Review set 15A – NON-CALCULATOR | 复习题集15A – 非计算器

📚 Review set 15A – NON-CALCULATOR | 复习题集15A – 非计算器

For IB Mathematics students, Review Set 15A brings together a wide range of non‑calculator skills that are essential for success in Paper 1. This article walks you through the key types of questions you will meet, from simplifying surds and solving exponential equations to differentiating from first principles and evaluating definite integrals—all without touching a calculator. Each section pairs a worked example with clear reasoning, helping you build the fluency and confidence needed for high‑stakes assessment.

对于 IB 数学学生而言，复习题集 15A 汇集了在试卷 1 中取得成功所必需的一系列非计算器技能。本文带你走过会遇到的典型题型，从根式化简、指数方程求解，到从第一原理求导、计算定积分——全程不碰计算器。每一部分都配对一个例题和清晰的推理，帮助你建立高水平评估所需的熟练度与信心。

1. Surds and Rationalisation | 根式与分母有理化

Many non‑calculator questions begin with manipulating surds. You are expected to simplify expressions such as √48 into the form a√b and to rationalise denominators like 1/(√3 − 1) by multiplying numerator and denominator by the conjugate.

许多非计算器题目从根式运算开始。你需要把 √48 化简为 a√b 的形式，并对类似 1/(√3 − 1) 的分母进行有理化，即分子分母同乘共轭根式。

Example: Write √48 + √27 in the form k√3.

示例：将 √48 + √27 写成 k√3 的形式。

√48 = √(16×3) = 4√3; √27 = √(9×3) = 3√3 → Sum = 7√3, k = 7

Similarly, for (5 + √2)/(√2 − 1), multiply top and bottom by (√2 + 1) to clear the radical from the denominator, then simplify the resulting integer‑plus‑surd form.

类似地，对于 (5 + √2)/(√2 − 1)，分子分母同乘 (√2 + 1) 消去分母中的根号，然后化简为整数加根式的形式。

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2. Exponent and Logarithm Laws | 指数与对数运算律

Questions in Set 15A often test simultaneous use of exponent rules and logarithm properties. You must convert between index form and log form fluently, and solve equations like 2^(x+1) = 8^(2x−3) by expressing both sides with the same base.

题集 15A 常考指数法则与对数性质的联合使用。你需要熟练地在指数式与对数式之间转换，并能通过化为同底求解如 2^(x+1) = 8^(2x−3) 的方程。

Example: Solve 2^(x+1) = 8^(2x−3).

示例：解方程 2^(x+1) = 8^(2x−3)。

8 = 2³, so RHS = (2³)^(2x−3) = 2^(6x−9) → x+1 = 6x−9 → 5x = 10 → x = 2

Log problems may ask you to evaluate log₂ 8 + log₃ (1/9) or to use the change‑of‑base formula. Always check the domain of the variable inside a logarithm — a common trap in non‑calculator papers.

对数题可能要求计算 log₂ 8 + log₃ (1/9) 或使用换底公式。永远要检查对数中变量的定义域——这是非计算器试卷中的常见陷阱。

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

Understanding the discriminant Δ = b² − 4ac is a recurring theme. You will be asked to find the number of real roots of a quadratic, determine values of k for which a quadratic has two equal roots, or prove that a quadratic is always positive by showing a > 0 and Δ < 0.

理解判别式 Δ = b² − 4ac 是一个反复出现的主题。你会被要求找出二次方程实根的个数、确定 k 使得二次方程有两个相等实根，或者通过证明 a > 0 且 Δ < 0 来论证一个二次式恒正。

Example: Find the range of m for which x² + (m+1)x + 4 = 0 has no real roots.

示例：求 m 的取值范围，使得 x² + (m+1)x + 4 = 0 无实根。

Δ = (m+1)² − 4·1·4 = m² + 2m + 1 − 16 = m² + 2m − 15 < 0 → (m+5)(m−3) < 0 → −5 < m < 3

In addition, you may need to complete the square to find the vertex of a parabola or to solve a quadratic by hand — classic no‑calculator territory.

此外，你可能需要通过配方求抛物线的顶点，或手解二次方程——典型的非计算器领域。

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4. Polynomial Division and Factor Theorem | 多项式除法与因式定理

Review Set 15A contains polynomial problems where you identify factors using the factor theorem and then perform polynomial long division (or synthetic division) to factorise a cubic completely. Being systematic with synthetic division saves time and reduces sign errors.

复习题集 15A 包含多项式问题，你需要先用因式定理识别因式，然后进行多项式长除（或综合除法）来完全分解一个三次多项式。有条不紊地使用综合除法能节省时间并减少符号错误。

Example: Given that (x−2) is a factor of P(x)=2x³ − 3x² − 3x + 2, factorise P(x) completely.

示例：已知 (x−2) 是 P(x)=2x³ − 3x² − 3x + 2 的因式，将 P(x) 完全分解。

Divide by (x−2): 2x³−3x²−3x+2 = (x−2)(2x² + x −1) = (x−2)(2x−1)(x+1)

When a polynomial is not fully factorised over the rationals, you may need to write it in the form (x − a)Q(x) + R and interpret the remainder theorem to find remainders without full division.

当多项式在有理数范围内不能完全分解时，你可能需要将其写成 (x − a)Q(x) + R 的形式，并运用余式定理求余数，而无需完整的除法过程。

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5. Trigonometric Identities and Equations | 三角恒等式与方程

This section tests your ability to solve trigonometric equations within a given interval, using exact values of sine, cosine and tangent for standard angles (30°, 45°, 60° etc. in degrees or π/6, π/4, π/3 in radians). You must also select the correct quadrants based on the sign of the ratio.

本节考察你在给定区间内求解三角方程的能力，要求使用标准角（30°、45°、60° 等，或弧度 π/6、π/4、π/3）的正弦、余弦和正切的精确值。你还必须根据三角比的正负选择正确的象限。

Example: Solve 2 sin² θ − sin θ − 1 = 0 for 0 ≤ θ ≤ 2π.

示例：在 0 ≤ θ ≤ 2π 内解方程 2 sin² θ − sin θ − 1 = 0。

Factorise: (2 sin θ + 1)(sin θ − 1) = 0 → sin θ = −½ or sin θ = 1 → θ = 7π/6, 11π/6, π/2

Proving simple identities, such as (sin θ + cos θ)² = 1 + sin 2θ, often appears as a warm‑up before solving. Familiarity with Pythagorean identities and double‑angle formulas is assumed.

证明简单恒等式，如 (sin θ + cos θ)² = 1 + sin 2θ，常作为解题前的热身。试卷默认你熟悉勾股恒等式和倍角公式。

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6. Radian Measure, Arc Length and Sector Area | 弧度制、弧长与扇形面积

IB non‑calculator papers regularly use radian measure. You need to interchange degrees and radians exactly, and apply the formulas l = rθ and A = ½ r²θ. A typical question provides the perimeter of a sector and asks for its area, requiring you to form a small system of equations.

IB 非计算器试卷经常使用弧度制。你需要精确换算角度与弧度，并应用公式 l = rθ 和 A = ½ r²θ。一个典型题目给出扇形的周长，要求计算面积，这需要你建立一个小型方程组。

Example: A sector has perimeter 20 cm and radius 6 cm. Find the area of the sector.

示例：一个扇形周长为 20 cm，半径为 6 cm。求扇形的面积。

Perimeter = 2r + rθ = 12 + 6θ = 20 → θ = 4/3 rad → Area = ½ × 6² × (4/3) = 24 cm²

Watch for angles given in terms of π — for example, a reflex angle of 4π/3 creates a major sector. The formula for area works equally well for any θ in radians.

注意以 π 表示的角——例如，优角 4π/3 构成一个优扇形。面积公式对任何以弧度表示的 θ 都同样适用。

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

Set 15A includes function notation drilled without a calculator: evaluating f(g(x)), finding the domain and range of a composite, and determining the inverse function f⁻¹(x) algebraically. Swapping x and y and then rearranging is the standard approach.

题集 15A 包含不依赖计算器的函数记号训练：计算 f(g(x))、求复合函数的定义域与值域，以及用代数方法求反函数 f⁻¹(x)。标准步骤是先交换 x 和 y，然后重新整理。

Example: Given f(x) = √(x − 2) for x ≥ 2, find f⁻¹(x) and state its domain.

示例：已知 f(x) = √(x − 2), x ≥ 2，求 f⁻¹(x) 并说明其定义域。

y = √(x−2) → x = √(y−2) → x² = y−2 → y = x²+2, with x ≥ 0 → f⁻¹(x) = x²+2, domain x ≥ 0

Be mindful of the condition that the domain of f⁻¹ is the range of f. Drawing a quick sketch of the original function helps catch domain restrictions, even without a calculator.

注意反函数的定义域正是原函数的值域。即使没有计算器，快速勾勒原函数的草图也有助于捕捉定义域限制。

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8. Limits and Differentiation from First Principles | 极限与导数第一原理

Non‑calculator differentiation from first principles is a hallmark of IB Mathematics Analysis and Approaches. You use the limit definition f'(x) = lim(h→0) [f(x+h) − f(x)]/h, expanding and simplifying algebraically so that the denominator cancels before taking the limit.

从第一原理求导是非计算器 IB 数学分析与方法的标志。你需要使用极限定义 f'(x) = lim(h→0) [f(x+h) − f(x)]/h，通过代数展开和化简，使得分母在取极限之前被消去。

Example: Differentiate f(x) = x³ from first principles.

示例：从第一原理对 f(x) = x³ 求导。

f(x+h) = (x+h)³ = x³ + 3x²h + 3xh² + h³ → f(x+h)−f(x) = 3x²h + 3xh² + h³ → divide by h: 3x² + 3xh + h² → limit as h→0 is 3x²

This method also appears for simple rational functions like f(x) = 1/x. Mastery of manipulating the difference quotient without arithmetic errors is essential for Paper 1.

此方法也会出现在如 f(x) = 1/x 的简单有理函数中。熟练操作差商且不出算术错误，对试卷 1 至关重要。

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9. Tangents, Normals and Stationary Points | 切线、法线与驻点

Once you have the derivative, Set 15A moves to applications: finding the equation of a tangent or normal at a point, and locating stationary points to classify maxima and minima. You should be quick at evaluating the derivative and substituting coordinates.

一旦求出导数，题集 15A 便进入应用：求给定点处的切线或法线方程，以及定位驻点以区分极大值与极小值。你需要能快速求导数值并代入坐标。

Example: Find the equation of the tangent to y = x² − 3x + 5 at x = 2.

示例：求 y = x² − 3x + 5 在 x = 2 处的切线方程。

dy/dx = 2x − 3 → m = 2(2) − 3 = 1 → y(2) = 4 − 6 + 5 = 3 → tangent: y − 3 = 1(x − 2) → y = x + 1

For normals, use the negative reciprocal of the gradient. When finding stationary points, set f'(x) = 0, solve, and then use the sign of f'(x) around the point or the second derivative to determine the nature.

对于法线，使用斜率的负倒数。找驻点时，令 f'(x) = 0，解方程，然后利用该点附近 f'(x) 的符号或二阶导数判定极值类型。

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10. Integration Basics — Indefinite and Definite Integrals | 积分基础——不定积分与定积分

Basic integration appears without a calculator, requiring you to reverse power rule, handle fractions such as 1/xⁿ written as x⁻ⁿ, and apply the constant of integration. A typical question asks you to find f(x) given f'(x) and a point on the curve.

基础积分也在非计算器环境下出现，你需要反向应用幂法则，处理如 1/xⁿ 写为 x⁻ⁿ 的分式，并加上积分常数。一个典型问题是已知 f'(x) 和曲线上一点，求 f(x)。

Example: Given f'(x) = 4x³ − 2x + 1 and f(1) = 3, find f(x).

示例：已知 f'(x) = 4x³ − 2x + 1 且 f(1) = 3，求 f(x)。

f(x) = ∫(4x³ − 2x + 1)dx = x⁴ − x² + x + C → 1⁴ − 1² + 1 + C = 3 → 1 + C = 3 → C = 2 → f(x) = x⁴ − x² + x + 2

Definite integrals are equally common. You must substitute upper and lower limits carefully, and remember that the area between a curve and the x‑axis can require splitting the interval when the function dips below the axis.

定积分同样常见。你需要仔细代入上下限，并记住曲线与 x 轴之间的面积在函数落入轴下方时，可能需要分割积分区间。

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11. Probability Without a Calculator | 概率非计算器题型

Review Set 15A often incorporates probability questions where arithmetic stays manageable. You might encounter tree diagrams with fractions, conditional probability formulas, and Venn diagram problems. The key is to keep probabilities in simplified fractional form and apply the formulas P(A|B) = P(A∩B)/P(B) without rounding.

复习题集 15A 通常包含算术可控的概率题。你可能会遇到带有分数的树状图、条件概率公式以及维恩图问题。关键是将概率保持为最简分数形式，并在不四舍五入的情况下应用公式 P(A|B) = P(A∩B)/P(B)。

Example: A bag contains 5 red and 3 blue marbles. Two marbles are drawn without replacement. Find the probability that the second is blue given that the first was red.

示例：一个袋中有 5 颗红弹珠和 3 颗蓝弹珠。先后取出两颗且不放回。已知第一颗是红色，求第二颗是蓝色的概率。

P(first red) = 5/8; after removal, P(second blue | first red) = 3/7

Also tested are expectation and simple discrete distributions. The arithmetic revolves around small integers and fractions, making it perfect for a non‑calculator assessment.

同样会考察期望和简单的离散分布。运算围绕小整数和分数展开，因此非常适合非计算器评估。

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12. Vectors — Dot Product and Angle Between Vectors | 向量——点积与向量夹角

Vector questions in Set 15A are designed for exact computation. You will compute the dot product a·b, find the magnitude |a|, and use cos θ = (a·b)/(|a||b|) to determine the angle between two vectors. All working should yield a neat cosine value, often linked to a standard angle.

题集 15A 中的向量题专为精确计算而设计。你将计算点积 a·b、求模长 |a|，并利用 cos θ = (a·b)/(|a||b|) 确定两向量的夹角。所有运算应得出一个整齐的余弦值，常与标准角对应。

Example: Let a = 2i − j + 3k and b = i + 2j − 2k. Find the angle between a and b.

示例：设 a = 2i − j + 3k, b = i + 2j − 2k。求 a 与 b 的夹角。

a·b = 2(1) + (−1)(2) + 3(−2) = 2 − 2 − 6 = −6; |a| = √(4+1+9) = √14; |b| = √(1+4+4) = √9 = 3 → cos θ = −6/(3√14) = −2/√14 → θ = arccos(−2/√14)

Often the answer is left in exact form or expressed as an angle whose cosine is a recognisable value, such as 1/2 or −√2/2. Beware of the sign to distinguish acute from obtuse angles.

通常答案会保留精确形式，或表示为一个其余弦值为可识别值的角，例如 1/2 或 −√2/2。注意正负号以区分锐角与钝角。

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