📚 AS Mathematics: Typical Example Questions with Step-by-Step Solutions | AS 数学：典型例题详解
Worked examples are one of the most effective ways to master AS-level Mathematics. This article presents carefully selected typical problems from the Pure Mathematics syllabus, each broken down into clear, logical steps. By following the reasoning and practicing similar questions, you will deepen your understanding of key concepts, improve your problem-solving technique, and gain confidence for your examinations.
典型例题是掌握 AS 数学最有效的方式之一。本文精选了纯数学大纲中的典型问题，每道题都拆解成清晰、有逻辑的步骤。通过跟随推理过程并练习类似题目，你能够加深对核心概念的理解，提高解题技巧，并为考试建立信心。
1. Solving Quadratic Equations | 解二次方程
Solve the equation 2x² – 3x – 2 = 0 by factorisation and verify using the quadratic formula.
用因式分解法解方程 2x² – 3x – 2 = 0，并用二次公式验证。
Factorising: look for two numbers that multiply to 2 × (-2) = -4 and add to -3. These are -4 and 1. Split the middle term: 2x² – 4x + x – 2 = 0. Group: 2x(x – 2) + 1(x – 2) = 0, giving (2x + 1)(x – 2) = 0.
因式分解：寻找两个数相乘得 2×(-2)=-4，相加得-3，即 -4 和 1。将中间项拆开：2x² – 4x + x – 2 = 0。分组：2x(x – 2) + 1(x – 2) = 0，得到 (2x + 1)(x – 2) = 0。
Set each factor equal to zero: 2x + 1 = 0 ⇒ x = -1/2, and x – 2 = 0 ⇒ x = 2.
令每个因式为零：2x + 1 = 0 ⇒ x = -1/2，以及 x – 2 = 0 ⇒ x = 2。
Quadratic formula check: a=2, b=-3, c=-2. x = [-(-3) ± √((-3)² – 4×2×(-2))] / (2×2) = (3 ± √(9 + 16)) / 4 = (3 ± √25) / 4 = (3 ± 5)/4. Thus x = 8/4 = 2 or x = -2/4 = -1/2.
二次公式验证：a=2, b=-3, c=-2。x = [-(-3) ± √((-3)² – 4×2×(-2))] / (2×2) = (3 ± √(9+16))/4 = (3 ± √25)/4 = (3±5)/4。因此 x = 8/4 = 2 或 x = -2/4 = -1/2。
Solutions: x = 2, x = -½
解：x = 2, x = -½
2. Quadratic Inequalities | 二次不等式
Find the set of values of x for which x² – 5x + 6 ≤ 0.
求满足 x² – 5x + 6 ≤ 0 的 x 的取值范围。
First factorise: x² – 5x + 6 = (x – 2)(x – 3). The critical values are x = 2 and x = 3.
首先因式分解：x² – 5x + 6 = (x – 2)(x – 3)。临界值为 x=2 和 x=3。
Sketch the quadratic: it opens upwards (coefficient of x² is positive), so it is ≤ 0 between the roots.
画二次函数草图：开口向上（x² 系数为正），因此在两根之间函数值 ≤ 0。
Thus the inequality holds when 2 ≤ x ≤ 3.
因此不等式在 2 ≤ x ≤ 3 时成立。
We can test a value, e.g. x=2.5: (2.5-2)(2.5-3) = (0.5)(-0.5) = -0.25 ≤ 0, confirming the interval.
可以取一个值测试，如 x=2.5：(2.5-2)(2.5-3) = (0.5)(-0.5) = -0.25 ≤ 0，证实该区间。
Solution set: {x ∈ ℝ : 2 ≤ x ≤ 3}
解集：{x ∈ ℝ : 2 ≤ x ≤ 3}
3. Simultaneous Equations (Linear and Quadratic) | 联立方程组（一次与二次）
Solve the simultaneous equations: y = 2x + 1 and y = x² + x – 1.
解联立方程组：y = 2x + 1 和 y = x² + x – 1。
Equate the expressions for y: 2x + 1 = x² + x – 1.
令两个 y 表达式相等：2x + 1 = x² + x – 1。
Rearrange to form a quadratic: 0 = x² + x – 1 – 2x – 1 ⇒ x² – x – 2 = 0.
移项得到二次方程：0 = x² + x – 1 – 2x – 1 ⇒ x² – x – 2 = 0。
Factorise: (x – 2)(x + 1) = 0, so x = 2 or x = -1.
因式分解：(x – 2)(x + 1) = 0，所以 x = 2 或 x = -1。
Substitute back: when x = 2, y = 2(2) + 1 = 5; when x = -1, y = 2(-1) + 1 = -1.
代回原式：当 x = 2 时，y = 2(2) + 1 = 5；当 x = -1 时，y = 2(-1) + 1 = -1。
Solutions: (2, 5) and (-1, -1)
解为：(2, 5) 和 (-1, -1)
4. Composite Functions | 复合函数
Given f(x) = 2x + 3 and g(x) = x² – 1, find fg(x) and gf(x), and evaluate fg(2).
已知 f(x) = 2x + 3 且 g(x) = x² – 1，求 fg(x) 和 gf(x)，并计算 fg(2)。
fg(x) means f(g(x)). Substitute g(x) into f: f(g(x)) = 2(g(x)) + 3 = 2(x² – 1) + 3 = 2x² – 2 + 3 = 2x² + 1.
fg(x) 表示 f(g(x))。将 g(x) 代入 f：f(g(x)) = 2(g(x)) + 3 = 2(x² – 1) + 3 = 2x² – 2 + 3 = 2x² + 1。
gf(x) means g(f(x)). Substitute f(x) into g: g(f(x)) = (f(x))² – 1 = (2x + 3)² – 1 = 4x² + 12x + 9 – 1 = 4x² + 12x + 8.
gf(x) 表示 g(f(x))。将 f(x) 代入 g：g(f(x)) = (f(x))² – 1 = (2x + 3)² – 1 = 4x² + 12x + 9 – 1 = 4x² + 12x + 8。
To evaluate fg(2): 2(2)² + 1 = 2×4 + 1 = 9.
计算 fg(2)：2(2)² + 1 = 2×4 + 1 = 9。
fg(x) = 2x² + 1, gf(x) = 4x² + 12x + 8, fg(2) = 9
fg(x) = 2x² + 1，gf(x) = 4x² + 12x + 8，fg(2) = 9
5. Inverse Functions | 反函数
Find the inverse function of f(x) = (3x – 2) / (x + 1), stating its domain.
求 f(x) = (3x – 2) / (x + 1) 的反函数，并说明其定义域。
Write y = (3x – 2)/(x + 1). To find the inverse, swap x and y: x = (3y – 2)/(y + 1).
令 y = (3x – 2)/(x + 1)。求反函数时交换 x 和 y：x = (3y – 2)/(y + 1)。
Multiply both sides by (y + 1): x(y + 1) = 3y – 2 ⇒ xy + x = 3y – 2.
两边乘以 (y + 1)：x(y + 1) = 3y – 2 ⇒ xy + x = 3y – 2。
Collect y terms: xy – 3y = -x – 2 ⇒ y(x – 3) = -x – 2 ⇒ y = (-x – 2)/(x – 3).
合并含 y 的项：xy – 3y = -x – 2 ⇒ y(x – 3) = -x – 2 ⇒ y = (-x – 2)/(x – 3)。
Thus f⁻¹(x) = (-x – 2)/(x – 3). The domain of f⁻¹ is all real numbers except where the denominator is zero, i.e. x ≠ 3.
因此 f⁻¹(x) = (-x – 2)/(x – 3)。反函数的定义域是所有实数，但分母为零的点除外，即 x ≠ 3。
f⁻¹(x) = (-x – 2)/(x – 3), domain: x ∈ ℝ, x ≠ 3
f⁻¹(x) = (-x – 2)/(x – 3)，定义域：x ∈ ℝ，x ≠ 3
6. Differentiation of Polynomials | 多项式微分
Differentiate y = 4x³ – 5x² + 2x – 7, and find the gradient of the curve at x = 2.
求 y = 4x³ – 5x² + 2x – 7 的导数，并求曲线在 x = 2 处的斜率。
Apply the power rule: d/dx (xⁿ) = n xⁿ⁻¹. Differentiate term by term.
应用幂法则：d/dx (xⁿ) = n xⁿ⁻¹。逐项求导。
dy/dx = 4 × 3x² – 5 × 2x¹ + 2 × 1x⁰ – 0 = 12x² – 10x + 2.
dy/dx = 4×3x² – 5×2x¹ + 2×1x⁰ – 0 = 12x² – 10x + 2。
At x = 2, substitute into dy/dx: 12(2)² – 10(2) + 2 = 12×4 – 20 + 2 = 48 – 20 + 2 = 30.
在 x = 2 处，代入 dy/dx：12(2)² – 10(2) + 2 = 12×4 – 20 + 2 = 48 – 20 + 2 = 30。
The gradient of the curve at x = 2 is 30.
曲线在 x = 2 处的斜率为 30。
dy/dx = 12x² – 10x + 2, gradient at x=2 is 30
dy/dx = 12x² – 10x + 2，x=2 处斜率为 30
7. Equation of Tangent and Normal | 切线与法线方程
Find the equations of the tangent and the normal to the curve y = x² + 3x at the point where x = -1.
求曲线 y = x² + 3x 在 x = -1 处的切线方程和法线方程。
First, find the y-coordinate: y = (-1)² + 3(-1) = 1 – 3 = -2, so the point is P(-1, -2).
首先求 y 坐标：y = (-1)² + 3(-1) = 1 – 3 = -2，因此该点为 P(-1, -2)。
Differentiate to find the gradient function: dy/dx = 2x + 3. At x = -1, m_tangent = 2(-1) + 3 = 1.
求导得斜率函数：dy/dx = 2x + 3。在 x = -1 处，切线斜率 m_t = 2(-1) + 3 = 1。
Tangent equation: y – y₁ = m(x – x₁) ⇒ y – (-2) = 1(x – (-1)) ⇒ y + 2 = x + 1 ⇒ y = x – 1.
切线方程：y – y₁ = m(x – x₁) ⇒ y – (-2) = 1(x – (-1)) ⇒ y + 2 = x + 1 ⇒ y = x – 1。
Normal gradient: m_normal = -1 / m_tangent = -1/1 = -1.
法线斜率：m_n = -1 / m_t = -1/1 = -1。
Normal equation: y – (-2) = -1(x – (-1)) ⇒ y + 2 = -x – 1 ⇒ y = -x – 3.
法线方程：y – (-2) = -1(x – (-1)) ⇒ y + 2 = -x – 1 ⇒ y = -x – 3。
Tangent: y = x – 1, Normal: y = -x – 3
切线：y = x – 1，法线：y = -x – 3
8. Integration of Polynomials | 多项式积分
Find the indefinite integral ∫ (6x² – 4x + 3) dx, and evaluate the definite integral ∫₁² (6x² – 4x + 3) dx.
求不定积分 ∫ (6x² – 4x + 3) dx，并计算定积分 ∫₁² (6x² – 4x + 3) dx。
Use the power rule for integration: ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, for n ≠ -1.
使用幂函数的积分法则：∫ xⁿ dx = xⁿ⁺¹/(n+1) + C，其中 n ≠ -1。
Integrate term by term: ∫ 6x² dx = 6 × (x³/3) = 2x³; ∫ -4x dx = -4 × (x²/2) = -2x²; ∫ 3 dx = 3x. So the indefinite integral is 2x³ – 2x² + 3x + C.
逐项积分：∫ 6x² dx = 6×(x³/3) = 2x³；∫ -4x dx = -4×(x²/2) = -2x²；∫ 3 dx = 3x。故不定积分为 2x³ – 2x² + 3x + C。
For the definite integral from 1 to 2: F(x) = 2x³ – 2x² + 3x. F(2) = 2(8) – 2(4) + 6 = 16 – 8 + 6 = 14; F(1) = 2(1) – 2(1) + 3 = 2 – 2 + 3 = 3. The value is F(2) – F(1) = 14 – 3 = 11.
计算从 1 到 2 的定积分：原函数 F(x) = 2x³ – 2x² + 3x。F(2) = 2(8) – 2(4) + 6 = 16 – 8 + 6 = 14；F(1) = 2(1) – 2(1) + 3 = 2 – 2 + 3 = 3。定积分的值为 F(2) – F(1) = 14 – 3 = 11。
∫ (6x² – 4x + 3) dx = 2x³ – 2x² + 3x + C; ∫₁² (…) dx = 11
∫ (6x² – 4x + 3) dx = 2x³ – 2x² + 3x + C；∫₁² (…) dx = 11
9. Arithmetic Sequences | 等差数列
In an arithmetic sequence, the 5th term is 22 and the 12th term is 57. Find the first term a and the common difference d, and the sum of the first 20 terms.
在一个等差数列中，第 5 项为 22，第 12 项为 57。求首项 a 和公差 d，以及前 20 项的和。
The nth term formula: uₙ = a + (n – 1)d. So u₅ = a + 4d = 22, and u₁₂ = a + 11d = 57.
第 n 项公式：uₙ = a + (n – 1)d。因此 u₅ = a + 4d = 22，u₁₂ = a + 11d = 57。
Subtract the first equation from the second: (a + 11d) – (a + 4d) = 57 – 22 ⇒ 7d = 35 ⇒ d = 5.
两式相减：(a + 11d) – (a + 4d) = 57 – 22 ⇒ 7d = 35 ⇒ d = 5。
Substitute d = 5 into a + 4(5) = 22 ⇒ a + 20 = 22 ⇒ a = 2.
代入 d = 5 到 a + 4(5) = 22 ⇒ a
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