📚 Miscellaneous Exercises 7: Mixed A-Level Maths Practice for AQA | AQA A-Level 数学杂项练习七综合解析
This revision article works through a set of mixed exercises suitable for the AQA A-Level Mathematics specification. Each section presents a typical question, a step-by-step solution, and a short exam tip to help you avoid common errors.
本篇复习文章精选一组适合 AQA A-Level 数学考纲的混合练习。每一节给出一道典型题、分步解答以及一条简短的考试提示，帮助你避开常见错误。
1. Algebraic Simplification and Surds | 代数化简与根式
Simplify (√48 - √27 + √75) / √3.
化简 (√48 - √27 + √75) / √3。
Write each surd in terms of √3: √48 = 4√3, √27 = 3√3, √75 = 5√3.
将每个根式写成 √3 的倍数：√48 = 4√3，√27 = 3√3，√75 = 5√3。
The numerator becomes 4√3 - 3√3 + 5√3 = 6√3, so dividing by √3 gives 6.
分子变为 4√3 - 3√3 + 5√3 = 6√3，再除以 √3 得 6。
Exam tip: always simplify surds before adding or subtracting them; this usually reveals a common factor.
考试提示：在进行根式加减前一定要先化简，这样通常能找出公因子。
2. Quadratic Functions and the Discriminant | 二次函数与判别式
Find the range of values of k for which x² + (k - 2)x + 9 = 0 has no real roots.
求使方程 x² + (k - 2)x + 9 = 0 无实根的 k 的取值范围。
For no real roots, the discriminant must be negative: Δ = b² - 4ac < 0.
无实根时，判别式必须小于零：Δ = b² - 4ac < 0。
Here a = 1, b = k - 2, c = 9, so Δ = (k - 2)² - 36.
这里 a = 1，b = k - 2，c = 9，因此 Δ = (k - 2)² - 36。
(k - 2)² - 36 < 0
This gives (k - 2)² < 36, so |k - 2| < 6.
由此得到 (k - 2)² < 36，所以 |k - 2| < 6。
Therefore -4 < k < 8.
因此 k 的取值范围是 -4 < k < 8。
Exam tip: the phrase 'no real roots' always points directly to Δ < 0, not Δ ≤ 0.
考试提示：题目中出现 'no real roots' 时，应直接用 Δ < 0，而不是 Δ ≤ 0。
3. Polynomial Division and the Factor Theorem | 多项式除法与因式定理
Given f(x) = 2x³ - 5x² + x + 2, show that (x - 1) is a factor and fully factorise f(x).
已知 f(x) = 2x³ - 5x² + x + 2，证明 (x - 1) 是一个因式，并将 f(x) 完全分解因式。
Use the factor theorem: evaluate f(1) = 2 - 5 + 1 + 2 = 0, so (x - 1) is a factor.
使用因式定理：计算 f(1) = 2 - 5 + 1 + 2 = 0，因此 (x - 1) 是一个因式。
Divide f(x) by (x - 1) using polynomial long division or synthetic division.
用多项式长除法或综合除法将 f(x) 除以 (x - 1)。
The quotient is 2x² - 3x - 2, which factorises as (2x + 1)(x - 2).
商式为 2x² - 3x - 2，可分解为 (2x + 1)(x - 2)。
f(x) = (x - 1)(2x + 1)(x - 2)
Exam tip: after factorising, expand the factors mentally to check the constant term and leading coefficient.
考试提示：完成因式分解后，可在心中展开核对常数项和首项系数是否正确。
4. Exponential and Logarithmic Equations | 指数方程与对数方程
Solve (3ˣ)² - 4 × 3ˣ + 3 = 0.
解方程 (3ˣ)² - 4 × 3ˣ + 3 = 0。
Let y = 3ˣ. Then the equation becomes y² - 4y + 3 = 0.
设 y = 3ˣ，则方程化为 y² - 4y + 3 = 0。
Factorise: (y - 1)(y - 3) = 0, so y = 1 or y = 3.
因式分解得 (y - 1)(y - 3) = 0，所以 y = 1 或 y = 3。
Since y = 3ˣ, solve 3ˣ = 1 giving x = 0, and 3ˣ = 3 giving x = 1.
由于 y = 3ˣ，解 3ˣ = 1 得 x = 0，解 3ˣ = 3 得 x = 1。
x = 0 or x = 1
Exam tip: hidden quadratic equations often appear in exponential form; substitution is the key first step.
考试提示：指数形式的隐性二次方程很常见，换元是关键的解题第一步。
5. Trigonometric Identities and Equations | 三角恒等式与三角方程
Solve 2sin²θ - 3cosθ = 0 for 0° ≤ θ ≤ 360°.
在 0° ≤ θ ≤ 360° 范围内解方程 2sin²θ - 3cosθ = 0。
Use sin²θ = 1 - cos²θ to rewrite the equation as 2(1 - cos²θ) - 3cosθ = 0.
利用 sin²θ = 1 - cos²θ，将方程改写为 2(1 - cos²θ) - 3cosθ = 0。
This simplifies to 2cos²θ + 3cosθ - 2 = 0.
化简得 2cos²θ + 3cosθ - 2 = 0。
Let u = cosθ, then 2u² + 3u - 2 = 0, so (2u - 1)(u + 2) = 0.
设 u = cosθ，则 2u² + 3u - 2 = 0，因式分解为 (2u - 1)(u + 2) = 0。
Thus u = ½ or u = -2. Since cosθ cannot be less than -1, reject u = -2.
因此 u = ½ 或 u = -2。由于 cosθ 不可能小于 -1，所以舍去 u = -2。
Solve cosθ = ½: θ = 60° or θ = 300°.
解 cosθ = ½ 得 θ = 60° 或 θ = 300°。
θ = 60°, 300°
Exam tip: always check the validity of trigonometric solutions, especially when factoring gives more than one value.
考试提示：当因式分解得到多个值时，一定要检验三角函数的解是否在允许范围内。
6. Differentiation Using the Product Rule | 乘积法则求导
Differentiate y = x² e²ˣ with respect to x.
对 y = x² e²ˣ 关于 x 求导。
Let u = x² and v = e²ˣ. Then u' = 2x and v' = 2e²ˣ.
设 u = x²，v = e²ˣ，则 u' = 2x，v' = 2e²ˣ。
Apply the product rule: dy/dx = u'v + uv'.
应用乘积法则：dy/dx = u'v + uv'。
So dy/dx = 2x e²ˣ + 2x² e²ˣ = 2x e²ˣ(1 + x).
因此 dy/dx = 2x e²ˣ + 2x² e²ˣ = 2x e²ˣ(1 + x)。
dy/dx = 2x e²ˣ(1 + x)
Exam tip: after differentiating, look for a common factor to present the answer in its simplest factorised form.
考试提示：求导后应寻找公因式，将结果写成最简的因式形式。
7. Integration by Substitution | 换元积分法
Find ∫ x√x² + 1 dx.
求 ∫ x√x² + 1 dx。
Let u = x² + 1. Then du/dx = 2x, so du = 2x dx, meaning x dx = ½ du.
设 u = x² + 1，则 du/dx = 2x，所以 du = 2x dx，即 x dx = ½ du。
The integral becomes ½ ∫ √u du.
原积分化为 ½ ∫ √u du。
Integrate: ½ × (2/3) u√u + C = ⅓ u√u + C.
积分得：½ × (2/3) u√u + C = ⅓ u√u + C。
Substitute back u = x² + 1 to obtain ⅓(x² + 1)√x² + 1 + C.
将 u = x² + 1 代回，得到 ⅓(x² + 1)√x² + 1 + C。
∫ x√x² + 1 dx = ⅓(x² + 1)√x² + 1 + C
Exam tip: when using substitution, always express the differential dx in terms of du before integrating.
考试提示：使用换元法时，一定要在积分前先把 dx 用 du 表示出来。
8. Sequences and Sigma Notation | 数列与求和符号
Evaluate Σ (from r = 1 to 20) (3r - 2).
求 Σ（r 从 1 到 20）(3r - 2) 的值。
Separate the sum: Σ(3r - 2) = 3Σr - Σ2.
拆分求和：Σ(3r - 2) = 3Σr - Σ2。
Use the standard results Σr = n(n+1)/2 and Σ2 = 2n with n = 20.
使用标准公式 Σr = n(n+1)/2 和 Σ2 = 2n，其中 n = 20。
Then the sum is 3 × (20 × 21 / 2) - 2 × 20.
因此和为 3 × (20 × 21 / 2) - 2 × 20。
This gives 630 - 40 = 590.
计算得 630 - 40 = 590。
Σ (r = 1 to 20) (3r - 2) = 590
Exam tip: sigma notation often breaks into separate sums, so identify standard arithmetic terms first.
考试提示：求和符号通常可以拆分为多个求和项，先识别出标准等差或常数项。
9. Vectors in Two and Three Dimensions | 二维与三维向量
Given a = 2i - j + 3k and b = i + 2j - k, find a·b and the angle between a and b.
已知 a = 2i - j + 3k，b = i + 2j - k，求 a·b 以及 a 与 b 的夹角。
Calculate the dot product: a·b = 2 × 1 + (-1) × 2 + 3 × (-1) = 2 - 2 - 3 = -3.
计算点积：a·b = 2 × 1 + (-1) × 2 + 3 × (-1) = 2 - 2 - 3 = -3。
Find the magnitudes: |a| = √(2² + (-1)² + 3²) = √14 and |b| = √(1² + 2² + (-1)²) = √6.
求模长：|a| = √(2² + (-1)² + 3²) = √14，|b| = √(1² + 2² + (-1)²) = √6。
Use cosθ = a·b / (|a||b|) = -3 / (√14 × √6) = -3 / √84.
利用公式 cosθ = a·b / (|a||b|) = -3 / (√14 × √6) = -3 / √84。
Since √84 = 2√21, cosθ = -3/(2√21). Therefore θ = arccos(-3/(2√21)) ≈ 109.1°.
因为 √84 = 2√21，所以 cosθ = -3/(2√21)，因此 θ = arccos(-3/(2√21)) ≈ 109.1°。
Exam tip: a negative dot product means the angle is obtuse, which is a useful quick check.
考试提示：点积为负意味着夹角为钝角，这是很有用的快速检验方法。
10. Proof by Contradiction |
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