📚 Year 12 CIE Maths: Unit Test Mock Paper Solutions | Year 12 CIE 数学：单元测试模拟卷解析
This article provides detailed solutions and examiner-style commentary for a mock unit test covering the Year 12 CIE AS Mathematics pure content. Each question is broken down step by step, with methods that reflect typical mark schemes and common pitfalls highlighted. Working through these problems will consolidate your algebra, calculus, and coordinate geometry skills ahead of internal assessments or the final exam.
本文针对一份覆盖 Year 12 CIE AS 数学纯数内容的单元测试模拟卷，提供详细的解答和评分风格评注。每道题都分步拆解，方法对应典型评分标准，并指出常见陷阱。通过练习这些问题，你可以在校内测评或最终考试前巩固代数、微积分和坐标几何技能。
1. Quadratic equation and discriminant | 二次方程与判别式
Question: Find the values of k for which the equation x² + (k – 3)x + 9 = 0 has no real roots.
题目：求 k 的值，使得方程 x² + (k – 3)x + 9 = 0 没有实数根。
For a quadratic ax² + bx + c = 0 to have no real roots, the discriminant must be negative: b² – 4ac < 0.
对于二次方程 ax² + bx + c = 0，没有实数根的条件是判别式小于零：b² – 4ac < 0。
Here a = 1, b = (k – 3), c = 9. So (k – 3)² – 4·1·9 < 0 → k² – 6k + 9 – 36 < 0 → k² – 6k – 27 < 0.
此处 a = 1，b = (k – 3)，c = 9。因此 (k – 3)² – 4·1·9 < 0 → k² – 6k + 9 – 36 < 0 → k² – 6k – 27 < 0。
Factorising the quadratic in k: (k – 9)(k + 3) < 0. The critical values are k = –3 and k = 9. A sign diagram shows the inequality holds for –3 < k < 9.
对 k 的二次式分解因式：(k – 9)(k + 3) < 0。临界值为 k = –3 和 k = 9。通过符号图可知不等式在 –3 < k < 9 时成立。
Answer: k ∈ (–3, 9). Common mistake: forgetting to reverse the inequality when dealing with a negative leading coefficient – here leading coefficient is 1, but always check the inequality direction after factorising.
答案：k ∈ (–3, 9)。常见错误：处理负的二次项系数时忘记反转不等号——此题中二次项系数为 1，但分解因式后仍需检查不等号方向。
2. Composite functions and domain | 复合函数与定义域
Question: f(x) = √(x – 2) for x ≥ 2, g(x) = x² + 1 for x ∈ ℝ. Find fg(x) and state its domain.
题目：f(x) = √(x – 2)，x ≥ 2；g(x) = x² + 1，x ∈ ℝ。求 fg(x) 并写出其定义域。
fg(x) = f(g(x)) = √(g(x) – 2) = √(x² + 1 – 2) = √(x² – 1).
fg(x) = f(g(x)) = √(g(x) – 2) = √(x² + 1 – 2) = √(x² – 1)。
For fg to be defined, the expression under the square root must be ≥ 0: x² – 1 ≥ 0 → |x| ≥ 1. Thus x ≤ –1 or x ≥ 1.
要使 fg 有意义，根号下的表达式需 ≥ 0：x² – 1 ≥ 0 → |x| ≥ 1。因此 x ≤ –1 或 x ≥ 1。
Notice that there is no further restriction from the inner function g, as its output is always ≥ 1, which automatically satisfies x ≥ 2 for the outer function? Wait – f requires its input ≥ 2. The input to f is g(x) = x² + 1, so we need x² + 1 ≥ 2 → x² ≥ 1, which gives the same condition. So domain = {x: x ≤ –1} ∪ {x: x ≥ 1}.
注意内层函数 g 无额外限制，因其输出始终 ≥ 1，而外层函数 f 要求输入 ≥ 2。f 的输入为 g(x) = x² + 1，故需 x² + 1 ≥ 2 → x² ≥ 1，得到相同条件。因此定义域为 {x: x ≤ –1} ∪ {x: x ≥ 1}。
Common pitfall: ignoring the domain restriction of the outer function. Always check the ‘input’ condition for the leftmost function in a composite.
常见陷阱：忽略外层函数的定义域限制。在复合函数中务必检查最左侧函数的“输入”条件。
3. Inequality with algebraic fraction | 分式不等式
Question: Solve (2x + 3)/(x – 1) ≤ 1.
题目：解不等式 (2x + 3)/(x – 1) ≤ 1。
Bring all terms to one side: (2x + 3)/(x – 1) – 1 ≤ 0 → (2x + 3 – (x – 1))/(x – 1) ≤ 0 → (x + 4)/(x – 1) ≤ 0.
将所有项移到一边：(2x + 3)/(x – 1) – 1 ≤ 0 → (2x + 3 – (x – 1))/(x – 1) ≤ 0 → (x + 4)/(x – 1) ≤ 0。
Critical values where numerator or denominator is zero: x = –4, x = 1. These divide the number line into intervals (–∞, –4), (–4, 1), (1, ∞). Test a point in each or use a sign table. The fraction is ≤ 0 in (–4, 1). But cannot include x = 1 as denominator would be zero. The inequality includes equality at x = –4 because numerator zero gives value 0, which satisfies ≤ 0.
分子或分母为零的临界值：x = –4，x = 1。这些点将数轴分为区间 (–∞, –4)，(–4, 1)，(1, ∞)。在各区间内取测试点或使用符号表。分式在 (–4, 1) 内 ≤ 0。但不能取 x = 1，因为分母为零。x = –4 时分子为零，分式值为 0，满足 ≤ 0，可包含。
Solution: –4 ≤ x < 1. Always remember to exclude values that make the denominator zero.
解：–4 ≤ x < 1。始终记住排除使分母为零的值。
4. Differentiation and tangent line | 求导与切线方程
Question: Find the equation of the tangent to the curve y = x³ – 3x² + 2 at x = 2.
题目：求曲线 y = x³ – 3x² + 2 在 x = 2 处的切线方程。
Differentiate: dy/dx = 3x² – 6x. At x = 2, gradient m = 3·4 – 6·2 = 12 – 12 = 0. So tangent is horizontal.
求导：dy/dx = 3x² – 6x。在 x = 2 处，梯度 m = 3·4 – 6·2 = 12 – 12 = 0。因此切线为水平线。
Find y-coordinate: y = (2)³ – 3·(2)² + 2 = 8 – 12 + 2 = –2. Point (2, –2).
求 y 坐标：y = 8 – 12 + 2 = –2。点 (2, –2)。
Equation of tangent: y – (–2) = 0·(x – 2) → y = –2.
切线方程：y – (–2) = 0·(x – 2) → y = –2。
Even when gradient is zero, the method is the same. Many students panic and think they’ve made an error – but horizontal tangents occur at turning points.
即便梯度为零，方法依然相同。许多学生惊慌失措，认为自己算错了——但实际上水平切线正出现在驻点处。
5. Integration and area under a curve | 积分与曲线下方面积
Question: Find the area enclosed by y = 4x – x² and the x-axis.
题目：求曲线 y = 4x – x² 与 x 轴围成的面积。
Find limits: set y = 0 → 4x – x² = 0 → x(4 – x) = 0 → x = 0 and x = 4.
求积分限：令 y = 0 → 4x – x² = 0 → x(4 – x) = 0 → x = 0 和 x = 4。
Integrate: ∫₀⁴ (4x – x²) dx = [2x² – (1/3)x³] from 0 to 4 = (2·16 – 64/3) – 0 = 32 – 64/3 = (96 – 64)/3 = 32/3 square units.
积分：∫₀⁴ (4x – x²) dx = [2x² – (1/3)x³]₀⁴ = (2·16 – 64/3) – 0 = 32 – 64/3 = (96 – 64)/3 = 32/3 平方单位。
Always draw a sketch; check that the curve is above the x-axis between the limits. Here the parabola opens downwards with vertex at (2,4), so the area is positive.
务必画出草图；检查在积分限内曲线是否位于 x 轴上方。此处抛物线开口向下，顶点在 (2,4)，因此面积为正。
6. Binomial expansion | 二项展开
Question: Expand (2 – 3x)⁴ up to the term in x³.
题目：将 (2 – 3x)⁴ 展开至 x³ 项。
Use the binomial expansion: (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴. Here a = 2, b = –3x.
使用二项展开：(a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴。此处 a = 2，b = –3x。
Term by term: a⁴ = 2⁴ = 16. 4a³b = 4·8·(–3x) = –96x. 6a²b² = 6·4·(9x²) = 216x². 4ab³ = 4·2·(–27x³) = –216x³.
逐项计算：a⁴ = 16；4a³b = 4·8·(–3x) = –96x；6a²b² = 6·4·(9x²) = 216x²；4ab³ = 4·2·(–27x³) = –216x³。
So (2 – 3x)⁴ = 16 – 96x + 216x² – 216x³ + … . Be careful with signs when b is negative.
因此 (2 – 3x)⁴ = 16 – 96x + 216x² – 216x³ + … 。当 b 为负时，注意符号的处理。
7. Vector magnitude and direction | 向量的模与方向
Question: Given vector a = 3i – 4j, find |a| and the angle a makes with the positive i-direction.
题目：已知向量 a = 3i – 4j，求 |a| 以及 a 与正 i 方向之间的夹角。
Magnitude: |a| = √(3² + (–4)²) = √(9 + 16) = √25 = 5.
模：|a| = √(3² + (–4)²) = √(9 + 16) = √25 = 5。
Angle θ from positive x-axis: tan θ = (–4)/3 → θ = arctan(–4/3). Since the vector is in the fourth quadrant, θ ≈ –53.1° or equivalently 306.9°. Usually answers are given as an acute angle with direction: 53.1° below the positive x-axis, or bearing 306.9°.
与正 x 轴的夹角 θ：tan θ = (–4)/3 → θ = arctan(–4/3)。向量位于第四象限，θ ≈ –53.1° 或等效为 306.9°。通常答案会以锐角加方向表示：与正 x 轴成 53.1° 下方，或方位角 306.9°。
Specify the angle clearly: 53.1° to the horizontal in the negative (clockwise) direction.
明确表示角度：与水平方向成 53.1° 向负方向（顺时针）。
8. Trigonometric equations | 三角方程
Question: Solve 2 sin² x – sin x – 1 = 0 for 0° ≤ x ≤ 360°.
题目：解方程 2 sin² x – sin x – 1 = 0，其中 0° ≤ x ≤ 360°。
Factorise the quadratic in sin x: (2 sin x + 1)(sin x – 1) = 0.
将 sin x 的二次式分解因式：(2 sin x + 1)(sin x – 1) = 0。
So sin x = –½ or sin x = 1.
因此 sin x = –½ 或 sin x = 1。
For sin x = 1: x = 90° (only solution in the interval).
对于 sin x = 1：x = 90°（区间内唯一解）。
For sin x = –½: reference angle = 30°. Sine negative in third and fourth quadrants → x = 180° + 30° = 210°, x = 360° – 30° = 330°.
对于 sin x = –½：参考角 = 30°。正弦在第三、第四象限为负 → x = 180° + 30° = 210°，x = 360° – 30° = 330°。
All solutions: x = 90°, 210°, 330°. Always check the domain; giving answers outside 0°–360° loses marks.
全部解：x = 90°，210°，330°。务必检查定义域；给出 0°–360° 以外的解会失分。
9. Coordinate geometry: circle equation | 坐标几何：圆的方程
Question: A circle has centre (2, –1) and radius 4. Write its equation and find the points where the circle cuts the y-axis.
题目：一个圆的圆心为 (2, –1)，半径为 4。写出圆的方程，并求该圆与 y 轴的交点。
Equation: (x – 2)² + (y + 1)² = 16.
方程：(x – 2)² + (y + 1)² = 16。
To find y-intercepts, set x = 0: (0 – 2)² + (y + 1)² = 16 → 4 + (y + 1)² = 16 → (y + 1)² = 12 → y + 1 = ±√12 = ±2√3 → y = –1 ± 2√3.
求 y 轴交点，令 x = 0：(0 – 2)² + (y + 1)² = 16 → 4 + (y + 1)² = 16 → (y + 1)² = 12 → y + 1 = ±√12 = ±2√3 → y = –1 ± 2√3。
Points: (0, –1 + 2√3) and (0, –1 – 2√3). Leave in simplified surd form unless otherwise instructed.
交点：(0, –1 + 2√3) 和 (0, –1 – 2√3)。除非另有说明，保留最简根式形式。
10. Rate of change and chain rule | 变化率与链式法则
Question: The radius of a circle is increasing at 3 cm s⁻¹. Find the rate of increase of the area when the radius is 5 cm.
题目：一个圆的半径以 3 cm s⁻¹ 的速率增加。求当半径为 5 cm 时面积的增加速率。
Area A = πr². Rate of change of area dA/dt = (dA/dr)·(dr/dt) = 2πr · 3 = 6πr.
面积 A = πr²。面积变化率 dA/dt = (dA/dr)·(dr/dt) = 2πr · 3 = 6πr。
When r = 5, dA/dt = 6π·5 = 30π cm² s⁻¹.
当 r = 5 时，dA/dt = 6π·5 = 30π cm² s⁻¹。
Always state units clearly. This is a standard connected rates problem requiring the chain rule.
务必清晰注明单位。这是一个标准的关联变化率问题，需要使用链式法则。
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