📚 IB Maths: Detailed Walkthrough of Typical Questions | IB 数学：典型例题详解
In IB Mathematics, mastering typical question types is the key to building confidence and achieving high marks. This article provides a step-by-step walkthrough of ten classic problems covering functions, trigonometry, calculus, probability, statistics, vectors, complex numbers, sequences, and exponentials. Each example is broken down into logical steps with paired explanations, helping you understand the reasoning behind every method. Whether you are preparing for Analysis & Approaches (AA) or Applications & Interpretation (AI), these worked solutions will sharpen your problem-solving skills and exam technique.
在 IB 数学中，掌握典型题型是建立信心和取得高分的关键。本文精选了十个经典例题，涵盖函数、三角学、微积分、概率、统计、向量、复数、数列和指数对数等板块，并以配对讲解的方式逐步拆解。无论你学习的是分析与方法 (AA) 还是应用与解释 (AI)，这些详细解答都将帮助你理清解题思路，提升应试能力。
1. Composite Function and Domain | 复合函数与定义域
Problem: Let f(x) = ln(x − 1) and g(x) = ex + 1. Find the expression for f(g(x)) and state its domain.
题目：已知 f(x) = ln(x − 1)，g(x) = ex + 1。求 f(g(x)) 的表达式并写出其定义域。
First, form the composite function: f(g(x)) = ln(g(x) − 1) = ln((ex + 1) − 1) = ln(ex) = x. This gives a surprisingly simple result.
首先构成复合函数：f(g(x)) = ln(g(x) − 1) = ln((ex + 1) − 1) = ln(ex) = x，化简后结果非常简单。
For the domain, recall that the input of the outer function ln requires its argument to be strictly positive: ex + 1 − 1 > 0 ⇒ ex > 0, which is true for all real x. Also g(x) must be within the domain of f, i.e., g(x) > 1 ⇒ ex + 1 > 1 ⇒ ex > 0, always satisfied. Hence the domain of f(g(x)) is all real numbers, (−∞, ∞).
关于定义域，外层函数 ln 要求其自变量严格大于零：ex + 1 − 1 > 0 ⇒ ex > 0，这对所有实数 x 都成立。同时 g(x) 必须落在 f 的定义域内，即 g(x) > 1 ⇒ ex + 1 > 1 ⇒ ex > 0，恒成立。因此复合函数 f(g(x)) 的定义域为全体实数 (−∞, ∞)。
Key point: Always check the domain restrictions of the innermost function and the outer function when dealing with compositions.
关键点：处理复合函数时务必检查内层和外层函数各自的定义域限制。
2. Solving a Trigonometric Equation | 解三角方程
Problem: Solve sin(2x) = ½ for 0 ≤ x ≤ 2π, giving answers in exact radian form.
题目：解方程 sin(2x) = ½，其中 0 ≤ x ≤ 2π，答案用精确弧度值表示。
Start by finding the reference angle for sin(θ) = ½: θ = π/6. Since sin is positive in the first and second quadrants, the principal solutions for 2x are θ = π/6 and θ = π − π/6 = 5π/6. Adding multiples of 2π, the general solutions are 2x = π/6 + 2kπ or 2x = 5π/6 + 2kπ, k ∈ ℤ.
先求 sin(θ) = ½ 的参考角：θ = π/6。正弦在第一象限和第二象限为正，因此 2x 的主值解为 θ = π/6 和 θ = π − π/6 = 5π/6。加上 2π 的整数倍，通解为 2x = π/6 + 2kπ 或 2x = 5π/6 + 2kπ，k ∈ ℤ。
Now divide by 2: x = π/12 + kπ or x = 5π/12 + kπ. Restrict to 0 ≤ x ≤ 2π. For k = 0: x = π/12, 5π/12. For k = 1: x = 13π/12, 17π/12. For k = 2: x = 25π/12 (out of range) and x = 29π/12 (out of range). Thus the solution set is {π/12, 5π/12, 13π/12, 17π/12}.
两边除以 2：x = π/12 + kπ 或 x = 5π/12 + kπ。限制在 0 ≤ x ≤ 2π。当 k = 0 时：x = π/12, 5π/12；k = 1 时：x = 13π/12, 17π/12；k = 2 时 x 值均超出范围。因此解集为 {π/12, 5π/12, 13π/12, 17π/12}。
Remember to adjust the period after finding the general solution for the doubled angle. Graphing the function can also help visualise the four intersections.
注意在求出倍角的通解后要调整周期。绘制函数图像也有助于直观看到四个交点。
3. Tangent Line Using Differentiation | 用导数求切线方程
Problem: Find the equation of the tangent to the curve y = x³ − 4x² + 7 at the point where x = 1.
题目：求曲线 y = x³ − 4x² + 7 在 x = 1 处的切线方程。
First compute the y-coordinate: y(1) = (1)³ − 4(1)² + 7 = 1 − 4 + 7 = 4. The point of tangency is (1, 4).
首先计算 y 坐标：y(1) = 1³ − 4·1² + 7 = 1 − 4 + 7 = 4。切点为 (1, 4)。
Differentiate: dy/dx = 3x² − 8x. At x = 1, the gradient m = 3(1)² − 8(1) = 3 − 8 = −5. So the tangent has slope −5.
求导：dy/dx = 3x² − 8x。当 x = 1 时，斜率 m = 3·1² − 8·1 = −5。因此切线斜率为 −5。
Using the point-slope form: y − 4 = −5(x − 1). Simplify to y = −5x + 5 + 4 = −5x + 9. The equation of the tangent is y = −5x + 9.
利用点斜式：y − 4 = −5(x − 1)。化简得 y = −5x + 5 + 4 = −5x + 9。切线方程为 y = −5x + 9。
A common mistake is misapplying the power rule when differentiating. Always check your derivative by expanding or using the power rule term by term.
常见错误是求导时幂法则运用不当。务必逐项检查导数，确保每一项的指数和系数正确。
4. Definite Integral for Area | 定积分求面积
Problem: Calculate the area enclosed between the curve y = sin x and the x-axis from x = 0 to x = π.
题目：计算曲线 y = sin x 与 x 轴之间在 x = 0 到 x = π 范围内围成的面积。
Since sin x ≥ 0 on [0, π], the area is simply the definite integral ∫0π sin x dx.
因为在 [0, π] 上 sin x ≥ 0，面积直接由定积分 ∫0π sin x dx 给出。
Find the antiderivative: ∫ sin x dx = −cos x. Apply the limits: Area = [−cos x]0π = (−cos π) − (−cos 0) = (−(−1)) − (−1) = 1 + 1 = 2. So the area is 2 square units.
求原函数：∫ sin x dx = −cos x。代入上下限：面积 = [−cos x]0π = (−cos π) − (−cos 0) = (−(−1)) − (−1) = 1 + 1 = 2。因此面积为 2 平方单位。
If the function were negative over part of the interval, we would need to split the integral or take absolute values. For this sine curve, the area matches the geometric intuition of one arch.
如果函数在区间内部分为负，则需要分段积分或取绝对值。对于这段正弦曲线，面积恰好符合一个拱形的几何直观。
5. Binomial Distribution | 二项分布
Problem: A fair die is rolled 5 times. Find the probability of getting exactly three 6s.
题目：一颗均匀骰子掷 5 次。求恰好得到三次点数 6 的概率。
Define X as the number of 6s obtained in 5 trials. Then X ~ B(5, p) with p = 1/6. The probability mass function is P(X = k) = C(n, k) pk (1−p)n−k.
设 X 为 5 次试验中出现 6 的次数。则 X ~ B(5, p)，其中 p = 1/6。概率质量函数为 P(X = k) = C(n, k) pk (1−p)n−k。
Plug in n = 5, k = 3, p = 1/6: P(X = 3) = C(5,3) × (1/6)³ × (5/6)². C(5,3) = 10. So P = 10 × (1/216) × (25/36) = 10 × 25 / (216 × 36) = 250 / 7776. Simplify dividing by 2: 125/3888. This is approximately 0.0322.
代入 n = 5, k = 3, p = 1/6：P(X = 3) = C(5,3) × (1/6)³ × (5/6)²。C(5,3) = 10。所以 P = 10 × (1/216) × (25/36) = 250 / 7776。约分为 125/3888，大约等于 0.0322。
Check that you have correctly distinguished between “exactly three” and “at least three”. Binomial pdf gives exact values, while cumulative probabilities handle ranges.
注意区分“恰好三次”和“至少三次”。二项概率密度给出精确值，累积概率用于区间计算。
6. Least Squares Regression Line | 最小二乘回归线
Problem: The table below shows five data pairs (x, y). Find the equation of the regression line y on x in the form y = a + bx.
| x | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| y | 2 | 4 | 5 | 7 | 8 |
题目：下表给出五组数据 (x, y)。求 y 对 x 的回归直线方程，形式为 y = a + bx。
First calculate means: x̄ = (1+2+3+4+5)/5 = 3, ȳ = (2+4+5+7+8)/5 = 26/5 = 5.2. Compute Sxx = Σ(x − x̄)² and Sxy = Σ(x − x̄)(y − ȳ).
先求均值：x̄ = 3，ȳ = 5.2。计算 Sxx = Σ(x − x̄)² 和 Sxy = Σ(x − x̄)(y − ȳ)。
Using a table: x − x̄: −2, −1, 0, 1, 2. y − ȳ: −3.2, −1.2, −0.2, 1.8, 2.8. Then (x−x̄)²: 4, 1, 0, 1, 4 → Sxx = 10. Products (x−x̄)(y−ȳ): (−2)(−3.2)=6.4, (−1)(−1.2)=1.2, 0×(−0.2)=0, 1×1.8=1.8, 2×2.8=5.6 → Sxy = 15.0.
列表计算：x − x̄: −2, −1, 0, 1, 2；y − ȳ: −3.2, −1.2, −0.2, 1.8, 2.8。(x−x̄)² 和为 10；积和为 15.0。
Then slope b = Sxy/Sxx = 15/10 = 1.5. Intercept a = ȳ − b x̄ = 5.2 − 1.5 × 3 = 5.2 − 4.5 = 0.7. Regression line: y = 0.7 + 1.5x.
斜率 b = Sxy/Sxx = 1.5。截距 a = ȳ − b x̄ = 0.7。回归直线为 y = 0.7 + 1.5x。
Make sure to identify which variable is dependent (y) and which is independent (x). The formula using Sxy and Sxx is standard for y-on-x regression.
注意明确因变量 (y) 和自变量 (x)。使用 Sxy 和 Sxx 的公式是 y 对 x 回归的标准方法。
7. Vector Equation of a Line and Angle Between Vectors | 向量直线方程与夹角
Problem: Given points A(1, 2, 3) and B(4, 0, −1). Find the vector equation of line AB and the acute angle between AB and the vector v = i + 2j + 2k.
题目：已知点 A(1, 2, 3) 和 B(4, 0, −1)。求直线 AB 的向量方程以及 AB 与向量 v = i + 2j + 2k 的锐角夹角。
Direction vector AB = (4−1, 0−2, −1−3) = (3, −2, −4). A vector equation of the line through A is r = (1, 2, 3) + t(3, −2, −4), t ∈ ℝ.
方向向量 AB = (3, −2, −4)。过点 A 的直线向量方程为 r = (1, 2, 3) + t(3, −2, −4)，t ∈ ℝ。
For the angle θ between AB and v = (1, 2, 2), use cosθ = |AB·v| / (|AB| |v|). Compute dot product: AB·v = 3×1 + (−2)×2 + (−4)×2 = 3 − 4 − 8 = −9. Magnitudes: |AB| = √(3² + (−2)² + (−4)²) = √(9+4+16) = √29; |v| = √(1+4+4) = √9 = 3.
求 AB 与 v 的夹角 θ，用 cosθ = |AB·v| / (|AB| |v|)。点积：AB·v = −9。模长：|AB| = √29，|v| = 3。
Thus cosθ = |−9|/(3√29) = 9/(3√29) = 3/√29. The acute angle is θ = arccos(3/√29) ≈ arccos(0.5578) ≈ 56.1° (to 1 d.p.).
于是 cosθ = 3/√29。锐角 θ = arccos(3/√29) ≈ 56.1°。
Remember to take the absolute value of the dot product to ensure you obtain the acute angle. The vector line equation can also be expressed symmetrically.
记得对标量积取绝对值以确保得到锐角。直线方程也可用参数形式或对称式表示。
8. Complex Numbers and Polar Form | 复数与极坐标形式
Problem: Solve the quadratic equation z² + 4z + 13 = 0, giving the roots in Cartesian form a + b i, and then express one root in polar form r cis θ.
题目：解二次方程 z² + 4z + 13 = 0，给出根的代数形式 a + b i，并将其中一个根表示为极坐标形式 r cis θ。
Discriminant Δ = 4² − 4×1×13 = 16 − 52 = −36. Roots: z = [−4 ± √(−36)] / 2 = (−4 ± 6i)/2 = −2 ± 3i. So the two roots are z₁ = −2 + 3i, z₂ = −2 − 3i.
判别式 Δ = 16 − 52 = −36。根为 z = (−4 ± 6i)/2 = −2 ± 3i。
For z₁ = −2 + 3i, the modulus r = √((−2)² + 3²) = √(4+9) = √13. The argument θ satisfies tan θ’ = opposite/adjacent = 3/2, giving reference angle arctan(1.5) ≈ 0.9828 rad. Since z₁ lies in the second quadrant (negative real part, positive imaginary part), θ = π − 0.9828 ≈ 2.1588 rad.
对于 z₁ = −2 + 3i，模长 r = √13。辐角参考值 arctan(1.5) ≈ 0.9828 rad。由于 z₁ 在第二象限，辐角 θ = π − 0.9828 ≈ 2.1588 rad。
Polar form: z₁ = √13 cis(2.159) or √13 (cos 2.159 + i sin 2.159). For exact expression, we can leave θ as π − arctan(3/2).
极坐标形式：z₁ = √13 cis(2.159) 或精确表达为 √13 (cos(π − arctan(3/2)) + i sin(π − arctan(3/2)))。
In IB exams, both Cartesian and modulus-argument forms are tested. Remember that the argument must be adjusted to the correct quadrant.
IB 考试中代数形式和模长-辐角形式都可能考查。务必根据象限修正辐角值。
9. Arithmetic Sequence and Summation | 等差数列与求和
Problem: In an arithmetic sequence, the 3rd term is 7 and the 8th term is 22. Find the first term u₁, the common difference d, and the sum of the first 20 terms.
题目：一个等差数列的第 3 项为 7，第 8 项为 22。求首项 u₁、公差 d 以及前 20 项的和。
Recall uₙ = u₁ + (n−1)d. Given u₃ = u₁ + 2d = 7 and u₈ = u₁ + 7d = 22. Subtract the first equation from the second: (u₁+7d) − (u₁+2d) = 22 − 7 ⇒ 5d = 15 ⇒ d = 3.
由通项公式得 u₃ = u₁ + 2d = 7，u₈ = u₁ + 7d = 22。两式相减得 5d = 15，故 d = 3。
Substitute d = 3 into u₁ + 2×3 = 7 ⇒ u₁ + 6 = 7 ⇒ u₁ = 1. So the sequence starts at 1, with common difference 3: 1, 4, 7, 10, …
代回得 u₁ = 1。数列为 1, 4, 7, 10, …
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