📚 AQA A Level Maths Unit 3 January 2022: Key Topics and Worked Solutions | AQA A Level 数学第三单元 2022年1月：核心考点与例题解析
The AQA A Level Maths Unit 3 paper from January 2022 is a pure mathematics assessment that tests core algebraic, trigonometric, logarithmic and calculus skills. It requires exact answers, clear reasoning and confident use of A-level methods under timed conditions.
AQA A Level 数学第三单元 2022 年 1 月试卷是一份纯数学评估，考查核心代数、三角、对数和微积分技能。它要求给出精确答案、清晰的推理过程，并能在限时条件下熟练运用 A-level 数学方法。
1. Paper Overview | 试卷概览
The paper typically lasts 1 hour 30 minutes and carries 75 marks. Questions are a mix of short structured items and longer problem-solving tasks, with an emphasis on proof, manipulation and exact values expressed in terms of π, e or ln 2.
试卷通常时长 1 小时 30 分钟，满分 75 分。题目包含短结构化小题和较长的综合应用题，重点考查证明、代数变形以及用 π、e 或 ln 2 等精确形式表示答案。
Many questions do not rely solely on calculator work. Students need to show full working because method marks are awarded even if the final answer contains an arithmetic slip.
很多题目不能仅靠计算器完成。学生需要展示完整步骤，因为即使最终答案出现计算失误，AQA 仍会给予方法分。
2. Algebra and Functions | 代数与函数
Typical questions test composition, inverse functions, range and domain, and modulus equations. For example, the function f(x) = 3 + 2/(x – 1) with domain x > 1 has range (3, ∞) because 2/(x – 1) tends to ∞ as x approaches 1 from the right, and it tends to 0 as x tends to ∞.
典型题目考查复合函数、反函数、值域与定义域以及绝对值方程。例如，函数 f(x) = 3 + 2/(x – 1) 在 x > 1 上的值域为 (3, ∞)，因为当 x 从右侧趋近 1 时，2/(x – 1) 趋近 ∞；当 x 趋近 ∞ 时，该分式趋近 0。
A modulus equation such as |2x – 1| = 3x + 2 must be solved by splitting into cases. Solving 2x – 1 = 3x + 2 gives x = -3, but this fails the original equation since the right-hand side is negative. Solving -(2x – 1) = 3x + 2 gives x = -1/5, which is the only valid solution.
例如绝对值方程 |2x – 1| = 3x + 2 必须分情况讨论。解 2x – 1 = 3x + 2 得 x = -3，但代入原方程右端为负，不满足条件。解 -(2x – 1) = 3x + 2 得 x = -1/5，这是唯一有效解。
3. Trigonometry | 三角学
Trigonometry questions often require solving equations over a given interval such as 0° ≤ x ≤ 360°. A standard example is 2sin²x – 3sinx + 1 = 0. Factorising gives (2sinx – 1)(sinx – 1) = 0, so sinx = 1/2 or sinx = 1. Within the interval, sinx = 1/2 yields x = 30° and 150°, while sinx = 1 yields x = 90°.
三角函数题通常要求在给定区间内解方程，例如 0° ≤ x ≤ 360°。一个典型例子是 2sin²x – 3sinx + 1 = 0。因式分解得 (2sinx – 1)(sinx – 1) = 0，所以 sinx = 1/2 或 sinx = 1。在区间内，sinx = 1/2 给出 x = 30° 和 150°，sinx = 1 给出 x = 90°。
Students must also be confident with exact trigonometric values, double angle identities and graph transformations. For example, y = 3sin(2x – 60°) has amplitude 3, period 180° and a phase shift of 30° to the right compared with y = 3sin(2x).
学生还需熟练掌握特殊角的精确三角函数值、倍角公式和图形变换。例如，y = 3sin(2x – 60°) 的振幅为 3，周期为 180°，与 y = 3sin(2x) 相比向右平移 30°。
4. Exponentials and Logarithms | 指数与对数
The January 2022 style paper frequently includes exponential growth or decay modelling and logarithmic manipulation. A typical question may ask to solve 3^(2x+1) = 5^(x-2). Taking natural logarithms on both sides gives (2x+1)ln3 = (x-2)ln5. This rearranges to x(2ln3 – ln5) = -2ln5 – ln3, so x = (-2ln5 – ln3)/(2ln3 – ln5).
2022 年 1 月风格试卷常包含指数增长或衰减建模以及对数运算。典型题目可能要求解 3^(2x+1) = 5^(x-2)。两边取自然对数得 (2x+1)ln3 = (x-2)ln5，整理后 x(2ln3 – ln5) = -2ln5 – ln3，因此 x = (-2ln5 – ln3)/(2ln3 – ln5)。
Logarithm rules are essential: ln(ab) = ln a + ln b, ln(a/b) = ln a – ln b and ln(aᵏ) = k ln a. Many students lose marks by changing bases incorrectly or failing to simplify expressions to the required exact form.
对数法则是关键：ln(ab) = ln a + ln b、ln(a/b) = ln a – ln b 以及 ln(aᵏ) = k ln a。许多学生因换底错误或未能将表达式化简为题目要求的精确形式而失分。
5. Differentiation | 微分
Unit 3 tests product rule, quotient rule and chain rule, often in combination with exponentials, logarithms and trigonometric functions. For y = x²e^(3x), the product rule gives dy/dx = 2xe^(3x) + 3x²e^(3x) = xe^(3x)(2 + 3x).
第三单元考查乘法法则、除法法则和链式法则，并常与指数、对数和三角函数结合。例如 y = x²e^(3x)，使用乘法法则得 dy/dx = 2xe^(3x) + 3x²e^(3x) = xe^(3x)(2 + 3x)。
Another standard skill is finding the gradient at a point and using it to form a tangent or normal equation. For y = ln(2x + 1), the derivative is dy/dx = 2/(2x + 1). At x = 1, the gradient is 2/3, so the tangent at (1, ln3) is y – ln3 = (2/3)(x – 1).
另一项基本技能是求某点的导数并用其建立切线或法线方程。例如 y = ln(2x + 1) 的导数为 dy/dx = 2/(2x + 1)。当 x = 1 时，梯度为 2/3，因此 (1, ln3) 处的切线为 y – ln3 = (2/3)(x – 1)。
6. Integration | 积分
Integration questions in this unit are often based on reverse differentiation, substitution and integration by parts. A common type asks to integrate ∫ (x² + 1)/(x³ + 3x) dx. Using u = x³ + 3x, du = (3x² + 3)dx = 3(x² + 1)dx, so the integral becomes (1/3)∫ du/u = (1/3)ln|u| + C = (1/3)ln|x³ + 3x| + C.
本单元的积分题常以逆微分、换元法和分部积分为基础。常见题型如求 ∫ (x² + 1)/(x³ + 3x) dx。令 u = x³ + 3x，du = (3x² + 3)dx = 3(x² + 1)dx，因此积分变为 (1/3)∫ du/u = (1/3)ln|u| + C = (1/3)ln|x³ + 3x| + C。
Definite integrals may require exact evaluation. For example, ∫₀¹ 2x/(1 + x²) dx can be solved by substituting u = 1 + x², giving [ln(1 + x²)]₀¹ = ln2 – ln1 = ln2. This is a typical AQA exact value question.
定积分可能要求精确求值。例如 ∫₀¹ 2x/(1 + x²) dx 可通过令 u = 1 + x² 求解，得到 [ln(1 + x²)]₀¹ = ln2 – ln1 = ln2。这是 AQA 常见的精确值题型。
7. Numerical Methods | 数值方法
The numerical methods section usually includes iterative schemes, sign-change root location and Simpson’s rule. A question may ask to show that x³ – 2x – 5 = 0 has a root between 2 and 3 by evaluating f(2) = -1 and f(3) = 16, noting the sign change.
数值方法部分通常包括迭代格式、用符号变化判断根的位置以及辛普森法则。题目可能要求通过计算 f(2) = -1 和 f(3) = 16，观察到符号改变，从而证明 x³ – 2x – 5 = 0 在 2 与 3 之间有一个根。
Simpson’s rule with 4 strips for ∫₀¹ e^(x²) dx uses h = 0.25. The ordinates are approximately 1, 1.0645, 1.2840, 1.7551 and 2.7183, giving an estimate of about 1.4637 to 4 decimal places. Accuracy in intermediate values is important because final answers are often required to a specified precision.
用 4 个条带的辛普森法则计算 ∫₀¹ e^(x²) dx 时，步长 h = 0.25。纵坐标约为 1、1.0645、1.2840、1.7551 和 2.7183，得到的估计值约为 1.4637（保留 4 位小数
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