Mastering Edexcel A-Level Pure Mathematics: Review Exercise 1 | 精通Edexcel A-Level纯数：复习练习1

📚 Mastering Edexcel A-Level Pure Mathematics: Review Exercise 1 | 精通Edexcel A-Level纯数：复习练习1

Review Exercise 1 in the Edexcel A Level Pure Mathematics course brings together foundational skills from the early chapters, including algebraic simplification, quadratic functions, inequalities, coordinate geometry, polynomial division, the binomial expansion, trigonometry, and graph transformations. A strong grasp of these topics is vital for tackling more advanced problems and succeeding in the final examinations. This bilingual guide walks through key techniques and worked examples, pairing each English explanation with a Chinese translation to reinforce understanding for learners who study in both languages.

Edexcel A Level纯数课程中的复习练习1汇集了前几章的基础技能，涉及代数化简、二次函数、不等式、坐标几何、多项式除法、二项式展开、三角学以及图像变换。扎实掌握这些主题对于攻克更复杂的问题并在终考中取得成功至关重要。本双语指南梳理了关键技巧和典型例题，将每一段英文解释与对应的中文翻译配对，帮助双语学习者加深理解。

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1. Simplifying Algebraic Expressions | 简化代数表达式

In Review Exercise 1, you are often asked to expand and simplify products of binomials or to collect like terms in polynomials. For example, (2x + 1)(x – 3) + 4x can be simplified by first expanding: 2x² – 6x + x – 3 + 4x, then combining like terms to obtain 2x² – x – 3. Always remember to apply the distributive law carefully and watch for negative signs.

在复习练习1中，经常需要展开并化简二项式的乘积，或合并多项式中的同类项。例如，(2x+1)(x-3)+4x 可以先展开：2x² – 6x + x – 3 + 4x，再合并同类项得到 2x² – x – 3。务必仔细运用分配律并留意负号。

Another common task is factorising expressions by taking out the highest common factor or by grouping. For instance, 6x³ – 4x² + 2x = 2x(3x² – 2x + 1). Recognising a difference of two squares such as 9x² – 16 = (3x – 4)(3x + 4) is also essential.

另一个常见任务是提取公因式或分组分解。例如，6x³ – 4x² + 2x = 2x(3x² – 2x + 1)。识别平方差公式如 9x² – 16 = (3x – 4)(3x + 4) 同样关键。

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2. Solving Quadratic Equations | 解二次方程

Quadratic equations in the form ax² + bx + c = 0 can be solved by factorising, completing the square, or using the quadratic formula. For factorisation, look for two numbers that multiply to ac and add to b. Example: x² – 5x + 6 = 0 factorises to (x-2)(x-3)=0, giving solutions x = 2 and x = 3.

形如 ax² + bx + c = 0 的二次方程可以通过因式分解、配方法或求根公式来求解。对于因式分解，找到两个乘起来等于 ac、加起来等于 b 的数。例如：x² – 5x + 6 = 0 可分解为 (x-2)(x-3)=0，解为 x = 2 和 x = 3。

When the quadratic does not factorise easily, use the quadratic formula:

x = [-b ± √(b² – 4ac)] / (2a)

Always check that the expression under the square root, the discriminant, is non-negative to obtain real solutions.

当二次方程不易因式分解时，使用求根公式：

x = [-b ± √(b² – 4ac)] / (2a)

务必检查根号内的判别式是否非负，以获得实数解。

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3. Completing the Square and the Vertex | 配方法与顶点

Completing the square rewrites a quadratic in the form a(x+p)² + q, which immediately reveals the vertex of its graph. For x² + 6x + 1, we take half of 6, square it: (6/2)² = 9, then write (x+3)² – 9 + 1 = (x+3)² – 8. Hence the vertex is at (-3, -8).

配方法将二次函数写成 a(x+p)² + q 的形式，从而直接显示出图像顶点的位置。对于 x² + 6x + 1，取6的一半平方得到 (6/2)² = 9，于是 (x+3)² – 9 + 1 = (x+3)² – 8。因此顶点坐标为 (-3, -8)。

If the coefficient of x² is not 1, factor it out first: 2x² – 8x + 5 = 2[x² – 4x] + 5 = 2[(x-2)² – 4] + 5 = 2(x-2)² – 8 + 5 = 2(x-2)² – 3. The vertex is (2, -3). This form is also useful for solving equations and sketching parabolas.

如果 x² 的系数不是1，先提取公因数：2x² – 8x + 5 = 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)。这种形式在解方程和绘制抛物线时也非常有用。

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4. The Discriminant and Nature of Roots | 判别式与根的性质

The discriminant of ax² + bx + c is Δ = b² – 4ac. It tells us how many real roots the quadratic has without solving it. The relationship is summarised below:

二次方程 ax² + bx + c 的判别式为 Δ = b² – 4ac。它能在不求解的情况下告诉我们实数根的个数。其关系总结如下：

Δ > 0	two distinct real roots	两个不等实根
Δ = 0	one repeated real root	一个重实根
Δ < 0	no real roots	无实根

For example, in 3x² + 2x – 1 = 0, Δ = 2² – 4(3)(-1) = 4 + 12 = 16 > 0, so there are two distinct real solutions. Problems may also ask you to find a value of k such that a quadratic has equal roots: set Δ = 0 and solve for k.

例如，对于 3x² + 2x – 1 = 0，Δ = 2² – 4(3)(-1) = 4 + 12 = 16 > 0，因此有两个不等实根。题目也可能要求找到使二次方程有等根的 k 值：令 Δ = 0 并解出 k。

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5. Inequalities and Set Notation | 不等式与集合记号

Linear and quadratic inequalities appear frequently in this review. To solve 2x – 3 < 5, we add 3 and divide by 2 to get x < 4. For quadratic inequalities such as x² - 4x - 5 ≤ 0, first find the critical values by solving x² - 4x - 5 = 0 giving x = -1 and x = 5. Then test intervals to obtain -1 ≤ x ≤ 5. Writing the solution in set notation: {x : -1 ≤ x ≤ 5}.

一次和二次不等式在本练习中频繁出现。求解 2x – 3 < 5，两边加3再除以2得 x < 4。对于如 x² - 4x - 5 ≤ 0 的二次不等式，先解方程 x² - 4x - 5 = 0 得到临界值 x = -1 和 x = 5，然后测试区间得到 -1 ≤ x ≤ 5。用集合记号表示为 {x : -1 ≤ x ≤ 5}。

Remember that multiplying or dividing by a negative number reverses the inequality sign. Graphical methods, such as sketching the parabola, can also help visualise the solution set.

记住，乘以或除以负数会改变不等号的方向。图形方法，如绘制抛物线，也有助于直观理解解集。

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6. Straight Line Graphs and Coordinate Geometry | 直线图与坐标几何

Essential skills include finding the gradient and equation of a line, and calculating distances and midpoints. Gradient m = (y₂ – y₁)/(x₂ – x₁). The equation of a line with gradient m passing through (x₁, y₁) is y – y₁ = m(x – x₁). The distance between two points is √[(x₂ – x₁)² + (y₂ – y₁)²]. The midpoint is ((x₁+x₂)/2, (y₁+y₂)/2).

基本技能包括求直线的斜率和方程，以及计算距离和中点。斜率 m = (y₂ – y₁)/(x₂ – x₁)。经过点 (x₁, y₁) 且斜率为 m 的直线方程为 y – y₁ = m(x – x₁)。两点间的距离为 √[(x₂ – x₁)² + (y₂ – y₁)²]。中点为 ((x₁+x₂)/2, (y₁+y₂)/2)。

Parallel lines share the same gradient; perpendicular lines have gradients whose product is -1. For instance, if a line has gradient 3, a perpendicular line’s gradient is -1/3. In the exam, you may need to prove a right angle or find the perpendicular bisector of a segment.

平行线具有相同的斜率；垂直线的斜率乘积为 -1。例如，若一直线的斜率为3，则与之垂直的直线斜率为 -1/3。考试中可能需要证明直角或求线段的垂直平分线。

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7. Polynomial Division and Factor Theorem | 多项式除法与因式定理

Dividing a cubic or quartic by a linear factor can be done by long division or synthetic division. For example, divide x³ – 2x² – 4 by (x – 3):

(x³ – 2x² + 0x – 4) ÷ (x – 3) = x² + x + 3 with remainder 5.

This means x³ – 2x² – 4 = (x – 3)(x² + x + 3) + 5.

多项式除法，如一个三次或四次多项式除以一个线性因式，可通过长除法或综合除法进行。例如，用 (x – 3) 除 x³ – 2x² – 4：

(x³ – 2x² + 0x – 4) ÷ (x – 3) = x² + x + 3 余 5。

这意味着 x³ – 2x² – 4 = (x – 3)(x² + x + 3) + 5。

The Factor Theorem states that if f(p)=0, then (x-p) is a factor of f(x). Combined with division, we can fully factorise a polynomial. For instance, given f(x)=x³ – x² – 4x + 4, since f(1)=0, (x-1) is a factor, and dividing gives (x-1)(x² – 4) = (x-1)(x-2)(x+2).

因式定理指出，若 f(p)=0，则 (x-p) 是 f(x) 的因式。结合除法即可将多项式完全分解。例如，已知 f(x)=x³ – x² – 4x + 4，由于 f(1)=0，故 (x-1) 为因式，除法后得 (x-1)(x² – 4) = (x-1)(x-2)(x+2)。

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8. The Binomial Expansion | 二项式展开

The binomial expansion for positive integer powers uses Pascal’s triangle or the nCr formula. The expansion of (a + b)ⁿ is given by:

Σ [nCr × aⁿ⁻ʳ × bʳ] for r = 0 to n

For (1 + x)ⁿ, this becomes 1 + nx + [n(n-1)/2!] x² + [n(n-1)(n-2)/3!] x³ + …

二项式定理对于正整数次幂可使用杨辉三角或组合数公式。(a + b)ⁿ 的展开式为：

∑ [nCr × aⁿ⁻ʳ × bʳ] （r 从 0 到 n）

对于 (1 + x)ⁿ，展开式为 1 + nx + [n(n-1)/2!] x² + [n(n-1)(n-2)/3!] x³ + …

A typical question asks for the first three terms in the expansion of (2 – 3x)⁴. Use a=2, b=-3x, n=4:

Term 1: ⁴C₀ × 2⁴ × (-3x)⁰ = 1 × 16 × 1 = 16

Term 2: ⁴C₁ × 2³ × (-3x)¹ = 4 × 8 × (-3x) = -96x

Term 3: ⁴C₂ × 2² × (-3x)² = 6 × 4 × 9x² = 216x²

So (2 – 3x)⁴ ≈ 16 – 96x + 216x².

典型的题目要求写出 (2 – 3x)⁴ 展开式的前三项。设 a=2, b=-3x, n=4：

第一项：⁴C₀ × 2⁴ × (-3x)⁰ = 16

第二项：⁴C₁ × 2³ × (-3x)¹ = -96x

第三项：⁴C₂ × 2² × (-3x)² = 216x²

因此 (2 – 3x)⁴ ≈ 16 – 96x + 216x²。

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9. Trigonometric Ratios and Equations | 三角比与三角方程

Exact values for sin, cos and tan at 30°, 45° and 60° must be memorised. For example, sin 30° = 1/2, cos 45° = 1/√2, tan 60° = √3. The sine and cosine rules are used in non-right-angled triangles:

a/sin A = b/sin B = c/sin C

a² = b² + c² – 2bc cos A

30°、45°、60° 的正弦、余弦和正切精确值必须牢记。例如 sin 30° = 1/2, cos 45° = 1/√2, tan 60° = √3。正弦定理和余弦定理适用于非直角三角形：

a/sin A = b/sin B = c/sin C

a² = b² + c² – 2bc cos A

Simple trigonometric equations like 2 sin x = 1 for 0° ≤ x ≤ 360° give sin x = 0.5, so x = 30° or 150°. Always consider the CAST diagram to find all solutions in the given interval.

简单的三角方程如 2 sin x = 1 在 0° ≤ x ≤ 360° 范围内，解得 sin x = 0.5，因此 x = 30° 或 150°。始终使用 CAST 图来寻找给定区间内的所有解。

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10. Graphs and Transformations of Functions | 函数图像与变换

You should be comfortable sketching basic functions like y = x², y = √x, y = 1/x, and y = sin x. Transformations include translations and reflections. For y = f(x), the graph of y = f(x) + a is a vertical translation by a; y = f(x + a) is a horizontal shift by -a; y = -f(x) reflects in the x-axis; y = f(-x) reflects in the y-axis. Stretches: y = a f(x) stretches vertically by factor a, and y = f(ax) stretches horizontally by factor 1/a.

你应能熟练绘制 y = x², y = √x, y = 1/x 和 y = sin x 等基本函数图像。变换包括平移和反射。对于 y = f(x)，y = f(x) + a 表示垂直平移 a 个单位；y = f(x + a) 表示水平平移 -a 个单位；y = -f(x) 关于 x 轴反射；y = f(-x) 关于 y 轴反射。伸缩变换：y = a f(x) 在竖直方向上拉伸因子 a，y = f(ax) 在水平方向上拉伸因子 1/a。

Applying combinations of transformations correctly is often tested. For instance, starting with y = x², the graph of y = 2(x – 3)² + 1 can be obtained by translating right by 3, stretching vertically by factor 2, then translating up by 1. The order of operations matters when horizontal transformations are combined with horizontal reflections or stretches.

复合变换的正确应用经常考到。例如，从 y = x² 出发，y = 2(x – 3)² + 1 的图像可以通过先向右平移3个单位，再垂直拉伸为原来的2倍，然后向上平移1个单位得到。当水平变换与水平反射或伸缩结合时，操作顺序很重要。

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