📚 Review Set 23A: Non-Calculator Practice for IB Mathematics | 复习集23A：IB数学非计算器练习
This review set is designed to strengthen your ability to handle IB Mathematics questions without relying on a calculator. Mastery of algebraic manipulation, exact trigonometric values, logarithmic properties, and analytical differentiation and integration is essential for success. Each problem in Set 23A targets a key skill area, and the worked solutions that follow illustrate clear manual techniques, reinforcing the importance of precision and logical reasoning in a non-calculator environment.
本复习集旨在强化你在不使用计算器的情况下解答 IB 数学题的能力。掌握代数变形、精确三角函数值、对数性质以及分析性求导与积分是成功的关键。23A 中的每一道题都针对一个核心技能领域，随后的解析通过清晰的笔算技巧展示，强调在非计算器环境中精确性与逻辑推理的重要性。
1. Algebraic Simplification | 代数化简
Simplify the expression fully, noting that x ≠ 2: (x³ – 8) / (x – 2).
化简下列表达式，需注意 x ≠ 2：(x³ – 8) / (x – 2)。
Recognise the numerator as a difference of cubes: x³ – 2³ = (x – 2)(x² + 2x + 4).
将分子看作立方差：x³ – 2³ = (x – 2)(x² + 2x + 4)。
Cancel the common factor (x – 2) since x ≠ 2, leaving x² + 2x + 4.
由于 x ≠ 2，可约去公因式 (x – 2)，得到 x² + 2x + 4。
The simplified expression is x² + 2x + 4. Always check for restrictions that avoid division by zero.
化简后的表达式为 x² + 2x + 4。务必检查避免分母为零的限制条件。
2. Solving Quadratic Equations | 解二次方程
Solve 2x² – 5x – 3 = 0 by factorising, giving exact values.
用因式分解法求解 2x² – 5x – 3 = 0，并给出精确值。
Multiply the coefficient of x² (2) and the constant (–3) to get –6. Find two numbers that multiply to –6 and add to –5: these are –6 and 1.
将 x² 的系数 2 与常数项 –3 相乘得 –6。寻找两个和为 –5、积为 –6 的数：–6 和 1。
Rewrite –5x as –6x + x: 2x² – 6x + x – 3 = 0. Factor by grouping: 2x(x – 3) + 1(x – 3) = 0, giving (x – 3)(2x + 1) = 0.
将 –5x 拆分为 –6x + x：2x² – 6x + x – 3 = 0。分组提取公因式：2x(x – 3) + 1(x – 3) = 0，得到 (x – 3)(2x + 1) = 0。
Set each factor to zero: x – 3 = 0 yields x = 3; 2x + 1 = 0 yields x = –1/2. The exact solutions are x = 3 and x = –½.
令每个因式为零：x – 3 = 0 得 x = 3；2x + 1 = 0 得 x = –1/2。精确解为 x = 3 和 x = –½。
3. Composite Functions and Domain | 复合函数与定义域
For f(x) = √x – 1 and g(x) = 1/x, evaluate f(g(2)) and state the domain of the composite function f(g(x)).
已知 f(x) = √x – 1 和 g(x) = 1/x，求 f(g(2)) 的值，并写出复合函数 f(g(x)) 的定义域。
First compute g(2) = 1/2. Then f(g(2)) = f(1/2) = √1/2 – 1 = √–1/2, which is not a real number. Therefore f(g(2)) is undefined in the reals.
先计算 g(2) = 1/2。然后 f(g(2)) = f(1/2) = √1/2 – 1 = √–1/2，这不是实数。因此 f(g(2)) 在实数范围内无定义。
For the composite function f(g(x)) = √1/x – 1, we require the expression inside the square root to be non-negative: 1/x – 1 ≥ 0 and x ≠ 0. Solve 1/x ≥ 1, which implies 0 < x ≤ 1.
对于复合函数 f(g(x)) = √1/x – 1，要求根号内的表达式非负：1/x – 1 ≥ 0 且 x ≠ 0。解 1/x ≥ 1，可得 0 < x ≤ 1。
The domain is therefore (0, 1]. Always check both the inner function's restrictions and the outer function's requirements.
因此定义域为 (0, 1]。务必同时检查内层函数的限制和外层函数的要求。
4. Exact Trigonometric Values | 精确三角函数值
Evaluate sin(π/3) + cos(π/6) – tan(π/4) without a calculator, giving an exact simplified result.
不使用计算器，求 sin(π/3) + cos(π/6) – tan(π/4) 的精确化简结果。
Recall the standard exact values: sin(π/3) = √3/2, cos(π/6) = √3/2, tan(π/4) = 1.
回忆标准精确值：sin(π/3) = √3/2，cos(π/6) = √3/2，tan(π/4) = 1。
Substitute into the expression: √3/2 + √3/2 – 1 = (2√3)/2 – 1 = √3 – 1.
代入表达式：√3/2 + √3/2 – 1 = (2√3)/2 – 1 = √3 – 1。
The exact value is √3 – 1. Memorising special angle ratios for π/6, π/4, π/3 is critical for non-calculator papers.
精确值为 √3 – 1。记住 π/6、π/4、π/3 的特殊角比值对于非计算器考试至关重要。
5. Logarithmic Equations | 对数方程
Solve for x: log₂(x) + log₂(x – 3) = 2.
解关于 x 的方程：log₂(x) + log₂(x – 3) = 2。
Combine the logs using the product rule: log₂[x(x – 3)] = 2.
利用乘积法则合并对数：log₂[x(x – 3)] = 2。
Rewrite in exponential form: x(x – 3) = 2² = 4, so x² – 3x – 4 = 0.
化为指数形式：x(x – 3) = 2² = 4，即 x² – 3x – 4 = 0。
Factorise: (x – 4)(x + 1) = 0, giving x = 4 or x = –1. Check the original domain: x > 0 and x – 3 > 0, so x > 3. Discard x = –1. The only valid solution is x = 4.
因式分解：(x – 4)(x + 1) = 0，得 x = 4 或 x = –1。检查原定义域：x > 0 且 x – 3 > 0，故 x > 3。舍去 x = –1。唯一有效解为 x = 4。
6. Differentiation from First Principles | 第一原理求导
Use the definition of the derivative to find f'(x) for f(x) = 3x² + 2x.
利用导数的定义求 f(x) = 3x² + 2x 的导数 f'(x)。
The definition: f'(x) = limh→0 [f(x + h) – f(x)] / h.
定义：f'(x) = limh→0 [f(x + h) – f(x)] / h。
f(x + h) = 3(x + h)² + 2(x + h) = 3(x² + 2xh + h²) + 2x + 2h = 3x² + 6xh + 3h² + 2x + 2h.
计算 f(x + h) = 3(x + h)² + 2(x + h) = 3(x² + 2xh + h²) + 2x + 2h = 3x² + 6xh + 3h² + 2x + 2h。
Then f(x + h) – f(x) = (3x² + 6xh + 3h² + 2x + 2h) – (3x² + 2x) = 6xh + 3h² + 2h.
于是 f(x + h) – f(x) = (3x² + 6xh + 3h² + 2x + 2h) – (3x² + 2x) = 6xh + 3h² + 2h。
Divide by h: (6xh + 3h² + 2h) / h = 6x + 3h + 2. Take the limit as h → 0: f'(x) = 6x + 2.
除以 h：(6xh + 3h² + 2h) / h = 6x + 3h + 2。令 h → 0 取极限：f'(x) = 6x + 2。
Therefore, the derivative is 6x + 2. First principles show how differentiation rules originate.
因此，导数为 6x + 2。第一原理揭示了求导法则的来源。
7. Indefinite Integration | 不定积分
Find ∫ (2x³ – 1/√x) dx, writing the result without negative or fractional exponents in the final answer.
求 ∫ (2x³ – 1/√x) dx，最终答案中不要出现负指数或分数指数。
Rewrite the integrand using exponents: 2x³ – x⁻¹/². Integrate term by term.
用指数形式重写被积函数：2x³ – x⁻¹/²。逐项积分。
∫ 2x³ dx = 2 * (x⁴/4) = x⁴/2.
∫ 2x³ dx = 2 * (x⁴/4) = x⁴/2。
∫ x⁻¹/² dx = x½ / (½) = 2x½ = 2√x. Note the sign: ∫ –x⁻¹/² dx = –2√x.
∫ x⁻¹/² dx = x½ / (½) = 2x½ = 2√x。注意符号：∫ –x⁻¹/² dx = –2√x。
Thus the integral is (1/2)x⁴ – 2√x + C. Avoid calculator dependency for basic integration patterns.
因此积分为 (1/2)x⁴ – 2√x + C。基本积分模式应避免依赖计算器。
8. Arithmetic Sequences and Series | 等差数列与求和
nce has first term a₁ = 5 and common difference d = 3. Find the sum of the first 20 terms.
一个等差数列的首项 a₁ = 5，公差 d = 3。求前 20 项的和。
Use the sum formula Sn = n/2 [2a₁ + (n – 1)d]. Here n = 20.
利用求和公式 Sn = n/2 [2a₁ + (n – 1)d]。此处 n = 20。
Substitute: S20 = 20/2 [2(5) + (20 – 1)(3)] = 10 [10 + 19×3] = 10 [10 + 57] = 10 × 67 = 670.
代入：S20 = 20/2 [2(5) + (20 – 1)(3)] = 10 [10 + 19×3] = 10 [10 + 57] = 10 × 67 = 670。
The sum of the first 20 terms is 670. Arithmetic series calculations should be done mentally or with simple pencil work.
前 20 项的和为 670。等差级数的计算应能通过心算或简单的笔算完成。
9. Complex Numbers in Standard Form | 复数的标准形式
Express (2 + 3i) / (1 – i) in the form a + bi, where a, b are real numbers.
将 (2 + 3i) / (1 – i) 表示为 a + bi 的形式，其中 a, b 为实数。
Multiply numerator and denominator by the conjugate of the denominator, which is (1 + i).
将分子和分母同乘分母的共轭复数 (1 + i)。
Numerator: (2 + 3i)(1 + i) = 2 + 2i + 3i + 3i² = 2 + 5i – 3 = –1 + 5i.
分子：(2 + 3i)(1 + i) = 2 + 2i + 3i + 3i² = 2 + 5i – 3 = –1 + 5i。
Denominator: (1 – i)(1 + i) = 1 – i² = 1 – (–1) = 2.
分母：(1 – i)(1 + i) = 1 – i² = 1 – (–1) = 2。
Divide: (–1 + 5i) / 2 = –1/2 + (5/2)i. Thus a = –½, b = 5/2.
相除：(–1 + 5i) / 2 = –1/2 + (5/2)i。因此 a = –½，b = 5/2。
10. Combinatorics and Counting | 组合与计数
A committee of 3 members is to be selected from 10 people. How many different committees are possible?
从 10 人中选出 3 人组成一个委员会。共有多少种不同的选法？
Use the combination formula C(n, k) = n! / [k!(n – k)!], where order does not matter. Here n = 10, k = 3.
使用组合公式 C(n, k) = n! / [k!(n – k)!]，此处顺序无关紧要。这里 n = 10, k = 3。
Compute C(10, 3) = 10! / (3!7!) = (10×9×8) / (3×2×1). Cancel efficiently: 10×9×8 = 720, denominator 6, 720/6 = 120.
计算 C(10, 3) = 10! / (3!7!) = (10×9×8) / (3×2×1)。约分计算：10×9×8 = 720，分母为 6，720/6 = 120。
simplify earlier: 10/(2)×9/(3)×8 = 5×3×8 = 120. So 120 distinct committees can be formed.
或者提前化简：10/2 × 9/3 × 8 = 5×3×8 = 120。因此可组成 120 个不同的委员会。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图