📚 Review Set 17A: Mastering Non-Calculator Skills | 复习题集 17A：掌握无计算器解题技巧
In IB Mathematics, the non-calculator paper demands fluency in algebraic manipulation, exact values, and strategic problem‑solving. Review Set 17A is designed to strengthen these skills through targeted exercises that mirror the style and difficulty of Paper 1 questions. This article walks you through the essential concepts covered in the set, offering worked examples, common pitfalls, and revision tips that will boost your confidence when you are not allowed to reach for a GDC.
在 IB 数学中，无计算器试卷要求学生具备熟练的代数操作能力、精确值计算能力以及策略性的解题思维。复习题集 17A 精心设计了一系列针对性练习，与 Paper 1 的题型和难度高度一致。本文将带领你逐一攻克题集中的核心概念，提供详细的例题解答、常见错误分析以及复习技巧，帮助你在没有图形计算器的情况下依然从容应对。
Working without a calculator rewards precision and a deep understanding of structure. Every simplification, every derivative, and every integral must be tackled with clear logic and neat working. Let’s explore the topics covered in Review Set 17A, turning potential weaknesses into strengths.
不使用计算器恰恰考验的是精确度和对数学结构的深刻理解。每一次化简、每一次求导、每一次积分都必须依靠清晰的逻辑和整洁的书写来完成。让我们一起走进复习题集 17A 所涉及的主题，把潜在的薄弱点变成你的得分武器。
1. Simplifying Algebraic Expressions | 代数表达式的化简
The foundation of many non-calculator problems is simplifying rational expressions, surds, and indices. You must be comfortable factorising, expanding, and cancelling without numerical assistance. For instance, when faced with (√3 + 1)/(√3 − 1), multiply numerator and denominator by the conjugate to achieve a rational denominator. The ability to spot a difference of two squares or a common factor quickly saves precious time.
许多无计算器题目的基础是化简有理式、根式和指数。你必须能在没有任何数值帮助的情况下，熟练地进行因式分解、展开和约分。例如，面对 (√3 + 1)/(√3 − 1) 时，分子分母同乘以共轭根式即可将分母有理化。快速识别平方差或公因式的能力会为你节省宝贵的时间。
Example: Simplify (x² − 4)/(x² + x − 6). First, factorise: (x − 2)(x + 2) / [(x + 3)(x − 2)]. Cancel the common factor (x − 2), leaving (x + 2)/(x + 3), provided x ≠ 2.
例题：化简 (x² − 4)/(x² + x − 6)。首先因式分解： (x − 2)(x + 2) / [(x + 3)(x − 2)]。约去公因式 (x − 2)，得到 (x + 2)/(x + 3)，其中 x ≠ 2。
When dealing with fractional indices, recall that am/n = ⁿ√(am). Simplify expressions like (27x⁶)⅓ to obtain the exact result 3x² without any decimal approximations.
在处理分数指数时，记住 am/n = ⁿ√(am)。将类似 (27x⁶)⅓ 的表达式化简得到精确结果 3x²，完全不需要小数近似。
2. Solving Exponential and Logarithmic Equations | 解指数与对数方程
Equations involving exponents and logarithms frequently appear in non-calculator sections. The key is to rewrite both sides of the equation with the same base, or to use the formal definition of a logarithm. For example, to solve 22x+1 = 32, recognise that 32 = 2⁵. Equating powers gives 2x + 1 = 5, so x = 2. No trial‑and‑error is needed.
涉及指数和对数的方程经常出现在无计算器试卷中。关键在于将方程两边化为同底数，或者使用对数的定义进行转换。例如，解 22x+1 = 32，识别出 32 = 2⁵。令指数相等得 2x + 1 = 5，则 x = 2。完全不需要试错。
When the unknown is in the logarithm, isolate the logarithmic term and convert to exponential form. For log₃(x − 1) = 2, rewrite as x − 1 = 3² = 9, giving x = 10. Always verify that the argument remains positive; here 10 − 1 > 0, so the solution is valid.
当未知数在对数符号内时，先孤立对数项，再还原为指数形式。对于 log₃(x − 1) = 2，改写为 x − 1 = 3² = 9，解得 x = 10。务必检验真数是否为正；这里 10 − 1 > 0，解答有效。
If the bases differ, apply the change‑of‑base formula or take natural logs. However, most non‑calculator exam questions are designed to yield integer or simple rational answers after careful rewriting.
如果底数不同，可以使用换底公式或取自然对数。不过，绝大多数无计算器考题经过恰当的表达式改写后，都会得到整数或简单有理数答案。
3. Finding Terms in Binomial Expansions | 二项式展开中的特定项
The binomial theorem allows you to find specific terms without expanding the whole expression. For (a + b)ⁿ, the general term is C(n, r) an−r br, where r = 0,1,…,n. In a non‑calculator setting, you must calculate combinations by hand, often using factorial notation or Pascal’s triangle. Patience with arithmetic is crucial.
二项式定理使你可以直接找出特定项而无需完整展开。对于 (a + b)ⁿ，通项为 C(n, r) an−r br，其中 r = 0,1,…,n。在无计算器的环境下，你必须手工计算组合数，通常使用阶乘记法或杨辉三角形。耐心处理算术运算至关重要。
Example: Find the coefficient of x³ in the expansion of (2x − 3)⁵. The general term is C(5, r) (2x)5−r (−3)r. We need the power of x to be 3, so 5 − r = 3 ⇒ r = 2. Substitute r = 2: term = C(5, 2) (2x)³ (−3)² = 10 × 8x³ × 9 = 720x³. The coefficient is 720.
例题：求 (2x − 3)⁵ 展开式中 x³ 的系数。通项为 C(5, r) (2x)5−r (−3)r。我们需要 x 的幂次为 3，所以 5 − r = 3 ⇒ r = 2。代入 r = 2：该项 = C(5, 2) (2x)³ (−3)² = 10 × 8x³ × 9 = 720x³。系数为 720。
Take care with signs: a negative b raised to an even r gives a positive contribution, while odd r yields a negative contribution. Write each factor explicitly to reduce sign errors.
注意正负号：当 b 为负且 r 为偶数时该项为正，r 为奇数时该项为负。明确写出每个因子，减少符号错误。
4. Trigonometric Identities and Exact Values | 三角恒等式与精确值
The non‑calculator paper expects you to know exact values for sin, cos, and tan of 0°, 30°, 45°, 60°, 90° and their radian equivalents. Practise deriving these from the special triangles, not by memorisation alone. Identities such as sin²θ + cos²θ = 1 and tanθ = sinθ/cosθ are tools for simplifying complicated trigonometric expressions.
无计算器试卷要求你熟知 0°、30°、45°、60°、90° 及其弧度制对应的正弦、余弦和正切精确值。练习从特殊三角形中推导这些数值，而不是单纯记忆。像 sin²θ + cos²θ = 1 和 tanθ = sinθ/cosθ 这样的恒等式是化简复杂三角表达式的利器。
To solve 2 sinθ = √3 for 0 ≤ θ ≤ 2π, first give sinθ = √3/2. Recognise that sinθ takes this value at θ = π/3 and θ = 2π/3. No calculator is needed—you simply recall the unit circle or the 30‑60‑90 triangle.
解方程 2 sinθ = √3 在 0 ≤ θ ≤ 2π 范围内，先得到 sinθ = √3/2。识别出 sinθ 在 θ = π/3 和 θ = 2π/3 时取得该值。完全无需计算器——你只需回忆单位圆或含有 30‑60‑90 的三角形即可。
When proving identities, work on one side only. Change everything into sines and cosines, factor, cancel, and apply Pythagorean identities. The final result must match the other side exactly. Never treat an identity as an equation to be solved by cross‑multiplication.
证明恒等式时，只处理等式的一边。将一切都转化为正弦和余弦，进行因式分解、约分，再运用毕达哥拉斯恒等式。最终结果必须与另一边严格相等。切勿将恒等式当作方程来交叉相乘求解。
5. Differentiation Without a Calculator | 无计算器求导技巧
Differentiation problems on a non‑calculator paper often involve polynomials, roots, and simple rational functions that can be rewritten using negative or fractional indices. Master the power rule: if y = xⁿ, then dy/dx = nxⁿ⁻¹. For example, to differentiate y = 2/√x, first write the function as y = 2x−½. Then dy/dx = 2 × (−½) x−³/² = −x−³/² = −1/(x√x). There is no need to compute decimal gradients.
无计算器试卷中的求导题通常涉及多项式、根式以及可以通过负指数或分数指数改写的简单有理函数。熟练掌握幂法则：若 y = xⁿ，则 dy/dx = nxⁿ⁻¹。例如，对 y = 2/√x 求导，先将函数写为 y = 2x−½。那么 dy/dx = 2 × (−½) x−³/² = −x−³/² = −1/(x√x)。无需计算任何小数斜率。
When finding equations of tangents, remember that the derivative at a point gives the gradient m. Then use y − y₁ = m(x − x₁). If asked to find the tangent to y = x³ − 3x at x = 2, first compute dy/dx = 3x² − 3. At x = 2, m = 9, y₁ = 2³ − 6 = 2. The tangent equation is y − 2 = 9(x − 2), which simplifies to y = 9x − 16.
当求切线方程时，记住函数在一点的导数给出斜率 m。然后利用点斜式 y − y₁ = m(x − x₁)。如需找出 y = x³ − 3x 在 x = 2 处的切线，先求导得 dy/dx = 3x² − 3。在 x = 2 处，m = 9，y₁ = 2³ − 6 = 2。切线方程为 y − 2 = 9(x − 2)，化简后为 y = 9x − 16。
Turning points are found where dy/dx = 0. Set the derivative to zero and solve the resulting equation by hand. Often the derivative factorises nicely, giving exact x‑coordinates. Determine their nature using a sign diagram or the second derivative.
驻点由 dy/dx = 0 确定。令导数为零并手工求解方程。导数通常能够很好地因式分解，从而给出精确的 x 坐标。利用符号表或二阶导数判断其性质。
6. Integration Techniques | 积分技巧
Non‑calculator integration relies on reversing the power rule: ∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + C, for n ≠ −1. Always add the constant of integration for indefinite integrals. Convert denominators and roots into negative or fractional powers before integrating.
无计算器积分依赖于逆用幂法则： ∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + C，其中 n ≠ −1。不定积分一定要记得加上积分常数。积分前先将分母和根式转化为负指数或分数指数。
Example: Evaluate ∫ (3x² + 2/x³) dx. Rewrite as ∫ (3x² + 2x⁻³) dx. Integrate term by term: 3 × (x³/3) + 2 × (x⁻²/(−2)) + C = x³ − x⁻² + C. The answer can be left as x³ − 1/x² + C.
例题：计算 ∫ (3x² + 2/x³) dx。改写为 ∫ (3x² + 2x⁻³) dx。逐项积分：3 × (x³/3) + 2 × (x⁻²/(−2)) + C = x³ − x⁻² + C。答案可保留为 x³ − 1/x² + C。
For definite integrals, such as ∫₁⁴ √x dx, write √x as x½. Integrate to get (2/3)x³/², then substitute the limits: (2/3)(4³/² − 1³/²) = (2/3)(8 − 1) = 14/3. Evaluate powers like 4³/² by thinking of √4 first, i.e. 2³ = 8.
对于定积分，例如 ∫₁⁴ √x dx，将 √x 写为 x½。积分得 (2/3)x³/²，然后代入上下限：(2/3)(4³/² − 1³/²) = (2/3)(8 − 1) = 14/3。计算 4³/² 时，先想 √4 = 2，再立方得 8。
When a question gives a boundary condition, use it to find the constant C. If f'(x) = 4x − 3 and f(1) = 5, integrate to f(x) = 2x² − 3x + C. Substitute x = 1: 2 − 3 + C = 5 ⇒ C = 6. So f(x) = 2x² − 3x + 6.
当题目给出边界条件时，用它来求出常数 C。若 f'(x) = 4x − 3 且 f(1) = 5，积分得 f(x) = 2x² − 3x + C。代入 x = 1：2 − 3 + C = 5 ⇒ C = 6。所以 f(x) = 2x² − 3x + 6。
7. Solving Equations and Inequalities | 解方程与不等式
Polynomial equations are the backbone of non‑calculator algebra. Always try to factorise. For a cubic like x³ − 4x² + x + 6 = 0, use the Rational Root Theorem to test possible factors. Once a root is found, perform polynomial division to reduce the degree and solve the remaining quadratic.
多项式方程是无计算器代数的核心。一定要尝试因式分解。对于像 x³ − 4x² + x + 6 = 0 这样的三次方程，使用有理根定理检验可能的因式。一旦找到一个根，就进行多项式除法降次，再解剩下的二次方程。
For inequalities such as (x − 2)(x + 1) < 0, sketch a sign diagram. The critical values are x = −1 and x = 2. Testing intervals gives the solution −1 < x < 2. Do not multiply by an expression containing x without considering its sign, as this can change the inequality direction.
对于像 (x − 2)(x + 1) < 0 这样的不等式，绘制符号图。临界值为 x = −1 和 x = 2。通过区间测试得到解集 −1 < x < 2。切勿在未考虑符号的情况下乘以含 x 的表达式，因为这会改变不等号的方向。
Simultaneous equations can be solved by substitution or elimination. In a non‑calculator context, the numbers are chosen so that the arithmetic remains manageable. For instance, solve y = x + 1 and x² + y² = 13 by substituting the linear equation into the quadratic. You obtain x² + (x+1)² = 13 ⇒ 2x² + 2x −12 = 0 ⇒ x² + x −6 = 0, which factorises to (x+3)(x−2)=0. The exact coordinate pairs follow quickly.
联立方程组可通过代入法或消元法求解。在无计算器环境下，数值经过精心设计，使得算术计算保持可控。例如，通过将 y = x + 1 代入 x² + y² = 13 来求解。得到 x² + (x+1)² = 13 ⇒ 2x² + 2x −12 = 0 ⇒ x² + x −6 = 0，该方程可分解为 (x+3)(x−2)=0。精确的坐标对随即得出。
8. Vectors and Magnitudes | 向量与模长
Vector questions without a calculator involve exact arithmetic with components. To find the magnitude of a vector v = 3i − 4j, compute |v| = √(3² + (−4)²) = √25 = 5. Keep such results in exact form. The dot product of a = 2i + j and b = i − 3j is (2)(1) + (1)(−3) = −1, and the angle between them can be found using cosθ = (a·b)/(|a||b|).
无计算器的向量题涉及分量的精确运算。要计算向量 v = 3i − 4j 的模，求 |v| = √(3² + (−4)²) = √25 = 5。将结果保留为精确值。向量 a = 2i + j 与 b = i − 3j 的点积为 (2)(1) + (1)(−3) = −1，它们之间的夹角可通过 cosθ = (a·b)/(|a||b|) 求出。
When asked whether two vectors are perpendicular, simply show that their dot product equals zero. For example, p = 5i + 2j and q = −2i + 5j gives a dot product of −10 + 10 = 0; hence they are perpendicular. No decimals or approximations are required.
当被问及两个向量是否垂直时，只需证明它们的点积为零。例如，p = 5i + 2j 与 q = −2i + 5j 的点积为 −10 + 10 = 0；因此它们相互垂直。完全不需要小数或近似值。
Vector equations of lines require you to write r = a + tb. If a line passes through (1, 2) and is parallel to 3i − j, its equation is r = (1, 2) + t(3, −1). You can then find the position vector for any t exactly, or solve for the intersection of two lines by equating components and handling the two equations manually.
直线的向量方程要求你写出 r = a + tb。若一直线经过 (1, 2) 且平行于 3i − j，其方程为 r = (1, 2) + t(3, −1)。然后你可以精确求得任意 t 对应的位置向量，或者通过令分量相等并手算两个方程来求两直线的交点。
9. Probability and Combinatorics | 概率与组合数学
Counting problems without a calculator demand careful use of factorial expressions and simplification. For example, the number of ways to choose 3 students from a group of 10 is C(10,3) = (10 × 9 × 8)/(3 × 2 × 1) = 120. Always cancel before multiplying to keep numbers small.
无计算器的计数问题要求仔细运用阶乘表达式并加以化简。例如，从 10 名学生中选出 3 人的方法数是 C(10,3) = (10 × 9 × 8)/(3 × 2 × 1) = 120。总是先约分再相乘，以保持数字较小。
When combining probabilities, remember that ‘AND’ means multiply, ‘OR’ means add, provided events are mutually exclusive. For independent events A and B, P(A ∩ B) = P(A) × P(B). Keep probabilities as fractions; a typical non‑calculator answer might be 5/36 or 7/18, never a rounded decimal.
在组合概率时，记住“并且”意味着相乘，“或者”意味着相加，前提是事件互斥。对于独立事件 A 和 B，P(A ∩ B) = P(A) × P(B)。将概率保持为分数；一道典型的无计算器题答案可能是 5/36 或 7/18，而绝不是四舍五入的小数。
Tree diagrams are your friend. They help organise outcomes and ensure no branches are missed. However, when drawing a tree, label branches with fractions, and calculate terminal probabilities by multiplying along the branches. Add the relevant terminal probabilities for ‘at least’ or ‘exactly one’ style questions.
树状图是你的好帮手。它有助于组织各种结果，确保不遗漏任何分支。然而，在绘制树状图时，要用分数标注分支，并通过沿分支相乘来计算终点概率。对于考察“至少”或“恰好一个”等题型，将相关的终点概率相加即可。
10. Function Transformations | 函数变换
Understanding how graphs shift, stretch, and reflect is critical in a non‑calculator environment. If f(x) is given, then f(x) + 2 shifts the graph up by 2 units; f(x + 2) shifts it left by 2 units. Be careful with horizontal transformations: they operate in the opposite direction to the sign inside the bracket.
在无计算器环境中，理解函数图像的平移、伸缩和反射至关重要。若给定 f(x)，则 f(x) + 2 将图像向上平移 2 个单位；f(x + 2) 将图像向左平移 2 个单位。注意水平变换：其作用方向与括号内符号相反。
When finding the inverse function, swap x and y and then solve for y. For f(x) = 2ex − 3, write y = 2ex − 3, swap to x = 2ey − 3, and isolate: 2ey = x + 3 ⇒ ey = (x+3)/2 ⇒ y = ln((x+3)/2). The domain of the inverse is the range of the original function, which should be stated in exact form.
求反函数时，交换 x 和 y 然后解出 y。对于 f(x) = 2ex − 3，令 y = 2ex − 3，交换为 x = 2ey − 3，然后分离变量：2ey = x + 3 ⇒ ey = (x+3)/2 ⇒ y = ln((x+3)/2)。反函数的定义域是原函数的值域，应以精确形式表述。
Composite functions f(g(x)) require careful substitution. If f(x) = √(x+1) and g(x) = 3x − 2, then f(g(x)) = √(3x − 2 + 1) = √(3x − 1). The domain of the composite is restricted to x ≥ 1/3 to keep the radicand non‑negative. Always define domains exactly.
复合函数 f(g(x)) 需要仔细代入。若 f(x) = √(x+1) 且 g(x) = 3x − 2，则 f(g(x)) = √(3x − 2 + 1) = √(3x − 1)。为保持被开方数非负，复合函数的定义域限制为 x ≥ 1/3。始终精确地给出定义域。
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