Review Set 21A – Non-Calculator Practice | 复习题集21A – 非计算器练习

📚 Review Set 21A – Non-Calculator Practice | 复习题集21A – 非计算器练习

This revision set covers a range of IB Mathematics topics that must be tackled without a calculator. You will practise algebraic manipulation, solving equations, logarithms, trigonometry, differentiation, integration, sequences, binomial expansion, and vector algebra – all using exact methods. Work through each problem step by step to strengthen your mental arithmetic and symbolic reasoning.

本复习题集涵盖了一系列必须在无计算器条件下完成的IB数学主题。你将练习代数运算、方程求解、对数、三角、微分、积分、数列、二项式展开和向量代数——全部使用精确方法。逐一完成每个问题，逐步强化你的心算能力和符号推理能力。

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1. Algebraic Expansion and Simplification | 代数展开与化简

Problem: Expand and simplify (2x – 3)(x + 4) – (x – 1)².

题目：展开并化简 (2x – 3)(x + 4) – (x – 1)²。

First expand the two binomials: (2x)(x) = 2x², (2x)(4) = 8x, (–3)(x) = –3x, (–3)(4) = –12. So (2x – 3)(x + 4) = 2x² + 5x – 12.

首先展开两个二项式：(2x)(x) = 2x²，(2x)(4) = 8x，(–3)(x) = –3x，(–3)(4) = –12。因此 (2x – 3)(x + 4) = 2x² + 5x – 12。

Now expand (x – 1)² = (x – 1)(x – 1) = x² – 2x + 1. Subtract this from the first result: (2x² + 5x – 12) – (x² – 2x + 1) = 2x² + 5x – 12 – x² + 2x – 1 = x² + 7x – 13.

接着展开 (x – 1)² = (x – 1)(x – 1) = x² – 2x + 1。从第一个结果中减去该式：(2x² + 5x – 12) – (x² – 2x + 1) = 2x² + 5x – 12 – x² + 2x – 1 = x² + 7x – 13。

The simplified expression is x² + 7x – 13.

化简后的表达式为 x² + 7x – 13。

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2. Solving a Quadratic Equation by Factorisation | 通过因式分解解二次方程

Problem: Solve x² – 5x + 6 = 0.

题目：解方程 x² – 5x + 6 = 0。

Look for two numbers that multiply to +6 and add to –5. These numbers are –2 and –3. So we factorise: (x – 2)(x – 3) = 0.

寻找两个数，它们相乘得 +6，相加得 –5。这两个数是 –2 和 –3。因此我们因式分解：(x – 2)(x – 3) = 0。

Set each factor equal to zero: x – 2 = 0 ⇒ x = 2; x – 3 = 0 ⇒ x = 3.

令每个因式等于零：x – 2 = 0 ⇒ x = 2；x – 3 = 0 ⇒ x = 3。

The solutions are x = 2 and x = 3.

解为 x = 2 和 x = 3。

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3. Logarithms and Exponents | 对数与指数

Problem: Evaluate log₂ 8 + log₂ (1/4).

题目：计算 log₂ 8 + log₂ (1/4)。

Recall that log₂ 8 asks “2 to what power gives 8?” Since 2³ = 8, log₂ 8 = 3.

回忆 log₂ 8 问的是“2的几次方等于8？”由于 2³ = 8，所以 log₂ 8 = 3。

Next, log₂ (1/4) = log₂ (4⁻¹) = –log₂ 4. Since 2² = 4, log₂ 4 = 2, so –log₂ 4 = –2. Alternatively, 1/4 = 2⁻², so log₂ (2⁻²) = –2.

接下来，log₂ (1/4) = log₂ (4⁻¹) = –log₂ 4。由于 2² = 4，log₂ 4 = 2，所以 –log₂ 4 = –2。或者，1/4 = 2⁻²，因此 log₂ (2⁻²) = –2。

Add the values: 3 + (–2) = 1.

将两个值相加：3 + (–2) = 1。

The result is 1.

结果为 1。

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4. Inverse Functions | 反函数

Problem: Given f(x) = 3x – 2, find f⁻¹(x) and evaluate f(f⁻¹(5)).

题目：已知 f(x) = 3x – 2，求 f⁻¹(x) 并计算 f(f⁻¹(5))。

To find the inverse, write y = 3x – 2. Swap x and y: x = 3y – 2. Solve for y: 3y = x + 2 ⇒ y = (x + 2)/3. Thus f⁻¹(x) = (x + 2)/3.

为求反函数，写出 y = 3x – 2。交换 x 和 y：x = 3y – 2。解出 y：3y = x + 2 ⇒ y = (x + 2)/3。因此 f⁻¹(x) = (x + 2)/3。

Now evaluate f(f⁻¹(5)): first f⁻¹(5) = (5 + 2)/3 = 7/3. Then f(7/3) = 3 × (7/3) – 2 = 7 – 2 = 5. Notice that f and f⁻¹ are inverses, so the composition returns the original input, 5.

现在计算 f(f⁻¹(5))：首先 f⁻¹(5) = (5 + 2)/3 = 7/3。然后 f(7/3) = 3 × (7/3) – 2 = 7 – 2 = 5。注意到 f 和 f⁻¹ 互为反函数，因此复合函数返回原输入 5。

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5. Trigonometric Simplification | 三角化简

Problem: Simplify (sin θ)/(1 + cos θ) + (1 + cos θ)/(sin θ).

题目：化简 (sin θ)/(1 + cos θ) + (1 + cos θ)/(sin θ)。

Find a common denominator: (sin θ)(1 + cos θ). The expression becomes [sin² θ + (1 + cos θ)²] / [sin θ (1 + cos θ)].

求公分母：(sin θ)(1 + cos θ)。表达式变为 [sin² θ + (1 + cos θ)²] / [sin θ (1 + cos θ)]。

Expand the numerator: sin² θ + (1 + 2cos θ + cos² θ) = (sin² θ + cos² θ) + 1 + 2cos θ. Using the identity sin² θ + cos² θ = 1, we get 1 + 1 + 2cos θ = 2 + 2cos θ = 2(1 + cos θ).

展开分子：sin² θ + (1 + 2cos θ + cos² θ) = (sin² θ + cos² θ) + 1 + 2cos θ。利用恒等式 sin² θ + cos² θ = 1，得到 1 + 1 + 2cos θ = 2 + 2cos θ = 2(1 + cos θ)。

Cancel the common factor (1 + cos θ) in the numerator and denominator, leaving 2 / sin θ = 2 csc θ.

约去分子和分母中的公因子 (1 + cos θ)，留下 2 / sin θ = 2 csc θ。

The simplified form is 2 csc θ (or 2/sin θ).

化简结果为 2 csc θ（或 2/sin θ）。

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6. Differentiation of Polynomials and Rational Terms | 多项式与有理项的微分

Problem: Differentiate y = 4x³ – 2x + 1/x with respect to x.

题目：对 y = 4x³ – 2x + 1/x 求关于 x 的导数。

Rewrite 1/x as x⁻¹. Then y = 4x³ – 2x + x⁻¹.

将 1/x 重写为 x⁻¹。那么 y = 4x³ – 2x + x⁻¹。

Apply the power rule d/dx (xⁿ) = n xⁿ⁻¹: derivative of 4x³ is 12x²; derivative of –2x is –2; derivative of x⁻¹ is –1 x⁻² = –1/x².

应用幂法则 d/dx (xⁿ) = n xⁿ⁻¹：4x³ 的导数是 12x²；–2x 的导数是 –2；x⁻¹ 的导数是 –1 x⁻² = –1/x²。

So dy/dx = 12x² – 2 – 1/x².

因此 dy/dx = 12x² – 2 – 1/x²。

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7. Indefinite Integration with Exponentials | 指数函数的不定积分

Problem: Find ∫ (2e²ˣ + 3) dx.

题目：求 ∫ (2e²ˣ + 3) dx。

Integrate term by term. For 2e²ˣ, recall that ∫ eᵏˣ dx = (1/k)eᵏˣ + c. Here k = 2, so ∫ 2e²ˣ dx = 2 × (1/2) e²ˣ = e²ˣ.

逐项积分。对于 2e²ˣ，回忆 ∫ eᵏˣ dx = (1/k)eᵏˣ + c。这里 k = 2，因此 ∫ 2e²ˣ dx = 2 × (1/2) e²ˣ = e²ˣ。

For the constant 3, ∫ 3 dx = 3x.

对于常数 3，∫ 3 dx = 3x。

Combine and add the constant of integration: ∫ (2e²ˣ + 3) dx = e²ˣ + 3x + C.

合并并加上积分常数：∫ (2e²ˣ + 3) dx = e²ˣ + 3x + C。

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8. Arithmetic Series Sum | 等差数列求和

Problem: Find the sum of the first 10 terms of an arithmetic sequence where the first term a = 3 and the common difference d = 2.

题目：求等差数列前10项的和，其中首项 a = 3，公差 d = 2。

Use the sum formula Sₙ = n/2 [2a + (n – 1)d]. Here n = 10, a = 3, d = 2.

使用求和公式 Sₙ = n/2 [2a + (n – 1)d]。这里 n = 10，a = 3，d = 2。

Calculate inside the brackets: 2a = 6; (n – 1)d = 9 × 2 = 18; sum = 24. Then S₁₀ = 10/2 × 24 = 5 × 24 = 120.

计算括号内的值：2a = 6；(n – 1)d = 9 × 2 = 18；和为 24。然后 S₁₀ = 10/2 × 24 = 5 × 24 = 120。

The sum of the first 10 terms is 120.

前10项的和为 120。

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9. Binomial Expansion Coefficient | 二项式展开系数

Problem: In the expansion of (1 + 2x)⁵, find the coefficient of x³.

题目：在 (1 + 2x)⁵ 的展开式中，求 x³ 的系数。

The general term in the binomial expansion of (a + b)ⁿ is given by Tₖ₊₁ = C(n, k) aⁿ⁻ᵏ bᵏ. Here a = 1, b = 2x, n = 5. We need the term with x³, so the power of b is 3, which means k = 3.

(a + b)ⁿ 的二项展开通项为 Tₖ₊₁ = C(n, k) aⁿ⁻ᵏ bᵏ。此处 a = 1，b = 2x，n = 5。我们需要 x³ 的项，因此 b 的幂次为 3，即 k = 3。

The term is C(5, 3) × (1)⁵⁻³ × (2x)³ = 10 × 1 × 8x³ = 80x³.

该项为 C(5, 3) × (1)⁵⁻³ × (2x)³ = 10 × 1 × 8x³ = 80x³。

So the coefficient of x³ is 80.

因此 x³ 的系数是 80。

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10. Vector Dot Product and Angle | 向量点积与夹角

Problem: Find the exact cosine of the angle between vectors a = (1, 2, 3) and b = (4, –1, 2).

题目：求向量 a = (1, 2, 3) 与 b = (4, –1, 2) 之间夹角的精确余弦值。

The dot product a·b = (1)(4) + (2)(–1) + (3)(2) = 4 – 2 + 6 = 8.

点积 a·b = (1)(4) + (2)(–1) + (3)(2) = 4 – 2 + 6 = 8。

The magnitudes: |a| = √(1² + 2² + 3²) = √(1 + 4 + 9) = √14; |b| = √(4² + (–1)² + 2²) = √(16 + 1 + 4) = √21.

模长：|a| = √(1² + 2² + 3²) = √(1 + 4 + 9) = √14；|b| = √(4² + (–1)² + 2²) = √(16 + 1 + 4) = √21。

Then cos θ = (a·b) / (|a||b|) = 8 / (√14 × √21) = 8 / √294. Simplify the radical: 294 = 49 × 6, so √294 = 7√6. Thus cos θ = 8 / (7√6). It can be rationalised if needed: (8√6) / 42 = (4√6) / 21.

于是 cos θ = (a·b) / (|a||b|) = 8 / (√14 × √21) = 8 / √294。化简根式：294 = 49 × 6，因此 √294 = 7√6。所以 cos θ = 8 / (7√6)。如有需要可分母有理化：(8√6) / 42 = (4√6) / 21。

The exact cosine is 8/(7√6) or equivalently (4√6)/21.

精确余弦值为 8/(7√6) 或等价地 (4√6)/21。

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