Answers to Selected IB Math Problems | IB数学精选问题解答

📚 Answers to Selected IB Math Problems | IB数学精选问题解答

This article provides detailed, step-by-step solutions to a curated set of IB Mathematics problems spanning algebra, functions, calculus, probability, vectors, sequences, and logarithms. Each solution is presented in a bilingual format, with English explanations immediately followed by Chinese translations to support learners at all levels. Work through these examples to reinforce your understanding and prepare confidently for your IB assessments.

本文提供了一组精选 IB 数学问题的详细分步解答，涵盖代数、函数、微积分、概率、向量、数列和对数。每道解答均采用双语形式，英文解释之后紧跟中文翻译，以帮助各层次的学习者。通过演练这些例子，你可以巩固理解，自信备战 IB 考核。

1. Solving a Quadratic Equation | 解二次方程

Problem: Solve the quadratic equation x² – 5x + 6 = 0.

问题：解二次方程 x² – 5x + 6 = 0。

We identify the coefficients: a = 1, b = –5, c = 6.

确定系数：a = 1，b = –5，c = 6。

The discriminant Δ = b² – 4ac = (–5)² – 4·1·6 = 25 – 24 = 1.

判别式 Δ = b² – 4ac = (–5)² – 4·1·6 = 25 – 24 = 1。

Since Δ > 0, there are two distinct real roots. Using the quadratic formula x = [–b ± √Δ] / (2a), we get x = [5 ± 1] / 2.

因为 Δ > 0，方程有两个不等实根。利用求根公式 x = [–b ± √Δ] / (2a)，得 x = [5 ± 1] / 2。

Thus, x₁ = (5+1)/2 = 3 and x₂ = (5–1)/2 = 2.

因此，x₁ = (5+1)/2 = 3，x₂ = (5–1)/2 = 2。

The solution set is {2, 3}.

解集为 {2, 3}。

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2. Composite Functions | 复合函数

Problem: Given f(x) = 2x + 1 and g(x) = x², find f(g(x)) and g(f(x)).

问题：已知 f(x) = 2x + 1，g(x) = x²，求 f(g(x)) 和 g(f(x))。

For f(g(x)), we substitute g(x) into f: f(g(x)) = 2·(g(x)) + 1 = 2·(x²) + 1 = 2x² + 1.

对于 f(g(x))，将 g(x) 代入 f：f(g(x)) = 2·(g(x)) + 1 = 2·(x²) + 1 = 2x² + 1。

For g(f(x)), substitute f(x) into g: g(f(x)) = (f(x))² = (2x + 1)² = 4x² + 4x + 1.

对于 g(f(x))，将 f(x) 代入 g：g(f(x)) = (f(x))² = (2x + 1)² = 4x² + 4x + 1。

Note that f(g(x)) ≠ g(f(x)) in general; composition of functions is not commutative.

注意，一般地 f(g(x)) ≠ g(f(x))；函数的复合不可交换。

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3. Differentiation of a Polynomial and Trigonometric Function | 多项式与三角函数的求导

Problem: Differentiate y = 3x⁴ – 2x² + sin x with respect to x.

问题：对 y = 3x⁴ – 2x² + sin x 关于 x 求导。

Apply the power rule and the derivative of sin x. The derivative of 3x⁴ is 3·4x³ = 12x³.

运用幂函数求导法则和 sin x 的导数。3x⁴ 的导数为 3·4x³ = 12x³。

The derivative of –2x² is –2·2x = –4x.

–2x² 的导数为 –2·2x = –4x。

The derivative of sin x is cos x.

sin x 的导数为 cos x。

Combine the results: dy/dx = 12x³ – 4x + cos x.

合并结果：dy/dx = 12x³ – 4x + cos x。

This is an exact derivative, no further simplification needed.

这就是精确的导数，无需进一步化简。

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4. Integration by Substitution | 换元积分法

Problem: Evaluate ∫ (2x + 1)³ dx.

问题：计算 ∫ (2x + 1)³ dx。

Use the substitution u = 2x + 1. Then du/dx = 2, so dx = du / 2.

使用换元 u = 2x + 1。则 du/dx = 2，从而 dx = du / 2。

The integral becomes ∫ u³ · (du/2) = (1/2) ∫ u³ du.

积分变为 ∫ u³ · (du/2) = (1/2) ∫ u³ du。

Integrate: (1/2) · (u⁴/4) + C = u⁴ / 8 + C.

积分得：(1/2) · (u⁴/4) + C = u⁴ / 8 + C。

Replace u with 2x + 1 to obtain the final answer: (1/8)(2x + 1)⁴ + C.

将 u 代回 2x + 1 得到最终答案：(1/8)(2x + 1)⁴ + C。

We can verify by differentiating the result, confirming it yields the original integrand.

可以通过对结果求导来验证，确认得到原来的被积函数。

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5. Probability without Replacement | 无放回概率

Problem: A bag contains 5 red balls and 3 blue balls. Two balls are drawn at random without replacement. Find the probability that both balls are red.

问题：袋中有 5 个红球和 3 个蓝球。随机无放回地抽取两个球。求两个球都是红色的概率。

Total number of balls = 8. Number of ways to choose 2 balls out of 8 is C(8,2) = 28.

总球数为 8。从 8 个球中选 2 个的组合数为 C(8,2) = 28。

Favorable outcomes: both red. Number of ways to choose 2 red out of 5 is C(5,2) = 10.

有利结果：两个都是红球。从 5 个红球中选 2 个的组合数为 C(5,2) = 10。

Thus, probability = 10/28 = 5/14.

因此，概率 = 10/28 = 5/14。

Alternatively, using conditional probability: P(first red) = 5/8, P(second red | first red) = 4/7, product = 5/8 × 4/7 = 20/56 = 5/14.

或者用条件概率：P(第一个红) = 5/8，P(第二个红 | 第一个红) = 4/7，乘积 = 5/8 × 4/7 = 20/56 = 5/14。

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6. Vectors – Dot Product and Angle | 向量 – 点积与夹角

Problem: Given vectors a = 3i – 2j + k and b = –i + 4j + 2k, calculate the dot product a·b and the angle between a and b.

问题：已知向量 a = 3i – 2j + k 和 b = –i + 4j + 2k，计算点积 a·b 以及 a 与 b 的夹角。

The dot product a·b = (3)(–1) + (–2)(4) + (1)(2) = –3 – 8 + 2 = –9.

点积 a·b = (3)(–1) + (–2)(4) + (1)(2) = –3 – 8 + 2 = –9。

Magnitudes: |a| = √(3² + (–2)² + 1²) = √9+4+1 = √14, |b| = √((–1)² + 4² + 2²) = √1+16+4 = √21.

模长：|a| = √(3² + (–2)² + 1²) = √9+4+1 = √14，|b| = √((–1)² + 4² + 2²) = √1+16+4 = √21。

Using a·b = |a||b| cos θ, we get cos θ = (–9) / (√14 · √21) = –9 / √294 = –9 / (7√6) ≈ –0.525.

利用 a·b = |a||b| cos θ，得 cos θ = (–9) / (√14 · √21) = –9 / √294 = –9 / (7√6) ≈ –0.525。

Then θ = arccos(–9 / (7√6 122° (to the nearest degree). The angle is obtuse, consistent with a negative dot product.

于是 θ = arccos(–9 / (7√6)) ≈ 122°（精确到度）。该角为钝角，与点积为负一致。

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7. Arithmetic Sequence | 等差数列

Problem: The 4th term of an arithmetic sequence is 14 and the 10th term is 38. Find the first term and the common difference.

问题：一个等差数列的第 4 项为 14，第 10 项为 38。求首项和公差。

In an arithmetic sequence, the nth term is given by uₙ = u₁ + (n – 1)d.

等差数列的第 n 项公式为 uₙ = u₁ + (n – 1)d。

For n=4: u₁ + 3d = 14. For n=10: u₁ + 9d = 38.

n=4 时：u₁ + 3d = 14。n=10 时：u₁ + 9d = 38。

Subtract the first equation from the second: (u₁+9d) – (u₁+3d) = 38 – 14 ⇒ 6d = 24 ⇒ d = 4.

用第二式减第一式：(u₁+9d) – (u₁+3d) = 38 – 14 ⇒ 6d = 24 ⇒ d = 4。

Substitute d = 4 into u₁ + 3·4 = 14 ⇒ u₁ + 12 = 14 ⇒ u₁ = 2.

将 d = 4 代入 u₁ + 3·4 = 14 ⇒ u₁ + 12 = 14 ⇒ u₁ = 2。

Thus, the first term is 2 and the common difference is 4.

因此，首项为 2，公差为 4。

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8. Logarithmic Equation | 对数方程

Problem: Solve log₂(x + 1) + log₂(x – 1) = 3.

问题：解方程 log₂(x + 1) + log₂(x – 1) = 3。

Combine the logarithms using the product rule: log₂[(x + 1)(x – 1)] = 3.

利用乘积法则合并对数：log₂[(x + 1)(x – 1)] = 3。

Simplify the argument: (x+1)(x–1) = x² – 1. So log₂(x² – 1) = 3.

化简括号内：(x+1)(x–1) = x² – 1。于是 log₂(x² – 1) = 3。

Rewrite in exponential form: x² – 1 = 2³ = 8.

改写成指数形式：x² – 1 = 2³ = 8。

Solve for x: x² = 9 ⇒ x = ±3.

解 x：x² = 9 ⇒ x = ±3。

Check the domain: the arguments x+1 and x–1 must be positive. For x = 3, 4 and 2 are positive. For x = –3, –2 and –4 are negative, which are not allowed. Hence, the only valid solution is x = 3.

检验定义域：真数 x+1 和 x–1 必须为正。x = 3 时，4 和 2 均为正。x = –3 时，–2 和 –4 为负，不符合要求。因此唯一有效解为 x = 3。

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