📚 Year 12 AQA Maths: Unit Test Mock Paper Walkthrough | AQA 数学：单元测试模拟卷解析
This article provides a complete walkthrough of a typical Year 12 AQA Mathematics Unit Test, covering key topics from pure mathematics. Each question is broken down step by step, highlighting effective solving techniques and common pitfalls. By studying these model solutions, you will strengthen your understanding of quadratics, differentiation, integration, exponentials, logarithms, trigonometry, binomial expansion and proof, all aligned with the AQA AS-level specification.
本文完整解析一套典型的 Year 12 AQA 数学单元测试卷，涵盖纯数学的重要考点。每个问题都配有详细的逐步解答，同时强调高效的解题技巧与常见误区。通过学习这些模拟解答，你将进一步巩固二次方程、微分、积分、指数与对数、三角函数、二项式展开以及证明等内容，完全贴合 AQA AS 阶段的考试要求。
1. Quadratic Equations | 二次方程
− 8x + 5, solve f(x) = 0 and find the coordinates of the vertex by completing the square.
题目：已知 f(x) = 2x² − 8x + 5，求解方程 f(x) = 0 并利用配方法求出顶点的坐标。
Step 1 – Identify coefficients: a = 2, b = −8, c = 5. Write the quadratic formula x = [−b ± √b² − 4ac] / (2a).
步骤 1 —— 识别系数：a = 2，b = −8，c = 5。写出求根公式 x = [−b ± √b² − 4ac] / (2a)。
Step 2 – Substitute: x = [8 ± √64 − 40] / 4 = [8 ± √24] / 4 = [8 ± 2√6] / 4 = 2 ± (√6)/2.
步骤 2 —— 代入计算：x = [8 ± √64 − 40] / 4 = [8 ± √24] / 4 = [8 ± 2√6] / 4 = 2 ± (√6)/2。
Step 3 – : 2x² − 8x + 5 = 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3.
步骤 3 —— 配方法：2x² − 8x + 5 = 2(x² − 4x) + 5 = 2[(x − 2)² − 4] + 5 = 2(x − 2)² − 8 + 5 = 2(x − 2)² − 3。
Step 4 – The vertex is therefore at (2, −3). The minimum value of the function is −3 when x = 2.
步骤 4 —— 因此顶点坐标为 (2, −3)。当 x = 2 时，函数取得最小值 −3。
Common mistake: Many students misapply the quadratic formula by writing the denominator as '2' instead of '2a', or they forget to take out the factor of 2 correctly when completing the square. Always factor the coefficient of x² from the first two terms.
常见错误：许多学生错误地将分母写成 '2' 而不是 '2a'，或者在配方时没有正确地提出 x² 的系数。务必先将二次项系数从前两项中提取出来。
2. Straight Lines and Perpendicular Bisectors | 直线与垂直平分线
of the perpendicular bisector of AB. Hence find its intersection with the line x + y = 4.
题目：已知点 A(1, 2) 和 B(5, 8)。求直线 AB 的方程（形式为 ax + by + c = 0），并确定线段 AB 的垂直平分线方程。然后求该平分线与直线 x + y = 4 的交点。
Step 1 – adient of AB: m₁ = (8 − 2)/(5 − 1) = 6/4 = 3/2. Hence the line AB is y − 2 = (3/2)(x − 1).
步骤 1 —— AB 的斜率：m₁ = (8 − 2)/(5 − 1) = 6/4 = 3/2。因此 AB 的方程为 y − 2 = (3/2)(x − 1)。
Step 2 – Rearranging: y − 2 = (3/2)x − 3/2 ⇒ multiply by 2: 2y − 4 = 3x − 3 ⇒ 3x − 2y + 1 = 0.
步骤 2 —— 整理方程：2y − 4 = 3x − 3 ⇒ 移项得 3x − 2y + 1 = 0。
Step 3 – Midpoint of AB: ((1+5)/2, (2+8)/2) = (3, 5). Perpendicular gradient m₂ = −1/(3/2) = −2/3.
步骤 3 —— AB 的中点：((1+5)/2, (2+8)/2) = (3, 5)。垂直平分线的斜率 m₂ = −1/(3/2) = −2/3。
Step 4 – Perpendicular bisector: y − 5 = (−2/3)(x − 3) ⇒ 3y − 15 = −2x + 6 ⇒ 2x + 3y − 21 = 0.
步骤 4 —— 垂直平分线：y − 5 = (−2/3)(x − 3) ⇒ 3y − 15 = −2x + 6 ⇒ 2x + 3y − 21 = 0。
Step 5 – 4: solve simultaneously. From x + y = 4, y = 4 − x. Substitute into 2x + 3(4 − x) − 21 = 0 ⇒ 2x + 12 − 3x − 21 = 0 ⇒ −x − 9 = 0 ⇒ x = −9, then y = 13. Intersection point: (−9, 13).
步骤 5 —— 与 x + y = 4 的交点：由 y = 4 − x 代入 2x + 3(4 − x) − 21 = 0，解得 x = −9，y = 13，交点坐标为 (−9, 13)。
ents; always check m₁ × m₂ = −1. Also watch for arithmetic errors when clearing fractions.
常见错误：混淆垂直斜率的负倒数关系；务必验证 m₁ × m₂ = −1。同时避免去分母时出现符号错误。
3. Differentiation and Tangents | 微分与切线
the equation of the tangent and the normal at x = 2, and determine the stationary points.
题目：已知 y = x³ − 3x² + 2，求 dy/dx、在 x = 2 处的切线和法线方程，并确定函数的驻点。
Step 1 – Differentiate: dy/dx = 3x² − 6x.
步骤 1 —— 求导：dy/dx = 3x² − 6x。
Step 2 – rizontal: y = y(2) = 8 − 12 + 2 = −2, so tangent is y = −2. The normal is vertical: x = 2.
步骤 2 —— 当 x = 2 时，斜率 mᵗ = 0，故切线为水平线 y = −2；法线为竖直线 x = 2。
Step 3 – Stationary points occur where dy/dx = 0: 3x² − 6x = 0 ⇒ 3x(x − 2) = 0 ⇒ x = 0 or x = 2. Find coordinates: (0, 2) and (2, −2).
步骤 3 —— 驻点：令 dy/dx = 0，得到 x = 0 或 x = 2，坐标为 (0, 2) 和 (2, −2)。
Step 4 – 6x − 6. At x = 0, d²y/dx² = −6 (< 0), so maximum; at x = 2, d²y/dx² = 6 (> 0), so minimum.
步骤 4 —— 利用二阶导数判定：d²y/dx² = 6x − 6。x = 0 处二阶导数为 −6，故为极大值；x = 2 处二阶导数为 6，故为极小值。
Common mistake: Forgetting to evaluate the original function to find the y-coordinate of stationary points, or misinterpreting a zero gradient as always being a maximum.
常见错误：忘记代入原函数求驻点的 y 坐标，或者误认为梯度为零就一定是极大值。
4. Integration and Area | 积分与面积
e area bounded by the curve y = 2x³ − 9x² + 12x, the x-axis and the lines x = 0 and x = 3.
题目：计算不定积分 ∫ (6x² − 2x + 1) dx，并求定积分 ∫₀² (6x² − 2x + 1) dx。然后计算曲线 y = 2x³ − 9x² + 12x 与 x 轴及直线 x = 0、x = 3 所围成的面积。
Step 1 – −2x dx = −x², ∫ 1 dx = x, plus constant C. So the indefinite integral is 2x³ − x² + x + C.
步骤 1 —— 逐项积分：∫ 6x² dx = 2x³, ∫ −2x dx = −x², ∫ 1 dx = x, 加上常数 C，得到 2x³ − x² + x + C。
Step 2 – Definite integral: [2x³ − x² + x]₀² = (16 − 4 + 2) − (0) = 14.
步骤 2 —— 定积分：代入上下限得 (16 − 4 + 2) − 0 = 14。
Step 3 – pt x = 0. Thus the curve does not cross the x-axis between 0 and 3, so the required area = |∫₀³ (2x³ − 9x² + 12x) dx|.
步骤 3 —— 对于 y = 2x³ − 9x² + 12x，因式分解得 x(2x² − 9x + 12)，判别式为负，故除 x = 0 外无实根。因此曲线在 [0,3] 内不穿过 x 轴，面积等于 |∫₀³ (2x³ − 9x² + 12x) dx|。
Step 4 – − 3x³ + 6x²]₀³ = (½×81 − 81 + 54) = 40.5 − 81 + 54 = 13.5. Thus area = 13.5 square units.
步骤 4 —— 计算定积分：原函数为 ½ x⁴ − 3x³ + 6x²，代入上下限得 13.5，故面积为 13.5。
Common mistake: Forgetting the constant of integration in indefinite integrals, or not checking whether the curve crosses the x-axis when finding areas. Always check the sign of the function over the interval.
常见错误：不定积分遗漏常数项，或在求面积时未检查曲线是否穿越 x 轴；务必确认区间内函数的正负。
5. Exponential and Logarithmic Equations | 指数与对数方程
Question: Solve the equation 3²ˣ = 81 and the logarithmic equation 2 log₃ x − log₃ (x − 2) = 2.
题目：解方程 3²ˣ = 81 以及对数方程 2 log₃ x − log₃ (x − 2) = 2。
Step 1 – For 3²ˣ = 81, write 81 as 3⁴. Then 2x = 4 ⇒ x = 2.
步骤 1 —— 将 81 写成 3⁴，得到 2x = 4，所以 x = 2。
Step 2 – e log equation, use the power rule: 2 log₃ x = log₃ (x²). So log₃ (x²) − log₃ (x − 2) = 2.
步骤 2 —— 运用幂法则：2 log₃ x = log₃ (x²)，于是原方程化为 log₃ (x²) − log₃ (x − 2) = 2。
Step 3 – Combine logs: log₃ [x²/(x − 2)] = 2 ⇒ x²/(x − 2) = 3² = 9.
步骤 3 —— 利用对数的减法规则：log₃ [x²/(x − 2)] = 2，即 x²/(x − 2) = 9。
Step 4 – Solve the rational equation: x² = 9x − 18 ⇒ x² − 9x + 18 = 0 ⇒ (x − 3)(x − 6) = 0 ⇒ x = 3 or x = 6.
步骤 4 —— 解分式方程得 x² − 9x + 18 = 0，因式分解为 (x − 3)(x − 6) = 0，故 x = 3 或 x = 6。
Step 5 – Check domain: for log₃ x and log₃ (x − 2), we need x > 0 and x − 2 > 0 ⇒ x > 2. Both solutions satisfy this, so final answer x = 3, 6.
步骤 5 —— 检验定义域：需满足 x > 0 且 x − 2 > 0，即 x > 2。两个解均符合，因此答案为 x = 3 和 6。
Common mistake: Forgetting to check the domain of logarithmic functions often leads to extraneous solutions being accepted. Always write down restrictions before solving.
常见错误：忽略对数函数的定义域检查，常常会保留增根。解题前务必先写出允许值范围。
6. Trigonometric Identities and Equations | 三角恒等式与方程
e exact values of cos θ and tan θ. Hence solve 2 sin² θ − cos θ − 1 = 0 for 0° ≤ θ ≤ 360°.
题目：已知 sin θ = 3/5 且 θ 为锐角，求 cos θ 和 tan θ 的精确值。然后解方程 2 sin² θ − cos θ − 1 = 0，区间为 0° ≤ θ ≤ 360°。
Step 1 – Using sin² θ + cos² θ ≡ , so cos θ = 4/5 (positive as θ is acute). Then tan θ = sin θ / cos θ = (3/5)/(4/5) = 3/4.
步骤 1 —— 由平方关系 sin² θ + cos² θ ≡ 1 得 cos² θ = 16/25，由于 θ 是锐角，cos θ = 4/5；tan θ = (3/5)/(4/5) = 3/4。
Step 2 – For the equation, replace sin² θ with 1 − cos² θ: 2(1 − cos² θ) − cos θ − 1 = 0 ⇒ 2 − 2 cos² θ − cos θ − 1 = 0 ⇒ −2 cos² θ − cos θ + 1 = 0. Multiply by
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