Year 9 Cambridge Further Mathematics: Unit Test Mock Paper Analysis | 九年级剑桥进阶数学：单元测试模拟卷解析

📚 Year 9 Cambridge Further Mathematics: Unit Test Mock Paper Analysis | 九年级剑桥进阶数学：单元测试模拟卷解析

Welcome to this detailed walkthrough of a Year 9 Cambridge Further Mathematics unit test mock paper. The mock paper comprises ten questions that cover core topics such as algebra, equations, inequalities, surds and indices. Each question is unpacked with step-by-step solutions and explanations, highlighting key techniques and common pitfalls.

欢迎来到这篇针对九年级剑桥进阶数学单元测试模拟卷的详细解析。这份模拟卷包含十道题目，覆盖了代数、方程、不等式、根式与指数等核心主题。每题都将进行逐步解答与阐明，突出关键技巧与常见易错点。

1. Question 1: Expanding and Simplifying Polynomials | 问题1：多项式展开与化简

Question: Expand and simplify (3x – 2)(2x + 5) – 4(x – 3)².

题目：展开并化简 (3x – 2)(2x + 5) – 4(x – 3)²。

First, expand the two parts separately. Begin with the product (3x – 2)(2x + 5). Multiply each term in the first bracket by each term in the second: 3x·2x + 3x·5 – 2·2x – 2·5. This yields 6x² + 15x – 4x – 10, which simplifies to 6x² + 11x – 10.

首先，分别展开两部分。先处理乘积 (3x – 2)(2x + 5)：将第一个括号每一项乘以第二个括号每一项，即 3x×2x + 3x×5 – 2×2x – 2×5，得到 6x² + 15x – 4x – 10，合并为 6x² + 11x – 10。

Next, expand the square (x – 3)² = x² – 6x + 9. Then multiply by -4: -4(x² – 6x + 9) = -4x² + 24x – 36.

接着，展开平方项 (x – 3)² = x² – 6x + 9，再乘以 -4 得 -4x² + 24x – 36。

Finally, combine the two results: (6x² + 11x – 10) + (-4x² + 24x – 36). Group like terms: 6x² – 4x² = 2x², 11x + 24x = 35x, -10 – 36 = -46.

最后，将两部分合并：(6x² + 11x – 10) + (-4x² + 24x – 36)。合并同类项：x² 项 6x² – 4x² = 2x²，x 项 11x + 24x = 35x，常数项 -10 – 36 = -46。

Simplified expression: 2x² + 35x – 46

化简后表达式：2x² + 35x – 46

Common mistake: forgetting to distribute the negative sign to all terms of the squared bracket. Always expand the bracket first, then apply the multiplier with its sign.

常见错误：忘记将负号分配给平方展开式的每一项。务必先展开括号，再将系数连同符号整个乘入。

2. Question 2: Factorising Quadratic Expressions | 问题2：二次三项式因式分解

Question: Factorise completely 6x² + x – 12.

题目：对 6x² + x – 12 进行完全因式分解。

We look for two binomials (px + q)(rx + s) such that pr = 6, qs = -12, and the cross terms sum to the coefficient of x, which is 1. By trial, using 6x² = 3x·2x and -12 = -4·3, we test (3x – 4)(2x + 3).

寻找两个二项式 (px + q)(rx + s)，使 pr = 6，qs = -12，且交叉项之和等于 x 的系数 1。尝试以 6x² = 3x·2x 且 -12 = -4×3，检验 (3x – 4)(2x + 3)。

Expand to verify: outer terms 3x·3 = 9x, inner terms -4·2x = -8x. Their sum is 9x – 8x = x. The constant is -12. Hence the factorisation is correct.

展开验证：外项 3x×3 = 9x，内项 -4×2x = -8x，和为 x。常数项 -12 成立，因此因式分解正确。

6x² + x – 12 = (3x – 4)(2x + 3)

Tip: always check your factors by expanding mentally or on paper to avoid sign errors.

提示：务必通过心算或纸面展开来验算，以免符号错误。

3. Question 3: Solving Quadratic Equations by Factorising | 问题3：用因式分解法解二次方程

Question: Solve 2x² – 7x + 3 = 0 by factorising.

题目：用因式分解法解方程 2x² – 7x + 3 = 0。

Find factors of 2x² and +3 that give a middle term of -7x. Using 2x and x, and -3 and -1, we form (2x – 1)(x – 3). Check: -1·x + 2x·(-3) = -x – 6x = -7x, constant 3.

寻找 2x² 和 +3 的因数组合使得中间项为 -7x。选用 2x 与 x，以及 -1 与 -3，构成 (2x – 1)(x – 3)。验证：-1×x + 2x×(-3) = -x – 6x = -7x，常数项为 3。

Set each factor to zero: 2x – 1 = 0 gives x = 1/2; x – 3 = 0 gives x = 3.

令每个因式为零：2x – 1 = 0 得 x = 1/2；x – 3 = 0 得 x = 3。

Solutions: x = 1/2 and x = 3

解：x = 1/2 和 x = 3

Always present solutions as a set or clearly separated values.

务必以集合形式或明确分开的形式给出所有根。

4. Question 4: Solving Using the Quadratic Formula | 问题4：用求根公式解二次方程

Question: Solve 3x² – 5x – 2 = 0, using the quadratic formula. Give exact answers.

题目：用求根公式解 3x² – 5x – 2 = 0，给出精确值。

Identify a = 3, b = -5, c = -2. The quadratic formula is:

识别系数：a = 3, b = -5, c = -2。求根公式为：

x = [ -b ± √(b² – 4ac) ] / (2a)

Substitute the values: b² – 4ac = (-5)² – 4(3)(-2) = 25 + 24 = 49. Then √(49) = 7.

代入数值：b² – 4ac = (-5)² – 4×3×(-2) = 25 + 24 = 49，√49 = 7。

Thus x = [ -(-5) ± 7 ] / (2×3) = [5 ± 7] / 6. This yields two solutions: (5 + 7)/6 = 12/6 = 2, and (5 – 7)/6 = -2/6 = -1/3.

因此 x = [5 ± 7] / 6。两个解：(5 + 7)/6 = 12/6 = 2；(5 – 7)/6 = -2/6 = -1/3。

Exact answers: x = 2 and x = -1/3

精确解：x = 2 和 x = -1/3

Remember: even if the discriminant is a perfect square, showing the formula steps demonstrates method mastery.

记住：即使判别式为完全平方，展示公式步骤也能体现对方法的掌握。

5. Question 5: Solving Simultaneous Equations | 问题5：解联立方程组

Question: Solve the simultaneous equations: 2x + y = 7, 3x – 2y = 1.

题目：解联立方程组：2x + y = 7，3x – 2y = 1。

Use elimination. Multiply the first equation by 2 to align coefficients of y: 4x + 2y = 14. Now add this to the second equation to eliminate y: (4x + 2y) + (3x – 2y) = 14 + 1 → 7x = 15.

采用消元法。将第一个方程乘以 2 使 y 系数对齐：4x + 2y = 14。与第二个方程相加消去 y：(4x + 2y) + (3x – 2y) = 14 + 1 → 7x = 15。

Solve for x: x = 15/7. Substitute back into the simpler original equation, 2x + y = 7, to find y: y = 7 – 2x = 7 – 2*(15/7) = 7 – 30/7 = 49/7 – 30/7 = 19/7.

解出 x = 15/7。代回较简的方程 2x + y = 7 求 y：y = 7 – 2×(15/7) = 7 – 30/7 = 49/7 – 30/7 = 19/7。

Solution: x = 15/7, y = 19/7

解：x = 15/7, y = 19/7

Check in the second equation: 3*(15/7) – 2*(19/7) = 45/7 – 38/7 = 7/7 = 1. Verified.

代入第二个方程验算：3×(15/7) – 2×(19/7) = 45/7 – 38/7 = 7/7 = 1，成立。

6. Question 6: Solving Linear Inequalities | 问题6：解一元一次不等式

Question: Solve the inequality 2(3x – 1) < 5x + 8, and illustrate the solution on a number line.

题目：解不等式 2(3x – 1) < 5x + 8，并在数轴上表示解集。

Expand the left side: 6x – 2 < 5x + 8. Subtract 5x from both sides: x - 2 < 8. Then add 2: x < 10.

展开左边：6x – 2 < 5x + 8。两边减 5x 得 x - 2 < 8，再加 2 得 x < 10。

Solution set: x < 10

解集：x < 10

On a number line, draw an open circle at 10 and shade all values to the left. The open circle indicates that 10 is not included.

在数轴上，在 10 处画空心圆，并向左画阴影。空心圆表示 10 不包含在内。

7. Question 7: Simplifying Surds | 问题7：化简根式

Question: Simplify √48 – √27 + 2√12.

题目：化简 √48 – √27 + 2√12。

Express each surd in terms of its simplest square factor. √48 = √(16×3) = 4√3; √27 = √(9×3) = 3√3; √12 = √(4×3) = 2√3. Therefore 2√12 = 2×2√3 = 4√3.

将每个根式化为最简二次根式：√48 = √(16×3) = 4√3；√27 = √(9×3) = 3√3；√12 = √(4×3) = 2√3，故 2√12 = 2×2√3 = 4√3。

Combine the terms: 4√3 – 3√3 + 4√3 = (4 – 3 + 4)√3 = 5√3.

合并同类根式：4√3 – 3√3 + 4√3 = (4 – 3 + 4)√3 = 5√3。

Simplified form: 5√3

化简结果：5√3

Always factor out the largest perfect square to minimise steps and errors.

务必提取最大的完全平方因子，以减少步骤和错误。

8. Question 8: Applying Laws of Indices | 问题8：指数律应用

Question: Simplify (2x³y⁻²)⁴ × (3x⁻¹y²)³ / (6x⁵y⁴). Express with positive indices.

题目：化简 (2x³y⁻²)⁴ × (3x⁻¹y²)³ / (6x⁵

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Year 9 Cambridge Further Mathematics: Unit Test Mock Paper Analysis | 九年级剑桥进阶数学：单元测试模拟卷解析