📚 SQA Year 12 Maths: Mock Unit Test Walkthrough | SQA 12 年级数学：单元测试模拟卷详解
This article provides a complete walkthrough of a mock unit test for SQA Year 12 Mathematics, carefully designed to reflect the style and difficulty of the official assessment. The questions cover straight line, differentiation, integration, trigonometry, circles, logarithms, tangents, and composite functions. Each solution is explained step by step, with English and Chinese commentary side by side to support bilingual learners.
本文为 SQA 12 年级数学单元测试模拟卷提供完整解析，题目严格参照官方评估的风格与难度设计，覆盖直线、微分、积分、三角学、圆、对数、切线以及复合函数等核心专题。每道题逐步讲解，配有中英对照注释，助力双语学习者深入理解。
1. Straight Line: Perpendicular Bisector | 直线：垂直平分线
Question: Find the equation of the perpendicular bisector of the line segment joining A(-2, 1) and B(4, 7).
题目：求连接 A(-2, 1) 与 B(4, 7) 的线段的垂直平分线方程。
First, calculate the midpoint M of AB. M = ( (-2+4)/2 , (1+7)/2 ) = (1, 4).
首先计算 AB 的中点 M：M = ( (-2+4)/2 , (1+7)/2 ) = (1, 4)。
The gradient of AB is m_AB = (7 – 1)/(4 – (-2)) = 6/6 = 1.
AB 的斜率 m_AB = (7 – 1)/(4 – (-2)) = 6/6 = 1。
The perpendicular gradient is the negative reciprocal: m_perp = -1/1 = -1.
垂直平分线的斜率为负倒数：m_perp = -1/1 = -1。
Using point-slope form with M(1, 4): y – 4 = -1(x – 1). Simplify to y = -x + 5.
利用点斜式，代入 M(1, 4)：y – 4 = -1(x – 1)。化简得 y = -x + 5。
Perpendicular bisector: y = -x + 5
垂直平分线方程：y = -x + 5
2. Differentiation: Increasing/Decreasing and Extrema | 微分：单调性与极值
Question: For f(x) = 2x³ – 9x² + 12x + 5, find the intervals where f is increasing or decreasing, and identify any local maxima or minima.
题目：已知 f(x) = 2x³ – 9x² + 12x + 5，求 f 的单调递增、递减区间，并指出局部极大值与极小值。
Differentiate: f'(x) = 6x² – 18x + 12 = 6(x² – 3x + 2) = 6(x – 1)(x – 2).
求导：f'(x) = 6x² – 18x + 12 = 6(x² – 3x + 2) = 6(x – 1)(x – 2)。
Set f'(x) = 0 → x = 1 or x = 2. Create a sign table for f'(x).
令 f'(x) = 0 → x = 1 或 x = 2。制作 f'(x) 的符号表。
| Interval | x < 1 | 1 < x < 2 | x > 2 |
|---|---|---|---|
| Sign of f'(x) | + | – | + |
Thus f is increasing on (-∞, 1), decreasing on (1, 2), increasing on (2, ∞).
因此 f 在 (-∞, 1) 递增，在 (1, 2) 递减，在 (2, ∞) 递增。
| 区间 | x < 1 | 1 < x < 2 | x > 2 |
|---|---|---|---|
| f'(x) 符号 | + | – | + |
Local maximum at x = 1: f(1) = 2 – 9 + 12 + 5 = 10. Local minimum at x = 2: f(2) = 16 – 36 + 24 + 5 = 9.
局部极大值在 x = 1：f(1) = 2 – 9 + 12 + 5 = 10。局部极小值在 x = 2：f(2) = 16 – 36 + 24 + 5 = 9。
3. Integration: Polynomial and Rational Functions | 积分：多项式与有理函数
Question: Evaluate ∫(6x² – 4/x² + 3) dx.
题目：计算 ∫(6x² – 4/x² + 3) dx。
Rewrite the integrand using negative exponents: ∫(6x² – 4x⁻² + 3) dx.
将被积函数用负指数改写：∫(6x² – 4x⁻² + 3) dx。
Integrate term by term: ∫6x² dx = 6·(x³/3) = 2x³; ∫-4x⁻² dx = -4·(x⁻¹/(-1)) = 4x⁻¹ = 4/x; ∫3 dx = 3x.
逐项积分：∫6x² dx = 6·(x³/3) = 2x³；∫-4x⁻² dx = -4·(x⁻¹/(-1)) = 4x⁻¹ = 4/x；∫3 dx = 3x。
Add the constant of integration C.
加上积分常数 C。
∫(6x² – 4/x² + 3) dx = 2x³ + 4/x + 3x + C
∫(6x² – 4/x² + 3) dx = 2x³ + 4/x + 3x + C
4. Trigonometric Equation: Quadratic in Cosine | 三角方程：余弦二次型
Question: Solve 2cos²x – 3cosx + 1 = 0 for 0 ≤ x ≤ 2π.
题目：解方程 2cos²x – 3cosx + 1 = 0，其中 0 ≤ x ≤ 2π。
Factor the quadratic in cosx: (2cosx – 1)(cosx – 1) = 0.
对 cosx 的二次式因式分解：(2cosx – 1)(cosx – 1) = 0。
Set each factor to zero: cosx = 1/2 or cosx = 1.
令每个因子为零：cosx = 1/2 或 cosx = 1。
For cosx = 1/2, the solutions in [0, 2π] are x = π/3 and x = 5π/3.
由 cosx = 1/2，在 [0, 2π] 中的解为 x = π/3 与 x = 5π/3。
For cosx = 1, the solutions are x = 0 and x = 2π.
由 cosx = 1，解为 x = 0 与 x = 2π。
Solution set: {0, π/3, 5π/3, 2π}
解集：{0, π/3, 5π/3, 2π}
5. Circle Equation from Centre and Point | 根据圆心和点求圆的方程
Question: Find the equation of the circle with centre C(3, -2) that passes through the point P(7, 1).
题目：求以 C(3, -2) 为圆心且经过点 P(7, 1) 的圆的方程。
The radius r is the distance from C to P: r = √((7 – 3)² + (1 – (-2))²) = √(4² + 3²) = √(16 + 9) = √25 = 5.
半径 r 是 C 到 P 的距离：r = √((7 – 3)² + (1 – (-2))²) = √(4² + 3²) = √(16 + 9) = √25 = 5。
The standard equation of a circle with centre (h, k) and radius r is (x – h)² + (y – k)² = r².
以 (h, k) 为圆心、半径为 r 的圆的标准方程为 (x – h)² + (y – k)² = r²。
Substitute h = 3, k = -2, r = 5: (x – 3)² + (y + 2)² = 25.
代入 h = 3, k = -2, r = 5：(x – 3)² + (y + 2)² = 25。
Circle equation: (x – 3)² + (y + 2)² = 25
圆的方程：(x – 3)² + (y + 2)² = 25
6. Logarithms: Properties and Evaluation | 对数：性质与求值
Question: Given logₐ(2) = 0.5 and logₐ(3) = 0.8, evaluate logₐ(48) and logₐ(√6).
题目：已知 logₐ(2) = 0.5 且 logₐ(3) = 0.8，求 logₐ(48) 与 logₐ(√6) 的值。
Express 48 as a product of prime factors: 48 = 2⁴ × 3.
将 48 表示为质因数的乘积：48 = 2⁴ × 3。
Then logₐ(48) = logₐ(2⁴) + logₐ(3) = 4 logₐ(2) + logₐ(3) = 4(0.5) + 0.8 = 2.0 + 0.8 = 2.8.
因此 logₐ(48) = logₐ(2⁴) + logₐ(3) = 4 logₐ(2) + logₐ(3) = 4(0.5) + 0.8 = 2.0 + 0.8 = 2.8。
For logₐ(√6): √6 = 6^(1/2). Write 6 = 2 × 3, so logₐ(√6) = (1/2)(logₐ2 + logₐ3) = 0.5 × (0.5 + 0.8) = 0.5 × 1.3 = 0.65.
对于 logₐ(√6)：√6 = 6^(1/2)。6 = 2 × 3，所以 logₐ(√6) = (1/2)(logₐ2 + logₐ3) = 0.5 × (0.5 + 0.8) = 0.5 × 1.3 = 0.65。
logₐ(48) = 2.8, logₐ(√6) = 0.65
logₐ(48) = 2.8，logₐ(√6) = 0.65
7. Tangent Condition: Discriminant | 切线条件：判别式
Question: The line y = 2x + 3 is tangent to the curve y = x² + k. Find the value of k.
题目：已知直线 y = 2x + 3 与曲线 y = x² + k 相切，求 k 的值。
At the point of tangency, the two equations are equal: x² + k = 2x + 3.
在切点处，两方程相等：x² + k = 2x + 3。
Rearrange into a quadratic equation: x² – 2x + (k – 3) = 0.
整理成二次方程：x² – 2x + (k – 3) = 0。
For tangency, there is exactly one solution, so the discriminant Δ must be zero. Here a = 1, b = -2, c = k – 3.
由于相切，方程有唯一解，故判别式 Δ 必须为零。此时 a = 1, b = -2, c = k – 3。
Δ = b² – 4ac = (-2)² – 4(1)(k – 3) = 4 – 4k + 12 = 16 – 4k.
Δ = b² – 4ac = (-2)² – 4(1)(k – 3) = 4 – 4k + 12 = 16 – 4k。
Set Δ = 0 → 16 – 4k = 0 → 4k = 16 → k = 4.
令 Δ = 0：16 – 4k = 0 → 4k = 16 → k = 4。
k = 4
k = 4
8. Composite Function: Domain of a Square Root | 复合函数：平方根的定义域
Question: Let h(x) = f(g(x)) where f(x) = √x and g(x) = x² – 9. Determine the domain of h(x).
题目：已知 h(x) = f(g(x))，其中 f(x) = √x 且 g(x) = x² – 9。求 h(x) 的定义域。
First, form the composite function: h(x) = √( g(x) ) = √(x² – 9).
首先写出复合函数：h(x) = √(x² – 9)。
The expression under the square root must be non-negative: x² – 9 ≥ 0.
根号内的表达式必须非负：x² – 9 ≥ 0。
Solve the inequality: (x – 3)(x + 3) ≥ 0. The critical points are x = -3 and x = 3.
解不等式：(x – 3)(x + 3) ≥ 0。
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