Year 10 CIE Additional Mathematics: Unit Test Mock Exam Analysis | Year 10 CIE 进阶数学：单元测试模拟卷解析

📚 Year 10 CIE Additional Mathematics: Unit Test Mock Exam Analysis | Year 10 CIE 进阶数学：单元测试模拟卷解析

This mock exam analysis breaks down a typical Year 10 CIE Additional Mathematics unit test, covering quadratic functions, inequalities, polynomials, binomial expansion, exponentials, logarithms, trigonometry and introductory differentiation. Each section works through a representative question with step-by-step guidance, highlighting common errors and essential exam techniques to help you build confidence and accuracy.

本模拟卷解析系统梳理了典型的 Year 10 CIE 进阶数学单元测试，涵盖二次函数、不等式、多项式、二项式展开、指数、对数、三角函数以及微分入门。每个小节通过一道代表性题目进行分步讲解，指出常见错误和关键的考试技巧，帮助你建立信心并提高解题准确性。

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1. Quadratic Function: Vertex Form and Range | 二次函数：顶点式与值域

Question: Express f(x) = 2x² – 8x + 5 in the form a(x – te the coordinates of the vertex, the equation of the axis of symmetry and the range of f.

题目：将 f(x) = 2x² – 8x + 5 写成 a(x – h)² + k 的形式，并由此写出顶点坐标、对称轴方程以及 f 的值域。

Step 1: Factor out the coefficient of x² from the first two terms: f(x) = 2(x² – 4x) + 5.

步骤 1：提取前两项中 x² 的系数：f(x) = 2(x² – 4x) + 5。

Step 2: To complete the square inside the bracket, take half of –4 (which is –2) and square it to get 4. Add and subtract 4 inside the parentheses: f(x) = 2(x² – 4x + 4 – 4) + 5.

步骤 2：在括号内进行配方，取 –4 的一半得 –2，平方得 4。在括号内加上并减去 4：f(x) = 2(x² – 4x + 4 – 4) + 5。

Step 3: Rewrite the perfect square trinomial and simplify: f(x) = 2[(x – 2)² – 4] + 5 = 2(x – 2)² – 8 + 5 = 2(x – 2)² – 3.

步骤 3：将完全平方式重写并化简：f(x) = 2[(x – 2)² – 4] + 5 = 2(x – 2)² – 8 + 5 = 2(x – 2)² – 3。

Hence the completed square form is:

因此，配方后的形式为：

f(x) = 2(x – 2)² – 3

Vertex: (h, k) = (2, –3). Axis of symmetry: x = 2.

顶点坐标为 (2, –3)，对称轴方程为 x = 2。

Since the coefficient a = 2 > 0, the parabola opens upwards, so the minimum value is –3. The range of f is f(x) ≥ –3.

由于系数 a = 2 > 0，抛物线开口向上，因此最小值为 –3，值域为 f(x) ≥ –3。

Common mistake: Forgetting to multiply the constant '–4' by the outside factor '2' before moving it out of the bracket.

常见错误：将括号内的常数 –4 移出时忘记先乘以外部的系数 2。

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2. Solving Quadratic Inequalities | 解二次不等式

Question: Solve the inequality 3x² + 5x – 2 > 0.

题目：解不等式 3x² + 5x – 2 > 0。

Step 1: Factorise the quadratic. Look for two numbers that multiply to 3 × (–2) = –6 and add to 5. The numbers are 6 and –1. Split the middle term: 3x² + 6x – x – 2 > 0. Factor by grouping: 3x(x + 2) – 1(x + 2) > 0 → (3x – 1)(x + 2) > 0.

步骤 1：对二次式进行因式分解。寻找两个数，其积为 3 × (–2) = –6，和为 5，这两个数是 6 和 –1。分解中项：3x² + 6x – x – 2 > 0。分组提取公因式：3x(x + 2) – 1(x + 2) > 0 → (3x – 1)(x + 2) > 0。

Step 2: Find the critical values where the expression equals zero: 3x – 1 = 0 gives x = 1/3; x + 2 = 0 gives x = –2.

步骤 2：找出使表达式等于零的临界值：3x – 1 = 0 得 x = 1/3；x + 2 = 0 得 x = –2。

Step 3: Place these values on a number line, dividing the line into three intervals: x < –2, –2 < x < 1/3, x > 1/3.

步骤 3：将这些值标在数轴上，将数轴分为三个区间：x < –2，–2 < x < 1/3，x > 1/3。

Step 4: Test a sample point from each interval in the factored inequality (3x – 1)(x + 2) > 0. For x < –2, pick x = –3: (negative)×(negative) = positive, so the inequality holds. For – ve) = negative, not > 0. For x > 1/3, pick x = 1: (positive)×(positive) = positive, holds.

步骤 4：在每个区间选取一个检测值代入因式分解后的不等式 (3x – 1)(x + 2) > 0。当 x < –2，取 x = –3：(负)×(负)= 正，不等式成立。当 –2 < x < 1/3，取 x = 0：(负)×(正)= 负，不成立。当 x > 1/3，取 x = 1：(正)×(正)= 正，成立。

Therefore, the solution set is x < –2 or x > 1/3.

因此，解集为 x < –2 或 x > 1/3。

Exam tip: Always sketch a quick sign diagram to avoid muddling greater than and less than signs.

考试技巧：始终快速画一个符号图，避免混淆大于号和小于号。

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3. Polynomials and the Factor Theorem | 多项式与因式定理

Question: Let f(x) = 2x³ – 3x² – 3x + 2. Show that (x – 2) is a factor, then factorise f(x) completely.

题目：设 f(x) = 2x³ – 3x² – 3x + 2。证明 (x – 2) 是一个因式，然后将 f(x) 彻底分解因式。

Step 1: Apply the Factor Theorem: if (x – 2) is a factor, then f(2) should equal 0. f(2) = 2(2)³ – 3(2)² – 3(2) + 2 = 16 – 12 – 6 + 2 = 0, so indeed (x – 2) is a factor.

步骤 1：应用因式定理：若 (x – 2) 为因式，则 f(2) 应等于 0。f(2) = 2(2)³ – 3(2)² – 3(2) + 2 = 16 – 12 – 6 + 2 = 0，因此 (x – 2) 确是因式。

Step 2: Divide f(x) by (x – 2) using polynomial long division or synthetic division. Performing the division yields quotient 2x² + x – 1.

步骤 2：用多项式长除法或综合除法除以 (x – 2)。完成除法后得到商式 2x² + x – 1。

Long division outline: 2x³ ÷ x = 2x²; multiply (x – 2) by 2x² → 2x³ – 4x²; subtract from f(x) to get x² – 3x + 2. Next, x² ÷ x = x; multiply → x² – 2x; subtract → –x + 2. Then –x ÷ x = –1; multiply → –x + 2; remainder 0.

长除法概略：2x³ ÷ x = 2x²；将 (x – 2) 乘以 2x² 得 2x³ – 4x²；相减得 x² – 3x + 2。接着 x² ÷ x = x；相乘得 x² – 2x；相减得 –x + 2。然后 –x ÷ x = –1；相乘得 –x + 2；余式为 0。

Step 3: Factorise the quadratic quotient. 2x² + x – 1 = (2x – 1)(x + 1).

步骤 3：将二次商式分解因式。2x² + x – 1 = (2x – 1)(x + 1)。

Thus, the complete factorisation is:

因此，完整因式分解为：

f(x) = (x – 2)(2x – 1)(x + 1)

Common mistake: Not fully factorising the quadratic quotient, especially when the leading coefficient is not 1.

常见错误：未能将二次商式彻底分解，尤其当首项系数不为 1 时。

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4. Binomial Expansion: Finding a Specific Coefficient | 二项式展开：求特定系数

Question: Find the coefficient of x³ in the expansion of (2 + 3x)⁵.

题目：求 (2 + 3x)⁵ 的展开式中 x³ 的系数。

Step 1: Use the binomial theorem: (a + b)ⁿ = Σ (n choose r) aⁿ⁻ʳ bʳ. Here a = 2, b = 3x, n = 5. The term containing x³ arises when r = 3.

步骤 1：使用二项式定理：(a + b)ⁿ = Σ (n choose r) aⁿ⁻ʳ bʳ。本题中 a = 2, b = 3x, n = 5。当 r = 3 时出现含有 x³ 的项。

Step 2: The general term for r = 3 is T₃₊₁ = ⁵C₃ · (2)⁵⁻³ · (3x)³.

步骤 2：r = 3 时的通项为 T₄ = ⁵C₃ · (2)² · (3x)³。

3: Calculate the binomial coefficient: ⁵C₃ = 10. Compute the powers: 2² = 4; (3x)³ = 27x³.

步骤 3：计算二项式系数：⁵C₃ = 10。计算幂：2² = 4；(3x)³ = 27x³。

ltiply these together: 10 × 4 × 27 x³ = 1080 x³. Therefore, the coefficient of x³ is 1080.

步骤 4：将这些值相乘：10 × 4 × 27 x³ = 1080 x³。因此，x³ 的系数为 1080。

Quick check: Ensure you correctly apply the power to both the constant 3 and the variable x inside (3x)³. A slip often produces 3x³ instead of 27x³.

快速检查：确保对括号内常数 3 和变量 x 都正确乘方。疏忽常导致误写为 3x³ 而非 27x³。

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5. Exponential Equations: Solving by a Common Base | 指数方程：化为同底求解

Question: Solve the equation 4^x+1 = 8^2x–3.

题目：解方程 4^x+1 = 8^2x–3。

Step 1: Express both sides with the same base. Since 4 = 2² and 8 = 2³, rewrite: (2²)^x+1 = (2³)^2x–3.

步骤 1：将两边表示为相同的底数。由 4 = 2²，8 = 2³，可重写为：(2²)^x+1 = (2³)^2x–3。

Step 2: Apply the power rule (aᵐ)ⁿ = aᵐⁿ: 2^(2(x+1)) = 2^(3(2x–3)).

步骤 2：应用幂的乘方法则 (aᵐ)ⁿ = aᵐⁿ：2^(2(x+1)) = 2^(3(2x–3))。

Step 3: Since the bases are equal, equate the exponents: 2(x + 1) = 3(2x – 3).

步骤 3：既然底数相同，令指数相等：2(x + 1) = 3(2x – 3)。

Step 4: Solve the linear equation: 2x + 2 = 6x – 9 → 2 + 9 = 6x – 2x → 11 = 4x → x = 11/4.

步骤 4：解一元一次方程：2x + 2 = 6x – 9 → 2 + 9 = 6x – 2x → 11 = 4x → x = 11/4。

Always verify by substitution: LHS = 4^{11/4 + 1} = 4^15/4 = (2²)^15/4 = 2^15/2; RHS = 8^(2×(11/4) – 3) = 8^{11/2 – 3} = 8^5/2 = (2³)^5/2 = 2^15/2. Confirmed.

务必代入验证：左式 = 4^15/4 = 2^15/2，右式 = 8^5/2 = 2^15/2，确认无误。

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6. Logarithmic Equations: Checking Domains | 对数方程：检查定义域

Question: Solve log₂(x + 1) + log₂(x – 1) = 3.

题目：解方程 log₂(x + 1) + log₂(x – 1) = 3。

Step 1: Combine the logs using the product rule: log₂[(x + 1)(x – 1)] = 3.

步骤 1：利用对数乘法法则合并：log₂[(x + 1)(x – 1)] = 3。

Step 2: Rewrite in exponential form: (x + 1)(x – 1) = 2³ = 8.

步骤 2：化为指数形式：(x + 1)(x – 1) = 2³ = 8。

Step 3: Expand and simplify: x² – 1 = 8 → x² = 9 → x = 3 or x = –3.

步骤 3：展开并化简：x² – 1 = 8 → x² = 9 → x = 3 或 x = –3。

Step 4: Check the domain restrictions. The original logs require x + 1 > 0 and x – 1 > 0, so x > –1 and x > 1, i.e. x > 1. x = –3 is invalid. x = 3 satisfies both conditions.

步骤 4：检查定义域限制。原对数要求 x + 1 > 0 且 x – 1 > 0，即 x > –1 且 x > 1，也就是 x > 1。x = –3 无效。x = 3 满足条件。

Therefore, the only valid solution is x = 3.

因此，唯一有效解为 x = 3。

Key reminder: Always state the domain before solving and reject extraneous solutions—losing marks here is very common.

重要提醒：解题前务必先声明定义域，并舍去增根；此处失分极为常见。

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7. Trigonometric Equations: Quadratic in sin x | 三角方程：关于 sin x 的二次方程

Question: Solve 2 sin²x – sin x – 1 = 0 for 0° ≤ x ≤ 360°.

题目：在 0° ≤ x ≤ 360° 范围内解方程 2 sin²x – sin x – 1 = 0。

Step 1: Treat as a quadratic in sin x. Let u = sin x, then 2u² – u – 1 = 0.

步骤 1：将此方程视作关于 sin x 的二次方程。令 u = sin x，则 2u² – u – 1 = 0。

Step 2: Factorise: (2u + 1)(u – 1) = 0, so u = –½ or u = 1.

步骤 2：因式分解：(2u + 1)(u – 1) = 0，因此 u = –½ 或 u = 1。

Step 3: Re-substitute: sin x = –½ or sin x = 1. Solve each in the given interval.

步骤 3：代回原函数：sin x = –½ 或 sin x = 1。在指定范围内分别求解。

For sin x = 1: the only solution in 0°–360° is x = 90° (the peak of the sine curve).

对于 sin x = 1：在 0°–360° 间唯一解为 x = 90°（正弦曲线的波峰）。

For sin x = – The reference angle for sin θ = ½ is 30°. In QIII, x = 180° + 30° = 210°; in QIV, x = 360° – 30° = 330°.

对于 sin x = –½：正弦在第三和第四象限为负。sin θ = ½ 的参考角为 30°。第三象限：x = 180° + 30° = 210°；第四象限：x = 360° – 30° = 330°。

All solutions: x = 90°, 210°, 330°.

全部解为：x = 90°、210°、330°。

so only one quadrant is needed. For the negative case, missing one of the quadrants loses marks.

常见错误：解 sin x = 正值时忘记第二象限，但此处 sin x = 1 是特例所以只需要一个象限。对负值情况，漏掉任一象限都会丢分。

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8. Differentiation: Equation of a Tangent | 微分：切线方程

Question: Find the equation of the tangent to the curve y = x³ – 3x + 2 at the point where x = 1.

题目：求曲线 y = x³ – 3x + 2 在 x = 1 处的切线方程。

Step 1: Find the y-coordinate when x = 1: y(1) = (1)³ – 3(1) + 2 = 0, so the point is (1, 0).

步骤 1：求 x = 1 时的 y 坐标：y(1) = 1 – 3 + 2 = 0，因此切点为 (1, 0)。

Step 2: Differentiate to find the gradient function: dy/dx = 3x² – 3.

步骤 2：求导得到梯度函数：dy/dx = 3x² – 3。

Step 3: Evaluate the gradient at x = 1: m = 3(1)² – 3 = 0. So the tangent is horizontal.

步骤 3：计算 x = 1 处的梯度：m = 3(1)² – 3 = 0。因此切线为水平线。

Step 4: Use the line equation y – y₁ = m(x – x₁): y – 0 = 0(x – 1) → y = 0.

步骤 4：使用直线方程 y – y₁ = m(x – x₁)：y – 0 = 0(x – 1) → y = 0。

The equation of the tangent is y = 0 (the x-axis).

切线方程为 y = 0（即 x 轴）。

This is a neat example where the gradient is zero—don’t be surprised by simple results; they often appear to test your understanding of stationary points.

这是一个梯度为零的简洁示例——不要对简单结果感到意外；这类题常用来检验你对驻点的理解。

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9. Coordinate Geometry: Intersection of Line and Circle | 坐标几何：直线与圆的交点

The line y = 2x + k is a tangent to the circle x² + y² = 5. Find the possible values of k.

题目：已知直线 y = 2x + k 与圆 x² + y² = 5 相切，求 k 的所有可能值。

Step 1: Substitute y = 2x + k into the circle equation: x² + (2x + k)² = 5.

步骤 1：将 y = 2x + k 代入圆的方程：x² + (2x + k)² = 5。

Step 2: Expand and simplify: x² + 4x² + 4kx + k² = 5 → 5x² + 4kx + (k² – 5) = 0.

步骤 2：展开并化简：x² + 4x² + 4kx + k² = 5 → 5x² + 4kx + (k² – 5) = 0。

atic must have exactly one solution, so its discriminant Δ = 0. Here a = 5, b = 4k, c = k² – 5.

步骤 3：要使直线与圆相切，二次方程必须有唯一解，因此判别式 Δ = 0。此处 a = 5, b = 4k, c = k² – 5。

Δ = b² – 4ac = (4k)² – 4×5×(k² – 5) = 16k² – 20(k² – 5) = 16k² – 20k² + 100 = –4k² + 100 = 0.

Δ = (4k)² – 4×5×(k² – 5) = 16k² – 20k² + 100 = –4k² + 100 = 0。

Step 4: Solve –4k² + 100 = 0 → 4k² = 100 → k² = 25 → k = ±5.

步骤 4：解 –4k² + 100 = 0 → k² = 25 → k = ±5。

Thus the two tangent lines are y = 2x + 5 and y = 2x – 5.

因此两条切线分别为 y = 2x + 5 和 y = 2x – 5。

Exam insight: Tangecy problems almost always reduce to solving 'discriminant = 0' after substitution—keep your algebra tidy.

考试洞察：相切问题几乎总是通过代入后解“判别式 = 0”来解决——保持代数演算整洁。

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10. Functions: Domain, Range and Inverse | 函数：定义域、值域与反函数

Question: The function f is defined by f(x) = √2x – 4 for x ≥ 2. Write down the range of f, then find

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