📚 A-Level CCEA Mathematics: Typical Worked Examples | A-Level CCEA 数学：典型例题详解
This article unpacks a selection of typical A-Level CCEA Mathematics problems, covering pure mathematics, mechanics and statistics. Each worked example illustrates key techniques and common question styles, helping you consolidate understanding and build confidence for the CCEA examination units. Use these step-by-step solutions to see how marks are earned and to sharpen your problem-solving approach.
本文精选 A-Level CCEA 数学考试中具有代表性的典型例题，涵盖纯数学、力学与统计。每个例题详解都展示了关键解题技巧和常见题型，帮助你在复习中巩固理解、增强信心。通过一步步的解题过程，你可以清楚地看到得分要点，并提升自己分析问题、组织解答的能力。
1. Quadratic Equations and the Discriminant | 二次方程与判别式
range of values of k for which the equation 2x² + kx + 8 = 0 has two distinct real roots.
例题：求 k 的取值范围，使方程 2x² + kx + 8 = 0 有两个不相等的实根。
greater than zero. We identify a = 2, b = k, c = 8 and set up the discriminant condition.
由于要有两个不相等的实根，判别式必须大于零。我们确定 a = 2，b = k，c = 8，并建立判别式条件。
Δ = b² − 4ac = k² − 4(2)(8) = k² − 64 > 0
quality opens upward, so the solution lies outside the roots of k² = 64, which are k = ±8.
解不等式 k² − 64 > 0，得 k² > 64。这个二次不等式开口向上，因此解集在方程 k² = 64 的根之外，即 k = ±8 的两侧。
Hence the required range is k < −8 or k > 8. In set notation, k ∈ (−∞, −8) ∪ (8, ∞).
因此所求的取值范围是 k < −8 或 k > 8。用集合符号表示为 k ∈ (−∞, −8) ∪ (8, ∞)。
Always remember to state the discriminant inequality explicitly and to interpret the solution of the quadratic inequality correctly. CCEA mark schemes reward clear reasoning, so include the condition Δ > 0 and the algebraic steps.
解答时务必明确写出判别式不等式，并正确解释二次不等式的解集。CCEA 的评分标准重视清晰的推理过程，因此要写出 Δ > 0 的条件及每一步代数变形。
2. Exponential and Logarithmic Equations | 指数与对数方程
Example: Solve the equation 32x+1 = 5x, giving your answer in the form x = ln a / ln b where a and b are integers.
例题：解方程 32x+1 = 5x，并将答案写成 x = ln a / ln b 的形式，其中 a 和 b 为整数。
Take natural logarithms on both sides to bring the powers down. The equation becomes ln(32x+1) = ln(5x). Then use the power law of logarithms.
对两边取自然对数以将指数降下来。方程变为 ln(32x+1) = ln(5x)，然后利用对数的幂运算法则。
(2x+1) ln 3 = x ln 5
Expand the left-hand side and collect the terms containing x on one side.
展开左边，并将含有 x 的项移到同一边。
2x ln 3 + ln 3 = x ln 5
2x ln 3 − x ln 5 = −ln 3
Factorise x and then divide by the coefficient. Note that the negative sign can be absorbed to reverse the fraction.
提取 x 的公因子，然后除以系数。注意负号可以吸收到分式中，以使结果符合要求。
x(2 ln 3 − ln 5) = −ln 3
x = −ln 3 / (2 ln 3 − ln 5) = ln 3 / (ln 5 − 2 ln 3)
b. We can also express the denominator as ln(5/9) with a=3, but 5/9 is not an integer. Let' ) / ln(9/5). That still gives non-integer. Wait, the question asked for a and b integers. Perhaps the equation can be solved using a different base or by writing 32x+1 = 3 × 9x, so 3 × 9x = 5x, then (9/5)x ybe the question meant expressing with integer a and b in the form x = ln a / ln b, and it's acceptable if b is a rational? e ? But the instruction says a and b are integers. Possibly I should solve a different equation. Let's change the example to 23x = 32x+1. Then we get integers. I'll use that instead to ensure an integer result.
I will modify the example to: Solve 23x = 32x+1 x = ln 3 / (3 ln 2 − 2 ln 3) = ln 3 / (ln 8 − ln 9) = ln 3 / ln(8/9). Still fraction. Let's try 5x+1 = 32x e a=5, b=9/5. Not integer. Probably the intended example in CCEA could be something like 2x = 3x+1 = ln 3 → x = ln 3 / (ln 2 − ln 3) = ln 3 / ln(2/3). So a=3, b=2/3, not integer. Another: 7x = 42x+1 ps the requirement of integer a and b might not be strict; they might accept rational b. I'll keep my original equation and note that b = 5/9, but it's not an integer. I can adjust the wording: "where a and b are integers" might be from a real CCEA question that gives 32x = 2x+1 /2, still fraction. To get integer b, the denominator would need to be of the form ln(integer). For instance, if we have 5x = 23x+1 → x ln 5 = 3x ln 2 + ln 2 → x(ln 5 − 3 ln 2) = ln 2 → x = ln 2 / ln(5/8). Not integer. It's hard to get an integer b unless the coefficient leads to cancellation. If we have 2x+1 = 3x e question expects an expression like x = ln 8 / ln 9, that would be integer a=8, b=9. Let's make that: Solve 23x+1 = 32x 2 / (2 ln 3 − 3 ln 2) = ln 2 / ln(9/8). So a=2, b=9/8. To get integer 8 and 9, we can do 23x = 32x+1 gave x = ln 3 / ln(8/9). So a=3, b=8/9. Not integer. What if we solve 8x = 9x+1? Then 23x = 3^(2(x+1)) = 32x+2 t integer b, we need denominator to be ln(integer). That means the coefficient of x on RHS minus LHS can result in ln(integer). For example, if original equation is 22x = 3x, then 2x ln 2 = x ln 3 → x(2 ln 2 − ln 3) = 0 → only solution x=0. Not good. I'll pick a slightly different problem: Solve 5x = 2x+3 a=8 if we write 3 ln 2 = ln 8, so x = ln 8 / ln(5/2). So a=8, b=5/2. Still fraction. So it's common to get a fraction. I'll change the requirement to "where a and b are integers" but maybe allow b to be a fraction? I' te that we can write it as x = ln 3 / ln(5/9), and if required to have integer a and b, it' To fix, I can choose an example where the final answer does yield integer a and b: e.g., 3x = 22x+1. Then x = ln 2 / ln(3/4) -> ln 2 / ln(0.75). Not integer. Or 2x = 3x+1 teger). That happens if the coefficient of x on one side is zero after moving terms, but that yields x constant. For example, 2x = 4x+1 gives 2x = 22x+2 → x = 2x+2 → x = -2, not logs. I think it's okay to have a rational b. I' / ln(5/9), and mention that it can be written as ln 3 / ln(5/9), with a=3 and b=5/9, but I' minator using log laws: x = ln(3) / ln(5/9) = ln(1/3) / ln(9/5) = ... still not integer. I'll adapt by saying "where a and b are integers" and adjust the equation to 23x = 32x, then (3x) ln 2 = (2x) ln 3 → x(3 ln 2 - 2 ln 3) = 0 → only trivial. Not working. I'll just use 32x+1 = 5x 3 and b = 5/9, but if the question demanded integers, we might have derived differently. I'll change the requirement to "giving your answer in the form x = ln p / ln q, where p and q are constants." Then I can avoid integer issue. I' in . I'll do that. So: giving your answer in the form x = ln p / ln q, where p and q are simplified constants. So the example works.
我们保留原方程，并把要求改为：将答案写成 x = ln p / ln q 的形式，其中 p 和 q 为化简后的常数。
5/9 cannot be reduced further. So p = 3, q = 5/9, or q = 9/5 with a negative sign absorbed? 9/5). Both are acceptable. The key is to show all steps and simplify using logarithm laws.
继续求解：x = ln 3 / ln(5/9)。分式 5/9 已经最简。因此 p = 3，q = 5/9。也可以吸收负号写成 x = −ln 3 / ln(9/5)。两种形式均可。重要的是展示所有步骤，并用对数法则进行化简。
3. Trigonometric Identities and Equations | 三角恒等式与方程
Example: Solve the equation sin 2θ = cos θ for 0° ≤ θ ≤ 360°.
例题：解方程 sin 2θ = cos θ，其中 0° ≤ θ ≤ 360°。
Begin by expressing sin 2θ in terms of sin θ and cos θ using the double-angle identity.
首先利用倍角公式将 sin 2θ 表示为 sin θ 和 cos θ 的形式。
sin 2θ = 2 sin θ cos θ
stitute into the equation. This yields 2 sin θ cos θ = cos θ. Bring all terms to one side.
代入原方程，得到 2 sin θ cos θ = cos θ。将所有项移到一边。
2 sin θ cos θ − cos θ = 0
Factorise by taking out the common factor cos θ.
提取公因式 cos θ 进行因式分解。
cos θ (2 sin θ − 1) = 0
Set each factor equal to zero. This gives two families of solutions.
令每个因式等于零，得到两组解。
First, cos θ = 0. For 0° ≤ θ ≤ 360°, this occurs at θ = 90° and θ = 270°.
首先，cos θ = 0。在区间 0° 到 360° 内，θ = 90° 和 θ = 270°。
Second, 2 sin θ − 1 = 0 ⇒ sine function is positive in the first and second quadrants, giving θ = 30° and θ = 150°.
其次，2 sin θ − 1 = 0 ⇒ sin θ = ½。正弦函数在第一象限和第二象限为正，因此 θ = 30° 和 θ = 150°。
0°. Always check for extraneous solutions by substituting back into the original equation.
汇总给定区间内的全部解：θ = 30°, 90°, 150°, 270°。务必代回原方程检验，排除增根。
4. Differentiation Techniques | 微分技巧
Example: Differentiate y = x³ ln(x) with respect to x.
例题：对 y = x³ ln(x) 关于 x 求导。
This is a product of two functions: u = x³ and v = ln x. Apply the product rule: dy/dx = u'v + uv'.
这是两个函数的乘积：u = x³，v = ln x。使用乘积法则：dy/dx = u'v + uv'。
Compute the derivatives: u' = 3x², v' = 1/x.
计算导数：u' = 3x²，v' = 1/x。
dy/dx = (3x²)(ln x) + (x³)(1/x)
Simplify the second term: x³ × (1/x) = x². Hence
化简第二项：x³ × (1/x) = x²。因此
dy/dx = 3x² ln x + x²
Factorise if required, e.g. x²(3 ln x + 1). Showing a simplified factored form can be useful in later integration or when setting the derivative to zero for stationary points.
必要时可以进行因式分解，例如写成 x²(3 ln x + 1)。化简后的乘积形式在后续积分或求驻点时非常有用。
5. Integration and Area | 积分与面积
Example: Find the area enclosed by the curve y = 4x − x² and the x-axis.
例题：求曲线 y = 4x − x² 与 x 轴所围成的区域面积。
First determine where the curve intersects the x-axis by setting y = 0: 4x − x² = 0 ⇒ x(4 − x) = 0 ⇒ x = 0 or x = 4.
首先通过设 y = 0 确定曲线与 x 轴的交点：4x − x² = 0 ⇒ x(4 − x) = 0 ⇒ x = 0 或 x = 4。
The area A is given by the definite integral from 0 to 4 of the function, which is always above the axis on this interval.
面积 A 由函数从 0 到 4 的定积分给出，该区间上曲线位于 x 轴上方。
A = ∫₀⁴ (4x − x²) dx
Integrate term by term: ∫ 4x dx = 2x², ∫ x² dx = ⅓ x³. So
逐项积分：∫ 4x dx = 2x²，∫ x² dx = ⅓ x³。于是
∫ (4x − x²) dx = 2x² − ⅓ x³
Evaluate the antiderivative between the limits 4 and 0.
计算原函数在上下限 4 和 0 的值。
A = [2(4)² − ⅓(4)³] − [0] = (2×16 − ⅓×64) = 32 − 64/3
Combine into a single fraction: 32 = 96/3, so A = 96/3 − 64/3 = 32/3 square units.
通分后相减：32 = 96/3，因此 A = 96/3 − 64/3 = 32/3 平方单位。
Always draw a quick sketch to confirm the curve lies above the axis between the roots. An inverted parabola with vertex at (2,4) confirms the area interpretation.
经常快速画图确认在两交点之间曲线位于 x 轴上方。该抛物线开口向下，顶点在 (2,4)，证实面积计算无误。
6. First Order Differential Equations | 一阶微分方程
Example: Solve the differential equation dy/dx = 2xy, given that y = 3 when x = 0.
例题：解微分方程 dy/dx = 2xy，已知 x = 0 时 y = 3。
Separate the variables: bring y terms to the left and x terms to the right.
分离变量：将含 y 的项移到左侧，含 x 的项移到右侧。
(1/y) dy = 2x dx
Integrate both sides. The integral of 1/y with respect to y is ln|y|, and the integral of 2x is x².
两边积分。1/y 关于 y 的积分是 ln|y|，2x 的积分是 x²。
ln|y| = x² + C
Exponentiate to solve for y. This gives |y| = ex² + C = eC · ex². Let A = ± eC, so y = A ex².
两边取指数解出 y。得到 |y| = ex² + C = eC · ex²。令 A = ± eC，则 y = A ex²。
Use the initial condition y(0) = 3 to find A: 3 = A e0 ⇒ A = 3.
利用初始条件 y(0) = 3 求 A：3 = A e0 ⇒ A = 3。
Thus the particular solution is y = 3 ex².
因此特解为 y = 3 ex²。
In CCEA exam questions, always clearly show the step where arbitrary constant is determined. Using absolute value in logarithmic integration is good practice, but the final answer with a positive exponential is correct for the given condition.
在 CCEA 考试题中，要清晰地展示确定任意常数的步骤。对数积分时使用绝对值是一种好习惯，但本题根据条件得到的最终指数函数形式是准确的。
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