Edexcel Maths: Worked Example Breakdown | Edexcel 数学：典型例题详解

📚 Edexcel Maths: Worked Example Breakdown | Edexcel 数学：典型例题详解

In this article, we walk through a selection of typical, exam-style questions from the Edexcel A Level Mathematics specification. Each worked example is chosen to highlight a core technique, common pitfall, or key piece of reasoning that students are expected to master. The explanations are given in a paired English–Chinese format, so that learners can follow the mathematical argument while building subject-specific vocabulary in both languages.

本文精选 Edexcel A Level 数学考试中的典型例题，逐一进行详细解析。每道例题都突出一种核心方法、常见易错点或关键推理，帮助学生牢固掌握必备技能。解析部分采用中英对照的形式，便于读者在理解数学逻辑的同时，积累双语的学科表达。

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1. Quadratic Discriminant | 二次方程的判别式

Problem: e : Find the range of values of k for which the quadratic equation x² + kx + 4 = 0 has no real roots.

题目：求使得二次方程 x² + kx + 4 = 0 没有实数根的 k 的取值范围。

For no real roots, the discriminant must be negative: b² – 4ac < 0.

没有实数根时，判别式必须小于零：b² – 4ac < 0。

Here a = 1, b = k, c = 4, so discriminant = k² – 16.

这里 a = 1, b = k, c = 4，所以判别式 = k² – 16。

Set up inequality: k² – 16 < 0 → (k – 4)(k + 4) < 0.

建立不等式：k² – 16 < 0 → (k – 4)(k + 4) < 0。

Critical values are k = – and k = 4. The quadratic in k opens upward, so the product is negative between the roots.

临界值为 k = –4 和 k = 4。关于 k 的二次函数开口向上，因此在两根之间乘积为负。

Thus the solution is –4 < k < 4.

因此解为 –4 < k < 4。

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2. Exponential and Logarithmic Equations | 指数与对数方程

Problem: Solve 3e²ˣ – 5eˣ – 2 = 0, giving your answer in exact form.

题目：解方程 3e²ˣ – 5eˣ – 2 = 0，答案用精确形式表示。

Let y = eˣ. Then the equation becomes 3y² – 5y – 2 = 0.

令 y = eˣ，则方程化为 3y² – 5y – 2 = 0。

Factorise: (3y + 1)(y – 2) = 0 → y = –1/3 or y = 2.

因式分解：(3y + 1)(y – 2) = 0 → y = –1/3 或 y = 2。

Since eˣ > 0 for all real x, we discard y = –1/3. Hence eˣ = 2 → x = ln 2.

因为对所有实数 x 都有 eˣ > 0，舍去 y = –1/3。因此 eˣ = 2 → x = ln 2。

The exact solution is x = ln 2.

精确解为 x = ln 2。

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3. Trigonometric Identities and Proof | 三角恒等式与证明

Problem: Prove that (sin θ + cos θ)² ≡ 1 + sin 2θ.

题目：证明 (sin θ + cos θ)² ≡ 1 + sin 2θ。

Expand the left-hand side: sin²θ + 2 sin θ cos θ + cos²θ.

展开左边：sin²θ + 2 sin θ cos θ + cos²θ。

Use the identity sin²θ + cos²θ ≡ 1, and 2 sin θ cos θ ≡ sin 2θ.

利用恒等式 sin²θ + cos²θ ≡ 1，以及 2 sin θ cos θ ≡ sin 2θ。

Then the expression simplifies to 1 + sin 2θ, which is the right-hand side.

式子化简为 1 + sin 2θ，即右边。

Hence the identity is proved.

恒等式得证。

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4. Chain Rule Differentiation | 链式法则求导

Problem: Differentiate f(x) = ln(3x² + 5) with respect to x.

题目：对 f(x) = ln(3x² + 5) 求导。

Let u = 3x² + 5, so f(x) = ln u. Then du/dx = 6x.

令 u = 3x² + 5，则 f(x) = ln u。du/dx = 6x。

By the chain rule, f'(x) = (1/u) × du/dx = (1/(3x² + 5)) × 6x.

由链式法则，f'(x) = (1/u) × du/dx = (1/(3x² + 5)) × 6x。

Simplify to f'(x) = 6x / (3x² + 5).

化简得 f'(x) = 6x / (3x² + 5)。

This is the required derivative; no further simplification is necessary.

这就是所求的导数，无需进一步化简。

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5. Integration by Substitution | 积分换元法

Problem: Find ∫ x√2x + 1 dx using the substitution u = 2x + 1.

题目：用换元 u = 2x + 1 求 ∫ x√2x + 1 dx。

If u = 2x + 1, then du/dx = 2, so dx = du/2. Also x = (u – 1)/2.

若 u = 2x + 1，则 du/dx = 2，所以 dx = du/2。同时 x = (u – 1)/2。

Substitute into the integral: ∫ ((u – 1)/2) × √u × (du/2) = (1/4) ∫ (u – 1) u^1/2 du.

代入积分：∫ ((u – 1)/2) × √u × (du/2) = (1/4) ∫ (u – 1) u^1/2 du。

Expand the integrand: u^3/2 – u^1/2.

展开被积函数：u^3/2 – u^1/2。

Integrate term by term: (1/4)[ (2/5) u^5/2 – (2/3) u^3/2 ] + C.

逐项积分：(1/4)[ (2/5) u^5/2 – (2/3) u^3/2 ] + C。

Simplify coefficients: = (1/10) u^5/2 – (1/6) u^3/2 + C.

化简系数：= (1/10) u^5/2 – (1/6) u^3/2 + C。

Substitute back u = 2x + 1 to obtain the final answer in terms of x.

将 u = 2x + 1 代回，得到用 x 表示的最终答案。

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6. Arithmetic and Geometric Sequences | 等差与等比数列

Problem: An arithmetic sequence has first term 7 and common difference –3. Find the sum of the first 20 terms.

题目：一等差数列首项为 7，公差为 –3。求前 20 项的和。

For an arithmetic series, Sₙ = n/2 [2a + (n – 1)d].

等差数列求和公式：Sₙ = n/2 [2a + (n – 1)d]。

Here a = 7, d = –3, n = 20.

这里 a = 7, d = –3, n = 20。

Calculate: S₂₀ = 20/2 [2(7) + (19)(–3)] = 10 [14 – 57] = 10 × (–43) = –430.

计算：S₂₀ = 20/2 [2(7) + (19)(–3)] = 10 [14 – 57] = 10 × (–43) = –430。

The sum of the first 20 terms is –430.

前 20 项的和为 –430。

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7. Vector Dot Product and Angle | 向量点积与夹角

Problem: Given vectors a = 3i – 2j + k and b = i + 4j – 2k, find the acute angle between them.

题目：已知向量 a = 3i – 2j + k 与 b = i + 4j – 2k，求它们之间的锐角。

Use the dot product formula: a • b = |a||b| cos θ.

使用点积公式：a • b = |a||b| cos θ。

Compute a • b = (3)(1) + (–2)(4) + (1)(–2) = 3 – 8 – 2 = –7.

计算 a • b = (3)(1) + (–2)(4) + (1)(–2) = 3 – 8 – 2 = –7。

Magnitudes: |a| = √(3² + (–2)² + 1²) = √9 + 4 + 1 = √14; |b| = √(1² + 4² + (–2)²) = √1 + 16 + 4 = √21.

模长：|a| = √(3² + (–2)² + 1²) = √14；|b| = √(1² + 4² + (–2)²) = √21。

Then cos θ = –7 / (√14 × √21) = –7 / √294 = –7 / (7√6) = –1/√6.

于是 cos θ = –7 / (√14 × √21) = –7 / √294 = –7 / (7√6) = –1/√6。

Since the cosine is negative, the angle is obtuse. The acute angle θ' is given by cos θ' = 1/√6, so θ' = arccos(1/√6).

余弦为负，说明该角为钝角。锐角 θ' 满足 cos θ' = 1/√6，因此 θ' = arccos(1/√6)。

The required acute angle is arccos(1/√6) (or approximately 65.9° if rounding is allowed).

所求锐角为 arccos(1/√6)（若允许近似，约为 65.9°）。

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8. Normal Distribution Probabilities | 正态分布概率计算

Problem: The random variable X ~ N(50, 4²). Find P(46 < X < 54).

题目：随机变量 X ~ N(50, 4²)。求 P(46 < X < 54)。

Standardise: Z = (X – μ) / σ = (X – 50) / 4.

标准化：Z = (X – μ) / σ = (X – 50) / 4。

When X = 46, Z = (46 – 50)/4 = –1; when X = 54, Z = (54 – 50)/4 = 1.

当 X = 46 时，Z = –1；当 X = 54 时，Z = 1。

Thus P(46 < X < 54) = P(–1 < Z < 1) = Φ(1) – Φ(–1).

因此 P(46 < X < 54) = P(–1 < Z < 1) = Φ(1) – Φ(–1)。

By symmetry of the standard normal distribution, Φ(–1) = 1 – Φ(1).

由标准正态分布的对称性，Φ(–1) = 1 – Φ(1)。

Hence probability = Φ(1) – (1 – Φ(1)) = 2Φ(1) – 1.

因此概率 = Φ(1) – (1 – Φ(1)) = 2Φ(1) – 1。

From tables, Φ(1) ≈ 0.8413, so P ≈ 2(0.8413) – 1 = 0.6826.

查表得 Φ(1) ≈ 0.8413，所以 P ≈ 2(0.8413) – 1 = 0.6826。

The probability that X lies within one standard deviation of the mean is approximately 0.683.

X 落在均值一个标准差范围内的概率约为 0.683。

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9. Newton's Second Law in One Dimension | 一维牛顿第二定律

Problem: A particle of mass 5 kg is pulled along a smooth horizontal surface by a force of 25 N inclined at 30° to the horizontal. Find the acceleration of the particle.

题目：一质量为 5 kg 的质点在光滑水平面上，被一个与水平方向成 30°、大小为 25 N 的力牵引。求质点的加速度。

Resolve the force horizontally: F_horizontal = 25 cos 30° = 25 × (√3/2) ≈ 21.65 N.

将力沿水平方向分解：F_水平 = 25 cos 30° = 25 × (√3/2) ≈ 21.65 N。

Using Newton's second law F = ma (horizontally), we have 21.65 = 5 × a.

根据牛顿第二定律 F = ma（水平方向），有 21.65 = 5 × a。

Thus a = 21.65 / 5 = 4.33 m s⁻² (exact value: a = (5√3) / 2 m s⁻²).

因此 a = 21.65 / 5 = 4.33 m s⁻²（精确值：a = (5√3) / 2 m s⁻²）。

The acceleration is approximately 4.33 m s⁻² in the direction of the horizontal component of the force.

加速度方向与拉力的水平分量方向相同，大小约为 4.33 m s⁻²。

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10. Binomial Expansion for Rational Powers | 有理数指数的二项展开式

Problem: Expand (1 + 2x)^1/2 in ascending powers of x up to and including the term in x³, stating the range of validity.

题目：将 (1 + 2x)^1/2 按 x 的升幂展开，包括 x³ 项，并写出成立范围。

Use the binomial series for (1 + u)ⁿ with n = 1/2 and u = 2x: (1 + u)^1/2 = 1 + (1/2)u + (1/2)(–1/2)/2! u² + (1/2)(–1/2)(–3/2)/3! u³ + …

利用二项级数展开 (1 + u)ⁿ，其中 n = 1/2, u = 2x：公式为 1 + (1/2)u + (1/2)(–1/2)/2! u² + (1/2)(–1/2)(–3/2)/3! u³ + …

Substitute u = 2x: first term: 1; second: (1/2)(2x) = x; third: (1/2)(–1/2)/2 × (2x)² = (–1/8) × 4x² = –x²/2; fourth: (1/2)(–1/2)(–3/2)/6 × (2x)³ = (3/8)/6 × 8x³ = (3/48) × 8x³ = x³/2.

代入 u = 2x：首项为 1；第二项：(1/2)(2x) = x；第三项：(1/2)(–1/2)/2 × (2x)² = (–1/8) × 4x² = –x²/2；第四项：(1/2)(–1/2)(–3/2)/6 × (2x)³ = (3/8)/6 × 8x³ = x³/2。

Hence (1 + 2x)^1/2 ≈ 1 + x – x²/2 + x³/2.

因此 (1 + 2x)^1/2 ≈ 1 + x – x²/2 + x³/2。

The expansion is valid for |u| < 1, i.e. |2x| < 1, giving |x| < 1/2.

展开式成立的条件是 |u| < 1，即 |2x| < 1，所以 |x| < 1/2。

Range of validity: –1/2 < x < 1/2.

成立范围：–1/2 < x < 1/2。

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