📚 Review Set 9A - Non-Calculator | 复习题集 9A - 非计算器
The IB Mathematics Review Set 9A is designed to be tackled without a calculator, reinforcing core algebraic manipulation, function analysis, and trigonometric understanding. Mastery of these non-calculator skills ensures a deep conceptual grasp that will benefit both Paper 1 and internal assessments. This article walks you through the key topics, providing worked examples and bilingual explanations.
IB 数学复习题集 9A 要求在无计算器条件下作答，旨在强化核心的代数运算、函数分析和三角理解。掌握这些非计算器技能能确保深刻的概念理解，对试卷一和内部评估都有帮助。本文将梳理关键主题，给出示例解答并附有中英双语解释。
1. Algebraic Simplification and Expansion | 代数化简与展开
When working without a calculator, it is essential to simplify rational expressions, factorise completely, and recognise standard expansions like the difference of two squares and perfect square trinomials.
在无计算器情况下，正确化简有理式、彻底因式分解以及识别平方差公式和完全平方三项式等标准展开式至关重要。
Simplify the expression (x² - 9) / (x² + 5x + 6).
化简表达式 (x² - 9) / (x² + 5x + 6)。
Factorise numerator as (x - 3)(x + 3) and denominator as (x + 2)(x + 3). Cancel the common factor (x + 3) to obtain (x - 3) / (x + 2), where x ≠ -3, -2.
分子因式分解为 (x - 3)(x + 3)，分母因式分解为 (x + 2)(x + 3)。约去公因子 (x + 3) 得到 (x - 3) / (x + 2)，其中 x ≠ -3, -2。
re: (x - 2)² = x² - 4x + 4. Subtract to get (2x² + 7x - 4) - (x² - 4x + 4) = x² + 11x - 8.
现在展开 (2x - 1)(x + 4) - (x - 2)²。先展开每一项：2x² + 8x - x - 4 = 2x² + 7x - 4。再展开平方：(x - 2)² = x² - 4x + 4。相减得 (2x² + 7x - 4) - (x² - 4x + 4) = x² + 11x - 8。
Remember that factorisation can reveal hidden simplifications, and expanding must be done with careful sign management - both are non-calculator fundamentals.
请记住，因式分解能揭示隐藏的化简机会，而展开时务必小心处理符号——这两者都是非计算器考试的基本功。
2. Solving Linear and Quadratic Equations | 解一次与二次方程
Non-calculator papers frequently test the ability to solve equations by factorising, completing the square, or using the quadratic formula and leaving surds in exact form.
非计算器试卷经常考察通过因式分解、配方法或使用求根公式来解方程，并保留根式精确形式的能力。
multiply to 12 and add to -7: -3 and -4. Hence (x - 3)(x - 4) = 0, giving x = 3 or x = 4.
用因式分解法解 x² - 7x + 12 = 0。寻找两个数，积为 12，和为 -7：即 -3 和 -4。因此 (x - 3)(x - 4) = 0，解得 x = 3 或 x = 4。
6x = -2. Add (6/2)² = 9 to both sides: x² - 6x + 9 = 7. Write as (x - 3)² = 7, so x = 3 ± √7.
当方程不能整齐因式分解时，使用配方法。对于 x² - 6x + 2 = 0，移项得 x² - 6x = -2。两边加上 (6/2)² = 9：x² - 6x + 9 = 7。写成 (x - 3)² = 7，于是 x = 3 ± √7。
: if Δ > 0, two distinct real roots; if Δ = 0, one repeated root; if Δ < 0, no real roots.
判别式 Δ = b² - 4ac 无需解方程即可判断根的性质：若 Δ > 0，有两个不同实根；Δ = 0，有一个重根；Δ < 0，无实根。
have two equal roots, set discriminant to zero: (-3)² - 4(2)(k) = 9 - 8k = 0, so k = 9/8.
要使方程 2x² - 3x + k = 0 有两个相等实根，令判别式为零：(-3)² - 4(2)(k) = 9 - 8k = 0，因此 k = 9/8。
3. Working with Surds and Rationalising Denominators | 根式运算与分母有理化
Manipulating surds accurately is a key non-calculator skill. You must simplify radicals, combine like surds, and rationalise denominators so that no surd appears in the denominator.
精确处理根式是重要的非计算器技能。你必须化简二次根式，合并同类根式，并使分母有理化，使分母中不出现根号。
Simplify √72. Since 72 = 36 × 2, we have √72 = √36 × √2 = 6√2. Always extract the largest square factor.
化简 √72。因为 72 = 36 × 2，所以 √72 = √36 × √2 = 6√2。始终提取最大的平方因子。
To rationalise the denominator of 5 / (√3 - 1), multiply numerator and denominator by the conjugate (√3 + 1): 5(√3 + 1) / ((√3)² - 1²) = 5(√3 + 1) / (3 - 1) = (5√3 + 5) / 2.
要将 5 / (√3 - 1) 的分母有理化，分子和分母同乘共轭根式 (√3 + 1)：5(√3 + 1) / ((√3)² - 1²) = 5(√3 + 1) / (3 - 1) = (5√3 + 5) / 2。
When adding or subtracting surds like √8 + √18, simplify each first: 2√2 + 3√2 = 5√2. Only like surds (same radicand) can be combined.
在进行根式加减法如 √8 + √18 时，先分别化简：2√2 + 3√2 = 5√2。只有同类根式（被开方数相同）才能合并。
4. Exponents and Logarithms – Laws and Simplification | 指数与对数 – 运算法则与化简
Fluency with the laws of exponents and logarithms allows you to simplify expressions and solve exponential equations without a calculator.
熟练掌握指数和对数运算法则，能让你在无计算器条件化简表达式并解指数方程。
| Exponent Law | 指数法则 |
|---|---|
| aᵐ × aⁿ = aᵐ⁺ⁿ | 同底数幂相乘，指数相加 |
| aᵐ ÷ aⁿ = aᵐ⁻ⁿ | 同底数幂相除，指数相减 |
| (aᵐ)ⁿ = aᵐⁿ | 幂的乘方，指数相乘 |
| a⁻ⁿ = 1/aⁿ | 负指数等于倒数 |
⁴)³/⁴ = 2³ = 8. The fraction exponent means take the fourth root then cube, or vice versa.
化简 16³/⁴。因为 16 = 2⁴，可改写为 (2⁴)³/⁴ = 2³ = 8。分数指数表示先开四次方再立方，或反过来。
(x/y) = logₐ x - logₐ y, and logₐ xⁿ = n logₐ x. Also know that logₐ a = 1 and logₐ 1 = 0.
对数的主要法则为：logₐ(xy) = logₐ x + logₐ y，logₐ(x/y) = logₐ x - logₐ y，以及 logₐ xⁿ = n logₐ x。还需知道 logₐ a = 1，logₐ 1 = 0。
) = 0. x = 4 is valid; x = -2 is extraneous because log of a negative number is undefined.
解方程 log₂(x) + log₂(x - 2) = 3。将对数合并：log₂[x(x - 2)] = 3，于是 x(x - 2) = 2³ = 8。得到 x² - 2x - 8 = 0，因式分解为 (x - 4)(x + 2) = 0。x = 4 有效；x = -2 为增根，因为负数的对数无定义。
5. Function Notation and Transformations | 函数符号与变换
You must be comfortable with function notation, composition, inverses, and transformations such as translations, stretches, and reflections applied to basic functions.
你必须熟悉函数符号、复合函数、反函数，以及平移、伸缩和反射等变换在基本函数上的应用。
1, find f(g(x)). Substitute g(x) into f: f(g(x)) = 2(x² + 1) - 3 = 2x² + 2 - 3 = 2x² - 1.
给定 f(x) = 2x - 3 和 g(x) = x² + 1，求 f(g(x))。将 g(x) 代入 f：f(g(x)) = 2(x² + 1) - 3 = 2x² + 2 - 3 = 2x² - 1。
y: x = (y + 4)/3, solve for y: y = 3x - 4. So f⁻¹(x) = 3x - 4. Verify that f(f⁻¹(x)) = x.
求 f(x) = (x + 4)/3 的反函数：设 y = (x + 4)/3，交换 x 与 y 得 x = (y + 4)/3，解出 y：y = 3x - 4。因此 f⁻¹(x) = 3x - 4。验证 f(f⁻¹(x)) = x。
left by 1 unit. y = -f(x) reflects it in the x-axis, and y = f(-x) reflects in the y-axis.
对于 y = f(x) + 2 的图像，将原图向上平移 2 个单位。y = f(x + 1) 向左平移 1 个单位。y = -f(x) 关于 x 轴反射，y = f(-x) 关于 y 轴反射。
6. Sketching Graphs Without Technology | 不借助技术绘制函数图像
Sketching requires you to identify intercepts, turning points, asymptotes, and basic shape from the equation, all by hand.
绘制函数图像需要你根据方程找出截距、极值点、渐近线以及基本形状，全部手工完成。
Sketch y = |x - 2| is 1, so vertex (2,1). The slopes are -1 to the left and +1 to the right, shifted up by 1.
绘制 y = |x - 2| + 1。这一 V 形绝对值图在内侧表达式为零的地方取得顶点，即 x = 2。y 坐标为 1，因此顶点 (2,1)。左侧斜率为 -1，右侧斜率为 +1，整体向上平移 1 个单位。
= 0 to get y = -1/2, set y = 0 to get x = -1. Sketch branches approaching the asymptotes.
对于有理函数 y = (x + 1)/(x - 2)，令分母为零得垂直渐近线 x = 2。水平渐近线为 y = 1（最高次项系数之比）。截距：令 x = 0 得 y = -1/2，令 y = 0 得 x = -1。画出两支分别趋近渐近线。
Always mark key points and label axes even in a rough sketch; this demonstrates understanding of the function's behaviour without numerical technology.
即使是草图，也要标注关键点并标记坐标轴；这能展现出你对函数行为的理解，而无需依赖数值工具。
7. Trigonometric Ratios and Exact Values | 三角比与精确值
In non-calculator trigonometry, you are expected to know the exact values of sine, cosine, and tangent for 0°, 30°, 45°, 60°, and 90°, and to use symmetry on the unit circle.
在非计算器的三角学中，你需要牢记 0°, 30°, 45°, 60° 和 90° 的正弦、余弦和正切的精确值，并会利用单位圆的对称性。
sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3
nt where sine is positive, and its reference angle is 30°. Hence sin 150° = sin 30° = 1/2.
利用单位圆求 sin 150°。150° 在第二象限，正弦为正，其参考角为 30°。因此 sin 150° = sin 30° = 1/2。
he third quadrant with reference 45°. Cosine is negative there, so cos 225° = -cos 45° = -√2/2. Tan 225° = tan 45° = 1 because tangent is positive in the third quadrant.
求 cos 225° 的精确值：注意到它在第三象限，参考角为 45°。余弦在第三象限为负，故 cos 225° = -cos 45° = -√2/2。tan 225° = tan 45° = 1，因为正切在第三象限为正。
8. Solving Trigonometric Equations | 解三角方程
ires finding all angles that satisfy the condition, using exact values and quadrant rules.
解如 2 sin x = 1（0 ≤ x ≤ 2π）这样的方程，需要找出所有满足条件的角，利用精确值和象限法则。
the first and second quadrants, the second solution is π - π/6 = 5π/6. Thus x = π/6, 5π/6.
由 2 sin x = 1 得 sin x = 1/2。主解为 x = π/6 (30°)。由于正弦在第一和第二象限为正，第二个解为 π - π/6 = 5π/6。因此 x = π/6, 5π/6。
o cos x = -1/2 or cos x = 1. cos x = 1 gives x = 0, 2π. cos x = -1/2 gives x = 2π/3, 4π/3.
对于 2 cos² x - cos x - 1 = 0，将其视为关于 cos x 的二次方程。因式分解得 (2 cos x +
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