📚 Year 11 AQA Further Maths: Mock Paper Walkthrough | AQA 进阶数学模拟卷精讲
This article breaks down a full unit test mock paper for the AQA Level 2 Certificate in Further Maths. Every key question is explained step by step, covering factor theorem, matrices, coordinate geometry, binomial expansion, trigonometry, and calculus. Use this walkthrough to identify common pitfalls and sharpen your exam technique before the real test.
本文详细解析一份针对 AQA 进阶数学（Level 2 Certificate）的单元测试模拟卷。内容涵盖因式定理、矩阵、坐标几何、二项式展开、三角学以及微积分初步，每一步都配有清晰讲解。通过这份模拟卷的逐题精讲，你可以及时发现易错点，在真正考试前提升解题能力与应试技巧。
1. Factor Theorem: Showing (x – 2) is a Factor | 因式定理：证明 (x-2) 为因式
The first question states: Show that (x – 2) is a factor of f(x) = 2x³ – 5x² – x + 6. To use the Factor Theorem, we evaluate f(2). Substituting x = 2 gives 2(8) – 5(4) – 2 + 6 = 16 – 20 – 2 + 6 = 0. Since the remainder is zero, (x – 2) is indeed a factor.
第一题要求证明 (x – 2) 是多项式 f(x) = 2x³ – 5x² – x + 6 的因式。根据因式定理，只需计算 f(2)。代入 x=2 得 2(8) – 5(4) – 2 + 6 = 16 – 20 – 2 + 6 = 0。余数为零，因此 (x – 2) 确实是一个因式。
2. Full Factorisation & Solving Cubic Equation | 完全因式分解与解三次方程
Next, we factorise f(x) completely. Divide f(x) by (x – 2) using algebraic long division or synthetic division. The quotient is 2x² – x – 3. This quadratic factorises further as (2x – 3)(x + 1). Hence f(x) = (x – 2)(2x – 3)(x + 1). Setting f(x) = 0 yields three solutions: x = 2, x = 3/2 and x = –1.
接下来需要完全分解 f(x)。用长除法或综合除法将 f(x) 除以 (x – 2)，商式为 2x² – x – 3。这个二次式可以继续分解为 (2x – 3)(x + 1)。因此 f(x) = (x – 2)(2x – 3)(x + 1)。令 f(x)=0 解得三个根：x = 2，x = 3/2 以及 x = –1。
3. Matrix Multiplication: Finding A² | 矩阵乘法：求 A²
In the matrix question, you are given A = [[2, –1], [3, 4]]. To find A², multiply A by itself:
矩阵题中给出 A = [[2, –1], [3, 4]]。计算 A² 即计算 A × A：
A² = A × A = [[2×2 + (–1)×3 , 2×(–1) + (–1)×4] , [3×2 + 4×3 , 3×(–1) + 4×4]] = [[1, –6], [18, 13]]
So the resulting matrix is:
所得矩阵为：
| 1 | –6 |
| 18 | 13 |
Always check that the inner dimensions match (2×2 × 2×2) and work row by column. The final matrix is also 2×2.
始终确保内维数匹配（2×2 乘 2×2），并按行乘列计算。结果仍为 2×2 矩阵。
4. Determinant and Inverse of a 2×2 Matrix | 2×2 矩阵的行列式与逆矩阵
For the same matrix A, the determinant is det A = (2 × 4) – (–1 × 3) = 8 + 3 = 11. Since the determinant is non‑zero, an inverse exists. The inverse is given by A⁻¹ = (1/det A) × [[4, 1], [–3, 2]], swapping a and d, changing signs of b and c.
对于同一矩阵 A，行列式为 det A = (2 × 4) – (–1 × 3) = 8 + 3 = 11。因为行列式非零，逆矩阵存在。公式为 A⁻¹ = (1/det A) × [[4, 1], [–3, 2]]，即交换 a 和 d，将 b 和 c 变号。
A⁻¹ = 1/11 ×
| 4 | 1 |
| –3 | 2 |
You can leave the answer with the scalar 1/11 outside, or multiply each entry by 1/11. This inverse can be used to solve matrix equations in later parts.
答案可以将标量 1/11 写在矩阵外，也可以将每个元素都乘以 1/11。此逆矩阵可用于后续求解矩阵方程。
5. Circle Centre and Radius from General Equation | 由一般方程求圆的圆心和半径
The mock paper gives the equation x² + y² – 6x + 8y = 0. To find the centre and radius, complete the square for x and y. Rewrite as (x² – 6x) + (y² + 8y) = 0. Half the coefficients, square them, and add to both sides: ( –6/2)² = 9 and (8/2)² = 16. This gives (x – 3)² + (y + 4)² = 25. The centre is (3, –4) and the radius is √25 = 5.
模拟卷给出方程 x² + y² – 6x + 8y = 0。为求圆心和半径，需对 x 和 y 配方。重写为 (x² – 6x) + (y² + 8y) = 0。取一次项系数一半的平方： (–6/2)² = 9，(8/2)² = 16，等式两边同时加上，得 (x – 3)² + (y + 4)² = 25。圆心为 (3, –4)，半径为 √25 = 5。
6. Tangent to a Circle at a Given Point | 圆在给定点处的切线方程
The next part asks for the equation of the tangent to the same circle at the point (0, –8). First verify the point lies on the circle: 0² + (–8)² – 6·0 + 8·(–8) = 64 – 64 = 0. The radius from centre (3, –4) to (0, –8) has gradient (–8 – (–4)) / (0 – 3) = (–4) / (–3) = 4/3. The tangent is perpendicular to this radius, so its gradient is the negative reciprocal: –3/4. Using point‑slope form: y + 8 = –3/4 (x – 0), which simplifies to y = –3/4 x – 8.
下一问要求写出同一圆在点 (0, –8) 处的切线方程。先验证点是否在圆上：0² + (–8)² – 6·0 + 8·(–8) = 64 – 64 = 0。从圆心 (3, –4) 到点 (0, –8) 的半径斜率为 (–8 + 4) / (0 – 3) = (–4)/(–3) = 4/3。切线与该半径垂直，因此切线斜率为其负倒数 –3/4。利用点斜式：y + 8 = –3/4 (x – 0)，化简得 y = –3/4 x – 8。
7. Binomial Expansion of (2 – 3x)⁴ | (2 – 3x)⁴ 的二项式展开
Expand (2 – 3x)⁴ up to the term in x². Use the binomial theorem with n = 4. The coefficients are ⁴C₀ = 1, ⁴C₁ = 4, ⁴C₂ = 6. The expansion is:
将 (2 – 3x)⁴ 展开至 x² 项。使用二项式定理，n = 4。系数为 ⁴C₀ = 1，⁴C₁ = 4，⁴C₂ = 6。展开式为：
(2 – 3x)⁴ = 1·2⁴ + 4·2³·(–3x) + 6·2²·(–3x)² + …
= 16 – 96x + 216x² + …
The terms in x³ and x⁴ are not required for the next part, but you should recognise the pattern of decreasing powers of 2 and increasing powers of (–3x).
后续题目不需要 x³ 和 x⁴ 项，但你应当能识别出 2 的幂递减、(–3x) 的幂递增的模式。
8. Binomial Approximation for (1.97)⁴ | 二项式近似计算 (1.97)⁴
We are asked to approximate (1.97)⁴ using the expansion. Set 2 – 3x = 1.97, which gives 3x = 0.03 so x = 0.01. Substitute x = 0.01 into the expansion up to x²: 16 – 96(0.01) + 216(0.01)² = 16 – 0.96 + 216 × 0.0001 = 15.04 + 0.0216 = 15.0616. To three decimal places, (1.97)⁴ ≈ 15.062. This matches the true value closely, demonstrating the power of binomial approximations for small x.
题目要求用展开式近似计算 (1.97)⁴。令 2 – 3x = 1.97，得 3x = 0.03，故 x = 0.01。将 x = 0.01 代入到 x² 项的展开式：16 – 96(0.01) + 216(0.01)² = 16 – 0.96 + 216 × 0.0001 = 15.04 + 0.0216 = 15.0616。精确到三位小数为 15.062。这与真实值非常接近，体现了二项式近似在小 x 时的威力。
9. Solving Trigonometric Equation: 2 sin x cos x = cos x | 解三角方程：2 sin x cos x = cos x
The equation is 2 sin x cos x = cos x for 0° ≤ x ≤ 360°. Bring all terms to one side and factorise: cos x (2 sin x – 1) = 0. This gives two cases: cos x = 0 or sin x = 1/2.
求解三角方程 2 sin x cos x = cos x，定义域为 0° ≤ x ≤ 360°。移项并提取公因式：cos x (2 sin x – 1) = 0。由此得两种情况：cos x = 0 或 sin x = 1/2。
-
cos x = 0 gives x = 90° and x = 270°.
cos x = 0 解得 x = 90° 和 x = 270°。
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sin x = 1/2 gives x = 30° and x = 150°.
sin x = 1/2 解得 x = 30° 和 x = 150°。
The full solution set is {30°, 90°, 150°, 270°}. Never divide by cos x without checking, or you risk losing solutions.
全部解为 {30°, 90°, 150°, 270°}。切勿不经检验就贸然除以 cos x，否则容易丢失解。
10. Proving a Trigonometric Identity | 证明三角恒等式
Prove that (sin x)/(1 + cos x) + (1 + cos x)/(sin x) ≡ 2 cosec x. Start by combining the left‑hand side over a common denominator:
证明恒等式 (sin x)/(1 + cos x) + (1 + cos x)/(sin x) ≡ 2 cosec x。先将左边通分：
LHS = [sin²x + (1 + cos x)²] / [sin x (1 + cos x)]
Expand the numerator: sin²x + 1 + 2 cos x + cos²x. Using sin²x + cos²x = 1, this simplifies to 1 + 1 + 2 cos x = 2 + 2 cos x. Factor out 2: 2(1 + cos x). The denominator is sin x (1 + cos x). Cancel the common factor (1 + cos x) provided cos x ≠ –1, giving 2 / sin x = 2 cosec x, which is the right‑hand side.
展开分子：sin²x + 1 + 2 cos x + cos²x。利用 sin²x + cos²x = 1，化简为 1 + 1 + 2 cos x = 2 + 2 cos x。提取公因子 2：2(1 + cos x)。分母为 sin x (1 + cos x)。约去公因子 (1 + cos x)（注意 cos x ≠ –1），得 2 / sin x 即 2 cosec x，等于右式。证毕。
11. Differentiation: Power Rule and Simplifying Expressions | 微分：幂法则与表达式化简
Differentiate y = 4x³ – 2/x + √x
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