📚 Answers to the Exercises in Support Pack 2 Page 143 | 支持包2第143页练习答案
This page provides full worked answers to the mixed exercises printed on page 143 of Support Pack 2 for AQA A-Level Mathematics. The pack covers the pure, mechanics and statistics content most commonly assessed under the AQA specification. Use these solutions to check your method, correct sign errors, and build confidence before timed practice.
本页提供 AQA A-Level 数学支持包 2 第 143 页混合练习的完整详细答案。该支持包涵盖 AQA 考试大纲中最常考查的纯数学、力学与统计内容。请使用这些答案检查解题方法、纠正符号错误，并在限时训练前建立信心。
1. Differentiation from First Principles | 从第一性原理求导
Question 1 asked you to differentiate f(x) = x² – 5x by using the limit definition f'(x) = lim[h→0] (f(x+h) – f(x))/h.
第 1 题要求使用极限定义 f'(x) = lim[h→0] (f(x+h) – f(x))/h 对函数 f(x) = x² – 5x 求导。
Answer: Expand f(x+h) = (x+h)² – 5(x+h) = x² + 2xh + h² – 5x – 5h. Subtract f(x) = x² – 5x, giving 2xh + h² – 5h. Divide by h to obtain 2x + h – 5. Let h → 0, so the derivative is f'(x) = 2x – 5.
答案：展开 f(x+h) = (x+h)² – 5(x+h) = x² + 2xh + h² – 5x – 5h。减去 f(x) = x² – 5x，得到 2xh + h² – 5h。除以 h 得到 2x + h – 5。令 h → 0，因此导数为 f'(x) = 2x – 5。
Common error: forgetting to subtract the original f(x) before dividing by h. Always write the full difference quotient first, then simplify.
常见错误：除以 h 之前忘记减去原来的 f(x)。务必先写出完整的差商，再进行化简。
2. Chain Rule and Product Rule | 链式法则与乘法法则
Question 2 asked you to differentiate y = (3x² + 1)⁵ and y = x² eˣ.
第 2 题要求对 y = (3x² + 1)⁵ 和 y = x² eˣ 求导。
For y = (3x² + 1)⁵, let u = 3x² + 1, so y = u⁵. Then dy/dx = 5u⁴ × 6x = 30x(3x² + 1)⁴.
对于 y = (3x² + 1)⁵，令 u = 3x² + 1，则 y = u⁵。因此 dy/dx = 5u⁴ × 6x = 30x(3x² + 1)⁴。
For y = x² eˣ, use the product rule: dy/dx = 2x eˣ + x² eˣ = x eˣ(2 + x).
对于 y = x² eˣ，使用乘法法则：dy/dx = 2x eˣ + x² eˣ = x eˣ(2 + x)。
Check: In the chain rule, the derivative of the inside function 3x² + 1 is 6x, and it must multiply the outer derivative. In the product rule, keep both original functions and multiply by the derivative of the other.
检查：在链式法则中，内层函数 3x² + 1 的导数是 6x，它必须与外层导数相乘。在乘法法则中，保留两个原函数并分别乘以另一个函数的导数。
3. Quotient Rule and Implicit Differentiation | 商法则与隐函数求导
Question 3 asked you to differentiate y = (x² + 1)/(x – 2) and to find dy/dx for x² + y² = 25.
第 3 题要求对 y = (x² + 1)/(x – 2) 求导，并求出 x² + y² = 25 中的 dy/dx。
Using the quotient rule: dy/dx = [(2x)(x – 2) – (x² + 1)(1)]/(x – 2)² = (2x² – 4x – x² – 1)/(x – 2)² = (x² – 4x – 1)/(x – 2)².
使用商法则：dy/dx = [(2x)(x – 2) – (x² + 1)(1)]/(x – 2)² = (2x² – 4x – x² – 1)/(x – 2)² = (x² – 4x – 1)/(x – 2)²。
For the implicit equation, differentiate both sides with respect to x: 2x + 2y dy/dx = 0, so dy/dx = -x/y.
对于隐函数方程，两边对 x 求导：2x + 2y dy/dx = 0，因此 dy/dx = -x/y。
Remember that when differentiating y² with respect to x, you must apply the chain rule and include dy/dx. This is the most common implicit differentiation mistake at A-Level.
请记住，对 y² 关于 x 求导时，必须应用链式法则并包含 dy/dx。这是 A-Level 隐函数求导中最常见的错误。
4. Integration by Substitution | 换元积分法
Question 4 asked you to evaluate ∫ 2x(x² + 3)⁴ dx and the definite integral ∫₀¹ 2x(x² + 3)⁴ dx.
第 4 题要求计算 ∫ 2x(x² + 3)⁴ dx 以及定积分 ∫₀¹ 2x(x² + 3)⁴ dx。
Let u = x² + 3, so du/dx = 2x, which gives du = 2x dx. The integral becomes ∫ u⁴ du = u⁵/5 + C = (x² + 3)⁵/5 + C.
令 u = x² + 3，则 du/dx = 2x，因此 du = 2x dx。积分变为 ∫ u⁴ du = u⁵/5 + C = (x² + 3)⁵/5 + C。
For the definite integral, substitute the limits: ∫₀¹ 2x(x² + 3)⁴ dx = [ (x² + 3)⁵/5 ]₀¹ = (4⁵ – 3⁵)/5 = (1024 – 243)/5 = 781/5.
对于定积分，代入上下限：∫₀¹ 2x(x² + 3)⁴ dx = [ (x² + 3)⁵/5 ]₀¹ = (4⁵ – 3⁵)/5 = (1024 – 243)/5 = 781/5。
Check: When changing variables in a definite integral, you may either convert the limits to u-values or return to x before substituting the original limits. Here returning to x is simpler.
检查：在定积分中换元时，可以将上下限转换为 u 值，也可以在代入原上下限之前回到 x。这里回到 x 更简单。
5. Integration by Parts | 分部积分法
Question 5 asked you to evaluate ∫ x eˣ dx and the definite integral ∫₀¹ x eˣ dx.
第 5 题要求计算 ∫ x eˣ dx 以及定积分 ∫₀¹ x eˣ dx。
Let u = x and dv/dx = eˣ. Then du/dx = 1 and v = eˣ. Applying integration by parts: ∫ x eˣ dx = x eˣ – ∫ eˣ dx = x eˣ – eˣ + C = eˣ(x – 1) + C.
令 u = x，dv/dx = eˣ。则 du/dx = 1，v = eˣ。应用分部积分：∫ x eˣ dx = x eˣ – ∫ eˣ dx = x eˣ – eˣ + C = eˣ(x – 1) + C。
For the definite integral: ∫₀¹ x eˣ dx = [eˣ(x – 1)]₀¹ = e¹(0) – e⁰(-1) = 0 + 1 = 1.
对于定积分：∫₀¹ x eˣ dx = [eˣ(x – 1)]₀¹ = e¹(0) – e⁰(-1) = 0 + 1 = 1。
Check: If you choose u and dv/dx the wrong way round, the new integral may become harder rather than easier. It is worth writing down u, v, du/dx and dv/dx before substituting.
检查：如果 u 和 dv/dx 的选择相反，新的积分可能会变得更复杂而不是更简单。在代入之前写下 u、v、du/dx 和 dv/dx 是值得的。
6. Trigonometric Identities and Equations | 三角恒等式与三角方程
Question 6 asked you to solve 2 sin² θ – cos θ = 1 for 0° ≤ θ ≤ 360°.
第 6 题要求在 0° ≤ θ ≤ 360° 范围内解方程 2 sin² θ – cos θ = 1。
Use sin² θ = 1 – cos² θ: 2(1 – cos² θ) – cos θ = 1 → 2 – 2 cos² θ – cos θ = 1 → 2 cos² θ + cos θ – 1 = 0.
使用 sin² θ = 1 – cos² θ：2(1 – cos² θ) – cos θ = 1 → 2 – 2 cos² θ – cos θ = 1 → 2 cos² θ + cos θ – 1 = 0。
Factor the quadratic: (2 cos θ – 1)(cos θ + 1) = 0. So cos θ = 1/2 or cos θ = -1. The solutions in the required interval are θ = 60°, 300°, 180°.
对二次式进行因式分解：(2 cos θ – 1)(cos θ + 1) = 0。因此 cos θ = 1/2 或 cos θ = -1。在指定区间内的解为 θ = 60°、300°、180°。
Check: Always verify all solutions inside the given interval. cos θ = 1/2 also gives θ = 300° because cosine is positive in the fourth quadrant. cos θ = -1 only gives θ = 180°.
检查：务必验证指定区间内的所有解。cos θ = 1/2 还给出 θ = 300°，因为余弦在第四象限为正。cos θ = -1 只给出 θ = 180°。
7. Exponential Growth and Decay | 指数增长与衰减
Question 7 stated that a population P satisfies dP/dt = 0.04P with P(0) = 250, and asked for the population after 10 years.
第 7 题给出人口 P 满足 dP/dt = 0.04P，且 P(0) = 250，要求 10 年后的人口。
Separate variables: ∫ 1/P dP = ∫ 0.04 dt → ln|P| = 0.04t + C. Write P = A e^(0.04t). Using P(0) = 250 gives A = 250.
分离变量：∫ 1/P dP = ∫ 0.04 dt → ln|P| = 0.04t + C。写成 P = A e^(0.04t)。利用 P(0) = 250 得 A = 250。
Therefore P(10) = 250 e^(0.04 × 10) = 250 e^0.4 ≈ 250 × 1.4918 ≈ 372.96, so the population is about 373.
因此 P(10) = 250 e^(0.04 × 10) = 250 e^0.4 ≈ 250 × 1.4918 ≈ 372.96，因此人口约为 373。
Check: In exponential growth, the rate of change is proportional to the current population. If the power is positive, the population increases without bound. A negative power would model exponential decay.
检查：在指数增长中，变化率与当前人口成正比。如果指数为正，人口无限增长。负指数则模拟指数衰减。
8. Sequences and Series: Arithmetic and Geometric | 数列与级数：等差与等比
Question 8 gave an arithmetic sequence with first term 7 and common difference 3, and a geometric sequence with first term 5 and common ratio 2. It asked for the 20th term and sum of the first 20 terms of the arithmetic sequence, and the 8th term and sum of the first 8 terms of the geometric sequence.
第 8 题给出首项为 7、公差为 3 的等差数列，以及首项为 5、公比为 2 的等比数列。要求等差数列的第 20 项和前 20 项和，以及等比数列的第 8 项和前 8 项和。
Arithmetic: aₙ = a + (n – 1)d, so a₂₀ = 7 + 19 × 3 = 64. The sum is Sₙ = n/2 (a + l), so S₂₀ = 20/2 (7 + 64) = 10 × 71 = 710.
等差数列：aₙ = a + (n – 1)d，因此 a₂₀ = 7 + 19 × 3 = 64。和为 Sₙ = n/2 (a + l)，因此 S₂₀ = 20/2 (7 + 64) = 10 ×
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