📚 8H Prelims Answers and Worked Solutions | 8H 预考答案解析
Welcome to this worked-solution guide for the 8H Prelims mathematics paper. The questions below have been selected to reflect the most common skills tested in a preliminary examination, including algebra, functions, coordinate geometry, trigonometry, calculus, probability, statistics, proof, vectors and sequences. Each section gives a full solution with clear reasoning so that you can check your own method and learn how marks are awarded.
欢迎使用本 8H 预考数学卷答案解析指南。以下题目选自预考中最常考查的核心技能，包括代数、函数、坐标几何、三角学、微积分、概率、统计、证明、向量与数列。每一节都给出完整解答和清晰推理，帮助你核对自己的方法，并了解阅卷给分点。
1. Algebraic Simplification | 代数化简
Question: Simplify (3x² - 2x + 1) - (x² - 4x - 5) + 2(3x - 1).
题目：化简 (3x² - 2x + 1) - (x² - 4x - 5) + 2(3x - 1)。
4. Therefore the simplified expression is 2x² + 8x + 4, which can also be written as 2(x² + 4x + 2).
解答：去括号时要仔细。第一个括号保持不变：3x² - 2x + 1。第二个括号前是负号，所以括号内各项符号全部改变：-x² + 4x + 5。第三项展开为 6x - 2。合并同类项：x² 项为 3x² - x² = 2x²；x 项为 -2x + 4x + 6x = 8x；常数项为 1 + 5 - 2 = 4。因此化简结果为 2x² + 8x + 4，也可写作 2(x² + 4x + 2)。
Common mistake: When subtracting a bracket, many students forget to change every sign inside the bracket. Always rewrite the expression with the signs flipped before collecting like terms.
常见错误：减去括号时，许多学生忘记改变括号内每一项的符号。在合并同类项之前，一定要先把括号内各项的符号改过来。
2. Solving Quadratic Equations | 解二次方程
Question: Solve 2x² - 5x - 3 = 0 by factorisation.
题目：用因式分解法解方程 2x² - 5x - 3 = 0。
+ 1 = 0 gives x = -½; x - 3 = 0 gives x = 3. Therefore the solutions are x = -½ or x = 3.
解答：我们需要找到两个数，它们的乘积为 2 × (-3) = -6，和为 -5。这两个数是 -6 和 1。拆分中间项：2x² - 6x + x - 3 = 0。分组分解：2x(x - 3) + 1(x - 3) = 0，即 (2x + 1)(x - 3) = 0。令每个因式为零：2x + 1 = 0 得 x = -½；x - 3 = 0 得 x = 3。因此方程的解为 x = -½ 或 x = 3。
Exam tip: Always check your factorisation by expanding the brackets back. If you get the original quadratic, your factors are correct. Also remember that a quadratic can have two, one or no real solutions.
应试技巧：分解因式后一定要把括号展开检查。如果得到原二次方程，说明因式正确。还要记住二次方程可能有两个、一个或没有实数解。
3. Functions and Inverse Functions | 函数与反函数
Question: Given f(x) = 3x - 2, find f⁻¹(x) and hence evaluate f⁻¹(7).
题目：已知 f(x) = 3x - 2，求 f⁻¹(x)，并求 f⁻¹(7)。
) ÷ 3. Thus f⁻¹(x) = (x + 2)/3. To evaluate f⁻¹(7), substitute x = 7: (7 + 2)/3 = 9/3 = 3.
解答：设 y = 3x - 2。求反函数时，交换 x 和 y：x = 3y - 2。解出 y：x + 2 = 3y，所以 y = (x + 2) ÷ 3。因此 f⁻¹(x) = (x + 2)/3。求 f
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