Year 10 Cambridge Further Maths: Unit Test Mock Paper Analysis | 剑桥10年级进阶数学：单元测试模拟卷解析

📚 Year 10 Cambridge Further Maths: Unit Test Mock Paper Analysis | 剑桥10年级进阶数学：单元测试模拟卷解析

This article provides a detailed walkthrough of a mock unit test for Year 10 Cambridge Further Mathematics. It covers typical topics such as quadratic equations, functions, logarithms, trigonometry, differentiation, and integration. Each question is solved step by step with clear explanations to help you prepare for your actual exam.

本文详细解析了一套针对剑桥10年级进阶数学的单元模拟测试卷。内容涵盖二次方程、函数、对数、三角学、微分和积分等常见主题。每道题均配有逐步解析，帮助同学们为考试做好充分准备。

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1. Quadratic Equations and Inequalities | 二次方程与不等式

Question uadratic equation 2x² - 5x - 3 = 0. (b) Hence, find the solution set for 2x² - 5x - 3 > 0.

问题：(a) 解二次方程 2x² - 5x - 3 = 0。 (b) 由此求出不等式 2x² - 5x - 3 > 0 的解集。

e are -6 and 1, so we split the middle term and factor by grouping, or directly recognise (2x+1)(x-3).

首先对二次表达式进行因式分解。寻找两个数，乘积为 2 × (-3) = -6，且和为 -5。这两个数是 -6 和 1，因此可以通过拆分中项分组分解，或直接写成 (2x+1)(x-3)。

2x² - 5x - 3 = (2x+1)(x-3)

Set each factor to zero to solve 2x² - 5x - 3 = 0.

令每个因子为零以解方程 2x² - 5x - 3 = 0。

2x+1 = 0 ⇒ x = -1/2;
x-3 = 0 ⇒ x = 3

For the inequality 2x² - 5x - 3 > 0, the roots -1/2 and 3 divide the number line into three intervals. Test a point in each interval or note that the parabola opens upward (coefficient of x² is positive). The product (2x+1)(x-3) is positive when x < -1/2 or x > 3.

对于不等式 2x² - 5x - 3 > 0，根 -1/2 和 3 将数轴分成三个区间。在每个区间内选点检验，或注意到抛物线开口向上（x² 系数为正）。乘积 (2x+1)(x-3) 在 x < -1/2 或 x > 3 时为正。

Hence, the solution set is {x : x < -1/2} ∪ {x : x > 3}.

因此解集为 {x : x < -1/2} ∪ {x : x > 3}。

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2. Composite Functions | 复合函数

Question (x) = 3x - 1 and g(x) = x² + 2, find (i) fg(x), (ii) gf(x), and (iii) solve fg(x) = gf(x).

问题：已知 f(x) = 3x - 1 和 g(x) = x² + 2，求 (i) fg(x)，(ii) gf(x)，以及 (iii) 解方程 fg(x) = gf(x)。

)). Substitute g(x) into f: f(g(x)) = 3(g(x)) - 1 = 3(x² + 2) - 1 = 3x² + 6 - 1 = 3x² + 5.

fg(x) 表示 f(g(x))。将 g(x) 代入 f： f(g(x)) = 3(g(x)) - 1 = 3(x² + 2) - 1 = 3x² + 6 - 1 = 3x² + 5。

gf(x) means g(f(x)). Substitute f(x) into g: g(f(x)) = (f(x))² + 2 = (3x - 1)² + 2.

gf(x) 表示 g(f(x))。将 f(x) 代入 g： g(f(x)) = (f(x))² + 2 = (3x - 1)² + 2。

Expand: (3x - 1)² = 9x² - 6x + 1, so gf(x) = 9x² - 6x + 1 + 2 = 9x² - 6x + 3

展开：(3x - 1)² = 9x² - 6x + 1，因此 gf(x) = 9x² - 6x + 1 + 2 = 9x² - 6x + 3

Set fg(x) = gf(x): 3x² + 5 = 9x² - 6x + 3.

令 fg(x) = gf(x)： 3x² + 5 = 9x² - 6x + 3。

Rearrange to bring all terms to one side: 0 = 9x² - 6x + 3 - 3x² - 5 ⇒ 0 = 6x² - 6x - 2.

移项将所有项移至一边： 0 = 9x² - 6x + 3 - 3x² - 5 ⇒ 0 = 6x² - 6x - 2。

Divide by 2: 3x² - 3x - 1 = 0. Use the quadratic formula: x = [3 ± √9 + 12] / (6) = [3 ± √21] / 6.

除以 2 得 3x² - 3x - 1 = 0。使用求根公式：x = [3 ± √9 + 12] / 6 = [3 ± √21] / 6。

Thus the solutions are x = (3 + √21)/6 and x = (3 - √21)/6.

因此解为 x = (3 + √21)/6 和 x = (3 - √21)/6。

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3. Exponential and Logarithmic Equations | 指数与对数方程

Question: Solve (a) 3²ˣ = 27, and (b) log₂(x+1) - log₂(x-1) = 2.

问题：解方程 (a) 3²ˣ = 27，以及 (b) log₂(x+1) - log₂(x-1) = 2。

Part (a): Recognise that 27 = 3³. So the equation becomes 3²ˣ = 3³. Equating exponents gives 2x = 3, so x = 3/2.

第 (a) 部分：注意到 27 = 3³，因此方程变为 3²ˣ = 3³。指数相等得 2x = 3，故 x = 3/2。

Part (b): Use the log subtraction law: log₂(x+1) - log₂(x-1) = log₂[(x+1)/(x-1)] = 2.

第 (b) 部分：使用对数减法法则：log₂(x+1) - log₂(x-1) = log₂[(x+1)/(x-1)] = 2。

Write in exponential form: (x+1)/(x-1) = 2² = 4.

写成指数形式：(x+1)/(x-1) = 2² = 4。

Multiply both sides by (x-1): x+1 = 4(x-1) ⇒ x+1 = 4x - 4.

两边同乘 (x-1)：x+1 = 4(x-1) ⇒ x+1 = 4x - 4。

Solve: 1+4 = 4x - x ⇒ 5 = 3x ⇒ x = 5/3.

求解得：1+4 = 4x - x ⇒ 5 = 3x ⇒ x = 5/3。

Check the domain: x+1 > 0 and x-1 > 0 ⇒ x > 1. x = 5/3 satisfies this, so it is valid.

检验定义域：x+1 > 0 且 x-1 > 0 ⇒ x > 1。x = 5/3 满足条件，解有效。

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4. Equation of a Tangent to a Circle | 圆的切线方程

Question: Find the equation of the tangent to the circle x² + y² = 25 at the point (3,4).

问题：求圆 x² + y² = 25 在点 (3,4) 处的切线方程。

The circle has centre (0,0) and radius 5. The gradient of the radius from the centre to the point (3,4) is m_rad = (4-0)/(3-0) = 4/3.

该圆圆心为 (0,0)，半径为 5。从圆心到点 (3,4) 的半径斜率为 m_rad = (4-0)/(3-0) = 4/3。

The tangent is perpendicular to the radius, so its gradient m_tan = -1 / (4/3) = -3/4.

切线与半径垂直，因此其斜率 m_tan = -1 / (4/3) = -3/4。

Using the point-slope form with point (3,4): y - 4 = (-3/4)(x - 3).

使用点斜式，点为 (3,4)：y - 4 = (-3/4)(x - 3)。

Multiply by 4: 4y - 16 = -3(x - 3) ⇒ 4y - 16 = -3x + 9.

两边乘 4：4y - 16 = -3(x - 3) ⇒ 4y - 16 = -3x + 9。

Rearrange to standard form: 3x + 4y - 25 = 0.

整理为标准形式：3x + 4y - 25 = 0。

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5. Trigonometric Identity Proving | 三角恒等式证明

Question: Prove that (sinθ + cosθ)² = 1 + sin2θ.

问题：证明 (sinθ + cosθ)² = 1 + sin2θ。

Start with the left-hand side (LHS): Expand (sinθ + cosθ)².

从左式 (LHS) 开始：展开 (sinθ + cosθ)²。

LHS = sin²θ + 2 sinθ cosθ + cos²θ

Use the Pythagorean identity sin²θ + cos²θ = 1.

使用恒等式 sin²θ + cos²θ = 1。

LHS = (sin²θ + cos²θ) + 2 sinθ cosθ = 1 + 2 sinθ cosθ

Recall the double-angle formula: sin2θ = 2 sinθ cosθ.

回忆二倍角公式：sin2θ = 2 sinθ cosθ。

Therefore, LHS = 1 + sin2θ = RHS, as required.

因此 LHS = 1 + sin2θ = RHS，证毕。

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6. Stationary Points via Differentiation | 利用微分求驻点

Question: Find the stationary points of the curve y = x³ - 3x + 2 and determine their nature.

问题：求曲线 y = x³ - 3x + 2 的驻点并判断其性质。

First derivative: dy/dx = 3x² - 3. Set dy/dx = 0 to find stationary points.

求一阶导数：dy/dx = 3x² - 3。令 dy/dx = 0 以寻找驻点。

3x² - 3 = 0 ⇒ 3(x² - 1) = 0 ⇒ x² = 1 ⇒ x = 1 or x = -1

s: For x = 1, y = (1)³ - 3(1) + 2 = 0. For x = -1, y = (-1)³ - 3(-1) + 2 = -1 + 3 + 2 = 4.

求相应的 y 坐标：当 x = 1 时，y = 1 - 3 + 2 = 0。当 x = -1 时，y = -1 + 3 + 2 = 4。

Stationary points are (1, 0) and (-1, 4).

驻点为 (1, 0) 和 (-1, 4)。

Determine nature using second derivative: d²y/dx² = 6x.

利用二阶导数判断性质：d²y/dx² = 6x。

At x = 1, d²y/dx² = 6 > 0 ⇒ minimum point. At x = -1, d²y/dx² = -6 < 0 ⇒ maximum point.

在 x = 1 处，d²y/dx² = 6 > 0 ⇒ 极小值点。在 x = -1 处，d²y/dx² = -6 < 0 ⇒ 极大值点。

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7. Definite Integration | 定积分计算

Question: Evaluate ∫₀² (2x³ - 3x² + 1) dx.

问题：计算定积分 ∫₀² (2x³ - 3x² + 1) dx。

vative term by term: ∫ 2x³ dx = (2/4)x⁴ = (1/2)x⁴; ∫ -3x² dx = (-3/3)x³ = -x³; ∫ 1 dx = x.

逐项求原函数：∫ 2x³ dx = (2/4)x⁴ = (1/2)x⁴；∫ -3x² dx = (-3/3)x³ = -x³；∫ 1 dx = x。

Thus, F(x) = (1/2)x⁴ - x³ + x.

因此 F(x) = (1/2)x⁴ - x³ + x。

Apply the limits: F(2) - F(0).

代入上下限：F(2) - F(0)。

F(2) = (1/2)(16) - 8 + 2 = 8 - 8 + 2 = 2. F(0) = 0 - 0 + 0 = 0.

F(2) = (1/2)(16) - 8 + 2 = 8 - 8 + 2 = 2。F(0) = 0 - 0 + 0 = 0。

The value of the definite integral is 2.

定积分的值为 2。

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8. Binomial Expansion Coefficient | 二项展开式系数

Question: Find the coefficient of x³ in the expansion of (1 + 2x)⁵.

问题：求 (1 + 2x)⁵ 的展开式中 x³ 的系数。

The general term in the binomial expansion is given by T_r+1 = C(5, r) * (1)^5-r * (2x)^r = C(5, r) * 2^r * x^r.

二项展开式的通项为 T_r+1 = C(5, r) * (1)^5-r * (2x)^r = C(5, r) * 2^r * x^r。

We need the term with x³, so r = 3.

我们需要 x³ 项，因此 r = 3。

The coefficient is C(5, 3) * 2³ = 10 * 8 = 80.

系数为 C(5, 3) * 2³ = 10 × 8 = 80。

The coefficient of x³ is 80.

x³ 的系数为 80。

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