📚 Year 12 WJEC Mathematics: Mock Unit Test Walkthrough | WJEC 12年级数学模拟单元测试详解
This step-by-step walkthrough breaks down a full mock unit test designed for the Year 12 WJEC Mathematics syllabus. The paper covers pure topics such as algebra, trigonometric equations, calculus, and sequences, alongside applied questions from mechanics and statistics. Working through these solutions will help you consolidate key techniques, avoid common pitfalls, and build confidence for the actual unit test.
本文逐步解析一份为 WJEC 12 年级数学课程设计的完整模拟单元测试卷。试卷涵盖纯数学内容，如代数、三角方程、微积分和数列，以及力学和统计方面的应用题。通过演练这些解答，你可以巩固关键技巧、避开常见错误，并为真正的单元测试建立信心。
1. Discriminant and Real Roots | 判别式与实根条件
Question: The quadratic equation x² + (k – 3)x + 4 = 0 has no real roots. Determine the range of values of k.
题目：二次方程 x² + (k – 3)x + 4 = 0 没有实根。求 k 的取值范围。
For a quadratic ax² + bx + c = 0, the condition for no real roots is that the discriminant is strictly less than zero: b² – 4ac < 0. Here a = 1, b = k - 3 and c = 4.
对于二次方程 ax² + bx + c = 0，没有实根的条件是判别式严格小于零：b² – 4ac < 0。此处 a = 1, b = k - 3, c = 4。
We compute the discriminant: (k – 3)² – 4 × 1 × 4 < 0. Expanding gives k² - 6k + 9 - 16 < 0, which simplifies to k² - 6k - 7 < 0.
计算判别式：(k – 3)² – 4 × 1 × 4 < 0。展开得 k² - 6k + 9 - 16 < 0，化简为 k² - 6k - 7 < 0。
Factorising the quadratic inequality: (k – 7)(k + 1) < 0. The critical values are k = -1 and k = 7. By testing intervals or considering the shape of the parabola, the product is negative when k lies between the two roots.
将二次不等式因式分解：(k – 7)(k + 1) < 0。临界值为 k = -1 和 k = 7。通过区间测试或考虑抛物线形状，可知当 k 介于两实根之间时乘积为负。
Therefore, the solution is -1 < k < 7. The quadratic will have no real roots for any value of k in this open interval.
因此，解集为 -1 < k < 7。当 k 取此开区间内任意值时，二次方程均无实根。
2. Polynomial Division and Factor Theorem | 多项式除法与因式定理
Question: f(x) = 2x³ – 3x² – 11x + 6. Show that (x – 3) is a factor of f(x) and hence factorise f(x) completely. Solve f(x) = 0.
题目：f(x) = 2x³ – 3x² – 11x + 6。证明 (x – 3) 是 f(x) 的一个因式，并由此将 f(x) 完全分解。解方程 f(x) = 0。
By the Factor Theorem, if (x – 3) is a factor then f(3) should be zero. Substituting x = 3: f(3) = 2(27) – 3(9) – 11(3) + 6 = 54 – 27 – 33 + 6 = 0, so the factor is confirmed.
根据因式定理，若 (x – 3) 是因式，则 f(3) 必为零。代入 x = 3：f(3) = 2×27 – 3×9 – 11×3 + 6 = 54 – 27 – 33 + 6 = 0，因此验证通过。
Now perform polynomial division of f(x) by (x – 3). Using algebraic long division or synthetic division, the quotient is 2x² + 3x – 2. Thus f(x) = (x – 3)(2x² + 3x – 2).
现在对 f(x) 除以 (x – 3)。使用代数长除法或综合除法，商式为 2x² + 3x – 2。因此 f(x) = (x – 3)(2x² + 3x – 2)。
Factorising the quadratic: 2x² + 3x – 2 = (2x – 1)(x + 2). So the fully factorised form is f(x) = (x – 3)(2x – 1)(x + 2).
对二次三项式进行分解：2x² + 3x – 2 = (2x – 1)(x + 2)。因此完全分解式为 f(x) = (x – 3)(2x – 1)(x + 2)。
Setting f(x) = 0 gives the three solutions: x = 3, x = ½, and x = -2. All roots are real and rational.
令 f(x) = 0 可得三个解：x = 3，x = ½ 和 x = -2。所有根均为实有理根。
3. Exponential and Logarithmic Equations | 指数与对数方程
Question: Solve the equation 3e²ˣ – 7eˣ + 2 = 0, giving your answers in exact form.
题目：解方程 3e²ˣ – 7eˣ + 2 = 0，答案保留精确形式。
Notice that e²ˣ = (eˣ)². This indicates a hidden quadratic. Let y = eˣ. The equation becomes 3y² – 7y + 2 = 0.
注意到 e²ˣ = (eˣ)²，这提示该方程是一个隐藏的二次方程。令 y = eˣ，原方程化为 3y² – 7y + 2 = 0。
Factorising: 3y² – 7y + 2 = (3y – 1)(y – 2) = 0. Hence y = ⅓ or y = 2.
因式分解：3y² – 7y + 2 = (3y – 1)(y – 2) = 0。因此 y = ⅓ 或 y = 2。
Re-substitute eˣ for y: eˣ = ⅓ gives x = ln(⅓) = -ln 3. eˣ = 2 gives x = ln 2. Both solutions are exact and valid since the exponential function is always positive.
将 y 替换回 eˣ：由 eˣ = ⅓ 得 x = ln(⅓) = -ln 3；由 eˣ = 2 得 x = ln 2。由于指数函数恒为正，两个解均为有效精确解。
4. Trigonometric Equations and Identities | 三角方程与恒等式
Question: Solve sin 2θ = cos θ for 0° ≤ θ ≤ 360°.
题目：在 0° ≤ θ ≤ 360° 范围内解方程 sin 2θ = cos θ。
Use the double-angle identity sin 2θ = 2 sin θ cos θ. The equation becomes 2 sin θ cos θ = cos θ.
利用倍角恒等式 sin 2θ = 2 sin θ cos θ，原方程化为 2 sin θ cos θ = cos θ。
Bring all terms to one side: 2 sin θ cos θ – cos θ = 0, then factor out cos θ: cos θ (2 sin θ – 1) = 0.
将所有项移至一边：2 sin θ cos θ – cos θ = 0，提取公因式 cos θ：cos θ (2 sin θ – 1) = 0。
Now apply the zero-product principle. Either cos θ = 0 or 2 sin θ – 1 = 0, which gives sin θ = ½.
应用零乘积原理，得 cos θ = 0 或 2 sin θ – 1 = 0，即 sin θ = ½。
For cos θ = 0 in the given interval, θ = 90°, 270°. For sin θ = ½, the principal solutions are θ = 30° and 150°, and these are the only ones within 0° to 360°.
在所给区间内，由 cos θ = 0 得 θ = 90°, 270°；由 sin θ = ½ 得主值 θ = 30° 和 150°，且这些是 0° 到 360° 内的全部解。
Therefore the full solution set is θ = 30°, 90°, 150°, 270°. Always check that no solutions have been lost by division — factoring avoids this risk.
因此全部解为 θ = 30°, 90°, 150°, 270°。务必检查是否存在因相除而丢失的解——采用因式分解可以避免这一风险。
5. Differentiation: Tangents and Stationary Points | 微分：切线与驻点
Question: A curve has equation y = x³ – 4x² + 5x + 1. Find the equation of the tangent to the curve at the point where x = 2. Determine the coordinates of any stationary points.
题目：曲线方程为 y = x³ – 4x² + 5x + 1。求曲线在 x = 2 处的切线方程，并确定所有驻点的坐标。
First differentiate: dy/dx = 3x² – 8x + 5. At x = 2, the gradient is m = 3(4) – 8(2) + 5 = 12 – 16 + 5 = 1.
首先求导：dy/dx = 3x² – 8x + 5。在 x = 2 处，梯度为 m = 3×4 – 8×2 + 5 = 12 – 16 + 5 = 1。
The y-coordinate when x = 2 is y = 8 – 16 + 10 + 1 = 3. So the tangent passes through (2, 3) with gradient 1. Its equation is y – 3 = 1(x – 2), or y = x + 1.
当 x = 2 时，y 坐标为 y = 8 – 16 + 10 + 1 = 3。因此切线过点 (2, 3) 且斜率为 1，其方程为 y – 3 = 1(x – 2)，即 y = x + 1。
For stationary points, set dy/dx = 0: 3x² – 8x + 5 = 0. Factorising gives (3x – 5)(x – 1) = 0, so x = 5/3 and x = 1.
对于驻点，令 dy/dx = 0：3x² – 8x + 5 = 0。因式分解得 (3x – 5)(x – 1) = 0，故 x = 5/3 和 x = 1。
Find the corresponding y-values: when x = 1, y = 1 – 4 + 5 + 1 = 3; when x = 5/3, substitute to get y = (125/27) – 4(25/9) + 5(5/3) + 1 = 125/27 – 100/9 + 25/3 + 1 = 125/27 – 300/27 + 225/27 + 27/27 = 77/27.
计算对应的 y 坐标：当 x = 1 时，y = 1 – 4 + 5 + 1 = 3；当 x = 5/3 时，代入得 y = (125/27) – 4×(25/9) + 5×(5/3) + 1 = 125/27 – 100/9 + 25/3 + 1 = 125/27 – 300/27 + 225/27 + 27/27 = 77/27。
The stationary points are (1, 3) and (5/3, 77/27). You can classify them using the second derivative or by testing the sign of the gradient change.
驻点为 (1, 3) 和 (5/3, 77/27)。可以利用二阶导数或检验梯度符号变化来对它们进行分类。
6. Integration and Area Under a Curve | 积分与曲线下面积
Question: Find the area bounded by the curve y = 4x – x² and the x-axis.
题目：求曲线 y = 4x – x² 与 x 轴所围成的面积。
First find the x-intercepts: solve 4x – x² = 0, factorising to x(4 – x) = 0, so x = 0 and x = 4. The region lies between these limits and is entirely above the x-axis because the quadratic opens downward with a positive intercept region.
首先确定 x 轴截距：解 4x – x² = 0，因式分解为 x(4 – x) = 0，得 x = 0 和 x = 4。该区域位于这两个界限之间，且由于二次函数开口向下，在此区间内曲线位于 x 轴上方。
Area = ∫₀⁴ (4x – x²) dx. Integrate term by term: ∫ 4x dx = 2x², ∫ x² dx = ⅓ x³. So the antiderivative is [2x² – ⅓ x³] evaluated from 0 to 4.
面积 = ∫₀⁴ (4x – x²) dx。逐项积分：∫ 4x dx = 2x²，∫ x² dx = ⅓ x³。因此原函数为 [2x² – ⅓ x³]₀⁴。
Upper limit: 2(16) – ⅓(64) = 32 – 64/3 = (96/3 – 64/3) = 32/3. Lower limit yields 0. So the exact area is 32/3 square units.
代入上限：2×16 – ⅓×64 = 32 – 64/3 = (96/3 – 64/3) = 32/3。下限结果为 0。因此所围面积为 32/3 平方单位。
Always provide the area as a positive quantity; a negative result would indicate the region is below the x-axis or that limits were reversed.
面积始终应取正值；负值结果则表明该区域位于 x 轴下方或积分上下限顺序颠倒。
7. Arithmetic and Geometric Sequences | 等差数列与等比数列
Question: The 5th term of an arithmetic sequence is 19 and the 9th term is 31. Find the first term and the common difference. Hence find the sum of the first 20 terms.
题目：一个等差数列的第 5 项为 19，第 9 项为 31。求首项和公差，并据此计算前 20 项的和。
Let the first term be a and the common difference d. The nth term is given by uₙ = a + (n – 1)d. For n = 5: a + 4d = 19. For n = 9: a + 8d = 31.
设首项为 a，公差为 d。第 n 项为 uₙ = a + (n – 1)d。代入 n = 5：a + 4d = 19；n = 9：a + 8d = 31。
Subtract the first equation from the second to eliminate a: (a + 8d) – (a + 4d) = 31 – 19, giving 4d = 12 so d = 3. Substituting back, a + 12 = 19 gives a = 7.
将第二式减去第一式消去 a：(a + 8d) – (a + 4d) = 31 – 19，得 4d = 12，故 d = 3。代回得 a + 12 = 19，因此 a = 7。
The sum of the first n terms of an arithmetic sequence is Sₙ = n/2 [2a + (n – 1)d]. For n = 20: S₂₀ = 10 [2×7 + 19×3] = 10 [14 + 57] = 10 × 71 = 710.
等差数列前 n 项和公式为 Sₙ = n/2 [2a + (n – 1)d]。代入 n = 20：S₂₀ = 10 [2×7 + 19×3] = 10 [14 + 57] = 10 × 71 = 710。
Always check the consistency of the sequence: terms are 7, 10, 13, 16, 19,… and the 9th is 31, which matches the given pattern.
始终检查数列的一致性：前几项为 7, 10, 13, 16, 19, …，第 9 项确实为 31，吻合给定的模式。
8. Mechanics: Constant Acceleration and SUVAT | 力学：匀加速运动与 SUVAT 方程
Question: A particle moves in a straight line from rest with constant acceleration 4 m s⁻² for 5 seconds. It then maintains constant speed for a further 10 seconds. Calculate the total distance travelled.
题目：一质点从静止开始以 4 m s⁻² 的恒定加速度直线运动 5 秒，然后保持匀速继续运动 10 秒。求通过的总路程。
Split the motion into two stages. Stage 1 (accelerating): initial velocity u = 0, a = 4, t = 5. Use s = ut + ½ at²: s₁ = 0 + ½ × 4 × 5² = 2 × 25 = 50 m. The velocity at the end of this stage is v = u + at = 0 + 4×5 = 20 m s⁻¹.
将运动分为两个阶段。第一阶段加速：初速 u = 0，a = 4，t = 5。使用 s = ut + ½ at²：s₁ = 0 + ½ × 4 × 5² = 2 × 25 = 50 m。该阶段结束时的速度 v = u + at = 0 + 4×5 = 20 m s⁻¹。
Stage 2 (constant speed): the particle travels at 20 m s⁻¹ for 10 seconds. Distance s₂ = speed × time = 20 × 10 = 200 m.
第二阶段匀速：质点以 20 m s⁻¹ 运动 10 秒。路程 s₂ = 速度 × 时间 = 20 × 10 = 200 m。
Total distance = s₁ + s₂ = 50 + 200 = 250 m. It is helpful to sketch a velocity–time graph: the area under the graph represents distance; a trapezium with base 15 s and heights 0 and 20 would give the same result.
总路程 = s₁ + s₂ = 50 + 200 = 250 m。画出速度–时间图会很有帮助：图线下面积代表距离，一个上底 0、下底 20、底边长 15 s 的梯形也会给出同样的结果。
9. Statistics: Binomial Distribution | 统计学：二项分布
Question: A fair die is rolled 8 times. Find the probability that exactly two of the rolls show a 6. Give your answer to three decimal places.
题目：抛掷一枚均匀骰子 8 次。求恰好出现 2 次 6 点的概率，答案保留三位小数。
This is a binomial setting: n = 8 independent trials, probability of success (rolling a 6) p = 1/6, and we need P(X = 2).
这是一个二项分布模型：n = 8 次独立试验，成功（掷得 6 点）的概率 p = 1/6，需要求 P(X = 2)。
The binomial probability formula is P(X = r) = ⁿCᵣ pʳ (1 – p)ⁿ⁻ʳ. Here ⁿCᵣ = C(8,2) = 28, p² = (1/6)² = 1/36, and (1 – p)⁶ = (5/6)⁶.
二项概率公式为 P(X = r) = ⁿCᵣ pʳ (1 – p)ⁿ⁻ʳ。此处 ⁿCᵣ = C(8,2) = 28，p² = (1/6)² = 1/36，(1 – p)⁶ = (5/6)⁶。
So P(X = 2) = 28 × (1/36) × (5/6)⁶. Compute (5/6)⁶ = 15625/46656. Then multiply: 28 × 1/36 = 28/36 = 7/9. Now (7/9) × (15625/46656) = 109375/419904 ≈ 0.260 (to 3 d.p.).
因此 P(X = 2) = 28 × (1/36) × (5/6)⁶。计算 (5/6)⁶ = 15625/46656。然后相乘：28 × 1/36 = 28/36 = 7/9。再 (7/9) × (15625/46656) = 109375/419904 ≈ 0.260（精确至三位小数）。
Be careful with the exponent; some students mistakenly use p⁶(1-p)². Setting up a binomial calculation systematically helps avoid sign errors.
注意指数勿用错；一些学生易错误地使用 p⁶(1-p)²。系统地建立二项计算有助于避免正误颠倒。
10. Vectors: Magnitude and Direction | 向量：模与方向
Question: Given vectors a = 3i – 4j and b = -i + 2j, find |a + 2b| and the angle that a + 2b makes with the positive i-direction.
题目：已知向量 a = 3i – 4j，b = -i + 2j，求 |a + 2b| 以及 a + 2b 与 i 正方向之间的夹角。
First compute a + 2b: 2b = -2i + 4j
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