📚 Edexcel A-Level Maths Review Exercise 2: Mixed Pure Topics | 爱德思A-Level数学复习练习2：纯数学综合主题
This review exercise brings together the core pure mathematics skills that regularly appear in Edexcel A-Level Maths Papers 1 and 2. It is designed to help you test your fluency across algebra, functions, trigonometry, calculus, sequences and vectors before moving on to timed past-paper practice.
本复习练习汇集了爱德思A-Level数学试卷1和试卷2中经常出现的核心纯数学技能。它旨在帮助你在进入限时真题训练之前，检验自己在代数、函数、三角、微积分、数列和向量等主题上的熟练程度。
1. Algebraic manipulation and surds | 代数运算与根式
Simplify expressions such as (√8 + √18)² and rationalise denominators like 5 / (2 - √3). These skills are frequently tested in the first few questions of Edexcel papers and often carry easy method marks.
化简诸如 (√8 + √18)² 的表达式，并将像 5 / (2 - √3) 这样的分母有理化。这些技能经常在爱德思试卷的前几题中出现，通常属于容易拿到的方法分。
To simplify a surd, factor out square numbers: √8 = √4 × 2 = 2√2 and √18 = √9 × 2 = 3√2. Then (√8 + √18)² = (2√2 + 3√2)² = (5√2)² = 50.
要化简根式，先提取平方因子：√8 = √4 × 2 = 2√2，√18 = √9 × 2 = 3√2。然后 (√8 + √18)² = (2√2 + 3√2)² = (5√2)² = 50。
When rationalising a denominator of the form a + b√c, multiply the numerator and denominator by a - b√c. This removes the surd from the denominator and leaves an integer or rational denominator.
当分母形如 a + b√c 时，将分子和分母同乘以 a - b√c。这样可以消去分母中的根式，留下整数或有理分母。
5 / (2 - √3) = 5(2 + √3) / [(2 - √3)(2 + √3)] = 10 + 5√3
2. Quadratics and discriminants | 二次函数与判别式
ether the quadratic has two distinct real roots, one repeated real root, or no real roots.
对于 ax² + bx + c = 0，判别式为 Δ = b² - 4ac。它告诉你二次方程有两个不同实根、一个重复实根还是没有实根。
l root; if Δ < 0 there are no real roots. You should also be able to complete the square to find the vertex of a quadratic graph.
如果 Δ > 0，则有两个不同实根；如果 Δ = 0，则恰好有一个实根；如果 Δ < 0，则没有实根。你还应该能够用配方法求出二次函数图像的顶点。
x² + 6x + 2 = (x + 3)² - 7, vertex = (-3, -7)
. The discriminant is k² - 16, so we require k² - 16 < 0, which gives -4 < k < 4.
例题：求使 x² + kx + 4 = 0 没有实根的 k 的取值范围。判别式为 k² - 16，因此需要 k² - 16 < 0，解得 -4 < k < 4。
3. Inequalities and modulus | 不等式与绝对值
Solving linear and quadratic inequalities requires careful sign diagrams. When you multiply or divide by a negative number, the inequality sign reverses.
解一次和二次不等式需要仔细使用符号表。当你乘以或除以一个负数时，不等号方向要反转。
For modulus equations such as |2x - 1| = 5, split into two cases: 2x - 1 = 5 or 2x - 1 = -5. For modulus inequalities like |x + 3| < 2, rewrite as -2 < x + 3 < 2.
对于像 |2x - 1| = 5 这样的绝对值方程，分成两种情况：2x - 1 = 5 或 2x - 1 = -5。对于像 |x + 3| < 2 这样的绝对值不等式，改写为 -2 < x + 3 < 2。
A common mistake is to square both sides of a modulus inequality without checking the original condition. Always draw a sign diagram or sketch the graph to confirm the final interval.
一个常见错误是在没有检查原始条件的情况下就对绝对值不等式两边平方。务必画出符号表或草图来确认最终区间。
4. Functions and inverse functions | 函数与反函数
ap x and y, and then rearrange to make y the subject. The domain of f⁻¹ is the range of f.
要求一一对应函数 f(x) 的反函数，先写 y = f(x)，交换 x 和 y，然后重新整理使 y 成为主项。f⁻¹ 的定义域就是 f 的值域。
ample, if f(x) = 2x + 1 and g(x) = x² - 3, then fg(x) = f(g(x)) = 2(x² - 3) + 1 = 2x² - 5.
复合函数通过将一个函数代入另一个函数来求值。例如，如果 f(x) = 2x + 1 且 g(x) = x² - 3，那么 fg(x) = f(g(x)) = 2(x² - 3) + 1 = 2x² - 5。
- 1) / (y + 2), then rearrange to y = (2x + 1) / (1 - x), so f⁻¹(x) = (2x + 1) / (1 - x).
例题：求 f(x) = (x - 1) / (x + 2) 的反函数 f⁻¹(x)。令 y = (x - 1) / (x + 2)，交换 x 和 y 得到 x = (y - 1) /
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