📚 Miscellaneous Exercises 5 | 混合练习五
This revision set pulls together core AQA A-level Mathematics techniques that often appear together in mixed papers. Each section below gives a short worked example and the key method you must be able to reproduce under timed conditions.
本复习练习将 AQA A-Level 数学考试中经常混合出现的核心技巧集中在一起。下面每一节给出一个简短例题以及你在限时条件下必须能够熟练掌握的关键方法。
1. Algebraic Simplification and Surds | 代数化简与根式
Surd questions test your ability to rewrite square roots in their simplest form. Always factor out the largest square number first.
根式题考查将平方根化为最简形式的能力。始终先提取最大的平方因数。
Example: Simplify (√8 + √18)².
示例：化简 (√8 + √18)²。
(√8 + √18)² = 50
Write √8 = 2√2 and √18 = 3√2, so the bracket becomes 5√2. Squaring gives 25 × 2 = 50.
因为 √8 = 2√2，√18 = 3√2，所以括号内为 5√2。平方后得到 25 × 2 = 50。
2. Quadratic Equations and the Discriminant | 二次方程与判别式
= b² - 4ac decides the nature of the roots. It is also used to find ranges of parameters.
对于 ax² + bx + c = 0，判别式 Δ = b² - 4ac 决定根的性质，也用于求参数的取值范围。
Example: Find the set of values of k for which x² + kx + 9 = 0 has real roots.
示例：求使 x² + kx + 9 = 0 有实根的 k 的取值范围。
Δ = b² - 4ac
Require Δ = k² - 36 ≥ 0, hence k² ≥ 36. The solution is k ≤ -6 or k ≥ 6.
需要 Δ = k² - 36 ≥ 0，因此 k² ≥ 36。解为 k ≤ -6 或 k ≥ 6。
3. Polynomial Division and the Factor Theorem | 多项式除法与因式定理
then (x - a) is a factor of f(x). Polynomial division then reduces a cubic to a quadratic.
因式定理指出，若 f(a) = 0，则 (x - a) 是 f(x) 的因式。多项式除法可将三次式降为二次式。
xample: Given f(x) = 2x³ - 3x² - 3x + 2, show that (x - 2) is a factor and solve f(x) = 0.
示例：已知 f(x) = 2x³ - 3x² - 3x + 2，证明 (x - 2) 是因式并解 f(x) = 0。
f(x) = (x - 2)(2x - 1)(x + 1)
a factor. Division gives (x - 2)(2x² + x - 1) = (x - 2)(2x - 1)(x + 1), so x = 2, 1/2, -1.
f(2) = 16 - 12 - 6 + 2 = 0，所以 (x - 2) 是因式。相除得 (x - 2)(2x² + x - 1) = (x - 2)(2x - 1)(x + 1)，因此 x = 2、1/2、-1。
4. Exponential and Logarithmic Equations | 指数与对数方程
Exponential equations often become quadratics after a substitution such as y = eˣ. Remember ln(eˣ) = x and eln x = x.
指数方程在代换 y = eˣ 后常化为二次方程。记住 ln(eˣ) = x 以及 eln x = x。
Example: Solve 2e²ˣ - 5eˣ + 2 = 0.
示例：解方程 2e²ˣ - 5eˣ + 2 = 0。
2e²ˣ - 5eˣ + 2 = 0
Let y = eˣ. Then 2y² - 5y + 2 = 0, giving y = 1/2 or y = 2. Hence eˣ = 1/2 gives x = ln(1/2) = -ln 2, and eˣ = 2 gives x = ln 2.
设 y = eˣ。则 2y² - 5y + 2 = 0，解得 y = 1/2 或 y = 2。因此 eˣ = 1/2 得到 x = ln(1/2) = -ln 2，eˣ = 2 得到 x = ln 2。
5. Trigonometric Identities and Equations | 三角恒等式与方程
n²θ + cos²θ = 1 and sin 2θ = 2 sinθ cosθ. Use them to rewrite equations in one trig ratio.
核心恒等式包括 sin²θ + cos²θ = 1 和 sin 2θ = 2 sinθ cosθ。利用它们将方程改写为只含一个三角比。
Example: Solve sin 2θ = 0.5 for 0 ≤ θ ≤ π.
示例：在 0 ≤ θ ≤ π 内解 sin 2θ = 0.5。
sin 2θ = 2 sinθ cosθ
u = 2θ. Since 0 ≤ u ≤ 2π, sin u = 0.5 gives u = π/6 or 5π/6. Therefore θ = π/12 or 5π/12.
令 u = 2θ。因为 0 ≤ u ≤ 2π，sin
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