📚 GCSE CIE Further Pure Maths Unit Test Mock Paper Analysis | GCSE CIE 进阶数学单元测试模拟卷解析
A well-structured mock paper helps students identify strengths and weaknesses before the actual examination. This analysis breaks down typical Unit Test questions, providing step-by-step solutions and highlighting common pitfalls. Mastering these techniques will boost your confidence and exam performance.
精心设计的模拟卷能帮助学生在正式考试前发现自己的强项与薄弱环节。本文剖析了典型的单元测试题，给出分步解答并指出常见误区。掌握这些技巧将提升你的信心和考试成绩。
1. Functions and Composite Functions | 函数与复合函数
Given f(x) = 2x + 1 and g(x) = x / (x – 1), find fg(x) and state its domain.
已知 f(x) = 2x + 1，g(x) = x / (x – 1)，求 fg(x) 并写出其定义域。
fg(x) means f(g(x)). Substitute g(x) into f: f(g(x)) = 2 * [x/(x – 1)] + 1 = (2x/(x – 1)) + 1. Combine over a common denominator: (2x + (x – 1)) / (x – 1) = (3x – 1)/(x – 1). The domain excludes x-values that make any denominator zero. g(x) is undefined when x – = 0, so x ≠ 1. The composite function fg(x) has the same restriction. Therefore, domain: x ∈ ℝ, x ≠ 1.
fg(x) 表示 f(g(x))。将 g(x) 代入 f：f(g(x)) = 2 * [x/(x – 1)] + 1 = (2x/(x – 1)) + 1。通分得：(2x + (x – 1)) / (x – 1) = (3x – 1)/(x – 1)。定义域要排除使任何分母为零的 x 值。g(x) 在 x – 1 = 0 时无定义，故 x ≠ 1。复合函数 fg(x) 也受此限制。因此定义域为 x ∈ ℝ, x ≠ 1。
2. Quadratic Functions and the Discriminant | 二次函数与判别式
Find the range of values of k for which the equation x² + kx + 9 = 0 has no real roots.
求使方程 x² + kx + 9 = 0 无实数根的 k 的取值范围。
For a quadratic ax² + bx + c = 0 to have no real roots, the discriminant Δ = b² – 4ac must be less than zero. Here a = 1, b = k, c = 9. So Δ = k² – 4 × 1 × 9 = k² – 36. Set Δ < 0 → k² – 36 < 0 → (k – 6)(k + 6) < 0. Solving the inequality gives –6 < k < 6.
对于二次方程 ax² + bx + c = 0 无实数根，判别式 Δ = b² – 4ac 必须小于零。此处 a = 1, b = k, c = 9。故 Δ = k² – 4×1×9 = k² – 36。令 Δ < 0 → k² – 36 < 0 → (k – 6)(k + 6) < 0。解不等式得 –6 < k < 6。
3. Exponential and Logarithmic Equations | 指数与对数方程
Solve 3²ˣ⁻¹ = 5ˣ, giving the answer in the form ln a / ln b.
解方程 3²ˣ⁻¹ = 5ˣ，答案以 ln a / ln b 的形式表示。
Take natural logs of both sides: ln(3²ˣ⁻¹) = ln(5ˣ). Use the power rule: (2x – 1) ln 3 = x ln 5. Expand: 2x ln 3 – ln 3 = x ln 5. Collect x terms: 2x ln 3 – x ln 5 = ln 3. Factor: x(2 ln 3 – ln 5) = ln 3. Therefore x = ln 3 / (2 ln 3 – ln 5). Using log rules, denominator = ln(3²) – ln 5 = ln(9/5). So x = ln 3 / ln(9/5), which is of the form ln a / ln b.
两边取自然对数：ln(3²ˣ⁻¹) = ln(5ˣ)。运用幂法则：(2x – 1) ln 3 = x ln 5。展开：2x ln 3 – ln 3 = x ln 5。合并含 x 项：2x ln 3 – x ln 5 = ln 3。提取公因式：x(2 ln 3 – ln 5) = ln 3
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