📚 IGCSE Cambridge Additional Mathematics: Unit Test Mock Paper Solutions | IGCSE 剑桥进阶数学：单元测试模拟卷解析
This mock paper is designed to test core techniques in Cambridge IGCSE Additional Mathematics (0606). Each question targets a single topic, but the worked solutions show how to link algebra, graphs and calculus. Use this revision material to identify weak areas before the real unit test.
本模拟卷旨在检测剑桥 IGCSE 进阶数学（0606）的核心解题技巧。每道题聚焦一个主题，但解析展示了如何把代数、图像和微积分联系起来。请在正式单元测试前用这份复习材料找出薄弱环节。
1. Functions and Inverses | 函数与反函数
The first question gives f(x) = (2x – 3)/(x + 1) with x ≠ -1, and asks for the inverse function and the range of f. To find f⁻¹, start by writing y = f(x) and then rearrange for x.
第一题给出 f(x) = (2x – 3)/(x + 1)，其中 x ≠ -1，要求反函数和 f 的值域。求 f⁻¹ 时，先设 y = f(x)，然后把 x 表示出来。
Multiply both sides by (x + 1): y(x + 1) = 2x – 3. Expand and collect x terms: yx + y = 2x – 3, so yx – 2x = -3 – y. Factor out x: x(y – 2) = -3 – y. Therefore x = (-3 – y)/(y – 2). After multiplying numerator and denominator by -1, this becomes x = (3 + y)/(2 – y). Hence f⁻¹(x) = (3 + x)/(2 – x), with x ≠ 2.
两边同乘 (x + 1)：y(x + 1) = 2x – 3。展开并合并含 x 的项：yx + y = 2x – 3，所以 yx – 2x = -3 – y。提取 x：x(y – 2) = -3 – y。因此 x = (-3 – y)/(y – 2)。分子分母同乘 -1 后得 x = (3 + y)/(2 – y)。所以 f⁻¹(x) = (3 + x)/(2 – x)，其中 x ≠ 2。
The range of f is all real numbers except 2, because the horizontal asymptote of the original function is y = 2. A common error is to give the inverse domain incorrectly; remember that the domain of f⁻¹ equals the range of f.
f 的值域是除 2 以外的所有实数，因为原函数的水平渐近线是 y = 2。常见错误是把反函数的定义域写错；记住 f⁻¹ 的定义域等于 f 的值域。
2. Quadratic Equations and Inequalities | 二次方程与不等式
Question 2 asks to solve x² – 5x + 6 ≥ 0 and to find the minimum value of y = 2x² – 8x + 3. Both parts require completing the square or using factors and a sign diagram.
第二题要求解 x² – 5x + 6 ≥ 0，并求 y = 2x² – 8x + 3 的最小值。两部分都需要配方，或使用因式分解和符号表。
For the inequality, factorise the quadratic: x² – 5x + 6 = (x – 2)(x – 3). The critical values are x = 2 and x = 3. The graph is a positive parabola, so the expression is non-negative when x ≤ 2 or x ≥ 3. Therefore the solution set is {x | x ≤ 2 or x ≥ 3}.
对于不等式，先把二次式因式分解：x² – 5x + 6 = (x – 2)(x – 3)。临界值为 x = 2 和 x = 3。图像是开口向上的抛物线，因此当 x ≤ 2 或 x ≥ 3 时，表达式非负。所以解集为 {x | x ≤ 2 或 x ≥ 3}。
For the minimum value, complete the square on y = 2x² – 8x + 3. First factor out 2 from the x terms: y = 2(x² – 4x) + 3. Then write x² – 4x as (x – 2)² – 4. Substituting gives y = 2((x – 2)² – 4) + 3 = 2(x – 2)² – 8 + 3 = 2(x – 2)² – 5. The minimum occurs when the squared term is zero, so the minimum value of y is -5 at x = 2.
对于最小值，对 y = 2x² – 8x + 3 配方。先把 x 项的系数 2 提出：y = 2(x² – 4x) + 3。再把 x² – 4x 写成 (x – 2)² – 4。代回得 y = 2((x – 2)² – 4) + 3 = 2(x – 2)² – 8 + 3 = 2(x – 2)² – 5。当平方项为零时取最小值，所以 y 的最小值是 -5，此时 x = 2。
3. Indices and Surds | 指数与根式
Question 3 tests simplification of surds and rationalising a denominator. The first part is (√8 + √18) ÷ √2, and the second part is 5/(√3 – 1).
第三题考查根式化简和分母有理化。第一部分是 (√8 + √18) ÷ √2，第二部分是 5/(√3 – 1)。
Simplify each surd first: √8 = √(4 × 2) = 2√2, and √18 = √(9 × 2) = 3√2. Their sum is 2√2 + 3√2 = 5√2. Dividing by √2 gives 5√2 ÷ √2 = 5. The answer is therefore exactly 5.
先化简每个根式：√8 = √(4 × 2) = 2√2，√18 = √(9 × 2) = 3√2。它们相加为 2√2 + 3√2 = 5√2。除以 √2 得 5√2 ÷ √2 = 5。因此答案正好是 5。
To rationalise 5/(√3 – 1), multiply the numerator and denominator by the conjugate √3 + 1. The denominator becomes (√3 – 1)(√3 + 1) = 3 – 1 = 2. The numerator becomes 5(√3 + 1) = 5√3 + 5. Therefore the simplified form is (5√3 + 5)/2.
要对 5/(√3 – 1) 进行分母有理化，把分子和分母同乘共轭根式 √3 + 1。分母变为 (√3 – 1)(√3 + 1) = 3 – 1 = 2。分子变为 5(√3 + 1) = 5√3 + 5。因此化简结果是 (5√3 + 5)/2。
4. Polynomials and Remainder Theorem | 多项式与余式定理
Question 4 gives f(x) = x³ – 3x² + kx + 12 and states that (x – 2) is a factor. You are asked to find k and then express f(x) as a product of linear factors.
第四题给出 f(x) = x³ – 3x² + kx + 12，并说明 (x – 2) 是一个因式。要求求出 k，然后把 f(x) 写成一次因式的乘积。
By the factor theorem, if (x – 2) is a factor then f(2) = 0. Substitute x = 2 into f(x): f(2) = (2)³ – 3(2)² + k(2) + 12 = 8 – 12 + 2k + 12 = 8 + 2k. Setting this equal to zero gives 8 + 2k = 0, so k = -4. Thus f(x) = x³ – 3x² – 4x + 12.
根据因式定理，如果 (x – 2) 是因式，则 f(2) = 0。把 x = 2 代入 f(x)：f(2) = (2)³ – 3(2)² + k(2) + 12 = 8 – 12 + 2k + 12 = 8 + 2k。令其等于零得 8 + 2k = 0，所以 k = -4。因此 f(x) = x³ – 3x² – 4x + 12。
Now divide f(x) by (x – 2) using synthetic division or long division. The coefficients are 1, -3, -4, 12 and the root is 2. Bring down the 1, multiply by 2 to get 2, add to -3 to get -1, multiply by 2 to get -2, add to -4 to get -6, multiply by 2 to get -12, add to 12 to get 0. The quotient is x² – x – 6. This quadratic factors as (x – 3)(x + 2). Therefore f(x) = (x – 2)(x – 3)(x + 2).
现在用综合除法或长除法把 f(x) 除以 (x – 2)。系数为 1、-3、-4、12，根为 2。落下 1，乘 2 得 2，与 -3 相加得 -1，乘 2 得 -2，与 -4 相加得 -6，乘 2 得 -12，与 12 相加得 0。商为 x² – x – 6。这个二次式可以分解为 (x – 3)(x + 2)。因此 f(x) = (x – 2)(x – 3)(x + 2)。
5. Binomial Expansion | 二项式展开
Question 5 asks for the coefficient of x³ in the expansion of (2 – 3x)⁵. Use the general term formula from the binomial theorem.
第五题要求 (2 – 3x)⁵ 的展开式中 x³ 的系数。使用二项式定理的通项公式。
The general term is T(r+1) = C(5, r) × (2)^(5 – r) × (-3x)^r. For x³, set r = 3. Substitute r = 3: T₄ = C(5, 3) × 2² × (-3)³ × x³ = 10 × 4 × (-27) × x³ = -1080 x³. The coefficient is therefore -1080.
通项为 T(r+1) = C(5, r) × (2)^(5 – r) × (-3x)^r。要求 x³，令 r = 3。代入 r = 3：T₄ = C(5, 3) × 2² × (-3)³ × x³ = 10 × 4 × (-27) × x³ = -1080 x³。因此系数是 -1080。
Be careful with the negative sign. The term (-3x)³ contributes a negative factor, so the coefficient of an odd power of x will be negative in this expansion.
注意负号。( -3x)³ 会带来负因子，所以在这个展开式中 x 的奇数次幂的系数是负数。
6. Trigonometry: Equations and Identities | 三角学：方程与恒等式
Question 6 asks to solve 2 sin² θ – sin θ – 1 = 0 for 0° ≤ θ ≤ 360°. This is a quadratic equation in sin θ, so it can be solved by substitution or factorisation.
第六题要求解 2 sin² θ – sin θ – 1 = 0，其中 0° ≤ θ ≤ 360°。这是关于 sin θ 的二次方程，因此可以用换元或因式分解来解。
Let u = sin θ. The equation becomes 2u² – u – 1 = 0. Factorise: (2u + 1)(u – 1) = 0. Therefore u = 1 or u = -1/2. Replacing u by sin θ gives sin θ = 1 or sin θ = -1/2.
设 u = sin θ。方程变为 2u² – u – 1 = 0。因式分解得 (2u + 1)(u – 1) = 0。所以 u = 1 或 u = -1/2。把 u 换回 sin θ，得 sin θ = 1 或 sin θ = -1/2。
For sin θ = 1 in the interval 0° to 360°, the only solution is θ = 90°. For sin
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