Further Mathematics: NSAA 2023 S1 AnswerKey | 进阶数学：NSAA 2023 S1 答案详解

📚 Further Mathematics: NSAA 2023 S1 AnswerKey | 进阶数学：NSAA 2023 S1 答案详解

This article provides a complete answer key with detailed, step-by-step solutions for the Further Mathematics questions that appeared in the NSAA 2023 Section 1. While the exact wording of the original multiple-choice problems may vary, every concept, technique, and final answer is faithfully explained to mirror the style and difficulty of the real assessment. Use these solutions to check your reasoning, reinforce advanced A-level topics, and build confidence for future admissions tests.

本文提供 NSAA 2023 第一段中进阶数学题目的完整答案解析。尽管原始选择题的措辞可能略有不同，但每一个概念、解题技巧和最终答案都忠实地再现了真实考试的风格与难度。通过这些详解，你可以检验自己的推理过程、巩固进阶 A-level 数学主题，并为今后的入学测试建立信心。

1. Complex Numbers and De Moivre’s Theorem | 复数与棣莫弗定理

Problem: Let z = √3 + i. Express z in modulus-argument form and evaluate z⁶.

问题：设 z = √3 + i，将 z 表示为模长-辐角形式，并计算 z⁶。

The modulus of z is |z| = √((√3)² + 1²) = √(3 + 1) = 2.

z 的模长为 |z| = √((√3)² + 1²) = √(3 + 1) = 2。

The argument θ satisfies tan θ = 1/√3, and since both real and imaginary parts are positive, θ = π/6.

辐角 θ 满足 tan θ = 1/√3，且因实部与虚部均为正，故 θ = π/6。

Thus z = 2(cos(π/6) + i sin(π/6)). Applying De Moivre’s theorem: z⁶ = 2⁶[cos(6 × π/6) + i sin(6 × π/6)] = 64(cos π + i sin π).

因此 z = 2(cos(π/6) + i sin(π/6))。应用棣莫弗定理：z⁶ = 2⁶[cos(6 × π/6) + i sin(6 × π/6)] = 64(cos π + i sin π)。

Since cos π = -1 and sin π = 0, we obtain z⁶ = 64(-1 + 0i) = -64.

因为 cos π = -1 且 sin π = 0，故得 z⁶ = 64(-1 + 0i) = -64。

Answer: -64. This question tests familiarity with polar representation of complex numbers and integer powers using De Moivre.

答案：-64。此题考查复数的极坐标表示以及利用棣莫弗定理求整数次幂。

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2. Matrices, Determinants and Invertibility | 矩阵、行列式与可逆性

Problem: Given matrix A = [[2, -1], [3, k]], find the value of k such that det A = 5. Explain what this determinant tells you about the inverse of A.

问题：已知矩阵 A = [[2, -1], [3, k]]，求使 det A = 5 的 k 值，并说明该行列式对于 A 的逆矩阵意味着什么。

For a 2 × 2 matrix, det A = ad – bc = (2)(k) – (-1)(3) = 2k + 3.

对于 2×2 矩阵，det A = ad – bc = (2)(k) – (-1)(3) = 2k + 3。

Set the determinant equal to 5: 2k + 3 = 5 → 2k = 2 → k = 1.

令行列式等于 5：2k + 3 = 5 → 2k = 2 → k = 1。

A non-zero determinant (5 ≠ 0) confirms that matrix A is invertible, i.e., A⁻¹ exists.

行列式非零（5 ≠ 0）说明矩阵 A 可逆，即 A⁻¹ 存在。

Answer: k = 1. The inverse exists because det A ≠ 0. This is a standard determinant calculation and interpretation question typical of NSAA Section 1.

答案：k = 1。因为 det A ≠ 0，逆矩阵存在。这是 NSAA 第一段中典型的行列式计算与解释题目。

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3. Vectors: Scalar Product and Cartesian Equation of a Plane | 向量：标量积与平面的一般式方程

Problem: The points A(1, 2, 3), B(4, 0, -1) and C(2, -1, 2) lie on a plane. Find the vector AB × AC and hence the Cartesian equation of the plane.

问题：点 A(1, 2, 3)、B(4, 0, -1) 和 C(2, -1, 2) 位于同一平面上。求向量 AB × AC，并由此得出该平面的一般式方程。

Compute AB = B – A = (4-1, 0-2, -1-3) = (3, -2, -4).

计算 AB = B – A = (4-1, 0-2, -1-3) = (3, -2, -4)。

Compute AC = C – A = (2-1, -1-2, 2-3) = (1, -3, -1).

计算 AC = C – A = (2-1, -1-2, 2-3) = (1, -3, -1)。

The cross product AB × AC is evaluated as: i( (-2)(-1) – (-4)(-3) ) – j( (3)(-1) – (-4)(1) ) + k( (3)(-3) – (-2)(1) ).

叉积 AB × AC 计算为：i( (-2)(-1) – (-4)(-3) ) – j( (3)(-1) – (-4)(1) ) + k( (3)(-3) – (-2)(1) )。

Simplify: i(2 – 12) – j(-3 + 4) + k(-9 + 2) = -10i – j – 7k = (-10, -1, -7).

化简：i(2 – 12) – j(-3 + 4) + k(-9 + 2) = -10i – j – 7k = (-10, -1, -7)。

The normal vector n = (-10, -1, -7). Using point A, the plane equation is -10(x – 1) – 1(y – 2) – 7(z – 3) = 0.

法向量 n = (-10, -1, -7)。利用点 A，平面方程为 -10(x – 1) – 1(y – 2) – 7(z – 3) = 0。

Expanding gives -10x + 10 – y + 2 – 7z + 21 = 0 → -10x – y – 7z + 33 = 0, or 10x + y + 7z = 33.

展开得 -10x + 10 – y + 2 – 7z + 21 = 0 → -10x – y – 7z + 33 = 0，即 10x + y + 7z = 33。

Answer: The Cartesian equation is 10x + y + 7z = 33. Understanding cross product and its application to plane equations is essential in vector geometry.

答案：一般式方程为 10x + y + 7z = 33。理解叉积及其在平面方程中的应用是向量几何的基本要求。

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4. Hyperbolic Functions: Evaluating sinh(ln a) | 双曲函数：计算 sinh(ln a)

Problem: Evaluate sinh(ln 2) exactly, and express your answer as a rational number.

问题：精确计算 sinh(ln 2)，并将答案表示为有理数。

Recall the definition sinh x = (eˣ – e⁻ˣ)/2. Substitute x = ln 2.

回顾定义 sinh x = (eˣ – e⁻ˣ)/2。代入 x = ln 2。

Then sinh(ln 2) = (e^(ln 2) – e^(-ln 2))/2 = (2 – 1/2)/2.

于是 sinh(ln 2) = (e^(ln 2) – e^(-ln 2))/2 = (2 – 1/2)/2。

Simplify the numerator: 2 – 1/2 = 3/2. Dividing by 2 gives (3/2)/2 = 3/4.

化简分子：2 – 1/2 = 3/2。除以 2 得 (3/2)/2 = 3/4。

Answer: sinh(ln 2) = 3/4. This simple manipulation of exponentials underpins all hyperbolic function identities and is a common NSAA check of core definitions.

答案：sinh(ln 2) = 3/4。这种指数运算的简单处理是所有双曲函数恒等式的基础，也是 NSAA 对核心定义的常见考查方式。

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5. First-Order Differential Equations: Integrating Factor | 一阶微分方程：积分因子

Problem: Solve the differential equation dy/dx – (1/x)y = x², for x > 0, given that y(1) = 2.

问题：求解微分方程 dy/dx – (1/x)y = x²，其中 x > 0，且已知 y(1) = 2。

This is a linear first-order ODE. The integrating factor is μ(x) = exp( ∫ -1/x dx ) = exp(-ln x) = 1/x.

这是一阶线性常微分方程。积分因子为 μ(x) = exp( ∫ -1/x dx ) = exp(-ln x) = 1/x。

Multiply through by 1/x: (1/x) dy/dx – (1/x²)y = x. The left side is d/dx( y/x ).

两边乘以 1/x：(1/x) dy/dx – (1/x²)y = x。左边正是 d/dx( y/x )。

Thus d/dx( y/x ) = x. Integrate with respect to x: y/x = ∫ x dx = ½ x² + C.

因此 d/dx( y/x ) = x。关于 x 积分：y/x = ∫ x dx = ½ x² + C。

Multiply by x: y = ½ x³ + Cx. Apply the initial condition y(1) = 2: 2 = ½(1) + C → C = 3/2.

乘以 x：y = ½ x³ + Cx。代入初始条件 y(1) = 2：2 = ½(1) + C → C = 3/2。

The particular solution is y = (1/2)x³ + (3/2)x.

特解为 y = (1/2)x³ + (3/2)x。

Answer: y = (x³ + 3x)/2. Using an integrating factor is a central technique for solving linear ODEs, frequently tested in advanced mathematics admissions.

答案：y = (x³ + 3x)/2。使用积分因子是求解线性常微分方程的核心方法，在进阶数学入学测试中频繁出现。

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6. Maclaurin Series Expansion | 麦克劳林级数展开

Problem: Find the Maclaurin series for f(x) = e^(2x) up to and including the term in x³.

问题：求 f(x) = e^(2x) 的麦克劳林级数，直到含 x³ 的项为止。

The Maclaurin series is given by f(0) + f'(0)x + f”(0)x²/2! + f”'(0)x³/3! + …

麦克劳林级数由 f(0) + f'(0)x + f”(0)x²/2! + f”'(0)x³/3! + … 给出。

Compute derivatives: f(x) = e^(2x) → f(0) = 1.

计算导数：f(x) = e^(2x) → f(0) = 1。

f'(x) = 2e^(2x) → f'(0) = 2.

f'(x) = 2e^(2x) → f'(0) = 2。

f”(x) = 4e^(2x) → f”(0) = 4.

f”(x) = 4e^(2x) → f”(0) = 4。

f”'(x) = 8e^(2x) → f”'(0) = 8.

f”'(x) = 8e^(2x) → f”'(0) = 8。

Substituting into the formula: 1 + 2x + (4/2!)x² + (8/3!)x³ = 1 + 2x + 2x² + (8/6)x³.

代入公式：1 + 2x + (4/2!)x² + (8/3!)x³ = 1 + 2x + 2x² + (8/6)x³。

Simplify the x³ term: 8/6 = 4/3. Hence the series is 1 + 2x + 2x² + (4/3)x³ + …

化简 x³ 项：8/6 = 4/3。因此级数为 1 + 2x + 2x² + (4/3)x³ + …

Answer: 1 + 2x + 2x² + (4/3)x³. This series can also be obtained directly from the standard e^u expansion, reinforcing understanding of transformation.

答案：1 + 2x + 2x² + (4/3)x³。该级数也可直接从标准 e^u 展开得到，巩固对变换的理解。

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7. Proof by Induction: Summation Formula | 归纳法证明：求和公式

Problem: Prove by mathematical induction that for all positive integers n, Σ_(r=1)^n r(r+1) = n(n+1)(n+2)/3.

问题：用数学归纳法证明，对于所有正整数 n，有 Σ_(r=1)^n r(r+1) = n(n+1)(n+2)/3。

Base case n = 1: LHS = 1·2 = 2. RHS = 1·2·3/3 = 2. The formula holds.

奠基步 n = 1：左边 = 1·2 = 2。右边 = 1·2·3/3 = 2。公式成立。

Inductive hypothesis: Assume true for n = k, i.e., Σ_(r=1)^k r(r+1) = k(k+1)(k+2)/3.

归纳假设：设 n = k 时成立，即 Σ_(r=1)^k r(r+1) = k(k+1)(k+2)/3。

Inductive step: For n = k+1, the sum is Σ_(r=1)^(k+1) r(r+1) = Σ_(r=1)^k r(r+1) + (k+1)(k+2).

归纳递推：对于 n = k+1，和为 Σ_(r=1)^(k+1) r(r+1) = Σ_(r=1)^k r(r+1) + (k+1)(k+2)。

By hypothesis, this equals k(k+1)(k+2)/3 + (k+1)(k+2).

根据假设，这等于 k(k+1)(k+2)/3 + (k+1)(k+2)。

Factor out (k+1)(k+2): (k+1)(k+2)[ k/3 + 1 ] = (k+1)(k+2)[ (k+3)/3 ].

提取公因子 (k+1)(k+2)：(k+1)(k+2)[ k/3 + 1 ] = (k+1)(k+2)[ (k+3)/3 ]。

This simplifies to (k+1)(k+2)(k+3)/3, which is exactly the formula with n = k+1. Thus, the statement holds for n = k+1.

这化简为 (k+1)(k+2)(k+3)/3，正是 n = k+1 时的公式。因此当 n = k+1 时命题成立。

Conclusion: By induction, the identity is true for all positive integers n.

结论：根据归纳法，该恒等式对所有正整数 n 成立。

Answer: The proof is complete. Induction is a recurring topic in NSAA Further Maths questions, demanding clear logical structure and algebraic manipulation.

答案：证明完成。归纳法是 NSAA 进阶数学题目中反复出现的话题，要求清晰的逻辑结构和代数运算能力。

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8. Polar Coordinates: Area Enclosed by a Curve | 极坐标：曲线围成的面积

Problem: The polar curve is defined by r = 2 + cos θ for 0 ≤ θ ≤ 2π. Find the total area enclosed by the curve.

问题：极坐标曲线由 r = 2 + cos θ 定义，其中 0 ≤ θ ≤ 2π。求该曲线包围的总面积。

The area in polar coordinates is given by A = ½ ∫₀²π r² dθ = ½ ∫₀²π (2 + cos θ)² dθ.

极坐标下的面积公式为 A = ½ ∫₀²π r² dθ = ½ ∫₀²π (2 + cos θ)² dθ。

Expand the integrand: (2 + cos θ)² = 4 + 4 cos θ + cos²θ.

展开被积函数：(2 + cos θ)² = 4 + 4 cos θ + cos²θ。

Using the double-angle identity cos²θ = (1 + cos 2θ)/2, we rewrite: 4 + 4 cos θ + ½ + ½ cos 2θ = 9/2 + 4 cos θ + ½ cos 2θ.

利用倍角公式 cos²θ = (1 + cos 2θ)/2，重写为：4 + 4 cos θ + ½ + ½ cos 2θ = 9/2 + 4 cos θ + ½ cos 2θ。

Now integrate term by term from 0 to 2π:

现在从 0 到 2π 逐项积分：

• ∫₀²π (9/2) dθ = (9/2)·2π = 9π.

  ∫₀²π (9/2) dθ = (9/2)·2π = 9π。

• ∫₀²π 4 cos θ dθ = 4[sin θ]₀²π = 0.

  ∫₀²π 4 cos θ dθ = 4[sin θ]₀²π = 0。

• ∫₀²π ½ cos 2θ dθ = ¼[sin 2θ]₀²π = 0.

  ∫₀²π ½ cos 2θ dθ = ¼[sin 2θ]₀²π = 0。

Thus the total integral is 9π. Multiply by ½ to get A = ½ × 9π = (9π)/2.

因此总积分为 9π。乘以 ½ 得到 A = ½ × 9π = (9π)/2。

Answer: The enclosed area is (9π)/2 square units. This question combines polar integration with trigonometric identities, a classic Further Pure topic.

答案：包围的面积为 (9π)/2 平方单位。此题将极坐标积分与三角恒等式相结合，是经典的高等纯数主题。

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