📚 Answers to the Exercises in Support Pack 2 | Support Pack 2 练习答案
This article provides worked solutions to the exercises from Support Pack 2, page 143, which form part of the AQA A-Level Mathematics revision materials. The questions cover core algebraic manipulation, differentiation, integration, trigonometry, sequences, vectors and coordinate geometry. Each solution is presented step by step to reinforce the method and to help you avoid common mistakes.
本文提供 AQA A-Level 数学复习资料 Support Pack 2 第143页练习的完整解答。这些题目涵盖代数运算、微分、积分、三角学、数列、向量与坐标几何等核心考点。每道题都给出分步演算，以强化解题方法，并帮助你避开常见错误。
1. Solving a Quadratic Equation | 解二次方程
Question: Solve the equation x² − 5x + 6 = 0.
问题： 解方程 x² − 5x + 6 = 0。
Answer: Factorise the quadratic into (x − 2)(x − 3) = 0. Hence x = 2 or x = 3.
答案： 将二次式因式分解为 (x − 2)(x − 3) = 0。因此 x = 2 或 x = 3。
Method check: Always expand your factors to verify the coefficient of x. (x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6 ✓.
方法检验： 务必展开因式以核对一次项系数。(x − 2)(x − 3) = x² − 3x − 2x + 6 = x² − 5x + 6 ✓。
2. Differentiation by Power Rule | 幂法则求导
Question: Find the derivative of f(x) = x³ − 4x with respect to x.
问题： 求函数 f(x) = x³ − 4x 关于 x 的导数。
Answer: Using the power rule, d/dx (xⁿ) = n xⁿ⁻¹, we obtain f′(x) = 3x² − 4.
答案： 利用幂法则 d/dx (xⁿ) = n xⁿ⁻¹，得到 f′(x) = 3x² − 4。
Note: The derivative of a constant is zero. Since −4x is a linear term, its derivative is simply −4.
注意： 常数的导数为零。由于 −4x 是一次项，其导数直接为 −4。
3. Indefinite Integration | 不定积分
Question: Evaluate ∫ (2x + 3) dx.
问题： 计算 ∫ (2x + 3) dx。
Answer: Integrate term by term: ∫ 2x dx = x², ∫ 3 dx = 3x. Therefore the general antiderivative is x² + 3x + C, where C is the constant of integration.
答案： 逐项积分：∫ 2x dx = x²，∫ 3 dx = 3x。因此原函数的一般形式为 x² + 3x + C，其中 C 为积分常数。
Common trap: Do not forget to add the constant C for indefinite integrals.
常见陷阱： 对于不定积分，切勿忘记加上常数 C。
4. Trigonometric Values in Right Triangles | 直角三角形中的三角函数值
Question: Given that sin θ = 3/5 and θ is acute, find cos θ.
问题： 已知 sin θ = 3/5，且 θ 为锐角，求 cos θ。
Answer: Use the identity sin²θ + cos²θ = 1. Then cos²θ = 1 − (3/5)² = 1 − 9/25 = 16/25. Since θ is acute, cos θ = √(16/25) = 4/5.
答案： 利用恒等式 sin²θ + cos²θ = 1。于是 cos²θ = 1 − (3/5)² = 1 − 9/25 = 16/25。因为 θ 为锐角，所以 cos θ = √(16/25) = 4/5。
Reasoning: The positive square root is chosen because Cos is positive in the first quadrant.
推理： 取正平方根是因为在第一象限余弦为正。
5. Simultaneous Equations | 联立方程
Question: Solve the simultaneous equations y = 2x + 1 and y = x² − 1.
问题： 解联立方程 y = 2x + 1 和 y = x² − 1。
Answer: Equate the two expressions for y: 2x + 1 = x² − 1. Rearranging gives x² − 2x − 2 = 0. Using the quadratic formula, x = [2 ± √(4 + 8)] / 2 = [2 ± √12] / 2 = 1 ± √3. Then substitute back to find y: when x = 1 + √3, y = 2(1 + √3) + 1 = 3 + 2√3; when x = 1 − √3, y = 3 − 2√3.
答案： 令两个 y 的表达式相等：2x + 1 = x² − 1。整理得 x² − 2x − 2 = 0。利用求根公式，x = [2 ± √(4 + 8)] / 2 = [2 ± √12] / 2 = 1 ± √3。代回求 y：当 x = 1 + √3 时，y = 2(1 + √3) + 1 = 3 + 2√3；当 x = 1 − √3 时，y = 3 − 2√3。
Check: Both points satisfy the line and the curve, so the solutions are consistent.
检验： 两组解均同时满足直线和曲线方程，因此结果一致。
6. Intersection with the x-axis | 与 x 轴的交点
Question: Find the x-coordinate of the point where the line y = 3x + 2 crosses the x-axis.
问题： 求直线 y = 3x + 2 与 x 轴交点的 x 坐标。
Answer: On the x-axis, y = 0. So solve 3x + 2 = 0, giving x = −2/3. The point is (−2/3, 0).
答案： 在 x 轴上，y = 0。因此解 3x + 2 = 0，得 x = −2/3。交点坐标为 (−2/3, 0)。
Graphical interpretation: This is the root of the linear function, where the line cuts the horizontal axis.
图像解释： 这是线性函数的零点，即直线与横轴相交的位置。
7. Logarithmic Simplification | 对数化简
Question: Simplify ln(e³).
问题： 化简 ln(e³)。
Answer: The natural logarithm and the exponential function are inverses. Therefore ln(e³) = 3.
答案： 自然对数与指数函数互为逆运算，因此 ln(e³) = 3。
General rule: ln(eᵏ) = k for any real k. This is also used when solving exponential equations by taking logs.
一般法则： 对任意实数 k，均有 ln(eᵏ) = k。这在通过取对数求解指数方程时也常用到。
8. Geometric Sequence Terms | 等比数列的项
Question: Find the 5th term of the geometric sequence 2, 6, 18, …
问题： 求等比数列 2, 6, 18, … 的第5项。
Answer: The first term is a = 2 and the common ratio is r = 6 ÷ 2 = 3. The n-th term is a rⁿ⁻¹, so the 5th term is 2 × 3⁴ = 2 × 81 = 162.
答案： 首项 a = 2，公比 r = 6 ÷ 2 = 3。第 n 项为 a rⁿ⁻¹，因此第5项为 2 × 3⁴ = 2 × 81 = 162。
Quick check: The sequence extends as 2, 6, 18, 54, 162, confirming the result.
快速验证： 数列依次为 2, 6, 18, 54, 162，确认结果正确。
9. Scalar Product of Vectors | 向量的数量积
Question: Given vectors a = 2i + 3j and b = 4i − j, compute a · b.
问题： 已知向量 a = 2i + 3j，b = 4i − j，计算 a · b。
Answer: For two-dimensional vectors a = a₁i + a₂j and b = b₁i + b₂j, the scalar product is a₁b₁ + a₂b₂. Hence a · b = (2)(4) + (3)(−1) = 8 − 3 = 5.
答案： 对于二维向量 a = a₁i + a₂j 和 b = b₁i + b₂j，数量积为 a₁b₁ + a₂b₂。因此 a · b = (2)(4) + (3)(−1) = 8 − 3 = 5。
Remember: The scalar product is a number, not a vector. Here the result is positive because the angle between the vectors is acute.
注意： 数量积是一个数，而不是向量。此处结果为正，说明两向量夹角为锐角。
10. Definite Integral | 定积分
Question: Evaluate ∫₁² x² dx.
问题： 计算定积分 ∫₁² x² dx。
Answer: First find the antiderivative: ∫ x² dx = x³/3. Then evaluate between the limits: [x³/3]₁² = (2³/3) − (1³/3) = 8/3 − 1/3 = 7/3.
答案： 先求原函数：∫ x² dx = x³/3。在上下限之间求值：[x³/3]₁² = (2³/3) − (1³/3) = 8/3 − 1/3 = 7/3。
Interpretation: This represents the exact area under y = x² between x = 1 and x = 2.
几何意义： 该定积分表示曲线 y = x² 在 x = 1 与 x = 2 之间与 x 轴围成的精确面积。
11. Circle Equation | 圆的方程
Question: Write down the equation of a circle with centre (2, −3) and radius 5.
问题： 写出圆心为 (2, −3)、半径为 5 的圆的方程。
Answer: The general formula is (x − h)² + (y − k)² = r², where (h, k) is the centre. Substituting gives (x − 2)² + (y + 3)² = 25.
答案： 圆的一般方程为 (x − h)² + (y − k)² = r²，其中 (h, k) 为圆心。代入得 (x − 2)² + (y + 3)² = 25。
Expanded form (optional): x² − 4x + 4 + y² + 6y + 9 = 25, which simplifies to x² + y² − 4x + 6y − 12 = 0.
展开形式（可选）： x² − 4x + 4 + y² + 6y + 9 = 25，化简得 x² + y² − 4x + 6y − 12 = 0。
12. Trigonometric Identity Proof | 三角恒等式证明
Question: Show that sin²θ + cos²θ = 1 for any angle θ.
问题： 证明对于任意角 θ，sin²θ + cos²θ = 1 成立。
Answer: In a right-angled triangle, sin θ = opposite/hypotenuse and cos θ = adjacent/hypotenuse. Let the sides be a (opposite), b (adjacent) and c (hypotenuse). Then sin²θ + cos²θ = a²/c² + b²/c² = (a² + b²)/c². By Pythagoras, a² + b² = c², so the expression equals c²/c² = 1.
答案： 在直角三角形中，sin θ = 对边/斜边，cos θ = 邻边/斜边。设三边分别为 a（对边）、b（邻边）、c（斜边）。则 sin²θ + cos²θ = a²/c² + b²/c² = (a² + b²)/c²。根据勾股定理，a² + b² = c²，因此该表达式等于 c²/c² = 1。
Extension: Dividing both sides by cos²θ gives tan²θ + 1 = sec²θ, another key identity.
拓展： 两边同时除以 cos²θ，可得 tan²θ + 1 = sec²θ，这是另一个重要恒等式。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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