📚 Practice Set from pdfjoiner_(4)-245 | pdfjoiner_(4)-245 练习题集
ilation, is tailored for students preparing for the Edexcel A-Level Mathematics examinations. Each question targets a core topic from the Pure Mathematics syllabus, including algebra, trigonometry, calculus, sequences, and vectors. By working through these carefully selected examples, you will reinforce your problem-solving skills and deepen your conceptual understanding.
这套习题选自 pdfjoiner_(4)-245 汇编，专为备考爱德思 A-Level 数学考试的学生设计。每道题都针对纯数大纲中的核心主题，涵盖代数、三角、微积分、数列和向量。通过练习这些精心挑选的例题，你将增强解题能力并加深对概念的理解。
1. Solving Quadratic Inequalities | 解二次不等式
Problem: Find the set of values of x for which x² – 5x + 6 > 0.
题目：求满足 x² – 5x + 6 > 0 的 x 的取值范围。
First, factorise the quadratic expression: x² – 5x + 6 = (x – 2)(x – 3).
首先，将二次式因式分解：x² – 5x + 6 = (x – 2)(x – 3)。
The critical values are x = 2 and x = 3, where the product equals zero.
临界点为 x = 2 和 x = 3，此时乘积等于零。
To determine the sign of the expression in each interval, choose test points. For the interval x < 2, pick x = 0: (0 – 2)(0 – 3) = 6 > 0, so the inequality holds.
为了判断各区间的符号，选取检验点。对于区间 x < 2，取 x = 0：(0 – 2)(0 – 3) = 6 > 0，不等式成立。
For 2 < x < 3, test x = 2.5: (2.5 – 2)(2.5 – 3) = (0.5)(–0.5) = –0.25 < 0, so the inequality is false.
对于 2 < x < 3，检验 x = 2.5：(2.5 – 2)(2.5 – 3) = (0.5)(-0.5) = -0.25 < 0，不等式不成立。
For x > 3, test x = 4: (4 – 2)(4 – 3) = 2 × 1 = 2 > 0, so the inequality holds.
对于 x > 3，检验 x = 4：(4 – 2)(4 – 3) = 2 × 1 = 2 > 0，不等式成立。
Therefore the solution set is {x | x < 2} ∪ {x | x > 3}. The graph of the quadratic opens upward, confirming that the parabola is above the x‑axis outside the interval between the roots.
因此解集为 {x | x < 2} ∪ {x | x > 3}。二次函数图像开口向上，这验证了抛物线在两根之间的区间外位于 x 轴上方。
2. Domain and Range of Functions | 函数的定义域与值域
Problem: Find the largest possible domain of the function f(x) = √x – 2 / (x – 5).
题目：求函数 f(x) = √x – 2 / (x – 5) 的最大可能定义域。
The square root requires x – 2 ≥ 0, hence x ≥ 2.
平方根要求 x – 2 ≥ 0，即 x ≥ 2。
The denominator cannot be zero, so x – 5 ≠ 0 ⇒ x ≠ 5.
分母不能为零，因此 x – 5 ≠ 0 ⇒ x ≠ 5。
Combining both conditions, the domain is {x ∈ ℝ | x ≥ 2, x ≠ 5}. In interval notation, this writes as [2, 5) ∪ (5, ∞).
综合两个条件，定义域为 {x ∈ ℝ | x ≥ 2, x ≠ 5}。用区间记号可表示为 [2, 5) ∪ (5, ∞)。
Note that near x = 5 the function has a vertical asymptote, and as x becomes large the value of f(x) tends to 0 from above. The range can be investigated by exploring the behaviour, but for the domain alone these two restrictions are sufficient.
注意在 x = 5 附近函数有垂直渐近线，且当 x 很大时 f(x) 的值从上方趋近于 0。值域可通过进一步分析行为求出，但仅就定义域而言，上述两个限制条件已足够。
3. Trigonometric Identities and Proof | 三角恒等式与证明
Problem: Show that (sinθ + cosθ)² = 1 + sin 2θ for all values of θ.
题目：证明对所有 θ 值，恒有 (sinθ + cosθ)² = 1 + sin 2θ。
Start by expanding the left-hand side: (sinθ + cosθ)² = sin²θ + 2 sinθ cosθ + cos²θ.
从左边展开：(sinθ + cosθ)² = sin²θ + 2 sinθ cosθ + cos²θ。
Use the Pythagorean identity sin²θ + cos²θ = 1, so the expression simplifies to 1 + 2 sinθ cosθ.
利用恒等式 sin²θ + cos²θ = 1，原式简化为 1 + 2 sinθ cosθ。
Recall the double-angle formula sin 2θ = 2 sinθ cosθ. Substituting this gives 1 + sin 2θ as required.
再根据二倍角公式 sin 2θ = 2 sinθ cosθ，代入即得 1 + sin 2θ，证毕。
This identity is often used in solving trigonometric equations or simplifying integrands, so recognising the connection between squares and double angles is very useful.
这一恒等式常用于解三角方程或化简被积函数，因此熟悉平方式与二倍角之间的联系十分有用。
4. Solving Trigonometric Equations | 解三角方程
Problem: Solve sin 2x = cos x for 0 ≤ x ≤ 2π, giving your answers in terms of π.
题目：求解 sin 2x = cos x，其中 0 ≤ x ≤ 2π，答案用 π 表示。
Apply the double-angle identity: sin 2x = 2 sin x cos x. The equation becomes 2 sin x cos x = cos x.
利用二倍角公式：sin 2x = 2 sin x cos x。方程化为 2 sin x cos x = cos x。
Bring all terms to one side: 2 sin x cos x – cos x = 0. Factor out cos x: cos x (2 sin x – 1) = 0.
移项得 2 sin x cos x – cos x = 0。提取公因式 cos x：cos x (2 sin x – 1) = 0。
Set each factor to zero. First, cos x = 0 gives x = π/2, 3π/2 within the interval.
令每个因子为零。首先 cos x = 0 在给定区间内得 x = π/2, 3π/2。
Second, 2 sin x – 1 = 0 ⇒ sin x = 1/2. The principal values are x = π/6 and x = 5π/6, both of which lie in the required range.
其次 2 sin x – 1 = 0 ⇒ sin x = 1/2，其主值解为 x = π/6 和 x = 5π/6，两者均在指定范围内。
Therefore the solution set is {π/6, π/2, 5π/6, 3π/2}. Always check all values satisfy the original equation, especially ensuring no division by zero was introduced (which did not occur here).
因此解集为 {π/6, π/2, 5π/6, 3π/2}。务必检查所有值是否满足原方程，尤其要确保未引入除以零的情况（本题未发生）。
5. Modelling with Exponential Functions | 指数函数模型
Problem: A population P grows according to P = 1000 e0.02t, where t is the time in years. Determine how many years it takes for the population to double.
题目：某种群数量按 P = 1000 e0.02t 增长，其中 t 为年数。求种群数量翻倍所需的年数。
Set P = 2000 (double of 1000). Then 2000 = 1000 e0.02t.
令 P = 2000（1000 的两倍），则 2000 = 1000 e0.02t。
Divide both sides by 1000: 2 = e0.02t.
两边除以 1000：2 = e0.02t。
Take the natural logarithm of both sides: ln 2 = 0.02t.
两边取自然对数：ln 2 = 0.02t。
Solve for t: t = ln 2 / 0.02. Using a calculator, ln 2 ≈ 0.6931, so t ≈ 34.66 years.
解出 t：t = ln 2 / 0.02。使用计算器得 ln 2 ≈ 0.6931，因此 t ≈ 34.66 年。
This is a standard exponential doubling model. Note that because the growth is continuous, the doubling time does not depend on the initial value – it is constant.
这是一个标准的指数翻倍模型。注意因为增长是连续的，翻倍时间与初始值无关，是恒定的。
6. Implicit Differentiation | 隐函数求导
Problem: Given the circle x² + y² = 25, find dy/dx in terms of x and y, and hence find the equation of the tangent at the point (3, 4).
题目：已知圆 x² + y² = 25，求用 x 和 y 表示的 dy/dx，并由此求出在点 (3, 4) 处的切线方程。
Differentiate both sides with respect to x, treating y as a function of x. For x² the derivative is 2x; for y² use the chain rule: d(y²)/dx = 2y (dy/dx).
两边对 x 求导，y 视为 x 的函数。对 x² 求导得 2x；对 y² 使用链式法则：d(y²)/dx = 2y (dy/dx)。
Thus 2x + 2y (dy/dx) = 0.
因此 2x + 2y (dy/dx) = 0。
Solve for dy/dx: 2y (dy/dx) = –2x ⇒ dy/dx = –x / y.
解出 dy/dx：2y (dy/dx) = –2x ⇒ dy/dx = –x / y。
At the point (3, 4), the gradient is m = –3/4.
在点 (3, 4) 处，斜率为 m = –3/4。
The equation of the tangent is y – 4 = m (x – 3), so y – 4 = –3/4 (x – 3). Multiplying through by 4 gives 4y – 16 = –3x + 9, or 3x + 4y = 25.
切线方程为 y – 4 = m (x – 3)，即 y – 4 = –3/4 (x – 3)。两边乘以 4 得 4y – 16 = –3x + 9，整理为 3x + 4y = 25。
ferentiation is essential when the function cannot be easily written in the form y = f(x).
当函数不易写成 y = f(x) 的显式时，隐函数求导至关重要。
7. Integration by Substitution | 换元积分法
Problem: Evaluate ∫ 2x √x² + 1 dx.
题目：求 ∫ 2x √x² + 1 dx。
Use the substitution u = x² + 1. Then du/dx = 2x, so du = 2x dx. Notice that the integrand contains exactly 2x dx.
作代换 u = x² + 1，则 du/dx = 2x，即 du = 2x dx。注意到被积式中恰好含有 2x dx。
The integral becomes ∫ √u du, which is ∫ u1/2 du.
积分变为 ∫ √u du，即 ∫ u1/2 du。
Integrate using the power rule: ∫ u1/2
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