📚 Year 11 CAIE Maths: Unit Test Mock Paper Walkthrough | Year 11 CAIE 数学：单元测试模拟卷解析
This article walks you through a carefully designed unit test mock paper for Year 11 CAIE IGCSE Mathematics. The selected questions cover essential topics from the Algebra, Functions and Graphs strand, including linear equations, quadratic factorisation, simultaneous equations, inequalities, composite functions, and direct proportion. Each question is followed by a step-by-step solution with paired English–Chinese explanations to help you master the methods, avoid common pitfalls, and build the confidence needed for your actual exam.
本文详细解析了一套为 Year 11 CAIE IGCSE 数学设计的单元测试模拟卷。所选题目覆盖了代数、函数与图像模块的核心考点，包括一元一次方程、二次因式分解、联立方程组、不等式、复合函数以及正比例关系。每道题都配有逐步求解过程，并采用英中双语解释，帮助你掌握解题方法、规避常见错误、建立迎战真实考试的信心。
1. Question 1: Solving Linear Equations | 第1题：解一元一次方程
Question 1: Solve 3(2x − 5) + 7 = 4x + 2.
First, expand the bracket. Multiply 3 by each term inside: 3 × 2x = 6x and 3 × (−5) = −15. The equation becomes 6x − 15 + 7 = 4x + 2.
首先展开括号。将 3 乘以括号内的每一项：3 × 2x = 6x，3 × (−5) = −15。方程变为 6x − 15 + 7 = 4x + 2。
Now simplify the left-hand side by combining the constant terms: −15 + 7 = −8. So we have 6x − 8 = 4x + 2.
化简左边，合并常数项：−15 + 7 = −8。得到 6x − 8 = 4x + 2。
Collect the x-terms on one side by subtracting 4x from both sides: 6x − 4x − 8 = 2, which gives 2x − 8 = 2. Then add 8 to both sides: 2x = 10. Finally, divide by 2: x = 5.
将含 x 的项移到一边，两边同时减去 4x：6x − 4x − 8 = 2，得到 2x − 8 = 2。然后两边加 8：2x = 10。最后除以 2 得 x = 5。
2. Question 2: Quadratic Factorisation | 第2题：二次因式分解
Question 2: Factorise x² − 7x + 12 fully.
We need two numbers that multiply to the constant term +12 and add up to the coefficient of x, which is −7. Since the product is positive and the sum is negative, both numbers must be negative.
我们需要寻找两个数，它们的乘积等于常数项 +12，和等于 x 的系数 −7。因为乘积为正、和为负，这两个数必定都是负数。
The pairs of factors of 12 are (1, 12), (2, 6), (3, 4). Testing with negative signs: (−3) + (−4) = −7, and (−3) × (−4) = 12. This works perfectly.
12 的因数对有 (1,12)、(2,6)、(3,4)。加上负号检验：(−3) + (−4) = −7，且 (−3) × (−4) = 12。完全符合。
Therefore, the factorised form is (x − 3)(x − 4). Always check by expanding: (x − 3)(x − 4) = x² − 4x − 3x + 12 = x² − 7x + 12. Correct.
因此，因式分解的结果是 (x − 3)(x − 4)。记得展开验证：(x − 3)(x − 4) = x² − 4x − 3x + 12 = x² − 7x + 12。正确。
3. Question 3: Solving a Quadratic Equation | 第3题：解二次方程
Question 3: Solve 2x² − 3x − 2 = 0.
First, factorise the quadratic expression. Look for two numbers that multiply to give the product of the coefficient of x² and the constant term: 2 × (−2) = −4, and add to −3 (the coefficient of x). The numbers are −4 and +1, because (−4) × 1 = −4 and (−4) + 1 = −3.
首先，对二次式进行因式分解。寻找两个数，它们的乘积等于二次项系数与常数项的积：2 × (−2) = −4，和等于 −3（x 的系数）。这两个数是 −4 和 +1，因为 (−4) × 1 = −4，(−4) + 1 = −3。
Split the middle term using these numbers: 2x² − 4x + x − 2 = 0. Then factor by grouping: 2x(x − 2) + 1(x − 2) = 0. Factor out the common bracket (x − 2): (x − 2)(2x + 1) = 0.
利用这两个数拆分中间项：2x² − 4x + x − 2 = 0。然后分组提取公因式：2x(x − 2) + 1(x − 2) = 0。提取公因式 (x − 2) 得到 (x − 2)(2x + 1) = 0。
Set each factor equal to zero: x − 2 = 0 gives x = 2; 2x + 1 = 0 gives x = −½. The solutions are x = 2 and x = −½.
令每个因式等于零：x − 2 = 0 得 x = 2；2x + 1 = 0 得 x = −½。方程的解为 x = 2 和 x = −½。
4. Question 4: Simultaneous Equations | 第4题：联立方程组
Question 4: Solve the simultaneous equations: y = 2x + 1 and 3x + 2y = 14.
One equation is already in the form y = …, so substitution is the most efficient method. Substitute y = 2x + 1 into the second equation: 3x + 2(2x + 1) = 14.
其中一个方程已经是 y = … 的形式，因此代入法最为高效。将 y = 2x + 1 代入第二个方程：3x + 2(2x + 1) = 14。
Expand and simplify: 3x + 4x + 2 = 14 → 7x + 2 = 14. Subtract 2 from both sides: 7x = 12. Divide by 7: x = 12/7.
展开并化简：3x + 4x + 2 = 14 → 7x + 2 = 14。两边减 2：7x = 12。除以 7 得 x = 12/7。
Now substitute this x-value back into y = 2x + 1: y = 2(12/7) + 1 = 24/7 + 7/7 = 31/7. The solution is x = 12/7, y = 31/7. You can leave it as an improper fraction.
将 x 的值代回 y = 2x + 1：y = 2(12/7) + 1 = 24/7 + 7/7 = 31/7。方程组的解为 x = 12/7, y = 31/7。答案可以保留为假分数。
5. Question 5: Equation of a Straight Line | 第5题：直线的方程
Question 5: Find the equation of the straight line that passes through (2, 5) and has gradient −3.
Use the point–gradient form of a linear equation: y − y1 = m(x − x1). Here m = −3, and (x1, y1) = (2, 5).
使用直线的点斜式方程：y − y1 = m(x − x1)。这里 m = −3, (x1, y1) = (2, 5)。
Substitute the values: y − 5 = −3(x − 2). Expand the right-hand side: y − 5 = −3x + 6. Then add 5 to both sides to isolate y: y = −3x + 11.
代入数值：y − 5 = −3(x − 2)。展开右边：y − 5 = −3x + 6。然后两边加 5 求出 y：y = −3x + 11。
The equation of the line is y = −3x + 11. Always check by substituting the given point: when x = 2, y = −3(2) + 11 = 5, so it passes through (2, 5). The gradient is indeed −3.
直线的方程为 y = −3x + 11。务必代入已知点检验：当 x = 2 时，y = −3(2) + 11 = 5，所以直线经过 (2,5)。斜率确实为 −3。
6. Question 6: Solving an Inequality | 第6题：解不等式
Question 6: Solve the inequality 4 − 3x > 2x − 11 and represent the solution on a number line.
First, collect the x-terms on one side. Add 3x to both sides: 4 > 5x − 11. Then add 11 to both sides: 15 > 5x. Alternatively, you could have moved the terms differently; the process is valid as long as you treat the inequality sign correctly.
首先，将含 x 的项移到一侧。两边加 3x：4 > 5x − 11。然后两边加 11：15 > 5x。当然，你也可以用其他移项方式，只要正确处理不等号即可。
Now divide both sides by 5: 3 > x, which is the same as x < 3. Note that we divided by a positive number, so the inequality direction stays unchanged.
两边除以 5：3 > x，等同于 x < 3。注意我们除以的是正数，因此不等号方向保持不变。
On a number line, this is shown as an open circle at 3, with an arrow pointing to the left, indicating all numbers less than 3.
在数轴上，解集表示为 3 处画一个空心圆，并向左画箭头，代表所有小于 3 的数。
7. Question 7: Composite Functions | 第7题：复合函数
Question 7: Given f(x) = 3x − 2 and g(x) = x² + 1, evaluate fg(−1) and find the expression for gf(x).
fg(−1) means f(g(−1)). Start with the inner function: g(−1) = (−1)² + 1 = 1 + 1 = 2. Then apply f: f(2) = 3(2) − 2 = 6 − 2 = 4.
fg(−1) 表示 f(g(−1))。先计算内层函数：g(−1) = (−1)² + 1 = 1 + 1 = 2。然后代入 f：f(2) = 3(2) − 2 = 6 − 2 = 4。
For gf(x), we substitute f(x) into g. Start with the outer function g and replace x with f(x): gf(x) = g(f(x)) = (f(x))² + 1.
对于 gf(x)，我们将 f(x) 代入 g。从外层函数 g 入手，把 x 替换为 f(x)：gf(x) = g(f(x)) = (f(x))² + 1。
Now replace f(x) with 3x − 2: gf(x) = (3x − 2)² + 1. Expand the square: (3x − 2)² = 9x² − 12x + 4. Then add 1: gf(x) = 9x² − 12x + 5.
将 f(x) 替换为 3x − 2：gf(x) = (3x − 2)² + 1。展开平方项：(3x − 2)² = 9x² − 12x + 4。再加 1 得：gf(x) = 9x² − 12x + 5。
Always remember the order: fg(x) means apply g first, then f; gf(x) means apply f first, then g.
务必牢记顺序：fg(x) 表示先作用 g 再作用 f；gf(x) 表示先作用 f 再作用 g。
8. Question 8: Direct Proportion with a Square | 第8题：含平方的正比例关系
Question 8: y is directly proportional to x². When x = 2, y = 16. Find y when x = 5.
Write the proportionality statement: y ∝ x², which means y = kx², where k is the constant of proportionality.
写出比例关系：y ∝ x²，即 y = kx²，其中 k 为比例常数。
Use the given pair to find k. Substitute x = 2 and y = 16: 16 = k × (2)² → 16 = 4k → k = 4. So the equation is y = 4x².
利用已知条件求 k。代入 x = 2, y = 16：16 = k × (2)² → 16 = 4k → k = 4。因此关系式为 y = 4x²。
Now find y when x = 5: y = 4 × (5)² = 4 × 25 = 100. The answer is y = 100.
现在求 x = 5 时的 y 值：y = 4 × (5)² = 4 × 25 = 100。答案为 y = 100。
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