Review Set 6A: Mastering IB Non-Calculator Skills | IB数学非计算器题型精练集6A

📚 Review Set 6A: Mastering IB Non-Calculator Skills | IB数学非计算器题型精练集6A

IB Mathematics non‑calculator papers demand fluency in algebra, manipulation, and exact reasoning. This article walks through a typical Review Set 6A, unpacking ten key question types that appear in Paper 1 of both Analysis & Approaches and Applications & Interpretation. By working through these solutions without any technological aid, you strengthen the foundational skills that separate a 5 from a 7.

IB数学非计算器考试对代数运算、恒等变形和精确推理能力提出了很高要求。本文沿着一个典型的复习题集6A，逐一拆解出现在分析与方法和应用与解释的试卷一中的十种核心题型。在不使用任何技术工具的情况下完成这些练习，能帮助你筑牢从5分迈向7分所必备的基本功。

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1. Simplifying Rational Expressions | 化简有理式

Simplify (x² – 4) / (x² – x – 6) for x ≠ –2, 3.

化简 (x² – 4) / (x² – x – 6)，其中 x ≠ –2, 3。

Factor both numerator and denominator: x² – 4 = (x – 2)(x + 2); x² – x – 6 = (x – 3)(x + 2). Cancel the common factor (x + 2) to obtain (x – 2) / (x – 3).

分别对分子和分母进行因式分解：x² – 4 = (x – 2)(x + 2)；x² – x – 6 = (x – 3)(x + 2)。约去公因式 (x + 2) 得到 (x – 2) / (x – 3)。

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2. Solving Logarithmic Equations | 解对数方程

Solve log₂ x + log₂ (x – 2) = 3.

解方程 log₂ x + log₂ (x – 2) = 3。

Combine the logs using the product law: log₂ [x(x – 2)] = 3. Rewrite in exponential form: x(x – 2) = 2³ = 8. This gives x² – 2x – 8 = 0, factoring as (x – 4)(x + 2) = 0. The possible solutions are x = 4 and x = –2. Since the logarithm argument must be positive, x = –2 is rejected. Hence x = 4.

利用乘积法则合并对数：log₂ [x(x – 2)] = 3。化为指数形式：x(x – 2) = 2³ = 8。得到 x² – 2x – 8 = 0，因式分解为 (x – 4)(x + 2) = 0。可能的解为 x = 4 和 x = –2。由于对数的真数必须为正，舍去 x = –2。最终解为 x = 4。

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

Find f ′(x) for f(x) = sin(2x) · eˣ.

求 f(x) = sin(2x) · eˣ 的导数 f ′(x)。

Apply the product rule: f ′(x) = [derivative of sin(2x)]·eˣ + sin(2x)·[derivative of eˣ sing the chain rule, derivative of sin(2x) is 2 cos(2x). Therefore, f ′(x) = 2 cos(2x) · eˣ + sin(2x) · eˣ = eˣ (2 cos 2x + sin 2x).

使用乘积法则：f ′(x) = [sin(2x) 的导数]·eˣ + sin(2x)·[eˣ 的导数]。根据链式法则，sin(2x) 的导数为 2 cos(2x)。因此，f ′(x) = 2 cos(2x) · eˣ + sin(2x) · eˣ = eˣ (2 cos 2x + sin 2x)。

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4. Integration by Recognition (Reverse Chain Rule) | 识别倒导数积分法

Evaluate ∫ (3x² + 1) / (x³ + x) dx.

计算不定积分 ∫ (3x² + 1) / (x³ + x) dx。

actly the numerator: d/dx (x³ + x) = 3x² + 1. This fits the form ∫ [f ′(x) / f(x)] dx = ln |f(x)| + C. Hence the integral equals ln |x³ + x| + C.

注意到分母的导数恰好是分子：d/dx (x³ + x) = 3x² + 1。这符合 ∫ [f ′(x) / f(x)] dx = ln |f(x)| + C 的形式。因此积分结果为 ln |x³ + x| + C。

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5. Solving Trigonometric Equations | 解三角方程

Solve 2 sin x = √3 for 0 ≤ x ≤ 2π. Give answers in exact radians.

在 0 ≤ x ≤ 2π 内解 2 sin x = √3，答案以精确弧度值表示。

Divide by 2: sin x = √3 / 2. From the unit circle, sin x = √3 at x = π/3 and x = 2π/3 within one full revolution. Thus the solution set is {π/3, 2π/3}.

两边除以 2：sin x = √3 / 2。由单位圆可知，在一周内 sin x = √3 / 2 的解为 x = π/3 和 x = 2π/3。因此解集为 {π/3, 2π/3}。

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6. Binomial Expansion | 二项式展开

Find the coefficient of x³ in the expansion of (1 + 2x)⁵.

求 (1 + 2x)⁵ 的展开式中 x³ 项的系数。

The general term is C(5, r) · 1⁵⁻ʳ · (2x)ʳ = C(5, r) · 2ʳ · xʳ. For the x³ term, set r = 3: coefficient = C(5, 3) · 2³ = 10 · 8 = 80.

通项为 C(5, r) · 1⁵⁻ʳ · (2x)ʳ = C(5, r) · 2ʳ · xʳ。令 r = 3 得到 x³ 项：系数 = C(5, 3) · 2³ = 10 × 8 = 80。

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7. Laws of Exponents | 指数运算律

Simplify (x⁻² y³)² / (x y⁻¹)³, expressing the answer with positive exponents.

化简 (x⁻² y³)² / (x y⁻¹)³，结果用正指数表示。

; denominator = x³ y⁻³. Divide by subtracting exponents: x⁻⁴⁻³ y⁶⁻⁽⁻³⁾ = x⁻⁷ y⁹ = y⁹ / x⁷.

利用幂的乘方法则：分子 = x⁻⁴ y⁶；分母 = x³ y⁻³。通过指数相减进行除法：x⁻⁴⁻³ y⁶⁻⁽⁻³⁾ = x⁻⁷ y⁹ = y⁹ / x⁷。

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8. Completing the Square | 完成平方

Express x² – 6x + 4 in the form (x – p)² + q, and state the coordinates of the vertex of the corresponding parabola.

将 x² – 6x + 4 写成 (x – p)² + q 的形式，并指出对应抛物线的顶点坐标。

Half the coefficient of x is –3, square it to get 9. Rewrite: x² – 6x + 9 – 9 + 4 = (x – 3)² – 5. So p = 3, q = –5. The vertex of y = (x – 3)² – 5 is (3, –5).

x 项系数的一半为 –3，平方得 9。重写表达式：x² – 6x + 9 – 9 + 4 = (x – 3)² – 5。故 p = 3，q = –5。抛物线 y = (x – 3)² – 5 的顶点为 (3, –5)。

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9. Probability Without Calculator | 无计算器的概率问题

A bag contains 5 red and 3 blue marbles. Two marbles are drawn at random without replacement. Find the probability that both are red.

一个袋子里有 5 颗红球和 3 颗蓝球。不放回地随机抽取两颗。求两颗都是红球的概率。

f second red = 4/7. Multiply along the branch: P(both red) = (5/8) × (4/7) = 20/56 = 5/14.

第一颗是红球的概率 = 5/8。抽走一颗红球后，剩下 4 红 3 蓝，第二颗为红球的概率 = 4/7。沿分支相乘：P(两红) = (5/8) × (4/7) = 20/56 = 5/14。

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10. Coordinate Geometry – Distance and Midpoint | 坐标几何 – 距离与中点

Given points A(2, 5) and B(–4, 1), find the exact distance AB and the midpoint of AB.

已知点 A(2, 5) 和 B(–4, 1)，求线段 AB 的精确长度及其中点坐标。

Distance AB = √[(–4 – 2)² + (1 – 5)²] = √[(–6)² + (–4)²] = √36 + 16 = √52 = 2√13. Midpoint = ((2 + (–4))/2, (5 + 1)/2) = (–1, 3).

距离 AB = √[(–4 – 2)² + (1 – 5)²] = √[(–6)² + (–4)²] = √36 + 16 = √52 = 2√13。中点 = ((2 + (–4))/2, (5 + 1)/2) = (–1, 3)。

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