📚 IB Maths Calculation Practice Drills: 12 Core Topics | IB 数学计算专项训练：12 大核心题型
Fluent calculation is the bedrock of high achievement in IB Mathematics, whether you are taking Analysis and Approaches (AA) or Applications and Interpretation (AI), and whether your school enters you for the IB Diploma or for a UK-style examination series. These targeted practice drills cover the most common computational question types, from algebra and functions through to calculus, vectors and statistics. Each step is presented in English first and then in Chinese, so equivalent, so you can strengthen both your mathematical reasoning and your bilingual command of the subject at the same time.
流利的计算是 IB 数学高分的基石，无论你选修分析与方法 (AA) 还是应用与解释 (AI)，也无论你参加的是 IB 文凭考试还是英式考试序列。本次专项训练涵盖了从代数、函数到微积分、向量、统计最常见的计算题型。每一步都先用英文给出，再配上中文对应内容，让你在强化数学推理的同时同步提升双语学科能力。
Work through each drill with a pen, paper and your graphic display calculator (GDC). Write every line down: examiners award method marks for correct intermediate steps even when the final answer slips, and the habit of writing full working is exactly what converts a 5 into a 7.
请用纸笔配合图形计算器 (GDC) 逐题演练。每一步都写下来：即使最后答案出错，考官也会给正确的中间步骤方法分；而完整书写过程的习惯，正是把 5 分变成 7 分的关键。
1. Algebraic Simplification and Equations | 代数化简与解方程
Simplify (3x² – 2x + 5) – (x² + 4x – 1). A minus sign in front of a bracket changes the sign of every term inside it.
化简 (3x² – 2x + 5) – (x² + 4x – 1)。括号前的负号会使括号内每一项都变号。
Distribute the minus sign: 3x² – 2x + 5 – x² – 4x + 1.
分配负号：3x² – 2x + 5 – x² – 4x + 1。
Group like terms: (3x² – x²) + (–2x – 4x) + (5 + 1).
合并同类项：(3x² – x²) + (–2x – 4x) + (5 + 1)。
Result: 2x² – 6x + 6.
结果：2x² – 6x + 6。
Now solve 2x² – 5x – 3 = 0 by factorising.
现在用因式分解法解 2x² – 5x – 3 = 0。
Multiply a × c = 2 × (–3) = –6. Find two numbers whose product is –6 and whose sum is –5: they are –6 and +1.
计算 a × c = 2 × (–3) = –6。找出乘积为 –6 且和为 –5 的两个数：–6 与 +1。
Rewrite: 2x² – 6x + x – 3 = 0, then group: 2x(x – 3) + 1(x – 3) = 0, giving (2x + 1)(x – 3) = 0.
改写：2x² – 6x + x – 3 = 0，然后分组：2x(x – 3) + 1(x – 3) = 0，得到 (2x + 1)(x – 3) = 0。
(2x + 1)(x – 3) = 0 → x = –1/2 or x = 3
Solutions: x = –1/2 and x = 3. Always substitute both roots back to confirm the sign of each term.
解得：x = –1/2 与 x = 3。务必把两个根代回原式，确认每一项的符号无误。
Two habits protect you here: check the discriminant b² – 4ac before factorising, and remember that a quadratic with no real roots cannot be factorised over the reals.
这里有两个好习惯：因式分解前先检验判别式 b² – 4ac；并记住没有实根的二次式在实数范围内无法分解。
2. Functions and Inverse Functions | 函数与反函数
Given f(x) = √3x – 2 for x ≥ 2/3, find its inverse f⁻¹(x) and state its domain.
已知 f(x) = √3x – 2，x ≥ 2/3，求反函数 f⁻¹(x) 并写出其定义域。
Write y = √3x – 2.
设 y = √3x – 2。
Swap x and y: x = √3y – 2.
交换 x 与 y：x = √3y – 2。
Square both sides: x² = 3y – 2. Rearrange: 3y = x² + 2, so y = (x² + 2)/3.
两边平方：x² = 3y – 2。整理得 3y = x² + 2，所以 y = (x² + 2)/3。
f⁻¹(x) = (x² + 2)/3, domain x ≥ 0
is never negative and f(2/3) = 0, the range of f is y ≥ 0, so the domain of f⁻¹ is x ≥ 0.
f⁻¹ 的定义域等于 f 的值域。由于平方根非负且 f(2/3) = 0，f 的值域为 y ≥ 0，因此 f⁻¹ 的定义域是 x ≥ 0。
Check by composition: f⁻¹(f(x)) = ((3x – 2) + 2)/3 = 3x/3 = x. The check confirms the algebra.
用复合函数检验：f⁻¹(f(x)) = ((3x – 2) + 2)/3 = 3x/3 = x。检验确认代数运算正确。
Three markers are almost always available in such a question: the correct swap, the correct rearrangement, and an explicit domain statement. Do not lose the last one.
这类题目通常有三个得分点：正确的变量交换、正确的变形整理，以及明确定义域。最后一点千万不要丢。
3. Indices, Surds and Rationalising | 指数、根式与有理化
Simplify (272/3 × 16–3/4) ÷ 81/3.
化简 (272/3 × 16–3/4) ÷ 81/3。
Evaluate each power separately. 272/3 = (271/3)² = 3² = 9.
分别计算各个幂。272/3 = (271/3)² = 3² = 9。
16–3/4 = 1 ÷ 163/4 = 1 ÷ 2³ = 1/8.
16–3/4 = 1 ÷ 163/4 = 1 ÷ 2³ = 1/8。
81/3 = 2. Therefore (9 × 1/8) ÷ 2 = 9/16.
81/3 = 2。因此 (9 × 1/8) ÷ 2 = 9/16。
Now simplify the surd expression √50 + √18. Write each surd with the largest possible square factor: √50 = √25 × 2 = 5√2 and √18 = √9 × 2 = 3√2.
再化简根式 √50 + √18。把每个根式写成含最大平方因子的形式：√50 = √25 × 2 = 5√2，√18 = √9 × 2 = 3√2。
So √50 + √18 = 5√2 + 3√2 = 8√2.
所以 √50 + √18 = 5√2 + 3√2 = 8√2。
Rationalise the denominator of 6/(√5 + 2) by multiplying numerator and denominator by the conjugate √5 – 2.
将 6/(√5 + 2) 分母有理化：分子分母同乘共轭式 √5 – 2。
6/(√5 + 2) × (√5 – 2)/(√5 – 2) = 6(√5 – 2)/(5 – 4) = 6√5 – 12
Key index laws to memorise: aᵐ × aⁿ = am + n, aᵐ ÷ aⁿ = am – n, (aᵐ)ⁿ = amn, a⁰ = 1, and a–n = 1/aⁿ.
必须记住的指数法则：aᵐ × aⁿ = am + n，aᵐ ÷ aⁿ = am – n，(aᵐ)ⁿ = amn，a⁰ = 1，以及 a–n = 1/aⁿ。
4. Logarithms and Exponential Equations | 对数与指数方程
Solve 32x + 1 = 81.
解 32x + 1 = 81。
Write 81 as a power of 3: 81 = 3⁴. Since the bases match, equate the indices: 2x + 1 = 4.
把 81 写成 3 的幂：81 = 3⁴。底数相同，故令指数相等：2x + 1 = 4。
Therefore 2x = 3 and x = 3/2 = 1.5.
因此 2x = 3，x = 3/2 = 1.5。
Next solve log₂(x + 5) + log₂(x – 1) = 4.
再解 log₂(x + 5) + log₂(x – 1) = 4。
Combine the logarithms: log₂[(x + 5)(x – 1)] = 4, so (x + 5)(x – 1) = 2⁴ = 16.
合并对数：log₂[(x + 5)(x – 1)] = 4，所以 (x + 5)(x – 1) = 2⁴ = 16。
Expand and solve: x² + 4x – 5 = 16 → x² + 4x – 21 = 0 → (x + 7)(x – 3) = 0.
展开求解：x² + 4x – 5 = 16 → x² + 4x – 21 = 0 → (x + 7)(x – 3) = 0。
Reject x = –7 because x – 1 must be positive, so the only solution is x = 3.
舍弃 x = –7，因为必须满足 x – 1 > 0，故唯一解为 x = 3。
When the base cannot be matched, take logs of both sides: 5ˣ = 40 gives x = ln 40 ÷ ln 5 ≈ 2.292 (3 s.f.).
当底数无法统一时，两边取对数：5ˣ = 40 得 x = ln 40 ÷ ln 5 ≈ 2.292（三位有效数字）。
Useful log rules: log(ab) = log a + log b, log(a/b) = log a – log b, log(aⁿ) = n log a, and the change-of-base formula log_a b = ln b ÷ ln a.
常用对数法则：log(ab) = log a + log b，log(a/b) = log a – log b，log(aⁿ) = n log a，以及换底公式 log_a b = ln b ÷ ln a。
5. Quadratics: Completing the Square and Inequalities | 二次式：配方与不等式
Complete the square for 2x² – 12x + 7 and state the coordinates of the vertex.
将 2x² – 12x + 7 配方，并写出顶点坐标。
Factor out the 2 from the first two terms: 2(x² – 6x) + 7.
从前两项提出 2：2(x² – 6x) + 7。
Complete the square inside: x² – 6x = (x – 3)² – 9.
括号内配方：x² – 6x = (x – 3)² – 9。
2x² – 12x + 7 = 2(x – 3)² – 11, vertex (3, –11)
he discriminant to find the value of k for which x² + kx + 9 = 0 has two equal real roots.
用判别式求使 x² + kx + 9 = 0 有两个相等实根的 k 值。
Equal roots mean Δ = b² – 4ac = 0, so k² – 4 × 1 × 9 = 0 → k² = 36 → k = ±6.
两根相等意味着 Δ = b² – 4ac = 0，故 k² – 4 × 1 × 9 = 0 → k² = 36 → k = ±6。
Now solve the inequality x² – x – 6 > 0. Factorise: (x – 3)(x + 2) > 0.
再解不等式 x² – x – 6 > 0。因式分解：(x – 3)(x + 2) > 0。
The critical values are x = –2 and x = 3, giving the solution x < –2 or x > 3.
临界值为 x = –2 与 x = 3，故解为 x < –2 或 x > 3。
Sketch the parabola whenever an inequality appears. The graph tells you instantly whether the answer lies outside or between the roots.
遇到不等式就画抛物线草图。图像能立刻告诉你答案在两根之外还是在两根之间。
6. Trigonometry: Identities and Equations | 三角恒等式与方程
Solve 2 sin²θ – 1 = 0 for 0° ≤ θ ≤ 360°.
解 2 sin²θ – 1 = 0，0° ≤ θ ≤ 360°。
Rearrange: sin²θ = 1/2, so sin θ = ±1/√2 = ±√2/2.
整理：sin²θ = 1/2，所以 sin θ = ±1/√2 = ±√2/2。
The four solutions in the required range are θ = 45°, 135°, 225° and 315°.
在给定区间内的四个解为 θ = 45°、135°、225°、315°。
Try a disguised quadratic: 2
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