📚 Review set 13A - NON-CALCULATOR | 复习题集13A - 无计算器
This review set targets the core skills tested without a calculator in IB Mathematics. You will practise algebraic manipulation, exact trigonometric values, logarithmic identities, differentiation from first principles, and definite integrals that yield rational or logarithmic results. Every question is designed to sharpen your mental arithmetic and symbolic reasoning so that you can move quickly and accurately on Paper 1.
本复习题集针对 IB 数学中不允许使用计算器的核心技能进行训练。你将练习代数变形、三角精确值、对数恒等式、导数第一原理以及得出有理数或对数结果的定积分。每道题都旨在强化你的心算和符号推理能力，让你在试卷一的答题中既快又准。
1. Algebraic manipulation and surds | 代数运算与根式化简
Simplify expressions such as (√3 + 2)(√3 − 5) without a calculator. Always expand brackets carefully and collect like terms. When rationalising denominators like 1/(2 − √3), multiply numerator and denominator by the conjugate 2 + √3. The result must be expressed in the form a + b√c where a, b, c are integers.
不要使用计算器化简 (√3 + 2)(√3 − 5) 这类式子。务必仔细展开括号并合并同类项。在对 1/(2 − √3) 这样的分母有理化时，分子分母同乘以共轭根式 2 + √3。结果必须写成 a + b√c 的形式，其中 a, b, c 为整数。
Example: Rationalise 1/(2 − √3) → (2 + √3)/(4 − 3) = 2 + √3
例题：有理化 1/(2 − √3) → (2 + √3)/(4 − 3) = 2 + √3
2. Laws of exponents and logarithms | 指数与对数运算法则
Rewrite 82/3 as (∛8)² = 2² = 4. Solve exponential equations like 2x+1 gₐM + logₐN and the change-of-base formula. Always check that the argument stays positive.
将 82/3 写成 (∛8)² = 2² = 4。解指数方程 2x+1 = 32 时，先把两边写成同底数幂。处理对数时，熟练运用 logₐ(MN) = logₐM + logₐN 和换底公式。始终注意真数必须为正。
Solve 2x+1 = 32 → 2x+1 = 2⁵ ⇒ x+1 = 5 ⇒ x = 4
解方程 2x+1 = 32 → 2x+1 = 2⁵ ⇒ x+1 = 5 ⇒ x = 4
3. Quadratic functions and the discriminant | 二次函数与判别式
ated root, and Δ < 0 no real roots. Use these facts to sketch graphs without a calculator.
通过配方法求出 y = x² − 6x + 5 的顶点，配方得 (x − 3)² − 4，顶点为 (3, −4)。判别式 Δ = b² − 4ac 指示实根个数：Δ > 0 有两个不同实根，Δ = 0 有一个重根，Δ < 0 无实根。利用这些结论可以在无计算器的情况下画函数草图。
y = x² − 6x + 5 → (x − 3)² − 4, axis x=3, vertex (3, −4)
y = x² − 6x + 5 → (x − 3)² − 4，对称轴 x=3，顶点 (3, −4)
4. Polynomials and the factor theorem | 多项式与因式定理
(x − k) is a factor. Use long division or synthetic division to factorise cubic expressions. In a non‑ usually a small integer divisor of the constant term. Once factorised, sketch the curve by finding axes intercepts and sign changes.
对于多项式 P(x)，若 P(k) = 0，则 (x − k) 是一个因式。可利用长除法或综合除法对三次多项式进行因式分解。在无计算器试卷中，所试的根通常为常数项的小整数因子。因式分解后，通过求截距和符号变化来绘制函数草图。
Factorise x³ − 4x² + x + 6: test x=2 → P(2)=0 ⇒ (x−2)(x²−2x−3) = (x−2)(x−3)(x+1)
分解 x³ − 4x² + x + 6：试根 x=2 → P(2)=0 ⇒ (x−2)(x²−2x−3) = (x−2)(x−3)(x+1)
5. Binomial expansion | 二项式展开
Use Pascal's triangle or the formula nCr = n!/(r!(n−r)!) to expand (a + b)ⁿ. For a non‑ e general term, set the powers equal, and compute the coefficient using simple arithmetic.
借助杨辉三角或组合数公式 nCr = n!/(r!(n−r)!) 来展开 (a + b)ⁿ。在无计算器题组中，通常会要求求特定项，例如 (2 + x)⁵ 中 x³ 的系数。写出通项公式，令指数相等，再用简单算术计算出系数。
In (2 + x)⁵, the x³ term is ⁵C₃ · 2² · x³ = 10·4·x³ = 40x³, so coefficient is 40.
在 (2 + x)⁵ 中，x³ 项为 ⁵C₃ · 2² · x³ = 10·4·x³ = 40x³，系数为 40。
6. Exact trigonometric values and the unit circle | 三角函数精确值与单位圆
x = 1/2 for 0 ≤ x ≤ 2π, read off the two solutions x = π/6 and 5π/6 without a calculator.
熟记 0°、30°、45°、60°、90°（及其弧度等价）的正弦、余弦和正切值。利用单位圆判断 120° 或 7π/6 等角的符号和函数值。求解 sin x = 1/2 在 0 ≤ x ≤ 2π 内的解时，直接从单位圆读取 x = π/6 和 5π/6，无需计算器。
| Angle (deg) | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| cos | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| tan | 0 | 1/√3 | 1 | √3 | undef. |
7. Trigonometric identities and equations | 三角恒等式与方程求解
verything in terms of cosθ or sinθ, form a quadratic, and solve within the given interval.
熟练运用勾股恒等式 sin²θ + cos²θ = 1 以及倍角公式 sin 2θ = 2 sinθ cosθ、cos 2θ = cos²θ − sin²θ。在解方程 2 sin²θ − cosθ = 1 时，先将所有项转化为 cosθ 或 sinθ，化为二次方程，再在给定区间内求解。
− cosθ = 1 → 2(1 − cos²θ) − cosθ = 1 → −2cos²θ − cosθ + 1 = 0 → (2cosθ − 1)(cosθ + 1) = 0
− cosθ = 1 → 2(1 − cos²θ) − cosθ = 1 → −2cos²θ − cosθ + 1 = 0 → (2cosθ − 1)(cosθ + 1) = 0
8. Differentiation from first principles | 导数第一原理
The derivative f'(x) is limh→0 ² or f(x) = 3x + 1. Expand, simplify, cancel h, then let h = 0. This proof is a common non‑calculator question that tests algebraic confidence.
导数 f'(x) 定义为 limh→0 [f(x+h) − f(x)]/h。利用此定义可求出 f(x) = x² 或 f(x) = 3x + 1 等简单函数的导数。先展开并化简，约去 h，再令 h 趋于 0。这个证明是常见的无计算器考题，旨在考察代数运算的扎实程度。
f(x) = x² → [ (x+h)² − x² ]/h = (2xh + h²)/h = 2x + h → limh→0 = 2x
f(x) = x² → [ (x+h)² − x² ]/h = (2xh + h²)/h = 2x + h → limh→0 = 2x
9. Integration and area under a curve | 积分与曲线下方面积
s [x³ − x² + x]₀² = (8 − 4 + 2) − 0 = 6. Remember to check whether the curve crosses the x‑ so, split the integral to find total area. Stay precise with fractions and negative signs.
计算定积分 ∫₀² (3x² − 2x + 1) dx 时，先求原函数，再代入上下限。结果为 [x³ − x² + x]₀² = (8 − 4 + 2) − 0 = 6。注意检查曲线是否穿过 x 轴；若穿过，则需将积分分段以求总面积。务必正确处理分数和负号。
∫₀² (3x² − 2x + 1) dx = [x³ − x² + x]₀² = 6
∫₀² (3x² − 2x + 1) dx = [x³ − x² + x]₀² = 6
10. Arithmetic and geometric sequences | 等差数列与等比数列
Sₙ = n/2 (u₁ + uₙ). For a geometric sequence, uₙ = u₁ rⁿ⁻¹ and Sₙ = u₁(1 − rⁿ)/(1 − r) for |r| ≠ 1. In non‑ l or simplifies beautifully. Be comfortable with infinite geometric series converging when |r| < 1: S∞ = u₁/(1 − r).
等差数列的通项为 uₙ = u₁ + (n−1)d，前 n 项和为 Sₙ = n/2 (u₁ + uₙ)。等比数列的通项为 uₙ = u₁ rⁿ⁻¹，前 n 项和为 Sₙ = u₁(1 − rⁿ)/(1 − r)（|r| ≠ 1）。无计算器考题中，n 通常较小或能巧妙化简。当 |r| < 1 时，无穷等比级数收敛，其和为 S∞ = u₁/(1 − r)。
Given u₁=3, r=1/2, find S∞: S∞ = 3/(1 − 1/2) = 6
已知 u₁=3，r=1/2，求 S∞：S∞ = 3/(1 − 1/2) = 6
11. Vectors and scalar product | 向量与数量积
alculate the angle between vectors a = 2i − j + k and b = i + 3j − 2k using cosθ = (a·b)/(|a||b|). Compute the dot product a·b = 2×1 + (−1)×3 + 1×(−2) = 2 − 3 − 2 = −3. Find magnitudes |a| = √4+1+1=√6, |b| = √1+9+4=√14. Then cosθ = −3/(√6√14) = −3/√84, and you can rationalise the denominator.
计算向量 a = 2i − j + k 与 b = i + 3j − 2k 的夹角，用公式 cosθ = (a·b)/(|a||b|)。点积 a·b = 2×1 + (−1)×3 + 1×(−2) = 2 − 3 − 2 = −3。模长 |a| = √4+1+1 = √6，|b| = √1+9+4 = √14。于是 cosθ = −3/(√6√14) = −3/√84，你可以对分母有理化。
a·b = −3, |a|=√6, |b|=√14 → cosθ = −3/√84 = −√84/28 (rationalised)
a·b = −3，|a|=√6，|b|=√14 → cosθ = −3/√84 = −√84/28（有理化后）
12. Counting and probability without replacement | 计数原理与无放回概率
Use nCr for selections where order does not matter. For probability questions like “ red and 3 blue marbles, two are drawn without replacement, find probability both are red,” e as 5/8 × 4/7 = 20/56 = 5/14. Always simplify fractions and show tree diagrams if needed.
当顺序无关时，用组合数 nCr 计算选法种数。对于“袋中有 5 个红球和 3 个蓝球，不放回地抽取两个，求均为红球的概率”这类问题，计算过程为 5/8 × 4/7 = 20/56 = 5/14。务必约简分数，必要时画出树状图。
P(both red) = (5/8)×(4/7) = 5/14
P(两个红球) = (5/8)×(4/7) = 5/14
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