Eduqas Year 11 Further Maths: Unit Test Mock Paper Walkthrough | Eduqas 11 年级进阶数学：单元测试模拟卷解析

📚 Eduqas Year 11 Further Maths: Unit Test Mock Paper Walkthrough | Eduqas 11 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 the Eduqas Level 2 Certificate in Additional Mathematics (often called Further Maths) for Year 11 students. The test covers key specification topics: algebra (quadratics, factor theorem, algebraic fractions, simultaneous equations), coordinate geometry of the circle, differentiation and integration, and matrix transformations. Working through each problem step by step will reinforce your understanding and sharpen your exam technique.

本文详细解析了一套为Eduqas 11年级进阶数学（Additional Mathematics，常被称为Further Maths）设计的单元模拟试卷。测试内容涵盖考纲核心主题：代数（二次方程、因式定理、代数分式、联立方程组）、圆的坐标几何、微分与积分以及矩阵变换。通过逐步演算每道题目，你将加深理解并提升应试技巧。

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1. Solving Quadratic Equations by Factorising | 因式分解解二次方程

Problem: Solve x² − 5x + 6 = 0.

问题：解方程 x² − 5x + 6 = 0。

iply to +6 and add to −5. These are −2 and −3. So the factorisation is (x − 2)(x − 3) = 0.

我们寻找两个数，乘积为 +6，和为 −5。这两个数是 −2 和 −3。因此因式分解为 (x − 2)(x − 3) = 0。

Setting each factor to zero gives x − 2 = 0 or x − 3 = 0, hence x = 2 or x = 3.

令每个因式等于零：x − 2 = 0 或 x − 3 = 0，于是 x = 2 或 x = 3。

Always check by substitution: 2² − 5(2) + 6 = 4 − 10 + 6 = 0, and 3² − 15 + 6 = 0.

代入检验：2² − 5(2) + 6 = 4 − 10 + 6 = 0，3² − 15 + 6 = 0，均成立。

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2. The Quadratic Formula and the Discriminant | 二次公式与判别式

lem: Find the range of values of k for which 2x² + kx + 8 = 0 has two distinct real roots.

问题：求使方程 2x² + kx + 8 = 0 有两个不同实数根的 k 的取值范围。

x² + bx + c = 0, the discriminant Δ = b² − 4ac. For two distinct real roots we need Δ > 0.

对于二次式 ax² + bx + c = 0，判别式 Δ = b² − 4ac。要有两个不同实数根，需要 Δ > 0。

Here a = 2, b = k, c = 8. So Δ = k² − 4(2)(8) = k² − 64.

这里 a = 2, b = k, c = 8，所以 Δ = k² − 4(2)(8) = k² − 64。

Set Δ > 0: k² − 64 > 0 ⇒ k² > 64 ⇒ k > 8 or k < −8.

令 Δ > 0：k² − 64 > 0 ⇒ k² > 64 ⇒ k > 8 或 k < −8。

Thus the required range is k < −8 or k > 8.

因此所求范围是 k < −8 或 k > 8。

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3. Algebraic Fractions Simplification | 代数分式化简

Problem: Simplify (x² − 4) ÷ (x² − x − 2).

问题：化简 (x² − 4) ÷ (x² − x − 2)。

denominator. Numerator: x² − 4 = (x − 2)(x + 2). Denominator: x² − x − 2 = (x − 2)(x + 1).

将分子分母因式分解。分子：x² − 4 = (x − 2)(x + 2)。分母：x² − x − 2 = (x − 2)(x + 1)。

The expression becomes ((x − 2)(x + 2)) / ((x − 2)(x + 1)). Cancel the common factor (x − 2), provided x ≠ 2.

表达式变为 ((x − 2)(x + 2)) / ((x − 2)(x + 1))。约去公因式 (x − 2)，但要注意 x ≠ 2。

Simplified form: (x + 2) / (x + 1), with restrictions x ≠ 2 and x ≠ −1 (the latter makes denominator zero).

化简结果为：(x + 2) / (x + 1)，且须满足 x ≠ 2 和 x ≠ −1（后者使分母为零）。

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4. Simultaneous Equations with One Quadratic | 含一个二次式的联立方程组

Problem: Solve the simultaneous equations y = x² − 3x + 4 and y = 2x + 2.

问题：解联立方程组 y = x² − 3x + 4 与 y = 2x + 2。

Equate the expressions for y: x² − 3x + 4 = 2x + 2 ⇒ x² − 5x + 2 = 0.

令两个 y 表达式相等：x² − 3x + 4 = 2x + 2 ⇒ x² − 5x + 2 = 0。

This quadratic does not factorise neatly, so use the quadratic formula: x = [5 ± √25 − 8] / 2 = (5 ± √17) / 2.

此二次式不易分解，使用求根公式：x = [5 ± √25 − 8] / 2 = (5 ± √17) / 2。

Corresponding y-values: y = 2x + 2 = 2((5 ± √17)/2) + 2 = 5 ± √17 + 2 = 7 ± √17.

对应的 y 值：y = 2x + 2 = 2((5 ± √17)/2) + 2 = 5 ± √17 + 2 = 7 ± √17。

The solutions are ((5 + √17)/2, 7 + √17) and ((5 − √17)/2, 7 − √17).

解为 ((5 + √17)/2, 7 + √17) 和 ((5 − √17)/2, 7 − √17)。

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5. Coordinate Geometry of a Circle | 圆的坐标几何

Problem: Find the equation of the circle with centre C(3, −2) and radius 5. Hence find the coordinates where the circle crosses the x-axis.

问题：求圆心为 C(3, −2)、半径为 5 的圆的方程，并由此求出圆与 x 轴的交点坐标。

x − a)² + (y − b)² = r². Substituting a = 3, b = −2, r = 5 gives (x − 3)² + (y + 2)² = 25.

圆的标准方程为 (x − a)² + (y − b)² = r²。代入 a = 3, b = −2, r = 5 得 (x − 3)² + (y + 2)² = 25。

To find x-axis intercepts, set y = 0: (x − 3)² + (0 + 2)² = 25 ⇒ (x − 3)² + 4 = 25 ⇒ (x − 3)² = 21.

求与 x 轴交点，令 y = 0：(x − 3)² + (0 + 2)² = 25 ⇒ (x − 3)² + 4 = 25 ⇒ (x − 3)² = 21。

So x − 3 = ±√21, giving x = 3 + √21 and x = 3 − √21. The intercepts are (3 + √21, 0) and (3 − √21, 0).

因此 x − 3 = ±√21，得 x = 3 + √21 和 x = 3 − √21。交点坐标为 (3 + √21, 0) 和 (3 − √21, 0)。

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

Problem: Differentiate f(x) = x³ using first principles.

问题：利用第一原理求 f(x) = x³ 的导数。

The derivative is given by f'(x) = lim_h→0 (f(x+h) − f(x)) / h.

导数定义为 f'(x) = lim_h→0 (f(x+h) − f(x)) / h。

Then the difference quotient is (x³ + 3x²h + 3xh² + h³ − x³) / h = (3x²h + 3xh² + h³) / h.

计算 f(x+h) = (x+h)³ = x³ + 3x²h + 3xh² + h³。差商为 (x³ + 3x²h + 3xh² + h³ − x³) / h = (3x²h + 3xh² + h³) / h。

Cancelling h gives 3x² + 3xh + h². Now take the limit as h → 0: f'(x) = lim_h→0 (3x² + 3xh + h²) = 3x².

约去 h 得到 3x² + 3xh + h²。当 h → 0 时取极限：f'(x) = lim_h→0 (3x² + 3xh + h²) = 3x²。

∴ f'(x) = 3x²

因此 f'(x) = 3x²

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7. Tangents and Normals | 切线与法线

m: Find the equation of the tangent to the curve y = x² − 4x + 1 at the point where x = 3.

问题：求曲线 y = x² − 4x + 1 在 x = 3 处的切线方程。

st find the y-coordinate: y(3) = 3² − 4(3) + 1 = 9 − 12 + 1 = −2. So the point is (3, −2).

先求 y 坐标：y(3) = 3² − 4(3) + 1 = 9 − 12 + 1 = −2。所以点为 (3, −2)。

Differentiate to find gradient: dy/dx = 2x − 4. At x = 3, gradient m = 2(3) − 4 = 2.

求导得斜率：dy/dx = 2x − 4。在 x = 3 处，斜率 m = 2(3) − 4 = 2。

Equation of tangent using y − y₁ = m(x − x₁): y − (−2) = 2(x − 3) ⇒ y + 2 = 2x − 6 ⇒ y = 2x − 8.

用点斜式 y − y₁ = m(x − x₁)：y − (−2) = 2(x − 3) ⇒ y + 2 = 2x − 6 ⇒ y = 2x − 8。

The tangent line is y = 2x − 8.

切线方程为 y = 2x − 8。

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8. Integration and Area Under a Curve | 积分与曲线下方面积

Problem: Find the area enclosed by the curve y = 4x − x² and the x-axis.

问题：求曲线 y = 4x − x² 与 x 轴所围成的面积。

First locate x-intercepts: set 4x − x² = 0 ⇒ x(4 − x) = 0, so x = 0 and x = 4. The region is above the x-axis between these points.

先求 x 轴交点：令 4x − x² = 0 ⇒ x(4 − x) = 0，得 x = 0 和 x = 4。该区间曲线在 x 轴上方。

Area = ∫₀⁴ (4x − x²) dx = [2x² − (x³/3)]₀⁴.

面积 = ∫₀⁴ (4x − x²) dx = [2x² − (x³/3)]₀⁴。

valuate at the upper limit: 2(4)² − (4³/3) = 2×16 − 64/3 = 32 − 64/3 = (96 − 64)/3 = 32/3.

代入上限：2(4)² − (4³/3) = 2×16 − 64/3 = 32 − 64/3 = (96 − 64)/3 = 32/3。

At the lower limit the value is 0, so the area is 32/3 square units.

下限处值为 0，因此面积为 32/3 平方单位。

Area = 32/3 units²

面积 = 32/3 单位²

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9. Matrix Transformations | 矩阵变换

Problem: A transformation is represented by the matrix M =

0	-1
1	0

. Find the image of the point P(2, 3) under this transformation and describe the transformation geometrically.

问题：变换矩阵为 M = [[0, -1], [1, 0]]。求点 P(2, 3) 在此变换下的像，并描述其几何意义。

Write P as

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