📚 Unit Test Mock Paper Walkthrough | 单元测试模拟卷解析
This mock paper is designed to mirror the style and depth of an SQA Year 11 (Higher) Mathematics unit test. Each question targets a core topic—from straight lines and polynomials to calculus and vectors—and is followed by a step‑by‑step bilingual solution. Working through these problems systematically will strengthen your conceptual understanding and exam technique, ensuring you are fully prepared for the real assessment.
本套模拟卷仿照 SQA 11 年级（进阶数学/Higher）单元测试的题型与难度命制。每道题聚焦一个核心考点——从直线与多项式到微积分与向量——并配有逐步的中英双语解析。系统演练这些题目将巩固你的概念理解与应试技巧，帮助你在真正的测评中游刃有余。
1. Straight Line and Perpendicular Bisector | 直线与垂直平分线
Question: ar bisector of AB. Give your answer in the form ax + by + c = 0 with integer coefficients.
题目：已知点A(2, 5) 和 B(8, 1)，求线段AB的垂直平分线方程，并将结果化为 ax + by + c = 0 的形式（系数为整数）。
AB: M = ((2+8)/2, (5+1)/2) = (5, 3). The midpoint will lie on the perpendicular bisector.
首先计算AB的中点M：M = ((2+8)/2, (5+1)/2) = (5, 3)。中点必然位于垂直平分线上。
8 - 2) = -4/6 = -2/3. The slope of any perpendicular line is the negative reciprocal, so mperp = 3/2.
直线AB的斜率为 (1 - 5) / (8 - 2) = -4/6 = -2/3。与之垂直的直线斜率为其负倒数，因此 m垂直 = 3/2。
Using the point‑slope form y - y1 = m(x - x1 M(5,3): y - 3 = (3/2)(x - 5). Multiply through by 2 to clear fractions: 2y - 6 = 3x - 15.
利用点斜式 y - y1 = m(x - x1) 代入中点M(5,3)：y - 3 = (3/2)(x - 5)。两边同乘2消去分母：2y - 6 = 3x - 15。
3x - 2y - 9 = 0. Always check that both A and B are equidistant from a point on this line.
整理为标准形式：3x - 2y - 15 + 6 = 0 → 3x - 2y - 9 = 0。务必检验A、B两点到该直线上任一点的距离是否相等。
2. Composite Functions and Inverses | 复合函数与反函数
Question: Let f(x) = 2x + 3 and g(x) = x2 - 1. (a) Find f(g(x)) and g(f(x)). (b) Determine f-1(x) and state its domain.
题目：设 f(x) = 2x + 3 和 g(x) = x2 - 1。(a) 求 f(g(x)) 和 g(f(x))。(b) 求 f-1(x) 并写出其定义域。
For f(g(x)), substitute g(x) into f: f(g(x)) = 2(g(x)) + 3 = 2(x2 - 1) + 3 = 2x2 - 2 + 3 = 2x2 + 1.
对于 f(g(x))，将 g(x) 代入 f：f(g(x)) = 2(g(x)) + 3 = 2(x2 - 1) + 3 = 2x2 - 2 + 3 = 2x2 + 1。
For g(f(x)), substitute f(x) into g: g(f(x)) = (f(x))2 - 1 = (2x + 3)2 - 1 = 4x2 + 12x + 9 - 1 = 4x2 + 12x + 8.
对于 g(f(x))，将 f(x) 代入 g：g(f(x)) = (f(x))2 - 1 = (2x + 3)2 - 1 = 4x2 + 12x + 9 - 1 = 4x2 + 12x + 8。
set y = 2x + 3. Swap x and y: x = 2y + 3. Solve for y: 2y = x - 3 → y = (x - 3)/2. Thus f-1(x) = (x - 3)/2.
为求 f 的反函数，令 y = 2x + 3。交换 x 和 y 得到 x = 2y + 3。解出 y：2y = x - 3 → y = (x - 3)/2。因此 f-1(x) = (x - 3)/2。
The domain of f-1 is all real numbers, because the original function f is a linear function with no restrictions on its output.
f-1 的定义域为全体实数，因为原函数 f 是一个线性函数，其值域没有限制。
3. Polynomial Division and the Factor Theorem | 多项式除法与因式定理
Question: f(x) = 2x3 - 3x2 - 3x + 2. Show that (x - 2) is a factor, and hence fully factorise f(x).
题目：f(x) = 2x3 - 3x2 - 3x + 2。证明 (x - 2) 是该多项式的一个因式，并由此完全分解 f(x)。
uate f(2): 2(8) - 3(4) - 3(2) + 2 = 16 - 12 - 6 + 2 = 0. Hence (x - 2) is indeed a factor.
根据因式定理，若 (x - 2) 是一个因式，必有 f(2) = 0。计算 f(2)：2(8) - 3(4) - 3(2) + 2 = 16 - 12 - 6 + 2 = 0。因此 (x - 2) 确实是一个因式。
Use synthetic division or algebraic long division to divide f(x) by (x - 2). The quotient is 2x2 + x - 1.
用综合除法或长除法将 f(x) 除以 (x - 2)，商式为 2x2 + x - 1。
Now factor the quadratic: 2x2 x - 1 = (2x - 1)(x + 1). Thus the fully factorised form is f(x) = (x - 2)(2x - 1)(x + 1).
接着分解二次式：2x2 + x - 1 = (2x - 1)(x + 1)。因此完全分解式为 f(x) = (x - 2)(2x - 1)(x + 1)。
4. Trigonometric Equations | 三角方程求解
Question: Solve the equation 2sin2x - sin x - 1 = 0 for 0° ≤ x ≤ 360°.
题目：解方程 2sin2x - sin x - 1 = 0，x 的范围为 0° ≤ x ≤ 360°。
Recognise this as a quadratic in sin x. Let u = sin x, giving 2u2 - u - 1 = 0. Factorise: (2u + 1)(u - 1) = 0. So u = -1/2 or u = 1.
这是一个关于 sin x 的二次方程。令 u = sin x，得 2u2 - u - 1 = 0。因式分解：(2u + 1)(u - 1) = 0。解得 u = -1/2 或 u = 1。
Case 1: sin x = 1. For 0° ≤ x ≤ 360°, x = 90°.
情形1：sin x = 1。在 0° 至 360° 范围内，x = 90°。
n x = -1/2. Sine is negative in the third and fourth quadrants. The reference angle is sin-1(1/2) = 30°. Thus x = 180° + 30° = 210° and x = 360° - 30° = 330°.
情形2：sin x = -1/2。正弦在第三、四象限为负。参考角为 sin-1(1/2) = 30°。因此 x = 180° + 30° = 210° 以及 x = 360° - 30° = 330°。
The full solution set is x = 90°, 210°, 330°.
完整的解集为 x = 90°、210°、330°。
5. Differentiation – Chain Rule and Tangents | 微分 – 链式法则与切线
Question: Differentiate y = (3x2 + 2)4 espect to x. Hence find the equation of the tangent to the curve at the point where x = 1.
题目：对 y = (3x2 + 2)4 求导。进而求出曲线在 x = 1 处的切线方程。
Apply the chain rule: let u = 3x2 + 2, so y = u4. Then dy/du = 4u3 and du/dx = 6x. Multiplying, dy/dx = 4u3 * 6x = 24x (3x2 + 2)3.
使用链式法则：令 u = 3x2 + 2，则 y = u4。dy/du = 4u3，du/dx = 6x。两者相乘得 dy/dx = 4u3 * 6x = 24x (3x2 + 2)3。
At x = 1, u = 3(1)2 + 2 = 5, so the gradient m = 24(1)(5)3 = 24 * 125 = 3000. The y‑coordinate is y = (5)4 = 625. The point is (1, 625).
当 x = 1 时，u = 3(1)2 + 2 = 5，梯度 m = 24(1)(5)3 = 24 × 125 = 3000。对应的 y 坐标为 y = 54 = 625，切点为 (1, 625)。
Equation of tangent: y - 625 = 3000(x - 1) → y = 3000x - 3000 + 625 → y = 3000x - 2375.
切线方程：y - 625 = 3000(x - 1) → y = 3000x - 3000 + 625 → y = 3000x - 2375。
6. Integration and Area Under a Curve | 积分与曲线下方面积
Question: Find the area enclosed between the curve y = x2 - 4x + 5, the x‑axis, and the lines x = 1 and x = 4.
题目：求曲线 y = x2 - 4x + 5、x 轴以及直线 x = 1 和 x = 4 所围成的面积。
upward with vertex at x = 2 (y = 1), it is positive over [1,4]. Area = ∫ from 1 to 4 of (x2 - 4x + 5) dx.
该曲线是开口向上的抛物线，顶点位于 x = 2 处（y = 1），在区间 [1,4] 上函数值为正。面积 = ∫14 (x2 - 4x + 5) dx。
Integrate term by term: ∫ x2 dx = x3/3, ∫ -4x dx = -2x2, ∫ 5 dx = 5x. So antiderivative F(x) = x3/3 - 2x2 + 5x.
逐项积分：∫ x2 dx = x3/3，∫ -4x dx = -2x2，∫ 5 dx = 5x。因此原函数 F(x) = x3/3 - 2x2 + 5x。
3 - 36/3 = 28/3. F(1) = 1/3 - 2 + 5 = 1/3 + 3 = 10/3. Difference = 28/3 - 10/3 = 18/3 = 6.
计算 F(4) - F(1)：F(4) = 64/3 - 32 + 20 = 64/3 - 12 = 28/3；F(1) = 1/3 - 2 + 5 = 10/3。差值为 28/3 - 10/3 = 18/3 = 6。
The exact area is 6 square units.
所求面积为 6 平方单位。
7. Equation of a Circle and Tangents | 圆的方程与切线
Question: A circle has centre C(2, -3) and radius 5 units. (a) Write its equation. (b) Show that the point P(6, 0) lies on the circle and find the equation of the tangent at P.
题目：已知圆的圆心为 C(2, -3)，半径为 5。(a) 写出圆的方程。(b) 证明点 P(6, 0) 在圆上，并求出过 P 点的切线方程。
(a) Standard form: (x - 2)2 + (y + 3)2 = 25.
(a) 标准方程为：(x - 2)2 + (y + 3)2 = 25。
(b) Substitute P(6,0): (6-2)2 + (0+3)2 = 42 + 32 = 16 + 9 = 25, so P lies on the circle.
(b) 代入 P(6,0)：(6-2)2 + (0+3)2 = 16 + 9 = 25，故 P 在圆上。
-3)) / (6 - 2) = 3/4. The tangent is perpendicular to the radius, so its gradient is -4/3.
半径 CP 的斜率为 (0 - (-3)) / (6 - 2) = 3/4。切线与半径垂直，因此切线斜率为 -4/3。
6) → y = (-4/3)x + 8. Multiply by 3 to avoid fractions: 3y = -4x + 24 → 4x + 3y - 24 = 0.
利用点 P 得出切线方程：y - 0 = (-4/3)(x - 6) → y = (-4/3)x + 8。两边乘 3：3y = -4x + 24 → 4x + 3y - 24 = 0。
8. Exponentials and Logarithms | 指数方程与对数方程
Question: Solve (a) 3e2x = 15, giving your answer in exact form, and (b) log2(x + 1) - log2(x - 2) = 3.
题目：(a) 解方程 3e2x = 15，答案保留精确值。(b) 解方程 log2(x + 1) - log2(x - 2) = 3。
(a) Divide both sides by 3: e2x = 5. Take the natural logarithm of both sides: 2x = ln 5 → x = (1/2) ln 5.
(a) 两边同除以 3：e2x = 5。取自然对数：2x = ln 5 → x = (1/2) ln 5。
(b) Combine the logs using the quotient rule: log2((x+1)/(x-2)) = 3. Rewrite in exponential form: (x+1)/(x-2) = 23 = 8.
(b) 利用差的对数性质合并：log2((x+1)/(x-2)) = 3。转化为指数形式：(x+1)/(x-2) = 23 = 8。
x - x → 17 = 7x → x = 17/7. Check domain: x+1>0 and x-2>0, so x>2. This solution is valid.
交叉相乘：x + 1 = 8(x - 2) → x + 1 = 8x - 16 → 17 = 7x → x = 17/7。检验定义域：x+1>0 且 x-2>0，即 x>2，解符合条件。
9. Vectors – Scalar Product and Angle | 向量 – 数量积与夹角
Question: i + 4j and b = -i + 2j, calculate the angle between a and b, correct to one decimal place.
题目：已知向量 a = 3i + 4j 和 b = -i + 2j，计算 a 与 b 的夹角，结果精确至一位小数。
Use the formula cos θ = (a·b) / (|a| |b|). Compute the dot product: a·b = (3)(-1) + (4)(2) = -3 + 8 = 5.
使用公式 cos θ = (a·b) / (|a| |b|)。计算数量积：a·b = (3)(-1) + (4)(2) = -3 + 8 = 5。
Magnitudes: |a| = √32 + 42 = √25 = 5. |b| = √((-1)2 + 22) = √5.
模长：|a| = √32 + 42 = √25 = 5；|b| = √((-1)2 + 22) = √5。
So cos θ = 5 / (5 √5) = 1/√5. Then θ = cos-1(1/√5) ≈ 63.4349...° → 63.4° to one decimal place.
因此 cos θ = 5 / (5 √5) = 1/√5。于是 θ = cos-1(1/√5) ≈ 63.4°（精确至一位小数）。
The acute angle between the vectors is 63.4°.
向量间的锐角为 63.4°。
10. Optimisation Using Differentiation | 利用微分解决最优化问题
Question: A farmer wishes to build a rectangular enclosure against a straight barn wall, using 60 m of fencing for the three remaining sides. Find the maximum possible area of the enclosure.
题目：一位农民想借助一面笔直的谷仓墙，用 60 米长的围栏围成一个矩形场地（另外三边）。求场地可能的最大面积。
rpendicular sides each be x metres. The total fencing used is 2x + y = 60, so y = 60 - 2x.
设平行于谷仓的边长为 y 米，两条垂直的边长各为 x 米。围栏总长为 2x + y = 60，得 y = 60 - 2x。
The area A = x * y = x(60 - 2x) = 60x - 2x2. This is a quadratic function opening downward. Find the maximum using differentiation or vertex.
面积 A = x × y = x(60 - 2x) = 60x - 2x2。这是一个
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