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Miscellaneous Exercises 2: Mixed AQA A-Level Maths Practice | AQA A-Level 数学杂项练习二

2026-08-21 20:42 · 本文部分内容由AI辅助创作

📚 Miscellaneous Exercises 2: Mixed AQA A-Level Maths Practice | AQA A-Level 数学杂项练习二

This revision set brings together typical AQA A-level Pure Mathematics skills in one mixed exercise. It covers algebra, quadratics, polynomials, exponentials, trigonometry, calculus, sequences and vectors, so you can test yourself under exam-style conditions.

本复习练习将 AQA A-Level 纯数学的常见技能整合为一套混合练习,涵盖代数、二次函数、多项式、指数、三角、微积分、序列与向量,帮助你以考试形式进行自测。


1. Algebraic Manipulation and Surds | 代数运算与根式

Simplifying surds and rationalising denominators appear regularly in AQA papers. A common type is to express (√a + √b)/(√a - √b) in the form p + q√c. To do this, multiply the numerator and denominator by the conjugate of the denominator.

在 AQA 试卷中,根式化简与分母有理化经常出现。常见类型是把 (√a + √b)/(√a - √b) 写成 p + q√c 的形式,方法是将分子分母同时乘以分母的共轭式。

Worked example: simplify (√5 + 1)/(√5 - 1). Multiply top and bottom by (√5 + 1). The numerator becomes (√5 + 1)² = 5 + 2√5 + 1 = 6 + 2√5; the denominator becomes (√5)² - 1² = 5 - 1 = 4. Therefore the expression simplifies to (6 + 2√5)/4 = (3 + √5)/2.

例题:化简 (√5 + 1)/(√5 - 1)。分子分母同乘 (√5 + 1)。分子变为 (√5 + 1)² = 5 + 2√5 + 1 = 6 + 2√5;分母变为 (√5)² - 1² = 5 - 1 = 4。因此原式化简为 (6 + 2√5)/4 = (3 + √5)/2。

(√5 + 1)/(√5 - 1) = (3 + √5)/2

When the denominator is a single surd such as √3, multiply by √3/√3; when it is a binomial such as √a + √b, multiply by its conjugate √a - √b. Never square the denominator without applying the same operation to the numerator.

当分母为单个根式如 √3 时,应乘以 √3/√3;当分母为二项式如 √a + √b 时,应乘以其共轭式 √a - √b。切勿只对方分母平方而不对分子进行相同运算。


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2. Quadratic Functions and the Discriminant | 二次函数与判别式

ct real roots; if Δ = 0 there is one repeated real root; if Δ < 0 there are no real roots.

对于二次方程 ax² + bx + c = 0,判别式为 Δ = b² - 4ac。若 Δ > 0,则有两个不相等实根;若 Δ = 0,则有一个重实根;若 Δ < 0,则没有实根。

o k² + 2k - 15 > 0, which factorises as (k + 5)(k - 3) > 0. The solution is k < -5 or k > 3.

例题:求使方程 x² + (k + 1)x + 4 = 0 有两个不相等实根的 k 的取值范围。这里 a = 1,b = k + 1,c = 4,所以 Δ = (k + 1)² - 16 = k² + 2k - 15。要求 Δ > 0,即 k² + 2k - 15 > 0,因式分解得 (k + 5)(k - 3) > 0。解为 k < -5 或 k > 3。

Δ = b² - 4ac

If the question asks for 'real roots' without saying 'distinct' d , then the repeated root case is allowed and the condition is Δ ≥ 0. A common error is to forget the equality sign in such questions.

如果题目只说“有实根”而未说“不相等”,则重根情况也允许,条件应为 Δ ≥ 0。常见错误是在此类题目中漏写等号。


3. Polynomial Division and the Factor Theorem | 多项式除法与因式定理

0. Once one factor is known, the remaining factors can be found by polynomial division or by inspection.

因式定理指出,当且仅当 f(a) = 0 时,(x - a) 是 f(x) 的因式。一旦知道一个因式,就可以通过多项式除法或观察法求出其余因式。

2) gives the quotient x² + 2x - 3, so f(x) = (x + 2)(x² + 2x - 3) = (x + 2)(x + 3)(x - 1).

例题:证明 (x + 2) 是 f(x) = x³ + 4x² + x - 6 的因式,并将 f(x) 完全因式分解。计算

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

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