📚 Year 12 Cambridge Maths: Unit Test Mock Exam Analysis | Year 12 Cambridge 数学：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test for Year 12 Cambridge Mathematics, covering core topics such as quadratics, functions, coordinate geometry, trigonometry, and calculus. Each question is solved step by step to reinforce understanding and exam technique.
本文详细解析一份为 Year 12 Cambridge 数学设计的单元测试模拟卷，涵盖二次函数、函数、坐标几何、三角函数和微积分等核心主题。通过逐步解答每道题目，巩固理解并提升应试技巧。
1. Solving Quadratic Equations | 解二次方程
Question: (a) Solve 3x² + x – 2 = 0 by factorisation. (b) Solve 2x² – 4x – 1 = 0 using the quadratic formula, giving answers in simplified surd form.
题目：(a) 用因式分解法解 3x² + x – 2 = 0。(b) 用求根公式解 2x² – 4x – 1 = 0，答案用最简根式表示。
Part (a): We look for two numbers that multiply to (3 × -2 = -6) and add to +1. The numbers are +3 and -2.
第(a)部分：寻找乘积为 3×(-2) = -6 且和为 +1 的两个数，它们是 +3 和 -2。
Split the middle term: 3x² + 3x – 2x – 2 = 0.
分解中间项：3x² + 3x – 2x – 2 = 0。
Factor by grouping: 3x(x + 1) – 2(x + 1) = 0, giving (3x – 2)(x + 1) = 0.
分组因式分解：3x(x + 1) – 2(x + 1) = 0，得到 (3x – 2)(x + 1) = 0。
Set each factor equal to zero: 3x – 2 = 0 → x = 2/3, and x + 1 = 0 → x = -1.
令各因式等于零：3x – 2 = 0 → x = 2/3，以及 x + 1 = 0 → x = -1。
Part (b): Identify a = 2, b = -4, c = -1. Write the quadratic formula:
第(b)部分：确定系数 a = 2，b = -4，c = -1。写出求根公式：
x = [ -b ± √(b² – 4ac) ] / (2a)
Substitute the values: x = [ -(-4) ± √((-4)² – 4(2)(-1)) ] / (2×2).
代入数值：x = [ -(-4) ± √((-4)² – 4(2)(-1)) ] / (2×2)。
Simplify inside the square root: x = [ 4 ± √(16 + 8) ] / 4 = [ 4 ± √24 ] / 4.
化简根号内：x = [ 4 ± √(16 + 8) ] / 4 = [ 4 ± √24 ] / 4。
Simplify the surd: √24 = √(4×6) = 2√6.
化简根式：√24 = √(4×6) = 2√6。
Thus x = (4 ± 2√6)/4 = 1 ± (√6)/2.
因此 x = (4 ± 2√6)/4 = 1 ± (√6)/2。
Final answers: x = 1 + √6/2 and x = 1 – √6/2.
最终答案：x = 1 + √6/2 和 x = 1 – √6/2。
2. Completing the Square | 配方法
Question: Write y = x² + 6x + 5 in the completed square form (x + p)² + q, and hence state the coordinates of the vertex.
题目：将 y = x² + 6x + 5 写成配方式 (x + p)² + q，并由此写出顶点坐标。
Take the coefficient of x, which is 6, halve it to get 3, then square it to obtain 9.
取 x 的系数 6，将其除以 2 得 3，再平方得 9。
Rewrite: y = (x² + 6x + 9) – 9 + 5.
重新表达：y = (x² + 6x + 9) – 9 + 5。
Factor the perfect square: y = (x + 3)² – 4.
因式分解完全平方式：y = (x + 3)² – 4。
The completed square form is y = (x + 3)² – 4, so p = 3 and q = -4.
配方式为 y = (x + 3)² – 4，因此 p = 3，q = -4。
The vertex of the parabola is at (-p, q), giving (-3, -4).
抛物线的顶点位于 (-p, q)，即 (-3, -4)。
3. Domain and Range of a Radical Function | 根式函数的定义域与值域
Question: For the function f(x) = √(x – 3), state the maximal domain and the corresponding range.
题目：对于函数 f(x) = √(x – 3)，写出最大定义域及相应的值域。
Inside the square root must be non-negative: x – 3 ≥ 0, so x ≥ 3. The domain is [3, ∞).
平方根内必须非负：x – 3 ≥ 0，因此 x ≥ 3。定义域为 [3, ∞)。
The square root function outputs only non-negative values. Hence f(x) ≥ 0, and the range is [0, ∞).
平方根函数只输出非负值，因此 f(x) ≥ 0，值域为 [0, ∞)。
4. Function Transformations | 函数变换
Question: The graph of y = f(x) is transformed to y = 2f(x – 1) + 3. Describe the sequence of transformations that maps y = f(x) onto the new graph.
题目：函数 y = f(x) 的图像变为 y = 2f(x – 1) + 3。描述将 y = f(x) 映射到新图像的变换顺序。
Start with x inside the function: (x – 1) indicates a horizontal translation 1 unit to the right.
从函数内部的 x 开始：(x – 1) 表示水平向右平移 1 个单位。
Next, the factor 2 multiplies the function, giving a vertical stretch by scale factor 2.
接着，系数 2 乘以函数值，产生垂直方向拉伸，缩放因子为 2。
Finally, the “+3” shifts the whole graph vertically upwards by 3 units.
最后，“+3”将整个图像垂直向上平移 3 个单位。
The correct sequence is: translate right by 1, stretch vertically by factor 2, translate up by 3.
正确的顺序是：向右平移 1 单位，垂直拉伸 2 倍，向上平移 3 单位。
5. Equations of Perpendicular Lines | 垂直直线的方程
Question: Find the equation of the line that passes through (4, -1) and is perpendicular to the line 3x – 2y = 7. Give your answer in the form ax + by + c = 0.
题目：求过点 (4, -1) 且与直线 3x – 2y = 7 垂直的直线方程。结果以 ax + by + c = 0 的形式给出。
First find the gradient of the given line. Rearrange to y = (3/2)x – 7/2, so gradient m₁ = 3/2.
首先求已知直线的斜率。整理得 y = (3/2)x – 7/2，因此斜率 m₁ = 3/2。
For perpendicular lines, m₁ × m₂ = -1, so m₂ = -2/3.
对于垂直直线，m₁ × m₂ = -1，所以 m₂ = -2/3。
Use point-slope form with (4, -1): y – (-1) = (-2/3)(x – 4).
用点斜式代入 (4, -1)：y – (-1) = (-2/3)(x – 4)。
Simplify: y + 1 = (-2/3)x + 8/3.
化简：y + 1 = (-2/3)x + 8/3。
Multiply through by 3: 3y + 3 = -2x + 8.
两边乘 3：3y + 3 = -2x + 8。
Bring all terms to one side: 2x + 3y – 5 = 0.
将所有项移到一侧：2x + 3y – 5 = 0。
6. Equation of a Circle | 圆的方程
Question: Find the centre and radius of the circle given by x² + y² – 6x + 4y – 12 = 0.
题目：求圆 x² + y² – 6x + 4y – 12 = 0 的圆心和半径。
Group x and y terms: (x² – 6x) + (y² + 4y) = 12.
将 x 和 y 的项分别组合：(x² – 6x) + (y² + 4y) = 12。
Complete the square for x: coefficient -6, half is -3, square is 9. Add 9 to both sides.
对 x 配方：系数为 -6，一半是 -3，平方得 9。两边同时加 9。
Complete the square for y: coefficient 4, half is 2, square is 4. Add 4 to both sides.
对 y 配方：系数为 4，一半是 2，平方得 4。两边同时加 4。
Equation becomes: (x – 3)² + (y + 2)² = 12 + 9 + 4 = 25.
方程变为：(x – 3)² + (y + 2)² = 12 + 9 + 4 = 25。
This is a circle with centre (3, -2) and radius √25 = 5.
这是圆心在 (3, -2)、半径为 √25 = 5 的圆。
7. Trigonometric Equation | 三角方程
Question: Solve 2sinθ = 1 for 0° ≤ θ ≤ 360°.
题目：在 0° ≤ θ ≤ 360° 范围内解方程 2sinθ = 1。
Divide both sides by 2: sinθ = 1/2.
两边除以 2：sinθ = 1/2。
The principal solution is θ = 30° (since sin30° = 1/2).
主解为 θ = 30°（因为 sin30° = 1/2）。
Sine is also positive in the second quadrant. The second solution is θ = 180° – 30° = 150°.
正弦在第二象限也为正，第二个解为 θ = 180° – 30° = 150°。
There are no other solutions in the interval, so the answer is θ = 30°, 150°.
在所给区间内没有其他解，因此答案为 θ = 30° 和 150°。
8. Differentiation from First Principles | 从第一原理求导
Question: Using first principles, differentiate f(x) = x² + 3x.
题目：利用第一原理求函数 f(x) = x² + 3x 的导数。
The derivative is defined as f'(x) = limh→0 [f(x+h) – f(x)] / h.
导数的定义为 f'(x) = limh→0 [f(x+h) – f(x)] / h。
Calculate f(x+h) = (x+h)² + 3(x+h) = x² + 2xh + h² + 3x + 3h.
计算 f(x+h) = (x+h)² + 3(x+h) = x² + 2xh + h² + 3x + 3h。
Now f(x+h) – f(x) = (x² + 2xh + h² + 3x + 3h) – (x² + 3x) = 2xh + h² + 3h.
于是 f(x+h) – f(x) = (x² + 2xh + h² + 3x + 3h) – (x² + 3x) = 2xh + h² + 3h。
Divide by h: (2xh + h² + 3h)/h = 2x + h + 3.
除以 h：(2xh + h² + 3h)/h = 2x + h + 3。
Take the limit as h → 0: f'(x) = limh→0 (2x + h + 3) = 2x + 3.
取 h → 0 时的极限：f'(x) = limh→0 (2x + h + 3) = 2x + 3。
9. Tangents and Normals | 切线与法线
Question: Find the equation of the tangent to the curve y = x³ – 2x at the point where x = 1.
题目：求曲线 y = x³ – 2x 在 x = 1 处的切线方程。
Differentiate to find the gradient function: dy/dx = 3x² – 2.
求导得到斜率函数：dy/dx = 3x² – 2。
At x = 1, the gradient m = 3(1)² – 2 = 1.
在 x = 1 处，斜率 m = 3(1)² – 2 = 1。
Find the y-coordinate when x = 1: y = (1)³ – 2(1) = -1. The point is (1, -1).
求 x = 1 时的 y 坐标：y = (1)³ – 2(1) = -1，点为 (1, -1)。
Use y – y₁ = m(x – x₁): y – (-1) = 1(x – 1).
利用点斜式 y – y₁ = m(x – x₁)：y – (-1) = 1(x – 1)。
Simplify: y + 1 = x – 1, so y = x – 2.
化简：y + 1 = x – 1，因此 y = x – 2。
10. Indefinite Integration | 不定积分
Question: Find ∫(4x³ – 6x + 2) dx.
题目：求 ∫(4x³ – 6x + 2) dx。
Integrate term by term using the power rule: increase the exponent by 1 and divide by the new exponent.
使用幂规则逐项积分：指数加 1，再除以新的指数。
∫ 4x³ dx = (4/4)x⁴ = x⁴.
∫ 4x³ dx = (4/4)x⁴ = x⁴。
∫ -6x dx = -6 × (x²/2) = -3x².
∫ -6x dx = -6 × (x²/2) = -3x²。
∫ 2 dx = 2x.
∫ 2 dx = 2x。
Don’t forget the constant of integration C. The final answer is x⁴ – 3x² + 2x + C.
别忘了积分常数 C，最终答案为 x⁴ – 3x² + 2x + C。
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