📚 Pre-U CCEA Mathematics: Unit Test Mock Paper Solutions | Pre-U CCEA 数学：单元测试模拟卷解析
This article provides detailed, step-by-step solutions to a mock unit test designed for the CCEA Pre-U Mathematics specification. The paper focuses on core pure-mathematics topics including algebra, calculus, vectors and complex numbers, typical of an end-of-unit assessment. Each question is broken down with clear reasoning, alternative approaches where helpful, and commentary on common pitfalls. Use this resource to refine your problem-solving technique and consolidate key concepts ahead of formal testing.
本文为 CCEA Pre-U 数学规范设计的单元模拟测试提供详细的分步解答。试卷聚焦于核心纯数学主题，包括代数、微积分、向量和复数，是单元结束评估的典型内容。每道题都拆解出清晰的推理过程，有用时还提供替代方法，并指出常见易错点。利用这份资源打磨解题技巧，在正式考试前巩固关键概念。
1. Question 1: Functions and Inverse Functions | 问题1：函数与反函数
The function f is defined by f(x) = (2x – 3) / (x + 1) for x ∈ ℝ, x ≠ -1. Find the inverse function f⁻¹(x), stating its domain. Hence solve the equation f(x) = f⁻¹(x).
函数 f 定义为 f(x) = (2x – 3) / (x + 1)，x ∈ ℝ，x ≠ -1。求反函数 f⁻¹(x)，并写出其定义域。由此求解方程 f(x) = f⁻¹(x)。
To find the inverse, write y = (2x – 3)/(x + 1), swap x and y, then solve for y. Cross-multiplying gives x(y + 1) = 2y – 3 → xy + x = 2y – 3 → xy – 2y = -x – 3 → y(x – 2) = -x – 3, so y = (-x – 3)/(x – 2) = (x + 3)/(2 – x). Thus f⁻¹(x) = (x + 3)/(2 – x), with domain x ≠ 2. For f(x) = f⁻¹(x), set (2x – 3)/(x + 1) = (x + 3)/(2 – x). Cross-multiplying gives (2x – 3)(2 – x) = (x + 3)(x + 1). Expand: -2x² + 4x + 3x – 6? Let’s compute: (2x)(2) = 4x, (2x)(-x) = -2x², (-3)(2) = -6, (-3)(-x) = 3x, so LHS = -2x² + 7x – 6. RHS: (x)(x)=x², (x)(1)=x, (3)(x)=3x, (3)(1)=3, so x² + 4x + 3. Equation: -2x² + 7x – 6 = x² + 4x + 3 → 0 = 3x² – 3x + 9 → divide by 3: x² – x + 3 = 0. Discriminant Δ = (-1)² – 4(1)(3) = 1 – 12 = -11 < 0, no real solutions. Therefore f(x) = f⁻¹(x) has no solution in the real numbers.
求反函数：令 y = (2x – 3)/(x + 1)，交换 x 和 y 并解出 y。交叉相乘得 x(y + 1) = 2y – 3，整理得 y(x – 2) = -x – 3，所以 y = (-x – 3)/(x – 2) = (x + 3)/(2 – x)。因此 f⁻¹(x) = (x + 3)/(2 – x)，定义域 x ≠ 2。令 f(x) = f⁻¹(x)，得到分式方程，交叉相乘展开后得到 -2x² + 7x – 6 = x² + 4x + 3，化简为 x² – x + 3 = 0。判别式为负数，无实数解。
2. Question 2: Differentiation from First Principles | 问题2：通过第一性原理求导
Using the limit definition, differentiate f(x) = √x at x = 4. Hence find the equation of the tangent to the curve y = √x at the point where x = 4.
利用极限定义求 f(x) = √x 在 x = 4 处的导数。由此求曲线 y = √x 在 x = 4 处的切线方程。
Derivative from first principles: f'(x) = lim(h→0) [f(x+h) – f(x)] / h. At x = 4, f'(4) = lim(h→0) [√(4+h) – 2] / h. Rationalise the numerator: multiply numerator and denominator by √(4+h) + 2. The numerator becomes (4+h) – 4 = h, so expression = h / [h(√(4+h) + 2)] = 1 / (√(4+h) + 2). As h→0, √(4+h) → 2, so limit = 1/(2+2) = 1/4. The gradient m = 1/4. At x = 4, y = 2. Tangent equation: y – 2 = (1/4)(x – 4) → y = (1/4)x + 1.
第一性原理求导：f'(4) = lim(h→0) [√(4+h) – 2]/h，有理化分子得 h / [h(√(4+h) + 2)] = 1/(√(4+h) + 2)，取极限得 1/4。切线斜率 m = 1/4，切点 (4, 2)，切线为 y = (1/4)x + 1。
3. Question 3: Integration by Substitution | 问题3：换元积分法
Evaluate the definite integral ∫₀¹ x/(1 + x²)² dx using the substitution u = 1 + x². Interpret the result geometrically.
利用代换 u = 1 + x² 计算定积分 ∫₀¹ x/(1 + x²)² dx，并从几何角度解释结果。
Let u = 1 + x², then du/dx = 2x → x dx = du/2. When x = 0, u = 1; when x = 1, u = 2. The integral becomes ∫₁² (1/u²)(du/2) = ½ ∫₁² u⁻² du = ½ [-u⁻¹]₁² = ½ (-1/2 + 1) = ½ × 1/2 = 1/4. The area under the curve y = x/(1 + x²)² from x = 0 to x = 1 is 0.25 square units.
令 u = 1 + x²，则 du/dx = 2x，积分限 x=0 → u=1，x=1 → u=2。原积分化为 ½ ∫₁² u⁻² du = ½[-1/u]₁² = ½( -1/2 + 1) = 1/4。曲线 y = x/(1 + x²)² 在 [0,1] 与 x 轴围成的面积为 0.25 平方单位。
4. Question 4: Vector Dot Product and Angle | 问题4：向量点乘与夹角
Given vectors a = 2i – j + 3k and b = -i + 4j – 2k, find the exact value of cos θ, where θ is the acute angle between a and b. Determine whether the vectors are perpendicular, parallel or neither.
已知向量 a = 2i – j + 3k，b = -i + 4j – 2k，求 a 与 b 夹角的余弦值精确值，并判断它们是垂直、平行还是其他关系。
Use the dot product: a·b = |a||b| cos θ. Compute a·b = (2)(-1) + (-1)(4) + (3)(-2) = -2 -4 -6 = -12. Magnitudes: |a| = √(4 + 1 + 9) = √14; |b| = √(1 + 16 + 4) = √21. Then cos θ = -12 / (√14 √21) = -12 / √294 = -12 / (7√6) because 294 = 49×6, √294 = 7√6. Rationalise: cos θ = -12√6 / (7×6) = -2√6 / 7. Since cos θ ≠ 0, vectors are not perpendicular. As a·b ≠ |a||b| and ratios of components are not equal (2/-1 ≠ -1/4), they are not parallel. They are skewed, obtuse angle because cos negative gives θ > 90°.
计算点乘：a·b = -12，模长 |a| = √14，|b| = √21，故 cos θ = -12/(√14 √21) = -12/√294 = -12/(7√6) = -2√6/7。cos θ ≠ 0，不垂直；对应分量不成比例，不平行；cos 为负值，夹角为钝角。
5. Question 5: Complex Numbers in Polar Form | 问题5：复数的极坐标形式
Express the complex number z = -1 + i√3 in modulus-argument form, with argument in the interval (-π, π]. Hence find z⁶ and all cube roots of z.
将复数 z = -1 + i√3 表示为模-辐角形式，辐角在 (-π, π] 内。由此求 z⁶ 和 z 的所有立方根。
Modulus: |z| = √((-1)² + (√3)²) = √(1+3) = 2. Argument: tan⁻¹(√3/-1) = tan⁻¹(-√3). The point is in the second quadrant, so principal argument θ = 2π/3. Thus z = 2(cos(2π/3) + i sin(2π/3)). By de Moivre, z⁶ = 2⁶[cos(6×2π/3) + i sin(6×2π/3)] = 64[cos(4π) + i sin(4π)] = 64(1 + 0i) = 64. For cube roots: z^(1/3) = 2^(1/3) [cos((2π/3 + 2kπ)/3) + i sin((2π/3 + 2kπ)/3)] for k = 0,1,2. So roots are: for k=0: ³√2 (cos(2π/9) + i sin(2π/9)); k=1: ³√2 (cos(8π/9) + i sin(8π/9)); k=2: ³√2 (cos(14π/9) + i sin(14π/9)), which is also ³√2 (cos(-4π/9) + i sin(-4π/9)).
模 |z|=2，辐角 2π/3，故 z = 2(cos(2π/3) + i sin(2π/3))。由棣莫弗定理，z⁶ = 64。立方根：z^(1/3) = ³√2 [cos((2π/3+2kπ)/3) + i sin(…)]，k=0,1,2 得三个根，辐角分别为 2π/9, 8π/9, 14π/9（或 -4π/9）。
6. Question 6: Series Expansion (Maclaurin) | 问题6：级数展开（麦克劳林级数）
Find the Maclaurin series for f(x) = ln(1 + sin x) up to and including the term in x³. Use this to approximate ln(1 + sin 0.2) and compare with the exact value.
求 f(x) = ln(1 + sin x) 的麦克劳林级数，包括到 x³ 项。用此近似式求 ln(1 + sin 0.2)，并与精确值比较。
We need derivatives at x = 0. f(0) = ln(1+0)=0. f'(x) = cos x / (1 + sin x) → f'(0)=1. f”(x) = derivative of cos x/(1+sin x): use quotient rule. Let u = cos x, v = 1+sin x, then u’ = -sin x, v’ = cos x. f” = (u’v – uv’)/v² = [(-sin x)(1+sin x) – cos x(cos x)] / (1+sin x)² = [-sin x – sin²x – cos²x]/(1+sin x)² = -(sin x + 1)/(1+sin x)² = -1/(1+sin x). So f”(0) = -1. f”'(x) = derivative of -1/(1+sin x) = (cos x)/(1+sin x)². At 0, f”'(0) = 1/(1)² = 1. The series: f(0) + f'(0)x + f”(0)x²/2! + f”'(0)x³/3! + … = 0 + 1·x + (-1)x²/2 + (1)x³/6 = x – x²/2 + x³/6. Approx for x=0.2: 0.2 – 0.04/2 + 0.008/6 = 0.2 – 0.02 + 0.001333… ≈ 0.18133. Exact: ln(1+sin0.2), sin0.2≈0.198669, 1+sin =1.198669, ln ≈0.18126. The approximation is accurate to 4 decimal places.
求导后在0处值：f(0)=0, f'(0)=1, f”(0)=-1, f”'(0)=1。麦克劳林展开为 x – x²/2 + x³/6。近似计算 x=0.2 得 0.18133，精确值约 0.18126，近似吻合良好。
7. Question 7: Parametric Equations and Tangent | 问题7：参数方程与切线
A curve is defined by x = t³ – 3t, y = t² + t. Find dy/dx in terms of t, and the equation of the tangent at the point where t = 2. Determine the nature of the stationary points.
曲线由参数方程 x = t³ – 3t, y = t² + t 定义。求 dy/dx 关于 t 的表达式，以及 t = 2 处的切线方程。确定驻点的性质。
Differentiate: dx/dt = 3t² – 3, dy/dt = 2t + 1. So dy/dx = (2t + 1) / (3t² – 3). At t = 2, dx/dt = 3(4)-3 = 9, dy/dt = 5, so gradient m = 5/9. Coordinates: x = 8 – 6 = 2, y = 4 + 2 = 6. Tangent: y – 6 = (5/9)(x – 2) → y = (5/9)x + 44/9? Wait: 6 – (5/9)*2 = 6 – 10/9 = 44/9. Yes, y = (5/9)x + 44/9. For stationary points, dy/dx = 0 → 2t + 1 = 0 → t = -½. Then d²y/dx² = d(dy/dx)/dt ÷ dx/dt. Compute d/dt (dy/dx) = [2(3t²-3) – (2t+1)(6t)] / (3t²-3)² = [6t²-6 -12t² -6t] / (3t²-3)² = (-6t² -6t -6)/(3t²-3)². At t=-½, dx/dt = 3(¼)-3 = -9/4, numerator of second derivative? Evaluate d(dy/dx)/dt at t=-½: (-6(¼) -6(-½) -6) = (-1.5 +3 -6) = -4.5. Over denominator (3(¼)-3)² = (-9/4)² = 81/16. So d(dy/dx)/dt = -4.5 / (81/16) = -4.5 × 16/81 = -72/81 = -8/9. Then d²y/dx² = (-8/9) / (-9/4) = (-8/9)×(-4/9) = 32/81 > 0, thus a minimum.
求导：dy/dx = (2t+1)/(3t²-3)。t=2 时斜率为 5/9，点 (2,6)，切线 y = (5/9)x + 44/9。驻点条件 dy/dx=0 → t = -½。求二阶导数判断，计算得 32/81 > 0，为极小值点。
8. Question 8: Differential Equation with Initial Condition | 问题8：带初始条件的微分方程
Solve the differential equation dy/dx = y² cos x, given that y = 1 when x = 0. Find the value of y when x = π/3.
求解微分方程 dy/dx = y² cos x，已知 x=0 时 y=1。求 x=π/3 时的 y 值。
Separate variables: dy / y² = cos x dx. Integrate both sides: ∫ y⁻² dy = ∫ cos x dx → -1/y = sin x + C. Use initial condition: x=0, y=1 → -1/1 = sin 0 + C → C = -1. So -1/y = sin x – 1 → 1/y = 1 – sin x → y = 1/(1 – sin x). At x = π/3, sin(π/3) = √3/2 ≈ 0.8660, so y = 1/(1 – √3/2) = 1 / ((2 – √3)/2) = 2/(2 – √3). Rationalise: multiply numerator and denominator by (2 + √3) → y = 2(2 + √3)/(4 – 3) = 4 + 2√3. So exact value is 4 + 2√3 (approx 7.464).
分离变量得 -1/y = sin x + C，代入初始条件得 C = -1，从而 y = 1/(1 – sin x)。x=π/3 时，y = 2/(2 – √3) = 4 + 2√3。
9. Question 9: Matrix Multiplication and Inverse | 问题9：矩阵乘法与逆矩阵
Consider the matrices A = [[2, 1], [3, 4]] and B = [[-1, 2], [0, 5]]. Find AB and A⁻¹. Verify that (AB)⁻¹ = B⁻¹A⁻¹.
给定矩阵 A = [[2, 1], [3, 4]]，B = [[-1, 2], [0, 5]]。求 AB 和 A⁻¹，并验证 (AB)⁻¹ = B⁻¹A⁻¹。
Compute AB = [[2×(-1)+1×0, 2×2+1×5], [3×(-1)+4×0, 3×2+4×5]] = [[-2, 4+5], [-3, 6+20]] = [[-2, 9], [-3, 26]]. Inverse of A: det A = 2×4 – 1×3 = 8-3=5. A⁻¹ = (1/5)[[4, -1], [-3, 2]] = [[4/5, -1/5], [-3/5, 2/5]]. Inverse of B: det B = (-1)×5 – 2×0 = -5. B⁻¹ = (-1/5)[[5, -2], [0, -1]] = [[-1, 2/5], [0, 1/5]]. Now B⁻¹A⁻¹ = [[-1, 2/5],[0, 1/5]] × [[4/5, -1/5],[-3/5, 2/5]] = first row: (-1)(4/5)+(2/5)(-3/5) = -4/5 -6/25 = -20/25 -6/25 = -26/25; second entry: (-1)(-1/5)+(2/5)(2/5) = 1/5 +4/25 = 5/25+4/25=9/25. Second row: (0)(4/5)+(1/5)(-3/5) = -3/25; (0)(-1/5)+(1/5)(2/5) = 2/25. So B⁻¹A⁻¹ = [[-26/25, 9/25], [-3/25, 2/25]]. Meanwhile det(AB) = (-2)×26 – 9×(-3) = -52 + 27 = -25. (AB)⁻¹ = (-1/25)[[26, -9], [3, -2]] = [[-26/25, 9/25], [-3/25, 2/25]], matching B⁻¹A⁻¹.
计算 AB 得 [[-2,9],[-3,26]]。A⁻¹=(1/5)[[4,-1],[-3,2]]，B⁻¹ 计算后相乘得到 B⁻¹A⁻¹ = [[-26/25, 9/25],[-3/25, 2/25]]，而 (AB)⁻¹ 为同样矩阵，验证正确。
10. Question 10: Probability – Binomial Distribution | 问题10：概率 – 二项分布
A biased coin is tossed 8 times. The probability of heads on any single toss is 0.35. Find the probability of obtaining exactly 2 heads. Find also the probability of at least 3 heads.
抛掷一枚偏斜硬币 8 次，单次正面概率为 0.35。求恰好出现 2 次正面的概率，以及至少 3 次正面的概率。
Use X ~ B(n=8, p=0.35). P(X = k) = C(8, k) p^k (1-p)^(8-k). Exactly 2 heads: C(8,2)=28, (0.35)²=0.1225, (0.65)⁶ ≈ 0.0754189? Let’s compute: 0.65⁶ = (0.65²)³ = 0.4225³ = 0.4225×0.4225×0.4225 = 0.1785×0.4225 ≈ 0.0754. So P(2) ≈ 28 × 0.1225 × 0.0754 ≈ 28 × 0.0092365 ≈ 0.2586. At least 3 heads: 1 – P(0) – P(1) – P(2). P(0) = 1 × 1 × 0.65⁸ = 0.65⁸ = (0.65⁴)². 0.65²=0.4225, 0.65⁴=0.1785, 0.65⁸=0.0319 approx. P(1) = 8 × 0.35 × 0.65⁷. 0.65⁷ = 0.65 × 0.65⁶ = 0.65 × 0.0754 ≈ 0.0490. So P(1) ≈ 8 × 0.35 × 0.0490 = 8 × 0.01715 = 0.1372. P(2) ≈ 0.2586. Thus P(at least 3) = 1 – (0.0319 + 0.1372 + 0.2586) = 1 – 0.4277 = 0.5723. So probability ≈ 0.572.
设 X~B(8,0.35)。P(X=2)=C(8,2)(0.35)²(0.65)⁶，计算约 0.2586。至少 3 次正面用互补事件：P(X≥3)=1-P(0)-P(1)-P(2) ≈ 1 – 0.0319 – 0.1372 – 0.2586 = 0.5723。
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