📚 Year 11 WJEC Further Mathematics: Unit Test Mock Paper Walkthrough | Year 11 WJEC 进阶数学：单元测试模拟卷解析
Welcome to this detailed walkthrough of a mock unit test designed for the Year 11 WJEC Further Mathematics specification. Each question is unpacked step by step, combining clear methodology with exam-focused commentary so that you can consolidate your understanding and build confidence in tackling similar problems under timed conditions.
欢迎阅读这份为 Year 11 WJEC 进阶数学设计的单元测试模拟卷详细解析。我们对每一道题都进行了逐步拆解，结合清晰的解题方法与针对考试的点评，帮助你巩固理解，并在限时条件下自信地解决同类问题。
1. Quadratic Inequalities | 二次不等式
Solve the inequality 2x² – 5x – 3 > 0.
解不等式 2x² – 5x – 3 > 0。
begin by factorising the quadratic expression. Look for two numbers that multiply to (2 × –3) = –6 and add to –5; these are –6 and +1. Split the middle term: 2x² – 6x + x – 3, then group: 2x(x – 3) + 1(x – 3) = (2x + 1)(x – 3). The inequality becomes (2x + 1)(x – 3) > 0.
我们首先对二次式进行因式分解。寻找两个数，乘积为 (2 × –3) = –6，且和为 –5；这两个数是 –6 和 +1。拆分中项：2x² – 6x + x – 3，然后分组：2x(x – 3) + 1(x – 3) = (2x + 1)(x – 3)。不等式化为 (2x + 1)(x – 3) > 0。
The critical values are found by setting each factor to zero: 2x + 1 = 0 gives x = –½; x – r line into three intervals. Test a value from each interval in the factored form. For x < –½, say x = –1, both factors are negative, giving a positive product. For –½ < x < 3, say x = 0, the factor (2x+1) is positive and (x– . For x > 3, say x = 4, both positive. The inequality requires > 0, so the solution is x < –½ or x > 3.
令每一个因式等于零求得临界值：2x + 1 = 0 得 x = –½；x – 3 = 0 得 x = 3。这两个值将数轴分成三个区间。在每个区间内取测试值代入因式积。当 x < –½，如 x = –1，两个因式均为负，积为正。当 –½ < x < 3，如 x = 0，(2x+1) 为正而 (x–3) 为负，积为负。当 x > 3，如 x = 4，两者皆正。原不等式要求 > 0，故解为 x < –½ 或 x > 3。
2. Polynomial Division and Factor Theorem | 多项式除法与因式定理
Given f(x) = 2x³ – 3x² – 3x + 2, show that (x – 2) is a factor and hence factorise f(x) completely.
已知 f(x) = 2x³ – 3x² – 3x + 2，证明 (x – 2) 是其因式，并由此将 f(x) 完全分解。
Use the Factor Theorem: if f(2) = 0 then (x – 2) is a factor. f(2) = 2(2)³ – 3(2)² – 3(2) + 2 = 2(8) – 3(4) – 6 + 2 = 16 – 12 – 6 + 2 = 0. Since the remainder is zero, (x – 2) is indeed a factor.
使用因式定理：若 f(2) = 0，则 (x – 2) 为因式。f(2) = 2(2)³ – 3(2)² – 3(2) + 2 = 16 – 12 – 6 + 2 = 0。余数为零，故 (x – 2) 确为因式。
Now perform algebraic division (or synthetic division). Dividing 2x³ – 3x² – 3x + 2 by (x – 2) yields the quadratic quotient 2x² + x – 1. Next, factorise this quadratic: 2x² + x – 1 = (2x – 1)(x + 1). Thus the full factorisation is f(x) = (x – 2)(2x – 1)(x + 1).
现在进行多项式除法（或用综合除法）。2x³ – 3x² – 3x + 2 除以 (x – 2) 得到二次商式 2x² + x – 1。接着分解该二次式：2x² + x – 1 = (2x – 1)(x + 1)。因此完全分解为 f(x) = (x – 2)(2x – 1)(x + 1)。
3. Differentiation and Tangent Equation | 微分与切线方程
Find the equation of the tangent to the curve y = 4x³ – 5x + 2 at the point where x = 1.
求曲线 y = 4x³ – 5x + 2 在 x = 1 处的切线方程。
First, find the derivative: dy/dx = 12x² – 5. At x = 1, the gradient of the tangent m = 12(1)² – 5 = 7. Next, find the y-coordinate of the point: y = 4(1)³ – 5(1) + 2 = 4 – 5 + 2 = 1. So the point is (1, 1). The tangent line has equation y – 1 = 7(x – 1), which simplifies to y = 7x – 6.
首先求导：dy/dx = 12x² – 5。在 x = 1 处，切线斜率 m = 12(1)² – 5 = 7。再求该点纵坐标：y = 4(1)³ – 5(1) + 2 = 1。所以点为 (1, 1)。切线方程为 y – 1 = 7(x – 1)，化简得 y = 7x – 6。
4. Matrix Multiplication and Inverse | 矩阵乘法与逆矩阵
Given A = [2, 1; 3, 2] and B = [1, –1; 0, 4], find AB and A⁻¹.
已知 A = [2, 1; 3, 2] 和 B = [1, –1; 0, 4]，求 AB 和 A⁻¹。
For matrix multiplication AB: multiply rows of A by columns of B. AB = [(2×1 + 1×0), (2×–1 + 1×4); (3×1 + 2×0), (3×–1 + 2×4)] = [(2+0), (–2+4); (3+0), (–3+8)] = [2, 2; 3, 5]. So AB = [2, 2; 3, 5].
计算矩阵乘法 AB：用 A 的行乘以 B 的列。AB = [(2×1 + 1×0), (2×–1 + 1×4); (3×1 + 2×0), (3×–1 + 2×4)] = [2, 2; 3, 5]。
To find the inverse of A, first calculate its determinant: det(A) = 2×2 – 1×3 = 4 – 3 = 1. Since det(A) ≠ 0, the inverse exists. A⁻¹ = (1/det) × [2, –1; –3, 2] = [2, –1; –3, 2].
求 A 的逆矩阵，先计算行列式：det(A) = 2×2 – 1×3 = 4 – 3 = 1。行列式不为零，故逆存在。A⁻¹ = (1/det) × [2, –1; –3, 2] = [2, –1; –3, 2]。
5. Vector Collinearity | 向量共线
ition vectors a = 2i + j, b = 5i + 3j and c = 8i + 5j. Show that A, B and C are collinear.
点 A、B、C 的位置向量分别为 a = 2i + j，b = 5i + 3j，c = 8i + 5j。证明 A、B、C 三点共线。
Find the vectors AB and BC. AB = b – a = (5 – 2)i + (3 – 1)j = 3i + 2j. BC = c – b = (8 – 5)i + (5 – , the vectors are parallel and share point B, so A, B and C lie on the same straight line.
计算向量 AB 和 BC。AB = b – a = (5–2)i + (3–1)j = 3i + 2j。BC = c – b = (8–5)i + (5–3)j = 3i + 2j。因 AB = BC，两向量平行且共点 B，故 A、B、C 三点共线。
6. Trigonometric Equations | 三角方程
Solve sin(2θ) = 0.5 for 0° ≤ θ ≤ 360°.
解 sin(2θ) = 0.5，其中 0° ≤ θ ≤ 360°。
sin X = 0.5. The principal value is X = 30° (since sin 30° = 0.5). Another solution in 0°–360° is X = 180° – °. k = 1: X = 390°, 510°. k = 2: X = 750° (too large). Thus X = 30°, 150°, 390°, 510°. Divide by 2: θ = 15°, 75°, 195°, 255°.
令 X = 2θ。解 sin X = 0.5。主值为 X = 30°（sin 30° = 0.5）。在 0°–360° 内另一个解为 X = 180° – 30° = 150°。加入周期得通解：X = 30° + 360°k 或 X = 150° + 360°k。要求 0° ≤ θ ≤ 360°，故 0° ≤ 2θ ≤ 720°。k=0 得 30°, 150°；k=1 得 390°, 510°。除以 2 得 θ = 15°, 75°, 195°, 255°。
7. Binomial Expansion | 二项式展开
Expand (2x – 3)⁴ and state the coefficient of x³.
展开 (2x – 3)⁴ 并写出 x³ 的系数。
Use the binomial theorem: (a + b)ⁿ = Σ (nCr) aⁿ⁻ʳ bʳ. Here a = 2x, b = –3, n = 4. The expansion up to all terms: (4C0)(2x)⁴(–3)⁰ + (4C1)(2x)³(–3)¹ + (4C2)(2x)²(–3)² + (4C3)(2x)¹(–3)³ + (4C4)(2x)⁰(–3)⁴. Compute coefficients: 1×16x⁴ + 4×8x³×(–3) + 6×4x²×9 + 4×2x×(–27) + 1×1×81 = 16x⁴ – 96x³ + 216x² – 216x + 81. The coefficient of x³ is –96.
使用二项式定理：(a + b)ⁿ = Σ (nCr) aⁿ⁻ʳ bʳ。这里 a = 2x，b = –3，n = 4。全部项为：1×(2x)⁴×1 + 4×(2x)³×(–3) + 6×(2x)²×9 + 4×(2x)×(–27) + 1×1×81 = 16x⁴ – 96x³ + 216x² – 216x + 81。x³ 的系数为 –96。
8. Composite Functions | 复合函数
f(x) = 2x + 1, g(x) = x² – 2. Find fg(x) and gf(x).
f(x) = 2x + 1，g(x) = x² – 2。求 fg(x) 和 gf(x)。
fg(x) means apply g first, then f: fg(x) = f(g(x)) = f(x² – 2) = 2(x² – 2) + 1 = 2x² – 4 + 1 = 2x² – 3. gf(x) means apply f first, then g: gf(x) = g(f(x)) = g(2x + 1) = (2x + 1)² – 2 = 4x² + 4x + 1 – 2 = 4x² + 4x – 1.
fg(x) 表示先 g 后 f：fg(x) = f(g(x)) = f(x² – 2) = 2(x² – 2) + 1 = 2x² – 3。gf(x) 表示先 f 后 g：gf(x) = g(f(x)) = g(2x+1) = (2x+1)² – 2 = 4x² + 4x – 1。
9. Algebraic Proof | 代数证明
Prove that the sum of any three consecutive integers is always a multiple of 3.
证明任意三个连续整数之和总是 3 的倍数。
an integer, n+1 is an integer, so S is clearly a multiple of 3. This completes the proof.
设这三个连续整数为 n，n+1 和 n+2，其中 n 为整数。它们的和 S = n + (n+1) + (n+2) = 3n + 3。提取公因数 3：S = 3(n + 1)。因为 n 是整数，n+1 也是整数，所以 S 显然是 3 的倍数。证明完毕。
10. Stationary Points and Nature | 驻点及其性质
For the curve y = x³ – 3x² + 2, find the coordinates of any stationary points and determine their nature.
对于曲线 y = x³ – 3x² + 2，求所有驻点的坐标并判断其性质。
Differentiate: dy/dx = 3x² – 6x. Set dy/dx = 0 to locate stationary points: 3x² – 6x = 0 → 3x(x – 2) = 0 → x = 0 or x = 2. Find corresponding y-values: when x = 0, y = 0³ – 3(0)² + 2 = 2; when x = 2, y = 8 – 12 + 2 = –2. So stationary points are (0, 2) and (2, –2).
求导：dy/dx = 3x² – 6x。令导数为零以确定驻点：3x² – 6x = 0 → 3x(x – 2) = 0 → x = 0 或 x = 2。求相应的 y 值：x = 0 时 y = 2；x = 2 时 y = –2。于是驻点为 (0, 2) 和 (2, –2)。
To classify, use the second derivative: d²y/dx² = 6x – 6. At x = 0, d²y/dx² = –6 < 0, so a local maximum. At x = 2, d²y/dx² = 6(2) – 6 = 6 > 0, so a local minimum. Hence, (0, 2) is a maximum point, (2, –2) is a minimum point.
为判断性质，求二阶导数：d²y/dx² = 6x – 6。在 x = 0 处，二阶导数为 –6 < 0，故为局部极大值；在 x = 2 处，二阶导数为 6 > 0，故为局部极小值。因此 (0, 2) 是极大点，(2, –2) 是极小点。
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