AS Maths Unit 1 Jan 2020 Paper Key Points | AS 数学 Unit 1 2020年1月试卷知识点精讲

📚 AS Maths Unit 1 Jan 2020 Paper Key Points | AS 数学 Unit 1 2020年1月试卷知识点精讲

The January 2020 AS Mathematics Unit 1 (Pure Mathematics) paper covered essential algebraic manipulation, coordinate geometry, trigonometry, calculus, sequences and binomial expansions. This revision article revisits the core concepts and typical question styles found in that exam, using bilingual explanations to help you build confidence and avoid common mistakes.

2020年1月的AS数学 Unit 1（纯数学）试卷涵盖了代数运算、坐标几何、三角函数、微积分、数列和二项式展开等核心内容。本文通过中英双语精讲这些常考知识点和典型题型，帮助你巩固理解，提升应试信心。

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1. Simplifying Surds and Rationalising Denominators | 根式化简与分母有理化

Surds such as √a are irrational numbers. When simplifying a surd, look for the largest square factor: √a²b = a√b for a ≥ 0 and b > 0. For example, √72 = √36×2 = 6√2.

根式如 √a 是无理数。化简时需提取最大平方因子：√a²b = a√b（a≥0，b>0）。例如 √72 = √36×2 = 6√2。

To add or subtract surds, first simplify each term so they share the same radicand. 5√3 + 2√12 becomes 5√3 + 2×2√3 = 9√3.

进行根式加减时，先将各项化为同根底数。5√3 + 2√12 化简为 5√3 + 4√3 = 9√3。

Rationalising a single-term denominator like 1/√a is done by multiplying numerator and denominator by √a. For a binomial denominator a + √b, multiply by its conjugate a – √b, and vice versa.

分母为单项根式如 1/√a，分子分母同乘 √a 即可有理化。若分母为二项式 a+√b，则乘以其共轭式 a–√b。

Example: rationalise 2/(√5 + 1) → multiply top and bottom by (√5 – 1) → [2(√5 – 1)]/(5 – 1) = (√5 – 1)/2.

例题：分母有理化 2/(√5 + 1)，上下同乘 (√5 – 1) 得到 [2(√5 – 1)]/4 = (√5 – 1)/2。

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2. Quadratic Functions – Completing the Square and the Discriminant | 二次函数 – 配方法与判别式

Writing a quadratic x² + bx + c in completed square form (x + p)² + q reveals the vertex (–p, q) and helps sketch the graph. For example, x² + 6x + 5 = (x + 3)² – 4 gives vertex (–3, –4).

将二次式 x²+bx+c 写成完全平方形式 (x+p)²+q，可读出顶点 (–p,q) 并帮助绘制图像。如 x²+6x+5 = (x+3)²–4，顶点为 (–3,–4)。

The discriminant Δ = b² – ot; Δ < 0: no real roots. This is frequently used to find conditions on unknown constants.

判别式 Δ = b²–4ac 决定根的性质。Δ>0：两个不等实根；Δ=0：一个重根；Δ<0：无实根。试卷常要求根据根的情况求未知参数范围。

Quadratic inequalities such as x² – 5x + 6 ≤ 0 should be factorised to (x – 2)(x – 3) ≤ 0, then solved using a sign diagram to give 2 ≤ x ≤ 3.

二次不等式如 x²–5x+6 ≤ 0，先因式分解为 (x–2)(x–3) ≤ 0，再用符号表判断得到 2 ≤ x ≤ 3。

Always check the critical values and the region that satisfies the inequality. Sketching the parabola or a number line helps avoid errors.

务必验证临界值并确定满足不等式的区间。画抛物线或数轴简图可防止出错。

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3. Graph Transformations | 函数图像变换

phs. y = f(x) + a shifts the graph vertically by a; y = f(x + a) shifts it horizontally by –a.

理解 y=f(x) 图像如何变换有助于写出新方程和匹配图形。y=f(x)+a 为纵向平移 a；y=f(x+a) 为横向平移 –a。

1/a. For example, y = 3sin x gives an amplitude of 3, while y = sin(2x) halves the period.

纵向伸缩：y=af(x) 纵坐标乘 a；横向伸缩：y=f(ax) 横坐标乘 1/a。例如 y=3sin x 振幅为3，y=sin(2x) 周期减半。

Reflections: y = –f(x) reflects in the x-axis; y = f(– them in the correct order: typically horizontal shifts and stretches first, then vertical.

反射：y=–f(x) 关于 x 轴对称；y=f(–x) 关于 y 轴对称。组合变换需按正确顺序进行，通常先处理水平方向再处理垂直方向。

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4. Coordinate Geometry – Straight Lines | 坐标几何 – 直线

The gradient between (x₁, y₁) and (x₂, y₂) is m = (y₂ – y₁)/(x₂ – x₁). A line's equation can be written as y – y₁ = m(x – x₁).

两点 (x₁,y₁) 和 (x₂,y₂) 间的斜率 m = (y₂–y₁)/(x₂–x₁)。直线方程可写为 y–y₁ = m(x–x₁)。

s are parallel if their gradients are equal (m₁ = m₂); they are perpendicular if m₁ × m₂ = –1. The perpendicular bisector of a segment passes through the midpoint and has slope –1/m.

两直线平行则斜率相等 (m₁=m₂)；垂直则 m₁×m₂ = –1。线段的垂直平分线经过中点且斜率为 –1/m。

Midpoint formula: M = ((x₁+x₂)/2, (y₁+y₂)/2). Distance formula: d = √[(x₂–x₁)² + (y₂–²]. These are often needed when finding the centre and radius of a circle from two points.

中点公式：M=((x₁+x₂)/2, (y₁+y₂)/2)。距离公式：d=√[(x₂–x₁)²+(y₂–y₁)²]。求圆的相关元素时常需要先算中点与距离。

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5. Circles – Centre, Radius and Tangents | 圆 – 圆心、半径与切线

The standard circle equation is (x – a)² + (y – the equation is given in expanded form, e.g. x² + y² + 2gx + 2fy + c = 0, the centre is (–g, –f) and the radius is √g² + f² – c.

圆的标准方程为 (x–a)²+(y–b)² = r²，圆心 (a,b)，半径 r。若给出一般式 x²+y²+2gx+2fy+c=0，则圆心为 (–g,–f)，半径为 √g²+f²–c。

ius is perpendicular to the tangent. Find the gradient of the radius, then use m_tangent = –1/m_radius, and apply the point-slope form.

求圆上一点 P 处的切线方程，可利用半径与切线垂直的性质。先求半径斜率，切线斜率取 –1/m_radius，再用点斜式写出方程。

When a circle is defined by a diameter or passes through given points, use distances and midpoints to set up the equation.

当圆由直径或经过定点定义时，可借助距离和中点建立方程。

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6. Trigonometric Equations and Identities | 三角方程与恒等式

ng the CAST diagram to locate all solutions within the given interval, e.g. 0° ≤ x ≤ 360°.

解基本三角方程 sin x = k 时，先用 arcsin 求主值，再借助 CAST 图找出给定区间（如 0°≤x≤360°）内的所有解。

he entire angle, then rearrange to find x, remembering to adjust the interval accordingly.

当角度为倍数如 sin(2x+30°)=0.5，先解整个角，再回代求 x，并注意将区间相应扩大。

The identity sin²θ + cos²θ ≡ 1 is essential for converting between sin and cos. For example, replace sin²θ with 1 – cos²θ to turn an equation into a quadratic in cosθ.

恒等式 sin²θ+cos²θ ≡ 1 是转换 sin 与 cos 的关键。例如用 1–cos²θ 替换 sin²θ，可将方程化为关于 cosθ 的二次方程。

Quadratic-type equations like 2sin²θ – bstituting u = sinθ, solving for u, and then finding θ. Always discard solutions outside [–1, 1].

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