IB Math AA Review Set 1C | IB数学AA复习题集1C

📚 IB Math AA Review Set 1C | IB数学AA复习题集1C

This integrated review set helps you consolidate the core algebra, functions, sequences, and introductory calculus topics commonly assessed in IB Mathematics Analysis and Approaches. Work through each example carefully to strengthen your problem-solving skills and deepen conceptual understanding.

本综合复习题集帮助你巩固IB数学分析与方法的必修代数、函数、数列及微积分入门内容。请仔细完成每个例题，以提升解题能力并加深概念理解。

1. Solving Linear Equations and Inequalities | 解线性方程和不等式

Linear equations require isolating the variable by performing inverse operations on both sides. When solving inequalities, remember to reverse the inequality sign if you multiply or divide by a negative number.

线性方程需要通过对方程两边进行逆运算来分离变量。解不等式时，若两边乘以或除以负数，必须反转不等号方向。

Example: Solve 3(x − 2) + 4 = 2x + 5 and represent the solution for the inequality 2 − 3x ≤ 7.

例题：解方程 3(x − 2) + 4 = 2x + 5，并求出不等式 2 − 3x ≤ 7 的解。

Expand the left side: 3x − 6 + 4 = 3x − 2, so the equation becomes 3x − 2 = 2x + 5. Subtract 2x from both sides: x − 2 = 5, hence x = 7.

展开左边：3x − 6 + 4 = 3x − 2，方程化为 3x − 2 = 2x + 5。两边减去 2x 得 x − 2 = 5，因此 x = 7。

For the inequality, subtract 2: −3x ≤ 5. Divide by −3 and flip the sign: x ≥ −5/3.

对于不等式，两边减2得 −3x ≤ 5。除以 −3 并反转不等号：x ≥ −5/3。

2. Quadratic Functions and the Discriminant | 二次函数与判别式

The discriminant Δ = b² − 4ac of a quadratic ax² + bx + c tells us the nature of the roots. If Δ > 0, two distinct real roots exist; if Δ = 0, one repeated real root; if Δ < 0, no real roots.

二次式 ax² + bx + c 的判别式 Δ = b² − 4ac 表明了根的性质。若 Δ > 0，有两个相异实根；若 Δ = 0，有一个重根；若 Δ < 0，无实根。

Example: Find the values of k for which x² + (k − 2)x + 4 = 0 has exactly one real solution.

例题：求使方程 x² + (k − 2)x + 4 = 0 恰有一个实根的 k 值。

Set the discriminant to zero: (k − 2)² − 4(1)(4) = 0. That gives (k − 2)² − 16 = 0, so (k − 2)² = 16, thus k − 2 = ±4, leading to k = 6 or k = −2.

令判别式为零：(k − 2)² − 4(1)(4) = 0。即 (k − 2)² − 16 = 0，得 (k − 2)² = 16，所以 k − 2 = ±4，解得 k = 6 或 k = −2。

3. Composite and Inverse Functions | 复合函数与反函数

A composite function (g ∘ f)(x) = g(f(x)) is formed by applying f first, then g. An inverse function f⁻¹(x) reverses the effect of f, satisfying f(f⁻¹(x)) = x. The domain of f⁻¹ is the range of f.

复合函数 (g ∘ f)(x) = g(f(x)) 是先应用 f 再应用 g。反函数 f⁻¹(x) 逆转 f 的作用，满足 f(f⁻¹(x)) = x。f⁻¹ 的定义域是 f 的值域。

Example: Given f(x) = 2x + 1 and g(x) = x² − 3, find (f ∘ g)(2) and determine f⁻¹(x).

例题：已知 f(x) = 2x + 1 和 g(x) = x² − 3，求 (f ∘ g)(2) 并确定 f⁻¹(x)。

First, g(2) = 2² − 3 = 1, then f(1) = 2(1) + 1 = 3, so (f ∘ g)(2) = 3.

先求 g(2) = 4 − 3 = 1，再求 f(1) = 2×1 + 1 = 3，所以 (f ∘ g)(2) = 3。

To invert f, write y = 2x + 1, swap x and y: x = 2y + 1, solve for y: y = (x − 1)/2, hence f⁻¹(x) = (x − 1)/2.

为求反函数，令 y = 2x + 1，交换 x 与 y：x = 2y + 1，解出 y = (x − 1)/2，因此 f⁻¹(x) = (x − 1)/2。

4. Exponential and Logarithmic Equations | 指数与对数方程

Exponential equations of the form aˣ = b can be solved by taking logarithms: x = logₐ b. Remember that ln eˣ = x and e^(ln x) = x. Log properties such as logₐ(xy) = logₐ x + logₐ y are essential.

形如 aˣ = b 的指数方程可通过取对数求解：x = logₐ b。请记住 ln eˣ = x 及 e^(ln x) = x。对数性质如 logₐ(xy) = logₐ x + logₐ y 至关重要。

Example: Solve 2^(x+1) = 5, giving your answer in terms of natural logarithms.

例题：解方程 2^(x+1) = 5，答案用自然对数表示。

Take ln on both sides: ln(2^(x+1)) = ln 5. Using the power rule, (x+1) ln 2 = ln 5, so x+1 = ln 5 / ln 2, therefore x = (ln 5 / ln 2) − 1.

两边取自然对数：ln(2^(x+1)) = ln 5。利用幂法则，(x+1) ln 2 = ln 5，得 x+1 = ln 5 / ln 2，所以 x = (ln 5 / ln 2) − 1。

5. Arithmetic Sequences and Series | 等差数列与级数

An arithmetic sequence has a common difference d. The nth term is uₙ = u₁ + (n−1)d, and the sum of the first n terms is Sₙ = n/2 [2u₁ + (n−1)d] = n/2 (u₁ + uₙ).

等差数列有公差 d。第 n 项为 uₙ = u₁ + (n−1)d，前 n 项和为 Sₙ = n/2 [2u₁ + (n−1)d] = n/2 (u₁ + uₙ)。

Example: The 5th term of an arithmetic sequence is 20 and the 12th term is 48. Find u₁ and d, then compute S₁₂.

例题：一个等差数列的第5项为20，第12项为48。求 u₁ 和 d，并计算 S₁₂。

Set up: u₅ = u₁ + 4d = 20, u₁₂ = u₁ + 11d = 48. Subtracting gives 7d = 28, so d = 4. Then u₁ = 20 − 4×4 = 4. Thus S₁₂ = 12/2 [2×4 + 11×4] = 6(8 + 44) = 6×52 = 312.

列式：u₅ = u₁ + 4d = 20, u₁₂ = u₁ + 11d = 48。相减得 7d = 28，所以 d = 4。进而 u₁ = 20 − 16 = 4。于是 S₁₂ = 6(8 + 44) = 312。

6. Geometric Sequences and Series | 等比数列与级数

A geometric sequence has a common ratio r (r ≠ 0). The nth term is uₙ = u₁ rⁿ⁻¹. The sum of the first n terms is Sₙ = u₁(1 − rⁿ)/(1 − r) for r ≠ 1. If |r| < 1, an infinite sum exists: S∞ = u₁/(1 − r).

等比数列有公比 r (r ≠ 0)。第 n 项为 uₙ = u₁ rⁿ⁻¹。前 n 项和为 Sₙ = u₁(1 − rⁿ)/(1 − r)（r ≠ 1）。若 |r| < 1，无穷级数收敛到 S∞ = u₁/(1 − r)。

Example: The 3rd term is 12 and the 6th term is 96 in a geometric sequence. Determine u₁ and r, then find the sum of the first 8 terms.

例题：在一等比数列中，第3项为12，第6项为96。求 u₁ 和 r，并求前8项的和。

We have u₁ r² = 12 and u₁ r⁵ = 96. Divide the second by the first: r³ = 8, so r = 2. Then u₁ = 12 / 4 = 3. S₈ = 3(1 − 2⁸)/(1 − 2) = 3(1 − 256)/(−1) = 3×255 = 765.

有 u₁ r² = 12，u₁ r⁵ = 96。两式相除得 r³ = 8，即 r = 2。于是 u₁ = 3。S₈ = 3(1 − 256)/(−1) = 3×255 = 765。

7. Introduction to Differentiation | 微分入门

The derivative f'(x) gives the gradient of the curve y = f(x). Using the power rule, if f(x) = xⁿ, then f'(x) = nxⁿ⁻¹. Derivatives are used to find rates of change and stationary points.

导数 f'(x) 给出曲线 y = f(x) 的斜率。根据幂法则，若 f(x) = xⁿ，则 f'(x) = nxⁿ⁻¹。导数用于求变化率及驻点。

Example: Differentiate f(x) = 3x⁴ − 5x² + 2x − 7 and evaluate f'(1).

例题：求 f(x) = 3x⁴ − 5x² + 2x − 7 的导数，并计算 f'(1)。

f'(x) = 12x³ − 10x + 2. Substituting x = 1 gives f'(1) = 12 − 10 + 2 = 4.

求导得 f'(x) = 12x³ − 10x + 2。代入 x = 1 得 f'(1) = 12 − 10 + 2 = 4。

8. Tangents and Normals | 切线与法线

The equation of the tangent at x = a is y − f(a) = f'(a)(x − a). The normal is perpendicular, so its gradient is −1/f'(a) (provided f'(a) ≠ 0).

在 x = a 处的切线方程为 y − f(a) = f'(a)(x − a)。法线与切线垂直，因此其斜率为 −1/f'(a)（前提 f'(a) ≠ 0）。

Example: Find the tangent and normal to the curve y = x² − 4x + 1 at the point where x = 3.

例题：求曲线 y = x² − 4x + 1 在 x = 3 处的切线和法线方程。

f(3) = 9 − 12 + 1 = −2. f'(x) = 2x − 4, so f'(3) = 2. Tangent: y − (−2) = 2(x − 3) → y = 2x − 8. Normal gradient = −1/2, equation: y + 2 = −½(x − 3) → y = −½x − 0.5.

计算得 f(3) = −2，f'(x) = 2x − 4，故 f'(3) = 2。切线：y + 2 = 2(x − 3)，即 y = 2x − 8。法线斜率为 −1/2，方程：y + 2 = −½(x − 3)，化简得 y = −½x − 0.5。

9. Basic Integration | 基本积分

Integration is the reverse of differentiation. The indefinite integral ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C, for n ≠ −1. Definite integrals compute the area under a curve between two limits.

积分是微分的逆运算。不定积分 ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C，其中 n ≠ −1。定积分计算曲线下两个界限之间的面积。

Example: Evaluate ∫₁³ (2x² − x + 4) dx.

例题：计算定积分 ∫₁³ (2x² − x + 4) dx。

Integrate term‑by‑term: ∫ 2x² dx = ⅔x³, ∫ −x dx = −½x², ∫ 4 dx = 4x. So the antiderivative is (2/3)x³ − (1/2)x² + 4x. Evaluate from 1 to 3:

逐项积分：∫ 2x² dx = ⅔x³，∫ −x dx = −½x²，∫ 4 dx = 4x。原函数为 (2/3)x³ − (1/2)x² + 4x。代入上下限：

At 3: (2/3)×27 − (1/2)×9 + 12 = 18 − 4.5 + 12 = 25.5. At 1: (2/3) − (1/2) + 4 = ⅔ − ½ + 4 = (4/6 − 3/6) + 4 = 1/6 + 4 = 25/6 ≈ 4.1667. Subtract: 25.5 − 25/6 = (51/2) − (25/6) = (153/6 − 25/6) = 128/6 = 64/3.

在 x=3：(2/3)×27 − (1/2)×9 + 12 = 18 − 4.5 + 12 = 25.5。在 x=1：⅔ − ½ + 4 = 1/6 + 4 = 25/6。相减：25.5 − 25/6 = 51/2 − 25/6 = (153 − 25)/6 = 128/6 = 64/3。

10. Trigonometric Equations | 三角方程

Trigonometric equations often require using identities such as sin²θ + cos²θ = 1 and the general solutions for sine and cosine. Always consider the periodic nature and the given domain.

三角方程常需使用恒等式如 sin²θ + cos²θ = 1，以及正弦、余弦的通解公式。务必考虑函数的周期性和给定定义域。

Example: Solve 2 sin x cos x = cos x for 0 ≤ x ≤ 2π.

例题：在 0 ≤ x ≤ 2π 内解方程 2 sin x cos x = cos x。

Move all terms to one side: 2 sin x cos x − cos x = 0 → cos x (2 sin x − 1) = 0. So either cos x = 0 or sin x = 1/2. In the domain, cos x = 0 gives x = π/2, 3π/2. sin x = 1/2 gives x = π/6, 5π/6. Solutions: x = π/6, π/2, 5π/6, 3π/2.

移项：2 sin x cos x − cos x = 0 → cos x (2 sin x − 1) = 0。于是 cos x = 0 或 sin x = 1/2。在给定区间内，cos x = 0 的解为 x = π/2, 3π/2；sin x = 1/2 的解为 x = π/6, 5π/6。全部解：x = π/6, π/2, 5π/6, 3π/2。

Published by TutorHao | IB Mathematics Analysis and Approaches Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

IB Math AA Review Set 1C | IB数学AA复习题集1C