Review Set 11A - Non-Calculator | 复习题集 11A - 非计算器

📚 Review Set 11A - Non-Calculator | 复习题集 11A - 非计算器

This review set is designed to help IB Mathematics students practise key non-calculator skills. It covers simplifying algebraic expressions, solving equations, working with functions, evaluating exact trigonometric values, solving logarithmic and exponential equations, differentiation, integration, and vector operations. Each problem is broken down into clear steps, with detailed explanations in both English and Chinese to deepen your understanding and boost your confidence before the exam.

本复习题集旨在帮助 IB 数学学生练习关键的非计算器技能。内容涵盖代数表达式化简、解方程、函数处理、精确三角函数值计算、对数和指数方程求解、微分、积分以及向量运算。每个问题都分解为清晰的步骤，并配有详细的中英文解释，以加深理解，增强考前信心。

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1. Simplifying Algebraic Expressions | 化简代数表达式

Problem: Simplify (2x³y⁻²)³ / (4x⁻¹y²)² without using a calculator.

题目：不用计算器化简 (2x³y⁻²)³ / (4x⁻¹y²)²。

Step 1: Apply the power to each factor in the numerator: (2x³y⁻²)³ = 2³ · (x³)³ · (y⁻²)³ = 8x⁹y⁻⁶.

步骤1：对分子中每个因式应用指数：(2x³y⁻²)³ = 2³ · (x³)³ · (y⁻²)³ = 8x⁹y⁻⁶。

Step 2: Do the same for the denominator: (4x⁻¹y²)² = 4² · (x⁻¹)² · (y²)² = 16x⁻²y⁴.

步骤2：对分母同样操作：(4x⁻¹y²)² = 4² · (x⁻¹)² · (y²)² = 16x⁻²y⁴。

Step 3: Simplify the fraction: 8/16 = 1/2; x⁹/x⁻² = x⁹⁺² = x¹¹; y⁻⁶/y⁴ = y⁻⁶⁻⁴ = y⁻¹⁰.

步骤3：化简分式：8/16 = 1/2；x⁹/x⁻² = x⁹⁺² = x¹¹；y⁻⁶/y⁴ = y⁻⁶⁻⁴ = y⁻¹⁰。

Result: (1/2)x¹¹y⁻¹⁰, which is often written as x¹¹/(2y¹⁰).

结果：(1/2)x¹¹y⁻¹⁰，通常写作 x¹¹/(2y¹⁰)。

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2. Solving a Quadratic Equation by Factoring | 因式分解解二次方程

Problem: Solve 3x² – 5x – 2 = 0 without using the quadratic formula or a calculator.

题目：不用求根公式或计算器，解方程 3x² – 5x – 2 = 0。

Step 1: add to -5. The pair -6 and 1 works: -6 × 1 = -6, -6 + 1 = -5. Rewrite the middle term: 3x² – 6x + x – 2 = 0.

步骤1：寻找两个数，乘积为 (3)(-2) = -6，和为 -5。 -6 和 1 满足条件：-6 × 1 = -6，-6 + 1 = -5。将中间项改写：3x² – 6x + x – 2 = 0。

Step 2: Factor by grouping: 3x(x – 2) + 1(x – 2) = 0 → (3x + 1)(x – 2) = 0.

步骤2：分组分解：3x(x – 2) + 1(x – 2) = 0 → (3x + 1)(x – 2) = 0。

Step 3: Set each factor equal to zero: 3x + 1 = 0 → x = -1/3; x – 2 = 0 → x = 2.

步骤3：令每个因式为零：3x + 1 = 0 → x = -1/3；x – 2 = 0 → x = 2。

Answer: x = -1/3 or x = 2.

答案：x = -1/3 或 x = 2。

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3. Exponential Equation (Hidden Quadratic) | 指数方程（隐藏二次型）

Problem: Solve e^2x – 5e^x + 6 = 0, giving exact answers.

题目：解方程 e^2x – 5e^x + 6 = 0，给出精确解。

Step 1: Recognise that e^2x = (e^x)². Let t = e^x. The equation becomes t² – 5t + 6 = 0.

步骤1：注意到 e^2x = (e^x)²。设 t = e^x，方程化为 t² – 5t + 6 = 0。

Step 2: Factor the quadratic: (t – 2)(t – 3) = 0, so t = 2 or t = 3.

步骤2：因式分解：(t – 2)(t – 3) = 0，得 t = 2 或 t = 3。

Step 3: Substitute back: e^x = 2 → x = ln 2; e^x = 3 → x = ln 3.

步骤3：代回原变量：e^x = 2 → x = ln 2；e^x = 3 → x = ln 3。

Answer: x = ln 2 or x = ln 3.

答案：x = ln 2 或 x = ln 3。

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4. Inverse Function and Its Domain | 反函数及其定义域

Problem: For f(x) = 2x + 3, find f⁻¹(x) and state its domain.

题目：已知 f(x) = 2x + 3，求 f⁻¹(x) 并写出其定义域。

Step 1: Write y = 2x + 3 and swap x and y: x = 2y + 3.

步骤1：写出 y = 2x + 3，交换 x 和 y：x = 2y + 3。

Step 2: Solve for y: 2y = x – 3 → y = (x – 3)/2.

步骤2：解出 y：2y = x – 3 → y = (x – 3)/2。

Step 3: Therefore f⁻¹(x) = (x – 3)/2. Since this is a linear function, it is defined for all real numbers. Domain: ℝ (all real numbers).

步骤3：因此 f⁻¹(x) = (x – 3)/2。因为这是一个一次函数，对所有实数都有定义。定义域：ℝ（所有实数）。

Answer: f⁻¹(x) = (x – 3)/2, domain: x ∈ ℝ.

答案：f⁻¹(x) = (x – 3)/2，定义域为一切实数。

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5. Exact Trigonometric Values | 精确三角函数值

Problem: Without using a calculator, find the exact value of sin(7π/6) + cos(5π/3).

题目：不用计算器，求 sin(7π/6) + cos(5π/3) 的精确值。

Step 1: le. 7π/6 = π + π/6, which is in the third quadrant. Sine is negative there, so sin(7π/6) = –sin(π/6) = –1/2.

步骤1：在单位圆上定位 7π/6。7π/6 = π + π/6，位于第三象限。该象限正弦为负，因此 sin(7π/6) = –sin(π/6) = –1/2。

Step 2: Now 5π/3 = 2π – which is in the fourth quadrant. Cosine is positive there, so cos(5π/3) = cos(π/3) = 1/2.

步骤2：而 5π/3 = 2π – π/3，位于第四象限。该象限余弦为正，因此 cos(5π/3) = cos(π/3) = 1/2。

Step 3: Add the two values: –1/2 + 1/2 = 0.

步骤3：两值相加：–1/2 + 1/2 = 0。

Answer: 0.

答案：0。

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6. Trigonometric Equation (Quadratic in Sine) | 三角方程（正弦的二次型）

Problem: Solve 2 sin² x – sin x – 1 = 0 for 0 ≤ x ≤ 2π without a calculator.

题目：在 0 ≤ x ≤ 2π 内，不用计算器解方程 2 sin² x – sin x – 1 = 0。

Step 1: Treat as a quadratic in sin x. Factor: (2 sin x + 1)(sin x – 1) = 0.

步骤1：视作关于 sin x 的二次方程。因式分解：(2 sin x + 1)(sin x – 1) = 0。

Step 2: Set each factor to zero: sin x = –1/2 or sin x = 1.

步骤2：令每个因式为零：sin x = –1/2 或 sin x = 1。

Step 3: For sin x = 1, the only solution in [0, 2π] is x =

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