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

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

In this article, we will work through a mock unit test designed for Year 10 AQA Further Maths, offering detailed, step-by-step solutions to each question. The topics covered include quadratic equations, inequalities, simultaneous equations, differentiation, matrices, trigonometric identities, sequences, and coordinate geometry. Use this walkthrough to sharpen your problem-solving skills and build confidence for your real assessment.

本文将逐题讲解一份专为 Year 10 AQA 进阶数学设计的单元模拟测试卷，提供详尽的逐步解答。内容涵盖二次方程、不等式、联立方程、微分、矩阵、三角恒等式、数列以及坐标几何。通过本解析，你可以磨炼解题技巧，为正式考试树立信心。

1. Quadratic Equation | 二次方程求解

Solve: 2x² − 5x − 3 = 0.

解方程：2x² − 5x − 3 = 0。

erm: 2x² − 6x + x − 3. Factor by grouping: 2x(x − 3) + 1(x − 3) = (2x + 1)(x − 3). Set each factor to zero: 2x + 1 = 0 ⇒ x = −½, or x − 3 = 0 ⇒ x = 3.

因式分解法：将 a 与 c 相乘得 2 × (−3) = −6。需找两个数，乘积为 −6 且和为 −5。这两个数是 −6 和 1。裂中项：2x² − 6x + x − 3。分组分解：2x(x − 3) + 1(x − 3) = (2x + 1)(x − 3)。令每个因式为零：2x + 1 = 0 ⇒ x = −½，或 x − 3 = 0 ⇒ x = 3。

Alternatively, using the quadratic formula x = [−b ± √b² − 4ac = 2, b = −5, c = −3. The discriminant b² − 4ac = 25 − 4(2)(−3) = 25 + 24 = 49. So x = [5 ± √49] / 4 = (5 ± 7) / 4. This yields x = 12/4 = 3 or x = −2/4 = −½.

亦可使用二次公式 x = [−b ± √b² − 4ac] / (2a)。此处 a = 2, b = −5, c = −3。判别式 b² − 4ac = 25 − 4×2×(−3) = 25 + 24 = 49。故 x = [5 ± √49] / 4 = (5 ± 7) / 4，解得 x = 12/4 = 3 或 x = −2/4 = −½。

2. Quadratic Inequality | 二次不等式求解

Solve x² − 4x − 5 > 0.

解不等式 x² − 4x − 5 > 0。

d x = −1. These split the number line into three intervals: x < −1, −1 < x < 5, and x > 5.

首先对二次式进行因式分解：x² − 4x − 5 = (x − 5)(x + 1)。令表达式的值为零，得到临界值 x = 5 和 x = −1。这将数轴分为三个区间：x < −1，−1 < x < 5，以及 x > 5。

= (−5)(1) = −5 < 0. For x = 6 (in x > 5): (6 − 5)(6 + 1) = (1)(7) = 7 > 0. The inequality is satisfied where the product is positive, i.e. x < −1 or x > 5.

在每个区间内选取一个测试点以判断乘积的符号。取 x = −2（区间 x < −1）：(−2 − 5)(−2 + 1) = (−7)(−1) = 7 > 0。取 x = 0（−1 < x < 5）：(0 − 5)(0 + 1) = (−5)(1) = −5 < 0。取 x = 6（x > 5）：(6 − 5)(6 + 1) = (1)(7) = 7 > 0。不等式要求乘积为正，故解集为 x < −1 或 x > 5。

rds and cuts the x-axis at −1 and 5. The graph lies above the x-axis for x < −1 and x > 5.

你还可以借助抛物线 y = x² − 4x − 5 的图像来理解：该抛物线开口向上，与 x 轴交于 −1 和 5。图像在 x < −1 和 x > 5 时位于 x 轴上方。

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3. Simultaneous Equations (Linear & Quadratic) | 联立方程（一次与二次）

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

解联立方程组：y = x² − 3x + 2 与 y = 2x − 2。

ch other: x² − 3x + 2 = 2x − 2. Rearrange to bring all terms to one side: x² − 5x + 4 = 0.

令两个关于 y 的表达式相等：x² − 3x + 2 = 2x − 2。将各项移至同侧并整理：x² − 5x + 4 = 0。

values. For x = 1: y = 2(1) − 2 = 0. For x = 4: y = 2(4) − 2 = 6. The solutions are (1, 0) and (4, 6).

因式分解二次式：(x − 1)(x − 4) = 0，得 x = 1 或 x = 4。将这些 x 值代入一次方程 y = 2x − 2 求对应的 y 值。当 x = 1：y = 2(1) − 2 = 0。当 x = 4：y = 2(4) − 2 = 6。解为 (1, 0) 和 (4, 6)。

0): LHS = 0, RHS = 1 − 3 + 2 = 0. For (4,6): LHS = 6, RHS = 16 − 12 + 2 = 6. Both satisfy.

始终将解代入原二次方程进行检验，以避免代数失误。(1,0)：左 = 0，右 = 1 − 3 + 2 = 0。(4,6)：左 = 6，右 = 16 − 12 + 2 = 6。两者均满足。

4. Differentiation & Tangent | 微分与切线

Given f(x) = 3x³ − 4x² + 2x − 7, find f'(x) and the equation of the tangent to the curve at the point where x = 1.

已知 f(x) = 3x³ − 4x² + 2x − 7，求 f'(x) 以及曲线在 x = 1 处的切线方程。

Differentiate term by term: f'(x) = 9x² − 8x + 2. To find the tangent at x = 1, first compute the gradient m = f' of contact: f(1) = 3(1)³ − 4(1)² + 2(1) − 7 = 3 − 4 + 2 − 7 = −6. So the point is (1, −6).

逐项微分：f'(x) = 9x² − 8x + 2。求 x = 1 处的切线时，先计算斜率 m = f'(1) = 9(1)² − 8(1) + 2 = 9 − 8 + 2 = 3。再求切点坐标：f(1) = 3(1)³ − 4(1)² + 2 − 7 = 3 − 4 + 2 − 7 = −6。切点为 (1, −6)。

Using the point-slope form y − y₁ = m(x − x₁): y − (−6) = 3(x − 1) ⇒ y + 6 = 3x − 3 ⇒ y = 3x − 9. This is the tangent equation.

利用点斜式 y − y₁ = m(x − x₁)：y − (−6) = 3(x − 1) ⇒ y + 6 = 3x − 3 ⇒ y = 3x − 9。此即切线方程。

Don't forget: the derivative of a constant is zero. Common mistakes include ignoring the constant term (−7) or misapplying the power rule (multiply by the old power and reduce the power by one).

牢记：常数的导数为零。常见错误包括遗漏常数项 −7 或误用幂法则（正确做法是将原指数乘到前面，再将指数减 1）。

5. Matrix Operations | 矩阵运算

Let A = ⎡2 1⎤ and B = ⎡0 −1⎤. Calculate A + B and A × B.

设 A = ⎡2 1⎤，B = ⎡0 −1⎤。计算 A + B 和 A × B。

A + B: Add corresponding elements. (2+0)=2, (1+(−1))=0, (−1+2)=1, (3+1)=4. So A + B = ⎡2 0⎤.

A + B：对应元素相加：(2+0)=2，(1+(−1))=0，(−1+2)=1，(3+1)=4。故 A + B = ⎡2 0⎤。

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