📚 Year 13 CIE Further Mathematics: Mock Unit Test Solutions | Year 13 CIE 进阶数学：单元测试模拟卷解析
This mock unit test covers key topics from the CIE Further Mathematics A-Level syllabus, including hyperbolic functions, Maclaurin series, polar coordinates, differential equations, rigid-body equilibrium, variable mass systems, continuous probability distributions, and hypothesis testing. Detailed step-by-step solutions are provided to reinforce understanding and exam technique.
本模拟单元测试涵盖 CIE 进阶数学 A-Level 大纲的核心主题，包括双曲函数、麦克劳林级数、极坐标、微分方程、刚体平衡、变质量系统、连续概率分布和假设检验。通过详细的步骤解析，帮助巩固理解并提升应考技巧。
1. Question 1: Hyperbolic Identities and Equations | 问题1：双曲恒等式与方程
Show that sinh 2x = 2 sinh x cosh x. Hence solve the equation 4 sinh x cosh x = 3 cosh x, giving your answers in logarithmic form.
证明 sinh 2x = 2 sinh x cosh x。由此求解方程 4 sinh x cosh x = 3 cosh x，答案以对数形式表示。
Using the definitions sinh x = (eˣ − e⁻ˣ)/2 and cosh x = (eˣ + e⁻ˣ)/2, the right-hand side 2 sinh x cosh x becomes 2 × (eˣ − e⁻ˣ)/2 × (eˣ + e⁻ˣ)/2 = (e²ˣ − e⁻²ˣ)/2 = sinh 2x.
利用定义 sinh x = (eˣ − e⁻ˣ)/2 和 cosh x = (eˣ + e⁻ˣ)/2，右式 2 sinh x cosh x 化为 2 × (eˣ − e⁻ˣ)/2 × (eˣ + e⁻ˣ)/2 = (e²ˣ − e⁻²ˣ)/2 = sinh 2x。
Replace 4 sinh x cosh x with 2 sinh 2x, giving 2 sinh 2x = 3 cosh x. Write in exponentials: (e²ˣ − e⁻²ˣ) = (3/2)(eˣ + e⁻ˣ). Multiply by 2e²ˣ: 2e⁴ˣ − 2 = 3e³ˣ + 3eˣ. Rearrange: 2e⁴ˣ − 3e³ˣ − 3eˣ − 2 = 0. Factor by grouping: (eˣ + 1)(2e³ˣ − 5e²ˣ + 5eˣ − 2) = 0. Since eˣ + 1 > 0, solve 2e³ˣ − 5e²ˣ + 5eˣ − 2 = 0. Let y = eˣ, then 2y³ − 5y² + 5y − 2 = 0. One root is y = 1, so factor (y − 1)(2y² − 3y + 2) = 0. The quadratic has discriminant 9 – 16 = -7, no real roots. Thus y = 1, giving x = ln 1 = 0. However, cosh x > 0 always, so the only solution is x = 0.
将 4 sinh x cosh x 替换为 2 sinh 2x，得到 2 sinh 2x = 3 cosh x。写成指数形式：(e²ˣ − e⁻²ˣ) = (3/2)(eˣ + e⁻ˣ)。两边乘 2e²ˣ：2e⁴ˣ − 2 = 3e³ˣ + 3eˣ。重新整理：2e⁴ˣ − 3e³ˣ − 3eˣ − 2 = 0。分组分解：(eˣ + 1)(2e³ˣ − 5e²ˣ + 5eˣ − 2) = 0。由于 eˣ + 1 > 0，需求解 2e³ˣ − 5e²ˣ + 5eˣ − 2 = 0。令 y = eˣ，得 2y³ − 5y² + 5y − 2 = 0。一根为 y = 1，分解得 (y − 1)(2y² − 3y + 2) = 0。二次项判别式为负，无实根。故 y = 1，得 x = ln 1 = 0。因 cosh x 恒正，唯一解为 x = 0。
2. Question 2: Maclaurin Series Expansion | 问题2：麦克劳林级数展开
Find the Maclaurin series for ln(1 + sin x) up to and including the term in x⁴.
求 ln(1 + sin x) 的麦克劳林级数，直至含 x⁴ 项。
First, recall sin x = x − x³/6 + … Let f(x) = ln(1 + sin x). Use the expansion ln(1 + u) = u − u²/2 + u³/3 − u⁴/4 + … where u = sin x. Substitute u = x − x³/6 and retain terms up to x⁴.
首先，sin x = x − x³/6 + … 记 f(x) = ln(1 + sin x)。利用展开式 ln(1 + u) = u − u²/2 + u³/3 − u⁴/4 + …，其中 u = sin x。代入 u = x − x³/6，并保留至 x⁴ 项。
Compute u² = (x − x³/6)² = x² − x⁴/3 + …; u³ = (x − x³/6)³ = x³ − x⁵/2 + … ; u⁴ = x⁴ + … Omit terms above x⁴. Then f(x) = (x − x³/6) − ½(x² − x⁴/3) + ⅓(x³) − ¼(x⁴) + … = x − x²/2 + (−x³/6 + x³/3) + (x⁴/6 − x⁴/4) + … = x − x²/2 + x³/6 + (−x⁴/12). So ln(1 + sin x) = x − ½ x² + ⅙ x³ − 1/12 x⁴ + …
计算 u² = (x − x³/6)² = x² − x⁴/3 + … ; u³ = (x − x³/6)³ = x³ − x⁵/2 + … ; u⁴ = x⁴ + … 舍去高于 x⁴ 的项。则 f(x) = (x − x³/6) − ½(x² − x⁴/3) + ⅓(x³) − ¼(x⁴) + … = x − x²/2 + (−x³/6 + x³/3) + (x⁴/6 − x⁴/4) + … = x − ½ x² + ⅙ x³ − 1/12 x⁴ + …
ln(1 + sin x) ≈ x − ½ x² + ⅙ x³ − 1/12 x⁴
3. Question 3: Polar Coordinates – Area | 问题3：极坐标 – 面积
The curve C has polar equation r = 2(1 + cos θ) for 0 ≤ θ ≤ 2π. Find the total area enclosed by C.
曲线 C 的极坐标方程为 r = 2(1 + cos θ)，0 ≤ θ ≤ 2π。求 C 所围总面积。
The polar area formula is A = ½ ∫ r² dθ. Here r² = 4(1 + 2 cos θ + cos² θ) = 4[1 + 2 cos θ + (1 + cos 2θ)/2] = 6 + 8 cos θ + 2 cos 2θ. Total area A = ½ ∫₀²π (6 + 8 cos θ + 2 cos 2θ) dθ = ½ [6θ + 8 sin θ + sin 2θ]₀²π = ½ (12π + 0 + 0) = 6π.
极坐标面积公式为 A = ½ ∫ r² dθ。这里 r² = 4(1 + 2 cos θ + cos² θ) = 4[1 + 2 cos θ + (1 + cos 2θ)/2] = 6 + 8 cos θ + 2 cos 2θ。总面积 A = ½ ∫₀²π (6 + 8 cos θ + 2 cos 2θ) dθ = ½ [6θ + 8 sin θ + sin 2θ]₀²π = ½ (12π + 0 + 0) = 6π。
4. Question 4: First-Order Linear Differential Equation | 问题4：一阶线性微分方程
Solve the differential equation dy/dx + (1/x) y = x², given that y = 1 when x = 1.
求解微分方程 dy/dx + (1/x) y = x²，已知 x = 1 时 y = 1。
The integrating factor is e^(∫(1/x)dx) = e^(ln x) = x. Multiply the equation by x: x dy/dx + y = x³. The left side is d/dx (x y). Integrate: x y = ∫ x³ dx = ¼ x⁴ + C. Thus y = ¼ x³ + C/x. Using y(1) = 1 gives 1 = ¼ + C → C = ¾. Hence the particular solution is y = ¼ x³ + 3/(4x).
积分因子为 e^(∫(1/x)dx) = e^(ln x) = x。方程两边乘 x 得：x dy/dx + y = x³。左侧即 d/dx (x y)。积分得 x y = ∫ x³ dx = ¼ x⁴ + C。因此 y = ¼ x³ + C/x。代入 y(1)=1 得 1 = ¼ + C → C = ¾。故特解为 y = ¼ x³ + 3/(4x)。
5. Question 5: Rigid Body Equilibrium – Ladder Problem | 问题5：刚体平衡 – 梯子问题
A uniform ladder of length 5 m and weight 200 N rests with one end on rough horizontal ground and the other end against a smooth vertical wall. The ladder makes an angle of 60° with the ground. Find the minimum coefficient of friction between the ladder and the ground that prevents slipping.
一架长 5 m、重 200 N 的均匀梯子，一端置于粗糙水平地面，另一端靠在光滑竖直墙上。梯子与地面夹角为 60°。求防止梯子滑动的梯子与地面间的最小摩擦系数。
For equilibrium, resolve horizontally and vertically, and take moments about the base. Let N₂ be the normal reaction from the wall (horizontal), N₁ from the ground (vertical), and F the friction at the ground (horizontal). Vertically: N₁ = 200 N. Horizontally: F = N₂. Moments about base: 200 × (2.5 cos 60°) = N₂ × (5 sin 60°). Thus 200 × 1.25 = N₂ × (5 × √3/2) → 250 = N₂ × (5√3/2) → N₂ = (250 × 2)/(5√3) = 100/√3 ≈ 57.7 N. Since F = μ N₁ and F = N₂, we have μ ≥ N₂ / N₁ = (100/√3) / 200 = 1/(2√3) ≈ 0.289.
平衡时，对水平和竖直方向分解力，并对梯脚取矩。设墙对梯子的法向反力为 N₂（水平），地面对梯子的法向反力为 N₁（竖直），地面摩擦力为 F（水平）。竖直方向：N₁ = 200 N。水平方向：F = N₂。对梯脚取矩：200 × (2.5 cos 60°) = N₂ × (5 sin 60°)。即 200 × 1.25 = N₂ × (5 × √3/2) → 250 = N₂ × (5√3/2) → N₂ = (250 × 2)/(5√3) = 100/√3 ≈ 57.7 N。因 F = μ N₁ 且 F = N₂，得 μ ≥ N₂ / N₁ = (100/√3) / 200 = 1/(2√3) ≈ 0.289。
6. Question 6: Variable Mass – Rocket Motion | 问题6：变质量系统 – 火箭运动
A rocket of initial mass 1000 kg ejects fuel at a constant rate of 20 kg s⁻¹ with a relative speed of 1500 m s⁻¹. Given that it starts from rest, find its speed after 5 seconds, neglecting gravity.
一枚初始质量 1000 kg 的火箭以恒定速率 20 kg s⁻¹ 喷射燃料，相对排放速度为 1500 m s⁻¹。忽略重力，求从静止开始 5 秒后火箭的速度。
Using the rocket equation m dv/dt = −u dm/dt, where u is the exhaust velocity relative to the rocket, and dm/dt = −20 kg s⁻¹ (mass decreasing). Thus dv/dt = (−u/m) dm/dt = (−1500/m)(−20) = 30000/m. The mass at time t is m = 1000 − 20t. So dv/dt = 30000/(1000 − 20t). Integrate from 0 to 5 s: v = ∫₀⁵ 30000/(1000 − 20t) dt. Let u_sub = 1000 − 20t, du = −20 dt, so dt = −du/20. v = ∫_{t=0}^{5} 30000/u × (−du/20) = −1500 ∫ du/u. Limits: u(0)=1000, u(5)=900. v = −1500 [ln u]₁₀₀₀⁹⁰⁰ = −1500 (ln 900 − ln 1000) = 1500 ln(1000/900) = 1500 ln(10/9) ≈ 1500 × 0.10536 = 158.0 m s⁻¹.
利用火箭方程 m dv/dt = −u dm/dt，其中 u 为燃料相对火箭的喷射速度，dm/dt = −20 kg s⁻¹。得 dv/dt = (−1500/m)(−20) = 30000/m。任意时刻 t 的质量为 m = 1000 − 20t。故 dv/dt = 30000/(1000 − 20t)。从 0 到 5 s 积分：v = ∫₀⁵ 30000/(1000 − 20t) dt。令 u = 1000 − 20t，则 du = −20 dt，dt = −du/20。v = ∫ 30000/u × (−du/20) = −1500 ∫ du/u。积分限：u(0)=1000，u(5)=900。v = −1500 [ln u]₁₀₀₀⁹⁰⁰ = −1500 (ln 900 − ln 1000) = 1500 ln(1000/900) = 1500 ln(10/9) ≈ 1500 × 0.10536 = 158.0 m s⁻¹。
7. Question 7: Continuous Random Variable – PDF and CDF | 问题7：连续随机变量 – 概率密度与分布函数
The continuous random variable X has probability density function f(x) = kx(2 − x) for 0 ≤ x ≤ 2, and 0 otherwise. Find the value of k, the cumulative distribution function F(x), and P(0.5 < X < 1.5).
连续随机变量 X 的概率密度函数为 f(x) = kx(2 − x)，0 ≤ x ≤ 2，其他为 0。求 k 值、累积分布函数 F(x) 以及 P(0.5 < X < 1.5)。
Total probability is 1: ∫₀² kx(2 − x) dx = k ∫₀² (2x − x²) dx = k [x² − x³/3]₀² = k (4 − 8/3) = k (4/3) = 1. Thus k = 3/4. For the CDF, F(x) = ∫₀ˣ (3/4)t(2 − t) dt = (3/4) [t² − t³/3]₀ˣ = (3/4)(x² − x³/3) = (3x²/4) − (x³/4). For P(0.5 < X < 1.5) = F(1.5) − F(0.5). F(1.5) = (3×2.25/4) − (3.375/4) = (6.75 − 3.375)/4 = 3.375/4 = 0.84375. F(0.5) = (3×0.25/4) − (0.125/4) = (0.75 − 0.125)/4 = 0.625/4 = 0.15625. Difference = 0.6875.
总概率为 1：∫₀² kx(2 − x) dx = k ∫₀² (2x − x²) dx = k [x² − x³/3]₀² = k (4 − 8/3) = k (4/3) = 1。故 k = 3/4。累积分布函数 F(x) = ∫₀ˣ (3/4)t(2 − t) dt = (3/4) [t² − t³/3]₀ˣ = (3/4)(x² − x³/3) = (3x²/4) − (x³/4)。P(0.5 < X < 1.5) = F(1.5) − F(0.5)。F(1.5) = (3×2.25/4) − (3.375/4) = (6.75 − 3.375)/4 = 0.84375。F(0.5) = (3×0.25/4) − (0.125/4) = (0.75 − 0.125)/4 = 0.15625。差值为 0.6875。
8. Question 8: Poisson Hypothesis Test | 问题8：泊松分布假设检验
The number of defects per metre of fabric follows a Poisson distribution with mean 2.5. A new process is tested on a 4-metre sample, and 6 defects are found. Test at the 5% significance level whether the mean has decreased.
每米织物的瑕疵数服从均值为 2.5 的泊松分布。在新工艺下检验 4 米样品，发现 6 个瑕疵。在 5% 显著性水平下检验均值是否降低。
Under the new process, the total number of defects in 4 metres follows Poisson(λ) where λ = 4 × 2.5 = 10 if the mean is unchanged. The alternative hypothesis is that the mean has decreased, so λ < 10. We use a one-tailed test, finding the largest c such that P(X ≤ c|λ=10) ≤ 0.05. From cumulative Poisson tables for λ=10, P(X≤4) = 0.0293, P(X≤5) = 0.0671. The critical value is 4. Observed defects = 6 > 4, so we do not reject H₀. There is insufficient evidence at the 5% level to say that the mean has decreased.
若均值未变，4 米样品总瑕疵数服从 Poisson(λ) 分布，λ = 4 × 2.5 = 10。备择假设为均值减小，故 λ < 10。采用单尾检验，寻找最大的 c 使得 P(X ≤ c|λ=10) ≤ 0.05。查 λ=10 的泊松累积表，P(X≤4) = 0.0293，P(X≤5) = 0.0671。临界值为 4。观测瑕疵数 6 > 4，故不拒绝 H₀。在 5% 显著性水平下，没有足够证据表明均值降低。
9. Question 9: Second-Order Differential Equation | 问题9：二阶微分方程
Solve d²y/dx² − 2 dy/dx + y = eˣ, with y(0)=0 and y'(0)=1. Find the particular solution.
解微分方程 d²y/dx² − 2 dy/dx + y = eˣ，满足 y(0)=0，y'(0)=1。求特解。
The auxiliary equation m² − 2m + 1 = 0 gives (m−1)² = 0, repeated root m=1. Complementary function: y_c = (A + Bx) eˣ. Particular integral: try y_p = Cx² eˣ (since eˣ and x eˣ are in the C.F.). Differentiate twice, substitute, and equate coefficients. y_p’ = C(2x + x²) eˣ, y_p” = C(2 + 4x + x²) eˣ. Substituting: C(2+4x+x²)eˣ − 2C(2x+x²)eˣ + Cx² eˣ = eˣ. Simplify: Ceˣ [(2+4x+x²) − (4x+2x²) + x²] = Ceˣ (2) = eˣ → C=½. Thus y_p = ½ x² eˣ. General solution: y = (A + Bx + ½ x²) eˣ. y(0)=0 → A = 0. y’ = (B + x) eˣ + (A + Bx + ½ x²) eˣ. At x=0: y'(0) = B + A = 1, with A=0 → B=1. Particular solution: y = (x + ½ x²) eˣ.
辅助方程 m² − 2m + 1 = 0 得 (m−1)² = 0，重根 m=1。余函数：y_c = (A + Bx) eˣ。特解尝试：由于 eˣ 和 x eˣ 已出现在余函数中，设 y_p = Cx² eˣ。微分两次并代入系数匹配。y_p’ = C(2x + x²) eˣ，y_p” = C(2 + 4x + x²) eˣ。代入：C(2+4x+x²)eˣ − 2C(2x+x²)eˣ + Cx² eˣ = eˣ。化简：Ceˣ [(2+4x+x²) − (4x+2x²) + x²] = Ceˣ (2) = eˣ → C=½。故 y_p = ½ x² eˣ。通解：y = (A + Bx + ½ x²) eˣ。代入 y(0)=0 → A=0。y’ = (B+x)eˣ + (A+Bx+½ x²)eˣ，x=0 时 y'(0)=B+A=1，因 A=0 得 B=1。特解为 y = (x + ½ x²) eˣ。
10. Question 10: Complex Numbers – De Moivre’s Theorem | 问题10：复数 – 棣莫弗定理
Use De Moivre’s theorem to find the two square roots of 8 − 6i, expressing them in the form a + bi where a and b are exact real numbers.
利用棣莫弗定理求 8 − 6i 的两个平方根，以精确实数 a + bi 形式表示。
First, express 8 − 6i in polar form. Modulus r = √(8² + (−6)²) = √100 = 10. Argument θ satisfies tan θ = −6/8 = −3/4, with θ in the fourth quadrant. Since cos θ = 8/10 = 0.8, sin θ = −6/10 = −0.6, θ ≈ −0.6435 rad. For square roots, modulus = √10, arguments = (θ + 2kπ)/2 for k = 0, 1. So θ/2 ≈ −0.32175 rad and θ/2 + π ≈ 2.81984 rad. However, exact rectangular form can be found by letting √(8−6i) = x + yi. Then (x+yi)² = x² − y² + 2xyi = 8 − 6i. Equate: x² − y² = 8, 2xy = −6 → xy = −3. Solve simultaneously: y = −3/x → x² − 9/x² = 8 → multiply x²: x⁴ − 8x² − 9 = 0 → (x² − 9)(x² + 1) = 0 → x² = 9 or x² = −1 (reject). Thus x = ±3. If x = 3, y = −1; if x = −3, y = 1. The two square roots are 3 − i and −3 + i.
先将 8 − 6i 化为极坐标形式。模 r = √(8² + (−6)²) = √100 = 10。辐角 θ 满足 tan θ = −6/8 = −3/4，位于第四象限。由 cos θ = 8/10 = 0.8，sin θ = −6/10 = −0.6，θ ≈ −0.6435 rad。平方根的模为 √10，辐角为 (θ + 2kπ)/2，k=0,1。但精确代数形式可通过设 √(8−6i) = x + yi 求得。则 (x+yi)² = x² − y² + 2xyi = 8 − 6i。比较实虚部：x² − y² = 8，2xy = −6 → xy = −3。联立消元：y = −3/x → x² − 9/x² = 8 → 两边乘 x² 得 x⁴ − 8x² − 9 = 0 → (x² − 9)(x² + 1) = 0 → x² = 9 或 x² = −1（舍）。从而 x = ±3。若 x = 3，y = −1；若 x = −3，y = 1。两平方根为 3 − i 和 −3 + i。
11. Further Practice and Key Takeaways | 拓展练习与核心要点
Mastering the solution methods demonstrated above—manipulating hyperbolic identities, methodically building Maclaurin series, applying polar area integrals, selecting appropriate integrating factors, resolving forces for rigid bodies, integrating the rocket equation, handling probability density functions, performing Poisson hypothesis tests, solving second-order ODEs with trial functions, and finding roots via De Moivre’s theorem—forms the backbone of success in CIE Further Mathematics. Always verify boundary conditions and check the validity domain of your answers.
掌握上述解题方法是 CIE 进阶数学成功的基础：灵活运用双曲恒等式、系统构建麦克劳林级数、应用极坐标面积积分、选取合适的积分因子、对刚体进行力的分解、积分火箭方程、处理概率密度函数、实施泊松假设检验、用试探函数解二阶常微分方程以及利用棣莫弗定理求根。务必验证边界条件并检查答案的有效域。
In your revision, attempt full mixed-topic papers under timed conditions, and use these solutions to diagnose weaknesses. Pay close attention to the precise wording of hypothesis tests and the justification of assumptions like smooth/wall or ideal rocket models.
在复习中，应在计时条件下练习完整的混合主题试卷，并利用本解析诊断薄弱点。特别注意假设检验的精确表述以及诸如光滑墙面或理想火箭模型等假设的合理性。
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