📚 Year 12 AQA Further Maths: Unit Test Mock Paper Walkthrough | Year 12 AQA 进阶数学：单元测试模拟卷解析
This article provides a detailed walkthrough of a Unit Test mock paper for Year 12 AQA Further Mathematics. Each section covers a typical exam-style question, breaking down the method, key formulas and common pitfalls. Use this revision resource to consolidate your understanding of core topics such as complex numbers, matrices, vectors, calculus, hyperbolic functions, polar coordinates and differential equations.
本文详细解析一份Year 12 AQA进阶数学单元测试模拟卷。每个小节涵盖一道典型考题，拆解解题方法、关键公式和常见易错点。利用这份复习资料巩固你对复数、矩阵、向量、微积分、双曲函数、极坐标和微分方程等核心专题的理解。
1. Complex Numbers: Simplifying Expressions | 复数：化简表达式
Question: Express z = (3 + 2i)(1 − i) + (4 − i) / (2 + i) in the form a + bi, where a, b ∈ ℝ.
题目：将 z = (3 + 2i)(1 − i) + (4 − i) / (2 + i) 写成 a + bi 的形式，其中 a, b ∈ ℝ。
1 = 2i, 2i × (−i) = −2i². Since i² = −1, −2i² = 2. So the sum is 3 − 3i + 2i + 2 = 5 − i.
首先展开乘积 (3 + 2i)(1 − i)。逐项相乘：3 × 1 = 3，3 × (−i) = −3i，2i × 1 = 2i，2i × (−i) = −2i²。因为 i² = −1，所以 −2i² = 2。因此和为 3 − 3i + 2i + 2 = 5 − i。
, (−i)×2 = −2i, (−i)×(−i) = i² = −1, giving 8 − 6i − 1 = 7 − 6i. Denominator: 2² − i² = 4 − (−1) = 5. Hence the fraction simplifies to (7 − 6i)/5 = 1.4 − 1.2i.
接着化简 (4 − i) / (2 + i)。分子分母同乘共轭 2 − i： (4 − i)(2 − i) / ( (2 + i)(2 − i) )。分子：4×2=8，4×(−i)=−4i，(−i)×2=−2i，(−i)×(−i)=i²=−1，得到 8−6i−1=7−6i。分母：2²−i²=4−(−1)=5。因此分式化简为 (7−6i)/5 = 1.4−1.2i。
Finally, add the two results: (5 − i) + (1.4 − 1.2i) = 6.4 − 2.2i. So a = 6.4, b = −2.2.
最后，将两部分相加：(5 − i) + (1.4 − 1.2i) = 6.4 − 2.2i。因此 a = 6.4，b = −2.2。
2. Matrix Operations: Multiplication and Determinants | 矩阵运算：乘法与行列式
Question: Given A =
| 2 | -1 |
| 0 | 3 |
| 1 | 4 |
| -2 | 0 |
题目：已知 A =
| 2 | -1 |
| 0 | 3 |
| 1 | 4 |
| -2 | 0 |
0 = 8. Row 2, column 1: 0×1 + 3×(−2) = 0 − 6 = −6. Row 2, column 2: 0×4 + 3×0 = 0. So AB =
| 4 | 8 |
| -6 | 0 |
计算 AB，将 A（2×2）乘以 B（2×2）。第1行第1列元素：2×1 + (−1)×(−2) = 2+2=4。第1行第2列：2×4 + (−1)×0 = 8+0=8。第2行第1列：0×1 + 3×(−2) = 0−6=−6。第2行第2列：0×4 + 3×0=0。因此 AB =
| 4 | 8 |
| -6 | 0 |
det(B) = 48. For AB, det(AB) = (4×0) − (8×−6) = 0 − (−48) = 48. The property is confirmed.
现在计算行列式。det(A) = (2×3) − (−1×0) = 6。det(B) = (1×0) − (4×−2) = 0 − (−8) = 8。因此 det(A) × det(B) = 48。对于 AB，det(AB) = (4×0) − (8×−6) = 0−(−48) = 48。性质得以验证。
3. Vectors: Scalar Product and Angles | 向量：数量积与夹角
d b = i + 4j − 2k, find the angle between a and b, giving your answer to the nearest 0.1°.
题目：已知向量 a = 3i − 2j + k 和 b = i + 4j − 2k，求 a 与 b 之间的夹角，答案精确至 0.1°。
Use the scalar product formula: a·b = |a||b| cos θ. First compute a·b = (3)(1) + (−2)(4) + (1)(−2) = 3 − 8 − 2 = −7.
使用数量积公式：a·b = |a||b| cos θ。先计算 a·b = (3)(1) + (−2)(4) + (1)(−2) = 3 − 8 − 2 = −7。
Next, find the magnitudes: |a| = √(3² + (−2)² + 1²) = √9 + 4 + 1 = √14. |b| = √(1² + 4² + (−2)²) = √1 + 16 + 4 = √21.
接着求模长：|a| = √(3² + (−2)² + 1²) = √9 + 4 + 1 = √14。|b| = √(1² + 4² + (−2)²) = √1 + 16 + 4 = √21。
Thus cos θ = −7 / (√14 √21) = −7 / √294. Simplify √294 = √49 × 6 = 7√6. Hence cos θ = −7 / (7√6) = −1/√6. Then θ = arccos(−1/√6) ≈ 114.1° (since cos θ negative, angle is obtuse).
因此 cos θ = −7 / (√14 √21) = −7 / √294。化简 √294 = √49×6 = 7√6。故 cos θ = −7/(7√6) = −1/√6。于是 θ = arccos(−1/√6) ≈ 114.1°（因 cos θ 为负，角为钝角）。
4. Further Calculus: Differentiating Inverse Trig Functions | 进阶微积分：反三角函数的微分
rentiate y = arctan(3x²) with respect to x. Hence find the gradient of the curve at x = ½.
题目：对 y = arctan(3x²) 求关于 x 的导数，并求曲线在 x = ½ 处的梯度。
u/dx. Here u = 3x², so du/dx = 6x. Therefore dy/dx = (1/(1+(3x²)²)) × 6x = 6x / (1 + 9x⁴).
回忆 arctan(u) 的导数为 (1/(1+u²)) × du/dx。此处 u = 3x²，所以 du/dx = 6x。因此 dy/dx = (1/(1+(3x²)²)) × 6x = 6x / (1 + 9x⁴)。
tor: 1 + 9/16 = 25/16. Numerator: 3. So gradient = 3 / (25/16) = 3 × 16/25 = 48/25 = 1.92.
在 x = ½ 处代入：dy/dx = 6(½) / (1 + 9(½)⁴)。计算 (½)⁴ = 1/16，9 × 1/16 = 9/16。分母：1 + 9/16 = 25/16。分子：3。因此梯度 = 3 / (25/16) = 3×16/25 = 48/25 = 1.92。
5. Hyperbolic Functions: Solving Equations | 双曲函数：解方程
Question: Solve the equation 4 sinh² x − 3 cosh x = 1, giving answers in logarithmic form.
题目：解方程 4 sinh² x − 3 cosh x = 1，用对数形式给出答案。
cosh² x − 1) − 3 cosh x = 1 → 4 cosh² x − 4 − 3 cosh x = 1 → 4 cosh² x − 3 cosh x − 5 = 0.
利用恒等式 sinh² x = cosh² x − 1。代入：(cosh² x − 1) − 3 cosh x = 1 → 4 cosh² x − 4 − 3 cosh x = 1 → 4 cosh² x − 3 cosh x − 5 = 0。
Let y = cosh x. Then 4y² − 3y − 5 = 0. Solve quadratic: y = [3 ± √9 + 80] / 8 = [3 ± √89] / 8. Since cosh x ≥ 1, we need y = (3 + √89)/8. Check: √89 ≈ 9.434, (3+9.434)/8 ≈ 1.554 > 1, valid. The other root (3 − √89)/8 is negative, so rejected.
令 y = cosh x。则 4y² − 3y − 5 = 0。解二次方程：y = [3 ± √9 + 80] / 8 = [3 ± √89] / 8。因为 cosh x ≥ 1，需取 y = (3 + √89)/8。验证：√89 ≈ 9.434，(3+9.434)/8 ≈ 1.554 > 1，有效。另一根 (3 − √89)/8 为负，舍去。
Now solve cosh x = y. Recall cosh x = (eˣ + e⁻ˣ)/2 = y → eˣ + e⁻ˣ = 2y → multiply by eˣ: e²ˣ − 2y eˣ + 1 = 0. This is quadratic in eˣ, so eˣ = y ± √y² − 1. Substituting y gives eˣ = (3 + √89)/8 ± √( ((3+√89)/8)² − 1 ). Simplify to obtain exact logarithmic forms, but generally x = ln( y + √y²−1 ) or x = ln( y − √y²−1 ). However cosh is even, both x and −x give same cosh. Thus x = ± ln( y + √y² − 1 ).
现在解 cosh x = y。已知 cosh x = (eˣ + e⁻ˣ)/2 = y → eˣ + e⁻ˣ = 2y → 两边同乘 eˣ：e²ˣ − 2y eˣ + 1 = 0。这是关于 eˣ 的二次方程，因此 eˣ = y ± √y² − 1。代入 y 得 eˣ = (3+√89)/8 ± √( ((3+√89)/8)² − 1 )。可化简得精确对数形式，但一般 x = ln( y + √y²−1 ) 或 x = ln( y − √y²−1 )。由于 cosh 是偶函数，x 和 −x 给出相同 cosh。所以 x = ± ln( y + √y² − 1 )。
Therefore, the solutions are x = ± ln( (3 + √89)/8 + √( ((3+√89)/8)² − 1 ) ).
因此解为 x = ± ln( (3 + √89)/8 + √( ((3+√89)/8)² − 1 ) )。
6. Polar Coordinates: Area Enclosed | 极坐标：围成面积
s given by r = 2 + 3 cos θ. Find the area enclosed by one loop of the curve for 0 ≤ θ ≤ π.
题目：极坐标曲线为 r = 2 + 3 cos θ。求该曲线在 0 ≤ θ ≤ π 内一个环所围成的面积。
e r² = (2 + 3 cos θ)² = 4 + 12 cos θ + 9 cos²θ. So A = ½ ∫₀^π (4 + 12 cos θ + 9 cos²θ) dθ.
使用面积公式 A = ½ ∫ r² dθ。这里 r² = (2 + 3 cos θ)² = 4 + 12 cos θ + 9 cos²θ。因此 A = ½ ∫₀^π (4 + 12 cos θ + 9 cos²θ) dθ。
Then the integrand becomes 4 + 12 cos θ + 4.5 + 4.5 cos 2θ = 8.5 + 12 cos θ + 4.5 cos 2θ.
将 cos²θ 替换为 (1 + cos 2θ)/2：9 cos²θ = (9/2)(1 + cos 2θ) = 4.5 + 4.5 cos 2θ。则被积函数变为 4 + 12 cos θ + 4.5 + 4.5 cos 2θ = 8.5 + 12 cos θ + 4.5 cos 2θ。
sin π + 2.25 sin 2π = 8.5π + 0 + 0. At 0: all zero. So A = ½ × 8.5π = 4.25π square units.
逐项积分：∫ 8.5 dθ = 8.5θ；∫ 12 cos θ dθ = 12 sin θ；∫ 4.5 cos 2θ dθ = 2.25 sin 2θ。从 0 到 π 求值。在 π 处：8.5π + 12 sin π + 2.25 sin 2π = 8.5π + 0 + 0。在 0 处全为 0。所以 A = ½ × 8.5π = 4.25π 平方单位。
7. Differential Equations: First Order Linear | 微分方程：一阶线性
Question: Solve the differential equation dy/dx + 2y tan x = sin x, given that y(0) = 1.
题目：解微分方程 dy/dx + 2y tan x = sin x，已知 y(0) = 1。
This is a linear first-order ODE with integrating factor IF = e∫ 2 tan x dx. Compute ∫ 2 tan x dx = 2 ln|sec x| = ln(sec² x). So IF = sec² x.
这是一阶线性常微分方程，积分因子 IF = e∫ 2 tan x dx。计算 ∫ 2 tan x dx = 2 ln|sec x| = ln(sec² x)。所以 IF = sec² x。
ides: y sec² x = ∫ sec² x sin x dx = ∫ (1/cos² x) sin x dx = ∫ tan x sec x dx = sec x + C.
方程两边同乘：sec² x dy/dx + 2 sec² x tan x y = sec² x sin x。左边是 d/dx (y sec² x)。于是两边积分：y sec² x = ∫ sec² x sin x dx = ∫ (1/cos² x) sin x dx = ∫ tan x sec x dx = sec x + C。
=1: at x=0, cos 0=1, so 1 + C × 1 = 1 → C = 0. Hence the particular solution is y = cos x.
因此 y = cos² x (sec x + C) = cos x + C cos² x。应用条件 y(0)=1：x=0 时 cos 0=1，得 1 + C×1 = 1 → C = 0。故特解为 y = cos x。
8. Roots of Polynomials: Complex Roots | 多项式根：复数根
hat 1 + 2i is a root of the cubic equation x³ − x² + 3x + 5 = 0, find the other two roots.
题目：已知 1 + 2i 是三次方程 x³ − x² + 3x + 5 = 0 的一个根，求其余两根。
(x − (1−2i)) = ( (x−1) − 2i)((x−1) + 2i) = (x−1)² + 4 = x² − 2x + 5 is a quadratic factor.
由于系数为实数，其共轭复数 1 − 2i 也是一个根。于是 (x − (1+2i))(x − (1−2i)) = ( (x−1) − 2i)((x−1) + 2i) = (x−1)² + 4 = x² − 2x + 5 是一个二次因式。
tiply: x(x² − 2x + 5) = x³ − 2x² + 5x. Subtract: (−x² + 2x²) = x², (3x − 5x) = −2x. Bring down +5: x² − 2x + 5. Next term: +1. Multiply: 1×(x² − 2x + 5) = x² − 2x + 5. Subtract gives 0. Hence quotient is x + 1. The third root is −1.
用该二次因式除三次多项式。进行多项式除法：(x³ − x² + 3x + 5) ÷ (x² − 2x + 5)。首项：x。乘：x(x² − 2x + 5) = x³ − 2x² + 5x。相减：(−x² + 2x²) = x²，(3x − 5x) = −2x。下移 +5：x² − 2x + 5。次项：+1。乘：1×(x² − 2x + 5) = x² − 2x + 5。相减得 0。故商式为 x + 1。第三个根为 −1。
Thus the roots are 1+2i, 1−2i, −1.
因此根为 1+2i，1−2i，−1。
9. Series: Method of Differences | 级数：差分法
Question: Use the method of differences to sum the series Σr=1n (1 / (r(r+2))). Express the result as a single fraction.
题目：使用差分法求级数 Σr=1n (1 / (r(r+2))) 的和，并将结果用单个分式表示。
e term in partial fractions: 1/(r(r+2)) = A/r + B/(r+2). Solving gives A = ½, B = −½. So ur = ½ (1/r − 1/(r+2)).
首先将项写成部分分式：1/(r(r+2)) = A/r + B/(r+2)。解得 A = ½，B = −½。所以 ur = ½ (1/r − 1/(r+2))。
Write out the sum Sn = ½ ∑r=1n (1/r − 1/(r+2)). Expand for first few terms to spot cancellation:
展开求和 Sn = ½ ∑r=1n (1/r − 1/(r+2))，列出前几项观察消去：
r=1: ½(1/1 − 1/3)
r=2: ½(1/2 − 1/4)
r=3: ½(1/3 − 1/5)
...
r=n−2: ½(1/(n−2) − 1/n)
r=n−1: ½(1/(n−1) − 1/(n+1))
r=n: ½(1/n − 1/(n+2))
Notice that all terms except 1/1, 1/2, −1/(n+1), −1/(n+2) cancel telescopically. Thus Sn = ½( 1 + 1/2 − 1/(n+1) − 1/(n+2) ).
观察发现除了 1/1、1/2、−1/(n+1)、−1/(n+2) 之外，所有项都通过错位相消。于是 Sn = ½( 1 + 1/2 − 1/(n+1) − 1/(n+2) )。
Simplify: Sn (n+1) ) / ((n+1)(n+2)) ) = ½( 3/2 − (2n+3)/((n+1)(n+2)) ). Combine to a single fraction: Sn n+3)) / (4(n+1)(n+2)). This can be expanded if needed, but leaving factored is acceptable.
化简：Sn = ½( 3/2 − ( (n+2)+(n+1) ) / ((n+1)(n+2)) ) = ½( 3/2 − (2n+3)/((n+1)(n+2)) )。合并为单个分式：Sn = (3(n+1)(n+2) − 2(2n+3)) / (4(n+1)(n+2))。如有需要可展开分子，保留因式分解亦可。
10. Matrices: Inverse and Linear Transformations | 矩阵：逆与线性变换
Question: A linear transformation T: ℝ² → ℝ² is represented by the matrix M =
| 2 | -3
Published by TutorHao | Year 12 进阶数学 Revision Series | aleveler.com
广告 / Sponsored ALevel国际高中数学课程 剑桥牛津培生 CAIE Edexcel AQA国际考试局 课本练习册复习指南 英文教材进… 数学 备考教辅 · 正版现货 · 与本文内容相关 查看详情 / View on Taobao参考价 ¥69.8更多咨询请联系16621398022（同微信） 备考资讯每日更新，关注不错过 考纲变化 · 科目分析 · 学习方法  在微信里长按二维码识别关注 或截图后用「扫一扫 → 相册」选择本图 公众号：tutorhao |