Answers to the Exercises in Support Pack 3 | 支持包3练习答案解析

📚 Answers to the Exercises in Support Pack 3 | 支持包3练习答案解析

This article provides detailed solutions and explanations for the exercises in AQA A-Level Mathematics Support Pack 3. Each question is worked through step by step, with key methods highlighted to reinforce your understanding of core topics.

本文为 AQA A-Level 数学 Support Pack 3 中的练习题提供详细解答与讲解。每道题都逐步演算，并重点标出关键方法，帮助大家巩固核心知识点的理解。

1. Algebraic Simplification and Factorisation | 代数化简与因式分解

Exercise 1.1 Simplify \(3x(2x – 5) – (x + 3)(x – 2)\). (Note: we use plain text notation here, but in the exam you may write fractions normally.)

练习1.1 化简 \(3x(2x – 5) – (x + 3)(x – 2)\)。（注意：此处使用纯文本记号，考试时可按常规书写分式。）

Expand both parts: \(3x(2x – 5) = 6x^2 – 15x\), and \((x + 3)(x – 2) = x^2 + x – 6\). Subtract: \(6x^2 – 15x – (x^2 + x – 6) = 5x^2 – 16x + 6\).

展开两部分：\(3x(2x – 5) = 6x^2 – 15x\)，\((x + 3)(x – 2) = x^2 + x – 6\)。相减得 \(6x^2 – 15x – (x^2 + x – 6) = 5x^2 – 16x + 6\)。

Exercise 1.2 Factorise completely: \(2x^3 – 8x\).

练习1.2 完全因式分解：\(2x^3 – 8x\)。

First take out the common factor \(2x\): \(2x(x^2 – 4)\). Then recognise the difference of squares: \(x^2 – 4 = (x – 2)(x + 2)\). Hence the complete factorisation is \(2x(x – 2)(x + 2)\).

先提取公因式 \(2x\)：\(2x(x^2 – 4)\)。再识别平方差：\(x^2 – 4 = (x – 2)(x + 2)\)。因此完整因式分解为 \(2x(x – 2)(x + 2)\)。

2. Solving Equations and Inequalities | 解方程与不等式

Exercise 2.1 Solve the simultaneous equations: \(y = 2x + 1\) and \(y = x^2 + x + 1\).

练习2.1 解联立方程：\(y = 2x + 1\) 和 \(y = x^2 + x + 1\)。

Equate the two expressions: \(2x + 1 = x^2 + x + 1\). Simplify: \(x^2 – x = 0\), so \(x(x – 1) = 0\). Thus \(x = 0\) or \(x = 1\). Substitute back: when \(x = 0\), \(y = 1\); when \(x = 1\), \(y = 3\). Solutions: \((0,1)\) and \((1,3)\).

令两式相等：\(2x + 1 = x^2 + x + 1\)。化简得 \(x^2 – x = 0\)，即 \(x(x – 1) = 0\)。因此 \(x = 0\) 或 \(x = 1\)。代回得：当 \(x = 0\) 时，\(y = 1\)；当 \(x = 1\) 时，\(y = 3\)。解为 \((0,1)\) 和 \((1,3)\)。

Exercise 2.2 Solve the inequality \(x^2 – 3x – 10 < 0\).

练习2.2 解不等式 \(x^2 – 3x – 10 < 0\)。

Factorise: \((x – 5)(x + 2) < 0\). The roots are \(x = -2\) and \(x = 5\). The parabola opens upwards, so the expression is negative between the roots. Hence the solution is \(-2 < x < 5\).

因式分解得 \((x – 5)(x + 2) < 0\)。根为 \(x = -2\) 和 \(x = 5\)。抛物线开口向上，故表达式在两根之间取负值。所以解为 \(-2 < x < 5\)。

3. Functions and Transformations | 函数与变换

Exercise 3.1 Given \(f(x) = 2x – 3\), find \(f^{-1}(x)\) and verify \(f(f^{-1}(x)) = x\).

练习3.1 已知 \(f(x) = 2x – 3\)，求 \(f^{-1}(x)\) 并验证 \(f(f^{-1}(x)) = x\)。

Let \(y = 2x – 3\). Swap \(x\) and \(y\): \(x = 2y – 3\). Solve for \(y\): \(y = \frac{x + 3}{2}\). So \(f^{-1}(x) = \frac{x + 3}{2}\). Verification: \(f(f^{-1}(x)) = 2 \times \frac{x + 3}{2} – 3 = x + 3 – 3 = x\).

设 \(y = 2x – 3\)。交换 \(x\) 与 \(y\)：\(x = 2y – 3\)。解出 \(y\)：\(y = \frac{x + 3}{2}\)。所以 \(f^{-1}(x) = \frac{x + 3}{2}\)。验证：\(f(f^{-1}(x)) = 2 \times \frac{x + 3}{2} – 3 = x + 3 – 3 = x\)。

Exercise 3.2 Describe the transformation that maps \(y = x^2\) to \(y = 3(x – 1)^2 + 2\).

练习3.2 描述将 \(y = x^2\) 映射为 \(y = 3(x – 1)^2 + 2\) 的变换。

Starting from \(y = x^2\): first translate 1 unit right (replacing \(x\) by \(x – 1\)) to get \(y = (x – 1)^2\). Then stretch vertically by factor 3 (multiplying by 3) to get \(y = 3(x – 1)^2\). Finally translate 2 units up (adding 2) to reach \(y = 3(x – 1)^2 + 2\). Order matters: translations in the \(y\)-direction can be interchanged with the vertical stretch, but the horizontal translation must be applied before any horizontal stretch if present.

从 \(y = x^2\) 开始：先向右平移1个单位（将 \(x\) 替换为 \(x – 1\)）得到 \(y = (x – 1)^2\)。然后纵向拉伸为原来的3倍（乘以3）得到 \(y = 3(x – 1)^2\)。最后向上平移2个单位（加2）得到 \(y = 3(x – 1)^2 + 2\)。顺序很重要：\(y\) 方向的平移可与纵向拉伸交换顺序，但水平平移必须在任何水平拉伸之前进行（如果存在）。

4. Differentiation | 微分法

Exercise 4.1 Differentiate \(y = 4x^3 – 2x^2 + 5x – 7\).

练习4.1 求 \(y = 4x^3 – 2x^2 + 5x – 7\) 的导数。

Apply the power rule term by term: \(\frac{dy}{dx} = 12x^2 – 4x + 5\). The constant term \(-7\) differentiates to zero.

逐项应用幂法则：\(\frac{dy}{dx} = 12x^2 – 4x + 5\)。常数项 \(-7\) 的导数为零。

Exercise 4.2 Find the gradient of the tangent to \(y = x^2 – 3x + 1\) at the point where \(x = 4\).

练习4.2 求曲线 \(y = x^2 – 3x + 1\) 在 \(x = 4\) 处的切线斜率。

The derivative is \(\frac{dy}{dx} = 2x – 3\). At \(x = 4\), this equals \(2(4) – 3 = 5\). So the gradient of the tangent is 5.

导数为 \(\frac{dy}{dx} = 2x – 3\)。在 \(x = 4\) 时，值为 \(2(4) – 3 = 5\)。因此切线斜率为5。

5. Integration | 积分法

Exercise 5.1 Evaluate \(\int (6x^2 + 4x – 3) \, dx\).

练习5.1 求 \(\int (6x^2 + 4x – 3) \, dx\)。

Integrate each term: \(\int 6x^2 \, dx = 2x^3\), \(\int 4x \, dx = 2x^2\), \(\int -3 \, dx = -3x\). Add the constant of integration: \(2x^3 + 2x^2 – 3x + C\).

逐项积分：\(\int 6x^2 \, dx = 2x^3\)，\(\int 4x \, dx = 2x^2\)，\(\int -3 \, dx = -3x\)。加上积分常数：\(2x^3 + 2x^2 – 3x + C\)。

Exercise 5.2 Calculate \(\int_{0}^{2} (x^3 – 2x) \, dx\).

练习5.2 计算 \(\int_{0}^{2} (x^3 – 2x) \, dx\)。

Find the antiderivative: \(\frac{x^4}{4} – x^2\). Evaluate from 0 to 2: \(\left(\frac{2^4}{4} – 2^2\right) – \left(\frac{0^4}{4} – 0^2\right) = (4 – 4) – 0 = 0\).

求原函数：\(\frac{x^4}{4} – x^2\)。从0到2计算：\(\left(\frac{2^4}{4} – 2^2\right) – \left(\frac{0^4}{4} – 0^2\right) = (4 – 4) – 0 = 0\)。

6. Trigonometric Identities | 三角恒等式

Exercise 6.1 Solve \(2 \cos \theta + \sqrt{3} = 0\) for \(0 \le \theta < 360°\).

练习6.1 解方程 \(2 \cos \theta + \sqrt{3} = 0\)，其中 \(0 \le \theta < 360°\)。

Rearrange: \(\cos \theta = -\frac{\sqrt{3}}{2}\). The reference angle is \(30°\) (since \(\cos 30° = \frac{\sqrt{3}}{2}\)). Cosine is negative in the second and third quadrants, so \(\theta = 180° – 30° = 150°\) and \(\theta = 180° + 30° = 210°\).

整理得 \(\cos \theta = -\frac{\sqrt{3}}{2}\)。参考角为 \(30°\)（因为 \(\cos 30° = \frac{\sqrt{3}}{2}\)）。余弦在第二和第三象限为负，所以 \(\theta = 180° – 30° = 150°\) 和 \(\theta = 180° + 30° = 210°\)。

Exercise 6.2 Prove the identity \(\frac{\sin^2 x}{1 – \cos x} = 1 + \cos x\).

练习6.2 证明恒等式 \(\frac{\sin^2 x}{1 – \cos x} = 1 + \cos x\)。

Use \(\sin^2 x = 1 – \cos^2 x\). Then \(\frac{1 – \cos^2 x}{1 – \cos x} = \frac{(1 – \cos x)(1 + \cos x)}{1 – \cos x} = 1 + \cos x\).

使用 \(\sin^2 x = 1 – \cos^2 x\)。于是 \(\frac{1 – \cos^2 x}{1 – \cos x} = \frac{(1 – \cos x)(1 + \cos x)}{1 – \cos x} = 1 + \cos x\)。

7. Exponentials and Logarithms | 指数与对数

Exercise 7.1 Solve \(\log_2 x + \log_2 (x – 2) = 3\).

练习7.1 解方程 \(\log_2 x + \log_2 (x – 2) = 3\)。

Combine the logs: \(\log_2 [x(x – 2)] = 3\). Convert to exponential form: \(x(x – 2) = 2^3 = 8\). This gives \(x^2 – 2x – 8 = 0\), factorising to \((x – 4)(x + 2) = 0\). Thus \(x = 4\) or \(x = -2\). Since the logarithms require \(x > 2\), the only valid solution is \(x = 4\).

合并对数：\(\log_2 [x(x – 2)] = 3\)。转换为指数形式：\(x(x – 2) = 2^3 = 8\)。整理得 \(x^2 – 2x – 8 = 0\)，因式分解为 \((x – 4)(x + 2) = 0\)。故 \(x = 4\) 或 \(x = -2\)。由于对数要求 \(x > 2\)，唯一有效解为 \(x = 4\)。

Exercise 7.2 Simplify \(e^{2 \ln 3}\).

练习7.2 化简 \(e^{2 \ln 3}\)。

Use the power property: \(e^{2 \ln 3} = (e^{\ln 3})^2 = 3^2 = 9\).

使用幂性质：\(e^{2 \ln 3} = (e^{\ln 3})^2 = 3^2 = 9\)。

8. Coordinate Geometry | 坐标几何

Exercise 8.1 Find the equation of the line through \((3, -2)\) and \((-1, 6)\).

练习8.1 求经过点 \((3, -2)\) 和 \((-1, 6)\) 的直线方程。

Gradient \(m = \frac{6 – (-2)}{-1 – 3} = \frac{8}{-4} = -2\). Using point-slope form with \((3, -2)\): \(y – (-2) = -2(x – 3)\), so \(y + 2 = -2x + 6\), giving \(y = -2x + 4\).

斜率 \(m = \frac{6 – (-2)}{-1 – 3} = \frac{8}{-4} = -2\)。用点斜式代入 \((3, -2)\)：\(y – (-2) = -2(x – 3)\)，即 \(y + 2 = -2x + 6\)，整理得 \(y = -2x + 4\)。

Exercise 8.2 The circle has equation \(x^2 + y^2 – 6x + 4y – 12 = 0\). Find its centre and radius.

练习8.2 圆 \(x^2 + y^2 – 6x + 4y – 12 = 0\)，求圆心和半径。

Complete the square: \((x^2 – 6x) + (y^2 + 4y) = 12\). \((x – 3)^2 – 9 + (y + 2)^2 – 4 = 12\), so \((x – 3)^2 + (y + 2)^2 = 25\). Centre is \((3, -2)\), radius \(r = 5\).

配方：\((x^2 – 6x) + (y^2 + 4y) = 12\)。\((x – 3)^2 – 9 + (y + 2)^2 – 4 = 12\)，所以 \((x – 3)^2 + (y + 2)^2 = 25\)。圆心为 \((3, -2)\)，半径 \(r = 5\)。

9. Sequences and Series | 数列与级数

Exercise 9.1 An arithmetic progression has first term 5 and common difference 3. Find the 20th term.

练习9.1 一个等差数列首项为5，公差为3。求第20项。

Use \(u_n = a + (n – 1)d\). Here \(a = 5\), \(d = 3\), \(n = 20\): \(u_{20} = 5 + 19 \times 3 = 5 + 57 = 62\).

使用 \(u_n = a + (n – 1)d\)。这里 \(a = 5\)，\(d = 3\)，\(n = 20\)：\(u_{20} = 5 + 19 \times 3 = 5 + 57 = 62\)。

Exercise 9.2 Find the sum of the first 10 terms of a geometric series with first term 2 and common ratio \(\frac{1}{2}\).

练习9.2 求首项为2、公比为 \(\frac{1}{2}\) 的等比级数前10项之和。

Use the sum formula \(S_n = \frac{a(1 – r^n)}{1 – r}\). Substitute \(a = 2\), \(r = \frac{1}{2}\), \(n = 10\): \(S_{10} = \frac{2\left(1 – \left(\frac{1}{2}\right)^{10}\right)}{1 – \frac{1}{2}} = 4\left(1 – \frac{1}{1024}\right) = \frac{4095}{256}\).

使用求和公式 \(S_n = \frac{a(1 – r^n)}{1 – r}\)。代入 \(a = 2\)，\(r = \frac{1}{2}\)，\(n = 10\)：\(S_{10} = \frac{2\left(1 – \left(\frac{1}{2}\right)^{10}\right)}{1 – \frac{1}{2}} = 4\left(1 – \frac{1}{1024}\right) = \frac{4095}{256}\)。

10. Exam-Style Mixed Exercise | 考试风格综合练习

Exercise 10.1 A curve has equation \(y = x^3 – 6x^2 + 9x\). Find the coordinates of the stationary points and determine their nature.

练习10.1 曲线方程为 \(y = x^3 – 6x^2 + 9x\)。求驻点坐标并判断其性质。

Differentiate: \(\frac{dy}{dx} = 3x^2 – 12x + 9 = 3(x – 1)(x – 3)\). Stationary points when \(\frac{dy}{dx} = 0\), so \(x = 1\) or \(x = 3\). The corresponding \(y\)-values are \(y(1) = 4\) and \(y(3) = 0\). To determine nature, use the second derivative: \(\frac{d^2y}{dx^2} = 6x – 12\). At \(x = 1\), \(\frac{d^2y}{dx^2} = -6 < 0\), so \((1,4)\) is a local maximum. At \(x = 3\), \(\frac{d^2y}{dx^2} = 6 > 0\), so \((3,0)\) is a local minimum.

求导：\(\frac{dy}{dx} = 3x^2 – 12x + 9 = 3(x – 1)(x – 3)\)。驻点条件为 \(\frac{dy}{dx} = 0\)，所以 \(x = 1\) 或 \(x = 3\)。对应的 \(y\) 值为 \(y(1) = 4\) 和 \(y(3) = 0\)。用二阶导数判断性质：\(\frac{d^2y}{dx^2} = 6x – 12\)。在 \(x = 1\) 处，\(\frac{d^2y}{dx^2} = -6 < 0\)，所以 \((1,4)\) 为局部极大值。在 \(x = 3\) 处，\(\frac{d^2y}{dx^2} = 6 > 0\)，所以 \((3,0)\) 为局部极小值。

Exercise 10.2 Evaluate \(\int_{0}^{1} \frac{2x + 1}{x^2 + x + 3} \, dx\).

练习10.2 计算 \(\int_{0}^{1} \frac{2x + 1}{x^2 + x + 3} \, dx\)。

Notice that the numerator is the derivative of the denominator: \(\frac{d}{dx}(x^2 + x + 3) = 2x + 1\). Therefore use the natural logarithm rule: \(\int \frac{f'(x)}{f(x)} \, dx = \ln |f(x)| + C\). So the integral equals \([\ln(x^2 + x + 3)]_0^1 = \ln 5 – \ln 3 = \ln \frac{5}{3}\).

注意分子恰好是分母的导数：\(\frac{d}{dx}(x^2 + x + 3) = 2x + 1\)。因此使用自然对数法则：\(\int \frac{f'(x)}{f(x)} \, dx = \ln |f(x)| + C\)。所以定积分为 \([\ln(x^2 + x + 3)]_0^1 = \ln 5 – \ln 3 = \ln \frac{5}{3}\)。

11. Common Mistakes and Exam Tips | 常见错误与应试技巧

One frequent error is forgetting the constant of integration when evaluating indefinite integrals. Always add \(+C\) unless the integral is definite. Another is misapplying the chain rule; for example, differentiating \(\sin(2x)\) must give \(2\cos(2x)\), not \(\cos(2x)\).

一个常见错误是求不定积分时忘记积分常数。除非是定积分，否则务必加上 \(+C\)。另一个常见错误是链式法则运用不当；例如，对 \(\sin(2x)\) 求导应得 \(2\cos(2x)\)，而不是 \(\cos(2x)\)。

When solving inequalities involving quadratics, always sketch the parabola or test intervals between roots to avoid sign errors. For logarithmic equations, always check the domain of the original expression; extraneous roots often arise when both sides are exponentiated.

解二次不等式时，务必画出抛物线草图或测试根之间的区间，避免符号错误。解对数方程时，始终检查原表达式的定义域；两边取指数时经常会产生增根。

12. Summary of Key Results | 关键结论总结

Below is a table of the main formulas and results that appear repeatedly in Support Pack 3 exercises.

下表总结了 Support Pack 3 练习中反复出现的主要公式与结论。

Topic	Key Result
Algebra	\((a+b)^2 = a^2 + 2ab + b^2\)
Quadratics	Discriminant \(\Delta = b^2 – 4ac\)
Differentiation	\(\frac{d}{dx}(x^n) = nx^{n-1}\)
Integration	\(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C, n \neq -1\)
Logarithms	\(\log_a (xy) = \log_a x + \log_a y\)

Master these core results and practise applying them in different contexts. Support Pack 3 is designed to consolidate exactly these skills, so working through every exercise carefully will prepare you well for the exam.

掌握这些核心结论，并练习在不同情境中加以应用。Support Pack 3 正是为了巩固这些技能而设计，因此认真完成每一道练习将使你为考试做好充分准备。

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Answers to the Exercises in Support Pack 3 | 支持包3练习答案解析