Year 10 Eduqas Further Maths: Unit Test Mock Paper Walkthrough | Year 10 Eduqas 进阶数学：单元测试模拟卷解析

📚 Year 10 Eduqas Further Maths: Unit Test Mock Paper Walkthrough | Year 10 Eduqas 进阶数学：单元测试模拟卷解析

This walkthrough breaks down a typical Year 10 Further Mathematics unit test mock paper for the Eduqas specification. We analyse common question types, step-by-step solutions, and the key skills each problem assesses. Use this guide to identify your strengths and target areas for improvement before your real exam.

本解析拆解了一份典型的 Year 10 进阶数学（Eduqas 考试局）单元测试模拟卷。我们剖析常见题型，逐步讲解解题过程以及每道题所考查的核心技能。利用本指南来明确自己的强项，并在正式考试前针对薄弱环节加以提升。

1. Simplifying Algebraic Fractions | 代数分式化简

Simplify fully (x² – 9) / (x² + 8x + 15). Remember to state any restrictions on x to ensure the denominator is not zero.

化简 (x² – 9) / (x² + 8x + 15)。记得注明 x 的限制条件以保证分母不为零。

Factorise both numerator and denominator. The numerator is a difference of squares: x² – 9 = (x – 3)(x + 3). The denominator factorises into (x + 3)(x + 5).

将分子和分母分别因式分解。分子是平方差：x² – 9 = (x – 3)(x + 3)。分母可分解为 (x + 3)(x + 5)。

(x² – 9)/(x² + 8x + 15) = (x – 3)(x + 3) / [(x + 3)(x + 5)] = (x – 3)/(x + 5)

Cancel the common factor (x + 3), leaving (x – 3)/(x + 5). The domain restriction is x ≠ -3, x ≠ -5, because these values make the original denominator zero.

约去公因式 (x + 3)，得到 (x – 3)/(x + 5)。定义域的限制为 x ≠ -3, x ≠ -5，因为当 x 取这些值时原分母为零。

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2. Quadratic Inequalities | 二次不等式

Solve the inequality x² – 4x > 5. Express your solution using set notation.

求解不等式 x² – 4x > 5，并用集合符号表示解集。

First, rearrange so one side is zero: x² – 4x – 5 > 0. Factorise the quadratic: (x – 5)(x + 1) > 0.

首先移项使一边为零：x² – 4x – 5 > 0。分解二次式：(x – 5)(x + 1) > 0。

The critical values are x = 5 and x = – ree intervals. Test a value from each interval to determine where the product is positive.

关键值为 x = 5 和 x = –1。它们将数轴分成三个区间。在每个区间内取测试值，判断乘积的正负。

• For x < –1, say x = –2: (–)(–) = +, so inequality holds.
• 对于 x < –1，例如 x = –2：负×负 = 正，不等式成立。
• For –1 < x < 5, say x = 0: (+)(–) = –, does not hold.
• 对于 –1 < x < 5，例如 x = 0：正×负 = 负，不成立。
• For x > 5, say x = 6: (+)(+) = +, holds.
• 对于 x > 5，例如 x = 6：正×正 = 正，成立。

The solution set is {x: x < –1} ∪ {x: x > 5}, or in interval notation x ∈ (–∞, –1) ∪ (5, ∞).

解集为 {x: x < –1} ∪ {x: x > 5}，或用区间表示为 x ∈ (–∞, –1) ∪ (5, ∞)。

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3. Matrix Multiplication & Determinants | 矩阵乘法与行列式

Given matrices A = [1 2; 3 4] and B = [2 0; –1 3], find AB and the determinant of A.

已知矩阵 A = [1 2; 3 4] 和 B = [2 0; –1 3]，求 AB 以及 A 的行列式。

For AB, multiply rows of A by columns of B:

计算 AB，将 A 的行与 B 的列相乘：

AB = [1×2 + 2×(–1)   1×0 + 2×3;
&nbsp;3×2 + 4×(–1)   3×0 + 4×3] = [0   6;
  2   12]

The determinant of A is det(A) = (1)(4) – (2)(3) = 4 – 6 = –2.

矩阵 A 的行列式为 det(A) = (1)(4) – (2)(3) = 4 – 6 = –2。

Remember, matrix multiplication is not commutative, so AB is generally different from BA.

注意，矩阵乘法不满足交换律，因此 AB 通常不等于 BA。

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4. Differentiation from First Principles | 从第一原理求导

Using first principles, find the derivative of f(x) = x².

利用第一原理，求 f(x) = x² 的导数。

The derivative is defined as f'(x) = lim_h→0 [f(x + h) – f(x)] / h.

导数的定义为 f'(x) = lim_h→0 [f(x + h) – f(x)] / h。

tute f(x) = x²: f(x + h) = (x + h)² = x² + 2xh + h². Then the difference quotient becomes:

代入 f(x) = x²：f(x + h) = (x + h)² = x² + 2xh + h²。差商变为：

[ (x² + 2xh + h²) – x² ] / h = (2xh + h²) / h = 2x + h

Take the limit as h → 0: f'(x) = lim_h→0 (2x + h) = 2x. This is the standard power rule result.

取 h → 0 时的极限：f'(x) = lim_h→0 (2x + h) = 2x。即为标准幂函数求导结果。

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5. Stationary Points & Nature | 驻点及其性质

Find the coordinates of the stationary point of y = x³ – 3x and determine its nature.

求曲线 y = x³ – 3x 的驻点坐标，并判断其性质。

Differentiate to find dy/dx = 3x² – 3. Set this equal to zero for stationary points: 3x² – 3 = 0 → x² = 1 → x = 1 or x = –1.

求导得 dy/dx = 3x² – 3。令其为零以求驻点：3x² – 3 = 0 → x² = 1 → x = 1 或 x = –1。

Find the second derivative: d²y/dx² = 6x. Evaluate at each stationary point.

求二阶导数：d²y/dx² = 6x。在每个驻点处计算二阶导数值。

• At x = 1, d²y/dx² = 6 > 0 → minimum point. y-coordinate: (1)³ – 3(1) = –2. Point (1, –2).
• 当 x = 1，二阶导数 = 6 > 0 → 极小值点。y 坐标：1 – 3 = –2。点 (1, –2)。
• At x = –1, d²y/dx² = –6 < 0 → maximum point. y-coordinate: (–1)³ – 3(–1) = –1 + 3 = 2. Point (–1, 2).
• 当 x = –1，二阶导数 = –6 < 0 → 极大值点。y 坐标：–1 + 3 = 2。点 (–1, 2)。

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6. Trigonometric Identities | 三角恒等式

Prove that (1 – cos²θ) / sin θ = sin θ, for sin θ ≠ 0.

证明 (1 – cos²θ) / sin θ = sin θ，其中 sin θ ≠ 0。

Use the Pythagorean identity sin²θ + cos²θ = 1, so 1 – cos²θ = sin²θ.

利用勾股恒等式 sin²θ + cos²θ = 1，因此 1 – cos²θ = sin²θ。

(1 – cos²θ) / sin θ = sin²θ / sin θ = sin θ

Thus the left-hand side simplifies to the right-hand side. This holds whenever sin θ ≠ 0.

于是左边可化简为右边。当 sin θ ≠ 0 时该等式恒成立。

This type of manipulation is common when simplifying trigonometric expressions or solving equations.

此类恒等变形在化简三角式或解三角方程时十分常见。

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7. Equation of a Circle from a Diameter | 由直径端点的圆的方程

The points P(2, –3) and Q(–4, 5) are endpoints of a diameter of a circle. Find the equation of the circle.

点 P(2, –3) 和 Q(–4, 5) 为圆某直径的两端点。求该圆的方程。

The centre is the midpoint of PQ: ((2 + (–4))/2, (–3 + 5)/2) = (–1, 1).

圆心为 PQ 的中点：((2 + (–4))/2, (–3 + 5)/2) = (–1, 1)。

The radius is half the length of PQ. First find PQ² = (–4 – 2)² + (5 – (–3))² = (–6)² + 8² = 36 + 64 = 100, so PQ = 10. Hence radius r = 5.

半径为 PQ 长度的一半。先求 PQ² = (–4 – 2)² + (5 – (–3))² = (–6)² + 8² = 36 + 64 = 100，故 PQ = 10。因此半径 r = 5。

Equation: (x + 1)² + (y – 1)² = 25

Expand if required: x² + 2x + 1 + y² – 2y + 1 = 25 → x² + y² + 2x – 2y – 23 = 0.

若需要展开：x² + 2x + 1 + y² – 2y + 1 = 25 → x² + y² + 2x – 2y – 23 = 0。

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8. Composite Functions & Domain | 复合函数与定义域

Let f(x) = 2x + 1 and g(x) = √x, for x ≥ 0. Find fg(x) and gf(x), and state the domain of each composite function.

设 f(x) = 2x + 1，g(x) = √x，其中 x ≥ 0。求 fg(x) 和 gf(x)，并写出每个复合函数的定义域。

fg(x) = f(g(x)) = f(√x) = 2√x + 1. The input to g must satisfy x ≥ 0, and the output of g (√x) is always defined for f, so domain of fg is x ≥ 0.

fg(x) = f(g(x)) = f(√x) = 2√x + 1。g 的输入需满足 x ≥ 0，且 g 的输出 √x 总能作为 f 的输入，故 fg 的定义域为 x ≥ 0。

gf(x) = g(f(x)) = g(2x + 1) = √2x + 1. The expression inside the square root must be non-negative: 2x + 1 ≥ 0 → x ≥ –½. Hence domain of gf is x ≥ –½.

gf(x) = g(f(x)) = g(2x + 1) = √2x + 1。根号内的式子必须非负：2x + 1 ≥ 0 → x ≥ –½。因此 gf 的定义域为 x ≥ –½。

Always check the domain restrictions when composing functions, especially with square roots, denominators, or logarithms.

在构造复合函数时，务必检查定义域的限制，特别是涉及平方根、分母或对数时。

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9. Vector Geometry & Ratio | 向量几何与比例

In triangle OAB, M is the midpoint of AB. OA = a, OB = b. Express OM in terms of a and b. If P divides OM in the ratio 2:1, find OP.

在三角形 OAB 中，M 为 AB 的中点。OA = a，OB = b。用 a 和 b 表示 OM。若点 P 将 OM 分成 2:1，求 OP。

The position vector of M is the average of A and B: OM = (a + b)/2.

M 的位置向量是 A 和 B 的平均值：OM = (a + b)/2。

in the ratio OP:PM = 2:1, meaning OP = (2/3) OM. Thus OP = (2/3) × (a + b)/2 = (a + b)/3.

P 将 OM 分成 OP:PM = 2:1，即 OP = (2/3) OM。于是 OP = (2/3) × (a + b)/2 = (a + b)/3。

Understanding section formulae in vector form is essential for harder geometry problems in the Eduqas Further Maths course.

掌握向量形式的定比分点公式，对攻克 Eduqas 进阶数学中较难的几何问题至关重要。

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10. Area Between a Curve and the x-axis | 曲线与 x 轴之间的面积

Find the area enclosed between the curve y = 4 – x² and the x-axis.

求曲线 y = 4 – x² 与 x 轴所围成的面积。

Find the x-intercepts: set y = 0 → 4 – x² = 0 → x = –2 and x = 2.

先求 x 轴截距：令 y = 0 → 4 – x² = 0 → x = –2 和 x = 2。

The required area is given by the definite integral from –2 to 2 of (4 – x²) dx.

所求面积由定积分 ∫_–2² (4 – x²) dx 给出。

Integrate: ∫ (4 – x²) dx = 4x – (1/3)x³. Evaluate from –2 to 2:

积分：∫ (4 – x²) dx = 4x – (1/3)x³。从 –2 到 2 计算：

[4(2) – (1/3)(8)] – [4(–2) – (1/3)(–8)] = (8 – 8/3) – (–8 + 8/3) = (16/3) – (–16/3) = 32/3

The area is 32/3 square units. This is an even function, so you could also compute 2 × ∫₀² (4 – x²) dx for a quicker calculation.

面积为 32/3 平方单位。该函数为偶函数，因此也可用 2 × ∫₀² (4 – x²) dx 来快速计算。

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