📚 Miscellaneous Exercises 4 | 综合练习四
This mixed revision set brings together key AQA A-Level Maths techniques from pure mathematics, including surds, quadratics, exponentials, trigonometry, calculus, binomial expansion, vectors, proof and modelling. Each section pairs a typical exam-style idea with a worked solution path so you can check both method and final answer quickly.
这套综合练习汇集了 AQA A-Level 数学纯数部分的关键题型，包括根式、二次函数、指数、三角、微积分、二项式展开、向量、证明和建模。每一节都配有一个典型考题思路和完整解答路径，帮助你快速检查方法与最终答案。
1. Algebraic Simplification with Surds | 代数化简与根式
Surd simplification questions often test whether you can identify square factors before adding, subtracting or dividing. Always express each surd in the form k√n where n has no square factor left.
根式化简题通常考查你是否能在加减或乘除前识别平方因子。务必先把每个根式写成 k√n 的形式，并保证 n 中不再含有平方因子。
Simplify (√12 + √27) ÷ √3
Write √12 as 2√3 and √27 as 3√3. Their sum is 5√3, and dividing by √3 gives 5.
把 √12 写成 2√3，把 √27 写成 3√3。两者之和为 5√3，再除以 √3，结果为 5。
(√12 + √27) ÷ √3 = (2√3 + 3√3) ÷ √3 = 5√3 ÷ √3 = 5
A second standard step is rationalising a denominator such as 1 ÷ (2 + √3). Multiply the numerator and denominator by the conjugate 2 - √3, then simplify using (a + b)(a - b) = a² - b².
第二个标准步骤是有理化分母，例如 1 ÷ (2 + √3)。将分子和分母同乘以共轭式 2 - √3，然后利用 (a + b)(a - b) = a² - b² 进行化简。
2. Quadratics and the Discriminant | 二次函数与判别式
distinct real roots, Δ > 0; for one repeated root, Δ = 0; for no real roots, Δ < 0.
判别式 Δ = b² - 4ac 决定二次函数 ax² + bx + c 的实根个数。若有两个不同实根，则 Δ > 0；若有一个重根，则 Δ = 0；若没有实根，则 Δ < 0。
n f(x) = 2x² - 3x + k, find the range of k for which f(x) = 0 has two distinct real roots.
已知 f(x) = 2x² - 3x + k，求使得 f(x) = 0 有两个不同实根时 k 的取值范围。
- 4(2)(k) = 9 - 8k. For two distinct real roots we need 9 - 8k > 0, giving k < 9/8.
这里 a = 2，b = -3，c = k，所以 Δ = (-3)² - 4(2)(k) = 9 - 8k。要有两个不同实根，需要 9 - 8k > 0，因此 k < 9/8。
Δ = 9 - 8k > 0 ⇒ k < 9/8
l roots, the inequality reverses to k > 9/8. A repeated root occurs exactly at k = 9/8.
如果题目要求没有实根，则不等式反转为 k > 9/8。当 k = 9/8 时，方程恰好有一个重根。
3. Exponential and Logarithmic Equations | 指数方程与对数方程
Exponential equations are solved by taking natural logarithms on both sides, while logarithmic equations are simplified by combining logs or using the definition of a logarithm.
指数方程通常通过对两边取自然对数求解，而对数方程则通过合并对数或使用对数的定义来化简。
Solve e³ˣ = 7, giving x in exact form.
解方程 e³ˣ = 7，并给出 x 的精确形式。
act and usually preferred in AQA mark schemes unless a decimal approximation is requested.
两边取自然对数得 3x = ln 7，所以 x = ln 7 ÷ 3。这是精确形式，除非题目要求近似小数，否则 AQA 评分标准通常优先接受精确值。
e³ˣ = 7 ⇒ 3x = ln 7 ⇒ x = ln 7 ÷ 3
) - log₂(x - 1) = 2 becomes log₂((x + 1)/(x - 1)) = 2, so (x + 1)/(x - 1) = 4 and x = 5/3.
对于对数方程，要记住 logₐ m - logₐ n = logₐ(m/n)。例如 log₂(x + 1) - log₂(x - 1) = 2 可化为 log₂((x + 1)/(x - 1)) = 2，因此 (x + 1)/(x - 1) = 4，解得 x = 5/3。
4. Trigonometric Equations and Identities | 三角方程与恒等式
Trigonometric equations often reduce to a quadratic in sin θ or cos θ. Factorise first, then use quadrant rules or the unit circle to find all solutions in the requested interval.
三角方程通常可以化为关于 sin θ 或 cos θ 的二次方程。先进行因式分解，然后利用象限规则或单位圆求出指定区间内的所有解。
Solve 2sin²θ - sinθ - 1 = 0 for 0° ≤ θ ≤ 360°.
解方程 2sin²θ - sinθ - 1 = 0，其中 0° ≤ θ ≤ 360°。
the given interval, sinθ = 1 gives θ = 90°, while sinθ = -1/2 gives θ = 210° and θ = 330°.
因式分解得 (2sinθ + 1)(sinθ - 1) = 0。因此 sinθ = -1/2 或 sinθ = 1。在给定区间内，sinθ = 1 得到 θ = 90°，而 sinθ = -1/2 得到 θ = 210° 和 θ = 330°。
θ = 90°, 210°, 330°
Always check that each result lies in the original interval and satisfies the original equation. A quick unit-circle sketch is a reliable way to avoid missing negative sine solutions in the third and fourth quadrants.
务必检查每个结果是否位于原始区间内，并满足原方程。快速画一个单位圆草图是避免遗漏第三、第四象限中负正弦解的可靠方法。
5. Differentiation Techniques | 微分技巧
Chain rule, product rule and quotient rule are core AQA differentiation tools. Identify the structure of the function before choosing a rule, and show the intermediate derivative clearly.
链式法则、乘法法则和商法则是 AQA 微分部分的核心工具。选择法则前要先判断函数结构，并清晰写出中间导数步骤。
Differentiate y = (3x² + 1)⁴ with respect to x.
求 y = (3x² + 1)⁴ 关于 x 的导数。
² + 1. Then y = u⁴, so dy/dx = 4u³ × du/dx. Since du/dx = 6x, the result is 24x(3x² + 1)³.
使用链式法则，设 u = 3x² + 1。则 y = u⁴，所以 dy/dx = 4u³ × du/dx。由于 du/dx = 6x，结果为 24x(3x² + 1)³。
dy/dx = 4(3x² + 1)³ × 6x = 24x(3x² + 1)³
For a product such as y = x²eˣ, use the product rule: dy/dx = 2xeˣ + x²eˣ = xeˣ(x + 2). For y = ln(x² + 1), the chain rule gives dy/dx = 2x/(x² + 1).
对于乘积函数，例如 y = x²eˣ，使用乘法法则：dy/dx = 2xeˣ + x²eˣ = xeˣ(x + 2)。对于 y = ln(x² + 1)，链式法则给出 dy/dx = 2x/(x² + 1)。
6. Integration and Definite Integrals | 积分与定积分
Definite integrals require an antiderivative, substitution of limits, and careful arithmetic with negative signs. Remember to include the constant of integration only for indefinite integrals.
定积分需要先求原函数，再代入上下限，并仔细处理负号运算。切记只有不定积分才需要加积分常数。
Evaluate ∫₀² (4x³ - 2x + 1) dx.
计算定积分 ∫₀² (4x³ - 2x + 1) dx。
egrate term by term to get x⁴ - x² + x. Substitute x = 2 and x = 0: (16 - 4 + 2) - 0 = 14.
逐项积分得到 x⁴ - x² + x。代入 x = 2 和 x = 0：(16 - 4 + 2) - 0 = 14。
∫₀² (4x³ - 2x + 1) dx = [x⁴ - x² + x]₀² = 14
This links integration with the natural logarithm and is often tested in mixed exercises.
另一种常见题型是 ∫₁ᵉ (1/x) dx = [ln x]₁ᵉ = ln e - ln 1 = 1。这把积分与自然对数联系起来，常在综合练习中出现。
7. Binomial Expansion | 二项式展开
For rational powers, binomial expansion is valid only within a specific range of x. For a positive integer power, the expansion is finite and the coefficients come from Pascal's triangle or nCr.
对于有理数次幂，二项式展开只在特定 x 范围内有效。对于正整数次幂，展开是有限的，系数来自帕斯卡三角或组合数 nCr。
Expand (1 - 2x)⁵ up to and including the x³ term.
展开 (1 - 2x)⁵，直到并包含 x³ 项。
1 + 5C1(-2x) + 5C2(-2x)² + 5C3(-2x)³ + ... This simplifies to 1 - 10x + 40x² - 80x³ + ...
使用二项式定理：(1 - 2x)⁵ = 1 + 5C1(-2x) + 5C2(-2x)² + 5C3(-2x)³ + ... 化简后为 1 - 10x + 40x² - 80x³ + ...
(1 - 2x)⁵ = 1 - 10x + 40x² - 80x³ + ...
For expansion of (1 + ax)ⁿ with rational n, convergence requires |ax| < 1. Here, if the power were rational, the validity
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