ > 0 and a perfect square, the equation has two distinct rational real roots.
示例：求 3x² + 2x − 5 = 0 的判别式。这里 a = 3，b = 2，c = −5。因此 Δ = 2² − 4 × 3 × (−5) = 4 + 60 = 64。因为 Δ > 0 且为完全平方数，方程有两个不相等的有理实根。
7. Simultaneous Equations | 联立方程
Solving simultaneous equations involves finding values of variables that satisfy all given equations. In A-Level, you must handle both linear-linear and linear-quadratic systems. The substitution method is often the most reliable: rearrange one equation to express one variable, then substitute into the other.
解联立方程需要找到满足所有给定方程的变量值。在 A-Level 中，你必须处理线性-线性与线性-二次的方程组。代入法通常最可靠：先整理一个方程表示一个变量，再代入另一个方程。
For a linear-quadratic system, substitution produces a quadratic equation in one variable. Solve this quadratic (by factorising or using the formula) to find the roots, then substitute back to obtain the corresponding y-values. The solution pairs correspond to intersection points of a line and a curve.
对于线性-二次方程组，代入后会得到一个一元二次方程。解出这个二次方程（用因式分解或求根公式）得到根，再代回求出对应的 y 值。每一组解对应直线与曲线的一个交点。
Alternatively, the elimination method works well for linear-linear systems. Add or subtract multiples of the equations to eliminate one variable, solve for the other, and back-substitute to find the remaining value. Whichever method you choose, always check your answers by substituting into both original equations.
另外，消元法在解线性-线性方程组时也很高效。将方程乘以适当倍数后相加或相减，消去一个未知数，解出另一个，然后代回求出剩下的未知数。无论使用哪种方法，都务必将答案代回原方程组进行验算。
Worked Example: Solve y = x² − 3 and y = 2x + 1. Substituting: 2x + 1 = x² − 3 → x² − 2x − 4 = 0. Then x = 1 ± √5. Substituting back yields the paired y-values.
示例：解方程组 y = x² − 3 与 y = 2x + 1。代入得 2x + 1 = x² − 3 → x² − 2x − 4 = 0。因此 x = 1 ± √5。代回即得对应的 y 值。
8. Inequalities Involving Quadratics | 含二次项的不等式
Solving quadratic inequalities requires a visual understanding of the parabola. The first step is always to rearrange to zero on one side. Factorise the quadratic and identify the critical values — the roots where the expression equals zero.
解二次不等式需要对抛物线有直观的理解。第一步始终是把不等式整理为一边为零的形式。将二次式因式分解，找出临界值——即表达式等于零的根。
Sketch a graph or use a sign table to determine which intervals satisfy the inequality. For example, if (x − 1)(x − 3) > 0, the valid x-values lie outside the interval [1, 3]. If the inequality is ≤ 0, then the solution is between the roots inclusive.
画出草图或使用符号表来确定哪些区间满足该不等式。例如，若 (x − 1)(x − 3) > 0，则满足条件的 x 值位于区间 [1, 3] 之外。若不等式为 ≤ 0，则解在两根之间且包含端点。
Pay careful attention to the inequality sign. A strict inequality (< or >) uses open intervals and excludes endpoints, while non-strict inequalities (≤ or ≥) include endpoints. In Exercise 1A, you should always express your answer using set notation or interval notation as required by the question.
请特别留意不等号方向。严格不等式 (< 或 >) 使用开区间并排除端点，而非严格不等式 (≤ 或 ≥) 则包含端点。在 1A 练习中，应根据题目要求使用集合符号或区间记号表达最终答案。
Worked Example: Solve x² − 5x + 6 < 0. Factorise: (x − 2)(x − 3) < 0. Drawing the graph, the curve lies below the x-axis between the roots, so 2 < x < 3 is the solution.
示例：解不等式 x² − 5x + 6 < 0。因式分解得 (x − 2)(x − 3) < 0。图像的曲线在两根之间位于 x 轴下方，所以解集为 2 < x < 3。
9. Function Notation and Evaluation | 函数记号与求值
Functions are a central theme in A-Level Mathematics. The notation f(x) denotes the output of the function f when the input is x. To evaluate a function, substitute the given x-value into every occurrence of x in the expression and simplify.
函数是 A-Level 数学的核心主题之一。记号 f(x) 表示函数 f 在输入为 x 时的输出。求函数值的方法是把给定的 x 值代入表达式中的每一个 x 并化简。
Exercise 1A often includes composite functions, where one function is applied after another. The composition f(g(x)) means first apply g to x, then apply f to the result. Work from the inside out, and be precise about the order — f(g(x)) is generally not equal to g(f(x)).
1A 练习常常包含复合函数，即先施加一个函数，再施加另一个。复合 f(g(x)) 表示先对 x 应用 g，然后将 f 应用于结果。从内向外求值，并注意顺序——f(g(x)) 通常不等于 g(f(x))。
Also be prepared for piecewise functions, which have different rules for different intervals of x. When evaluating a piecewise function, first determine which interval the given x-value falls into, then apply the corresponding formula.
你还需要准备分段函数，它在不同的 x 区间上有不同的规则。求分段函数的值时，先确定给定的 x 属于哪个区间，再应用相应的公式。
Worked Example: If f(x) = x² + 1, then f(3) = 3² + 1 = 10. If further g(x) = 2x, then g(f(3)) = g(10) = 20. Note the order: first evaluate f(3), then feed it into g.
示例：若 f(x) = x² + 1，则 f(3) = 3² + 1 = 10。若再设 g(x) = 2x，那么 g(f(3)) = g(10) = 20。注意顺序：先求 f(3)，再把它代入 g。
10. Domain and Range | 定义域与值域
The domain of a function is the set of all permitted input values x, while the range is the set of all possible output values f(x). In Exercise 1A, you must identify restrictions on the domain, such as avoiding division by zero or taking the square root of a negative number.
函数的定义域是所有允许的输入值 x 的集合，而值域是所有可能的输出值 f(x) 的集合。在 1A 练习中，你需要识别定义域的限制，例如避免除以零，或者对负数开平方根。
For linear and quadratic functions, the domain is usually all real numbers, ℝ. However, functions involving fractions or square roots require careful analysis. For example, f(x) = √(x − 2) requires x ≥ 2.
对于线性和二次函数，定义域通常是全体实数，即 ℝ。然而，涉及分式或平方根的函数需要仔细分析。例如，f(x) = √(x − 2) 要求 x ≥ 2。
To determine the range, consider the graph or use algebraic reasoning. A quadratic in the form f(x) = a(x − h)² + k has a minimum value (a > 0) or maximum value (a < 0) at y = k. Thus the range is either [k, ∞) or (−∞, k].
要确定值域，可以借助函数图像或代数推理。形式为 f(x) = a(x − h)² + k 的二次函数在 y = k 处有最小值（a > 0）或最大值（a < 0）。因此值域为 [k, ∞) 或 (−∞, k]。
Worked Example: Find the domain and range of f(x) = 1/(x − 3). The denominator cannot be zero, so x ≠ 3. As x approaches 3, f(x) approaches infinity; therefore the range is all real numbers except 0.
示例：求 f(x) = 1/(x − 3) 的定义域和值域。分母不能为零，所以 x ≠ 3。当 x 趋近于 3 时，f(x) 趋向无穷大；因此值域为除 0 以外的所有实数。
11. Sketching Quadratic Graphs | 绘制二次函数图像
Sketching a quadratic graph y = ax² + bx + c requires key features: the y-intercept, the x-intercepts (roots), and the turning point (vertex). The y-intercept is simply c, the point where the graph crosses the y-axis. The x-intercepts are the roots found by solving the equation.
绘制 y = ax² + bx + c 的二次函数图像需要确定几个关键特征：y 截距、x 截距（根）和转折点（顶点）。y 截距就是 c，即图像与 y 轴的交点。x 截距通过解方程得到，即根。
The turning point can be found by completing the square or by using x = −b/(2a) for the x-coordinate. The sign of a determines the direction: if a > 0, the parabola opens upwards (a “smile”); if a < 0, it opens downwards (a "frown").
转折点可以通过配方法求得，也可用 x = −b/(2a) 求横坐标。二次项系数 a 的符号决定开口方向：若 a > 0，抛物线开口向上（笑脸形）；若 a < 0，开口向下（倒脸形）。
In an exam, a correct sketch should clearly label all intercepts and the vertex. Use a smooth curve and avoid drawing ruler-straight lines. Mark the axis of symmetry with a dashed line where appropriate, as this demonstrates your understanding of the graph’s structure.
在考试中，正确的草图应清晰标注所有截距和顶点。用平滑曲线绘制，避免画出笔直的折线。在适当位置用虚线标出对称轴，以展示你对图像结构的理解。
Worked Example: Sketch y = x² − 2x − 3. First, y-intercept = −3. Roots: (x − 3)(x + 1) = 0, so x = 3 and x = −1. Vertex: x = 1 (midpoint of roots), giving y = 1 − 2 − 3 = −4. The graph opens upwards.
示例：绘制 y = x² − 2x − 3 的草图。首先，y 截距 = −3。根：由 (x − 3)(x + 1) = 0，得 x = 3 和 x = −1。顶点：x = 1（两点的中点），代回得 y = 1 − 2 − 3 = −4。图像开口向上。
12. Practice Question Mix and Exam Tips | 综合练习与考试技巧
Exercise 1A culminates in mixed exercises that integrate multiple topics. To excel, you should practice a wide range of questions and time yourself under exam conditions. Review any mistakes thoroughly to avoid repeating them.
1A 练习以综合题收尾，整合多个知识点。为了取得好成绩，你应该练习各种类型的题目，并在模拟考试条件下计时。仔细复盘每一个错误，避免重蹈覆辙。
Show every step of your working clearly. In AQA mark schemes, method marks are awarded even when the final answer is incorrect. Algebraic mistakes can often be spotted early if you write neatly and check each line for continuity.
清晰地写下每一步过程。在 AQA 评分方案中，即使最终答案错误，也能获得方法分。整齐书写并检查每一步的连贯性，可以帮助你在早期发现代数错误。
Memorise key formulas from the AQA formula booklet and know when to apply them. Practice recognising which factoring technique to use, when to complete the square, and when the quadratic formula is necessary. Speed and accuracy come from repeated practice.
熟记 AQA 公式手册中的关键公式，并知道何时应用。练习判断何时使用哪种因式分解技巧、何时配方、何时需要求根公式。速度和准确性来自于反复练习。
Finally, use the exam checklist: read the question twice, identify the required topic, plan your approach, execute the algebra, and verify your solution by substituting back into the original equation. This systematic routine will maximise your marks.
最后，请遵循答题清单：阅读题目两遍、确定所涉及的知识点、规划解题思路、执行代数运算，并将解代回原式验证。这套系统化的步骤将帮助你最大化得分。
Here is a summary table of key concepts and the strategies you should employ:
下表汇总了关键概念及应采用的解题策略：
| Topic/主题 | Key Strategy/核心策略 |
| Index Laws 指数法则 | Combine exponents; apply to fractions 合并指数；应用于分式 |
| Surds 根式 | Extract squares; rationalise denominators 提取平方因子；有理化分母 |
| Quadratics 二次式 | Factorise; complete square; use formula 因式分解；配方；求根公式 |
| Discriminant 判别式 | Evaluate b² − 4ac; state root nature 计算 b² − 4ac；说明根的性质 |
| Inequalities 不等式 | Find roots; sketch sign chart 求根；绘制符号图 |
Remember, the foundation you build in Exercise 1A will support every subsequent topic in A-Level Mathematics. Take your time, practice diligently, and do not hesitate to revisit this material before your exams.
请记住，你在 1A 练习中打下的基础将支撑 A-Level 数学中的所有后续主题。请分配充足的时间、勤加练习，并在考试前随时回顾本材料。
In conclusion, mastering Exercise 1A is non-negotiable for success in AQA A-Level Mathematics. By working through these twelve key areas, you will develop both the technical fluency and the conceptual depth required. Always prioritise accuracy in algebra, clarity of method, and rigorous checking.
总而言之，熟练掌握 1A 练习是 AQA A-Level 数学成功的关键所在。通过攻克这 12 个核心领域，你将在技术操作和概念理解上双双达标。请始终把代数准确性、方法清晰度和严谨验算放在首位。
We hope this bilingual guide has clarified every aspect of the chapter. Use it alongside your textbook, and attempt every question with confidence. Your journey to an A* begins here.
我们希望这份双语指南已帮助你透彻理解了本章的各个方面。请配合教材使用本文，并自信地尝试每一道题。你通往 A* 的旅程就从这里开始。
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