IB Mathematics: Analysis and Approaches HL Practice Paper 1 Solutions | IB数学：分析与方法HL练习卷一解析

📚 IB Mathematics: Analysis and Approaches HL Practice Paper 1 Solutions | IB数学：分析与方法HL练习卷一解析

This article provides a detailed walkthrough of a complete IB Mathematics: Analysis and Approaches Higher Level (AA HL) Practice Paper 1. Paper 1 is taken without a calculator, so the solutions emphasise algebraic manipulation, exact values and conceptual understanding.

本文详细解析一套完整的IB数学：分析与方法高级水平（AA HL）练习卷一。卷一不允许使用计算器，因此解答强调代数运算、精确值以及概念理解。

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1. Functions and Transformations | 函数与变换

Consider the function f(x)=2x-1x+3, for x≠ -3. Find the inverse function f^-1(x) and state its domain.

考虑函数 f(x)=2x-1x+3，其中 x≠ -3。求反函数 f^-1(x) 并写出其定义域。

Let y=2x-1x+3. Swap x and y: x=2y-1y+3. Then x(y+3)=2y-1, so xy+3x=2y-1. Rearranging: xy-2y=-3x-1, hence y(x-2)=-3x-1.

设 y=2x-1x+3。交换 x 与 y：x=2y-1y+3。于是 x(y+3)=2y-1，即 xy+3x=2y-1。整理得 xy-2y=-3x-1，从而 y(x-2)=-3x-1。

f^-1(x)=-3x-1x-2=3x+12-x, x≠ 2

The domain of the inverse is the range of the original function. Since f(x) has a horizontal asymptote y=2, the range is y≠ 2. Therefore the domain of f^-1 is x≠ 2.

反函数的定义域是原函数的值域。由于 f(x) 有水平渐近线 y=2，其值域为 y≠ 2。因此 f^-1 的定义域为 x≠ 2。

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2. Complex Numbers | 复数

Find the square roots of the complex number z=-3+4i, giving your answers in the form a+bi, where a,b∈ℝ.

求复数 z=-3+4i 的平方根，写出形如 a+bi 的答案，其中 a,b∈ℝ。

Let (a+bi)²=-3+4i. Expanding gives a²-b²+2abi=-3+4i. Equating real and imaginary parts:

设 (a+bi)²=-3+4i。展开得 a²-b²+2abi=-3+4i。比较实部和虚部：

• a²-b²=-3
• 2ab=4, so ab=2

From ab=2, we have b=2a. Substitute: a²-⁴⁄_a²=-3. Multiplying by a²: a⁴+3a²-4=0. Factor: (a²+4)(a²-1)=0. Since a²=-4 is not real, a=± 1.

由 ab=2 得 b=2a。代入：a²-⁴⁄_a²=-3。两边乘以 a²：a⁴+3a²-4=0。因式分解得 (a²+4)(a²-1)=0。由于 a²=-4 非实数，所以 a=± 1。

a=1, b=2 或 a=-1, b=-2，故平方根为 1+2i 和 -1-2i

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3. Vectors and Geometry | 向量与几何

The points A(1,2,3), B(2,-1,4) and C(-1,0,2) are given. Find the area of triangle ABC.

已知点 A(1,2,3)、B(2,-1,4) 和 C(-1,0,2)。求三角形 ABC 的面积。

Form vectors: AB⃗ = (1,-3,1) and AC⃗ = (-2,-2,-1). The area of the triangle is half the magnitude of the cross product.

构造向量：AB⃗ = (1,-3,1)，AC⃗ = (-2,-2,-1)。三角形面积为叉积模长的一半。

AB⃗×AC⃗ = (i, j, k; 1, -3, 1; -2, -2, -1) = (3+2, -1+1?) Let us compute carefully: i((-3)(-1)-1(-2)) - j(1(-1)-1(-2)) + k(1(-2)-(-3)(-2)).

AB⃗×AC⃗ = (i, j, k; 1, -3, 1; -2, -2, -1) = i((-3)(-1)-1(-2)) - j(1(-1)-1(-2)) + k(1(-2)-(-3)(-2))。

This gives (3+2)i - (-1+2)j + (-2-6)k = 5i -1j -8k. The magnitude is √25+1+64=√90=3√10.

计算得 (3+2)i - (-1+2)j + (-2-6)k = 5i -1j -8k。模长为 √25+1+64=√90=3√10。

面积 = 12× 3√10 = 3√102

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4. Calculus: Differentiation | 微积分：微分法

Differentiate y=ln(x²+1)+arctan x. Hence find the exact value of the derivative at x=1.

求 y=ln(x²+1)+arctan x 的导数，并由此计算 x=1 处导数的精确值。

Using standard derivatives: ddxln(x²+1)=^2x⁄_x²+1, and ddxarctan x=¹⁄_x²+1.

利用标准导数公式：ddxln(x²+1)=^2x⁄_x²+1，ddxarctan x=¹⁄_x²+1。

dydx=^2x+1⁄_x²+1

At x=1, dydx=32.

当 x=1 时，dydx=32。

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5. Calculus: Integration | 微积分：积分法

Evaluate the definite integral ∫₀¹ xe^2xdx, giving your answer exactly.

计算定积分 ∫₀¹ xe^2xdx，给出精确答案。

Use integration by parts: let u=x, dv=e^2xdx. Then du=dx, v=12e^2x.

使用分部积分法：令 u=x，dv=e^2xdx。则 du=dx，v=12e^2x。

∫ xe^2xdx = x2e^2x - ∫ 12e^2xdx = x2e^2x - 14e^2x + C.

∫ xe^2xdx = x2e^2x - ∫ 12e^2xdx = x2e^2x - 14e^2x + C。

Evaluate from 0 to 1: [x2e^2x - 14e^2x]₀¹ = (12e² - 14e²) - (0 - 14) = 14e² + 14.

从0到1计算：[x2e^2x - 14e^2x]₀¹ = (12e² - 14e²) - (0 - 14) = 14e² + 14。

∫₀¹ xe^2xdx = ^e2+1⁄₄

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6. Sequences and Series | 数列与级数

The sum of the first n terms of a geometric series is S_n = 3(1-4^-n). Find the first term u₁ and the common ratio r.

一个等比数列前 n 项和为 S_n = 3(1-4^-n)。求首项 u₁ 和公比 r。

The first term is u₁ = S₁ = 3(1-4^-1) = 3(1-14) = 94.

首项为 u₁ = S₁ = 3(1-4^-1) = 3(1-14) = 94。

The sum formula for a geometric series is S_n = ^a(1-rn)⁄_1-r. Comparing, we have 1-rⁿ matching 1-4^-n = 1-(4^-1)ⁿ, so r=4^-1=14. Also a1-r=3, giving a=3(1-14)=94.

等比数列求和公式为 S_n = ^a(1-rn)⁄_1-r。比较可知，1-rⁿ 对应 1-4^-n = 1-(4^-1)ⁿ，因此 r=4^-1=14。同时 a1-r=3，得 a=3(1-14)=94。

u₁=94, r=14

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7. Probability and Statistics | 概率与统计

A discrete random variable X has probability distribution given by P(X=x)=k(3-x) for x=0,1,2, and P(X=3)=0. Find k and E(X).

离散型随机变量 X 的概率分布为 P(X=x)=k(3-x)，x=0,1,2，且 P(X=3)=0。求 k 和 E(X)。

The probabilities must sum to 1: k(3-0)+k(3-1)+k(3-2)=3k+2k+k=6k=1. Hence k=16.

概率总和必须为1：k(3-0)+k(3-1)+k(3-2)=3k+2k+k=6k=1。因此 k=16。

Then E(X)=0· 36+1· 26+2· 16 = 0+26+26=46=23.

于是 E(X)=0· 36+1· 26+2· 16 = 0+26+26=46=23。

k=16, E(X)=23

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8. Proof and Logic | 证明与逻辑

Prove that the sum of any three consecutive integers is divisible by 3.

证明任意三个连续整数的和能被3整除。

Let the three consecutive integers be n, n+1, n+2, where n∈ℤ. Their sum is n+(n+1)+(n+2)=3n+3=3(n+1). Since n+1 is an integer, the sum is divisible by 3.

设三个连续整数为 n、n+1、n+2，其中 n∈ℤ。它们的和为 n+(n+1)+(n+2)=3n+3=3(n+1)。因为 n+1 是整数，所以该和能被3整除。

This proof uses direct proof, one of the fundamental methods in AA HL Paper 1. It is essential to write full sentences and clearly state what each variable represents.

该证明使用了直接证明法，这是AA HL卷一中的基本方法之一。书写时必须写出完整句子，并清楚说明每个变量代表的含义。

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9. Common Mistakes and Strategies | 常见错误与解题策略

Several errors frequently appear in Paper 1. Avoiding them can substantially increase your score.

卷一中有几个常见错误。避免它们可以显著提高分数。

• Forgetting the constant of integration C in indefinite integrals.
• 忽略不定积分中的积分常数 C。
• Dropping absolute values when simplifying √x².
• 化简 √x² 时漏掉绝对值。
• Incorrectly applying the chain rule, especially in composite functions.
• 错误应用链式法则，尤其是复合函数情形。
• Mixing up the conditions for using the binomial expansion on (a+x)ⁿ vs (1+x)ⁿ.
• 混淆 (a+x)ⁿ 与 (1+x)ⁿ 二项式展开的适用条件。

When reviewing solutions, always check the domain of inverse functions, the sign of square roots, and whether your final answer is exact as required.

复习解答时，务必检查反函数的定义域、平方根的符号，以及最终答案是否符合题目要求的精确值。

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