Self-tests: Selected Answers | 自测题答案选编

📚 Self-tests: Selected Answers | 自测题答案选编

Mastering IB Mathematics requires consistent self-assessment, and working through self-tests is one of the most effective revision strategies. This article presents selected answers and worked solutions from a comprehensive self-test covering core topics in the IB Mathematics: Analysis and Approaches and Applications and Interpretation courses. For each question, we provide the final answer alongside key steps and common pitfalls. Use these to verify your own work, deepen your understanding, and sharpen your problem-solving skills before exams.

掌握IB数学需要持续的自我评估，而进行自测题练习是最有效的复习策略之一。本文从涵盖IB数学“分析与方法”及“应用与解释”课程核心主题的综合自测题中，节选了一些答案和解题步骤。每个问题都提供了最终答案、关键步骤以及常见错误，用于帮助你核对答案、加深理解，并在考试前提升解题能力。

1. Algebraic Simplification | 代数化简

Question: Simplify the expression (3x²y)² × (2xy³) ÷ (6x⁴y⁵).

题目：化简表达式 (3x²y)² × (2xy³) ÷ (6x⁴y⁵)。

². Multiply by 2xy³ to obtain 18x⁵y⁵. Finally, divide by 6x⁴y⁵: (18/6) · x⁵⁻⁴ · y⁵⁻⁵ = 3x.

解答：首先将幂分配到括号内的每个因子：(3x²y)² = 3² · (x²)² · y² = 9x⁴y²。乘以 2xy³ 得到 18x⁵y⁵。最后除以 6x⁴y⁵：(18/6) · x⁵⁻⁴ · y⁵⁻⁵ = 3x。

Answer: 3x

答案：3x

2. Solving Exponential Equations | 解指数方程

Question: Solve for x: 2^2x+1 = 8^x-3.

题目：解方程 2^2x+1 = 8^x-3。

ion: Express both sides with the same base. Since 8 = 2³, the right-hand side becomes (2³)^x-3 = 2^3x-9. Equate the exponents: 2x+1 = 3x-9. Rearranging gives x = 10.

解答：两边化为同底数的幂。因为 8 = 2³，右边等于 (2³)^x-3 = 2^3x-9。令指数相等：2x+1 = 3x-9，移项解得 x = 10。

Answer: x = 10

答案：x = 10

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3. Composite Functions | 复合函数

Question: Given f(x) = 1/(x-1) and g(x) = √x+2, find f(g(x)) and state its domain.

题目：已知 f(x) = 1/(x-1) 与 g(x) = √x+2，求 f(g(x)) 并写出其定义域。

Solution: f(g(x)) = 1/( √x+2 in requires the radicand x+2 ≥ 0, so x ≥ -2. Additionally, the denominator cannot be zero: √x+2 - 1 ≠ 0 ⇒ √x+2 ≠ 1 ⇒ x+2 ≠ 1 ⇒ x ≠ -1. Thus domain: x ≥ -2, x ≠ -1.

解答：f(g(x)) = 1/( √x+2 - 1 )。定义域需满足被开方数 x+2 ≥ 0，即 x ≥ -2；同时分母不为零：√x+2 - 1 ≠ 0 ⇒ √x+2 ≠ 1 ⇒ x+2 ≠ 1 ⇒ x ≠ -1。因此定义域为 x ≥ -2 且 x ≠ -1。

Answer: f(g(x)) = 1/(√x+2 - 1), domain: [-2, -1) ∪ (-1, ∞)

答案：f(g(x)) = 1/(√x+2 - 1)，定义域：[-2, -1) ∪ (-1, ∞)

4. Differentiation Using the Product Rule | 乘积法则求导

Question: Differentiate y = x² ln x with respect to x.

题目：求 y = x² ln x 的导数。

Solution: Let u = x², v' = ln x. Then u' = 2x, v' = 1/x. Using the product rule y' = u'v + uv', we get y' = 2x ln x + x²·(1/x) = 2x ln x + x. Factor out x: y' = x(2 ln x + 1).

解答：设 u = x², v = ln x，则 u' = 2x, v' = 1/x。根据乘积法则 y' = u'v + uv'，得 y' = 2x ln x + x²·(1/x) = 2x ln x + x。提取公因式 x：y' = x(2 ln x + 1)。

Answer: dy/dx = x(2 ln x + 1)

答案：dy/dx = x(2 ln x + 1)

5. Integration by Substitution | 换元积分法

Question: Evaluate the indefinite integral ∫ 2x e^x² dx.

题目：计算不定积分 ∫ 2x e^x² dx。

Solution: Use the substitution u = x², so du = 2x dx. The integral becomes ∫ e^u du = e^u + C. Replace u with x² to obtain the final result.

解答：令 u = x²，则 du = 2x dx。积分化为 ∫ e^u du = e^u + C。将 u 代回 x²，即得最终结果。

Answer: e^x² + C

答案：e^x² + C

6. Trigonometric Equations | 三角方程

Question: Solve 2 sin²θ − cosθ − 1 = 0 for 0 ≤ θ ≤ 2π.

题目：解方程 2 sin²θ − cosθ − 1 = 0，其中 0 ≤ θ ≤ 2π。

− cosθ − 1 = 0, and then to −2cos²θ − cosθ + 1 = 0. Multiply by −1: 2cos²θ + cosθ − 1 = 0

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