📚 CIE Pure Mathematics: May 2018 Paper 1 Worked Solutions | CIE 纯数学：2018年5月真题解析（试卷1）
In this article, we provide detailed worked solutions for selected questions from the CIE Pure Mathematics Paper 1 (9709/12) May/June 2018 exam. The paper covers algebra, coordinate geometry, differentiation, integration, trigonometry, vectors and more. Each solution is broken down step by step to illustrate the methods required for full marks.
本文提供 CIE 纯数学试卷1（9709/12）2018年5月考试中部分题目的详细解析。试卷涵盖代数、坐标几何、微分、积分、三角学、向量等内容。每道题的解答均逐步拆解，展示获得满分所需的解题方法。
1. Question 1: Binomial Expansion | 问题1：二项式展开
The first question tests binomial expansion and algebraic manipulation. We are asked to find the coefficient of x³ in the expansion of (2 − x)⁵ and then determine the coefficient of x³ in the expansion of (1 + 2x)(2 − x)⁵.
第一题考查二项式展开与代数运算。题目要求找出 (2 − x)⁵ 展开式中 x³ 的系数，再求 (1 + 2x)(2 − x)⁵ 展开式中 x³ 的系数。
For part (a), use the general term Tᵣ₊₁ = ⁵Cᵣ × 2⁵⁻ʳ × (−x)ʳ. We need the power of x to be 3, so r = 3. Then T₄ = ⁵C₃ × 2² × (−x)³ = 10 × 4 × (−1)³ x³ = −40x³. Hence the coefficient is −40.
对于 (a) 部分，通项公式为 Tᵣ₊₁ = ⁵Cᵣ × 2⁵⁻ʳ × (−x)ʳ。我们需要 x 的指数为3，因此 r = 3。于是 T₄ = ⁵C₃ × 2² × (−x)³ = 10 × 4 × (−1)³ x³ = −40x³，故系数为 −40。
In part (b), the product (1 + 2x)(2 − x)⁵ requires us to combine terms. The x³ term arises in two ways: taking 1 from the first factor and the x³ term from the binomial, or taking 2x from the first factor and the x² term from the binomial. The x² term in (2 − x)⁵ corresponds to r = 2: T₃ = ⁵C₂ × 2³ × (−x)² = 10 × 8 × x² = 80x². Multiplying by 2x gives 160x³. Adding the contribution from the x³ term (−40x³) gives a total coefficient of 120.
在 (b) 部分，乘积 (1 + 2x)(2 − x)⁵ 需要合并同类项。x³ 项可以通过两种方式产生：从第一个因式中取 1 再取二项式中的 x³ 项，或从第一个因式中取 2x 再取二项式中的 x² 项。(2 − x)⁵ 中 x² 项对应 r = 2：T₃ = ⁵C₂ × 2³ × (−x)² = 10 × 8 × x² = 80x²。乘以 2x 得到 160x³。再加上来自 x³ 项 (−40x³) 的贡献，总系数为 120。
2. Question 2: Tangents and Normals | 问题2：切线与法线
A curve has equation y = 2x³ − 5/x. We need the equation of the normal to the curve at the point where x = 1.
某曲线的方程为 y = 2x³ − 5/x。需要求出该曲线在 x = 1 处法线的方程。
First rewrite y as 2x³ − 5x⁻¹. Differentiate to get dy/dx = 6x² + 5x⁻². When x = 1, dy/dx = 6(1)² + 5/(1)² = 11. The y-coordinate is y = 2(1)³ − 5/1 = −3. The gradient of the tangent is 11, so the gradient of the normal is −1/11.
首先将 y 写为 2x³ − 5x⁻¹。求导得 dy/dx = 6x² + 5x⁻²。当 x = 1 时，dy/dx = 6(1)² + 5/(1)² = 11。y 坐标为 y = 2(1)³ − 5/1 = −3。切线斜率为 11，因此法线斜率为 −1/11。
Using the point (1, −3) and the normal gradient, the equation is y − (−3) = −1/11 (x − 1). Multiply by 11: 11y + 33 = −x + 1, and rearranging gives x + 11y + 32 = 0.
利用点 (1, −3) 与法线斜率，方程为 y − (−3) = −1/11 (x − 1)。两边乘以 11：11y + 33 = −x + 1，整理得 x + 11y + 32 = 0。
3. Question 3: Completing the Square | 问题3：配方法
This question asks us to express f(x) = 3x² − 12x + 7 in the form a(x + b)² + c and then find the coordinates of the vertex and solve an inequality.
本题要求将 f(x) = 3x² − 12x + 7 写成 a(x + b)² + c 的形式，并求顶点坐标和解不等式。
Factor out the coefficient of x²: f(x) = 3(x² − 4x) + 7. Complete the square inside: x² − 4x becomes (x − 2)² − 4. So f(x) = 3[(x − 2)² − 4] + 7 = 3(x − 2)² − 12 + 7 = 3(x − 2)² − 5. Thus a = 3, b = −2, c = −5.
提取 x² 的系数：f(x) = 3(x² − 4x) + 7。在括号内配方：x² − 4x 变为 (x − 2)² − 4。所以 f(x) = 3[(x − 2)² − 4] + 7 = 3(x − 2)² − 12 + 7 = 3(x − 2)² − 5。因此 a = 3，b = −2，c = −5。
The vertex of the parabola is at (2, −5). For the inequality f(x) ≤ 11, we set 3(x − 2)² − 5 ≤ 11 ⇒ 3(x − 2)² ≤ 16 ⇒ (x − 2)² ≤ 16/3. Taking square roots gives −4/√3 ≤ x − 2 ≤ 4/√3, so the solution set is 2 − 4/√3 ≤ x ≤ 2 + 4/√3.
抛物线顶点为 (2, −5)。对于不等式 f(x) ≤ 11，我们列出 3(x − 2)² − 5 ≤ 11 ⇒ 3(x − 2)² ≤ 16 ⇒ (x − 2)² ≤ 16/3。开平方得 −4/√3 ≤ x − 2 ≤ 4/√3，因此解集为 2 − 4/√3 ≤ x ≤ 2 + 4/√3。
4. Question 4: Intersection of Line and Circle | 问题4：直线与圆的交点
We need to find the points where the line y = 2x − 1 meets the circle x² + y² = 10.
需要求直线 y = 2x − 1 与圆 x² + y² = 10 的交点。
Substitute y into the circle equation: x² + (2x − 1)² = 10. Expand: x² + 4x² − 4x + 1 = 10 ⇒ 5x² − 4x − 9 = 0. This quadratic factorises as (5x − 9)(x + 1) = 0, giving x = 9/5 or x = −1.
将 y 代入圆的方程：x² + (2x − 1)² = 10。展开得 x² + 4x² − 4x + 1 = 10 ⇒ 5x² − 4x − 9 = 0。该二次方程因式分解为 (5x − 9)(x + 1) = 0，解得 x = 9/5 或 x = −1。
For x = 9/5, y = 2(9/5) − 1 = 13/5. For x = −1, y = 2(−1) − 1 = −3. The intersection points are (9/5, 13/5) and (−1, −3).
当 x = 9/5 时，y = 2(9/5) − 1 = 13/5。当 x = −1 时，y = 2(−1) − 1 = −3。交点坐标为 (9/5, 13/5) 和 (−1, −3)。
5. Question 5: Differentiation and Integration of Logarithmic Functions | 问题5：对数函数的微分与积分
This question involves differentiating x² ln x and then using the result to integrate a related function.
本题要求对 x² ln x 进行微分，并利用结果积分一个相关函数。
Using the product rule: d/dx [x² ln x] = 2x ln x + x² × (1/x) = 2x ln x + x. Therefore, ∫ (2x ln x + x) dx = x² ln x + C. We can rearrange to find ∫ x ln x dx. From the above, ∫ 2x ln x dx = x² ln x − ∫ x dx = x² ln x − x²/2 + C. Dividing by 2 gives ∫ x ln x dx = ½ x² ln x − ¼ x² + C.
使用乘积法则：d/dx [x² ln x] = 2x ln x + x² × (1/x) = 2x ln x + x。因此，∫ (2x ln x + x) dx = x² ln x + C。我们可以重新整理来求 ∫ x ln x dx。由上式，∫ 2x ln x dx = x² ln x − ∫ x dx = x² ln x − x²/2 + C。除以 2 得到 ∫ x ln x dx = ½ x² ln x − ¼ x² + C。
Thus the derivative result enables us to perform integration by recognition, a common CIE technique.
这样，通过微分的逆运算就能完成积分，这是 CIE 考试中常用的技巧。
6. Question 6: Trigonometric Equations | 问题6：三角方程
Solve the equation sin 2x = 1/2 for 0 ≤ x ≤ π, giving answers in terms of π.
解方程 sin 2x = 1/2，其中 0 ≤ x ≤ π，答案用 π 表示。
The basic solutions for sin θ = 1/2 are θ = π/6 and θ = 5π/6. Since θ = 2x, we have 2x = π/6 or 5π/6, plus multiples of 2π. But because x is restricted to [0, π], the range for 2x is [0, 2π]. Within this interval, the possible values of 2x are π/6 and 5π/6. Adding the period 2π gives 13π/6 and 17π/6, which exceed 2π and are therefore rejected.
sin θ = 1/2 的基本解为 θ = π/6 和 θ = 5π/6。令 θ = 2x，则 2x = π/6 或 5π/6，再加上 2π 的整数倍。但由于 x 的范围是 [0, π]，2x 的范围为 [0, 2π]。在此区间内，2x 可能的取值为 π/6 和 5π/6。加上周期 2π 得 13π/6 和 17π/6，超出 2π，故舍去。
Dividing by 2 gives x = π/12 and x = 5π/12. Both solutions lie within the required interval.
除以 2 得 x = π/12 和 x = 5π/12。这两个解均在给定区间内。
7. Question 7: Integration Techniques | 问题7：积分技巧
Find ∫ (e²x + 1)² dx, simplifying your answer.
求 ∫ (e²x + 1)² dx，并化简答案。
First expand the integrand: (e²x + 1)² = e⁴x + 2e²x + 1. Now integrate term by term: ∫ e⁴x dx = (1/4)e⁴x, ∫ 2e²x dx = 2 × (1/2)e²x = e²x, and ∫ 1 dx = x. So the integral is (1/4)e⁴x + e²x + x + C.
先将被积函数展开：(e²x + 1)² = e⁴x + 2e²x + 1。然后逐项积分：∫ e⁴x dx = (1/4)e⁴x，∫ 2e²x dx = 2 × (1/2)e²x = e²x，∫ 1 dx = x。因此积分结果为 (1/4)e⁴x + e²x + x + C。
As a second part, evaluate ∫ sin² x dx using the double-angle identity cos 2x = 1 − 2sin² x, which gives sin² x = ½(1 − cos 2x). Then ∫ sin² x dx = ∫ ½(1 − cos 2x) dx = ½(x − ½ sin 2x) + C = ½ x − ¼ sin 2x + C.
作为第二部分，利用倍角公式 cos 2x = 1 − 2sin² x 求 ∫ sin² x dx，即 sin² x = ½(1 − cos 2x)。于是 ∫ sin² x dx = ∫ ½(1 − cos 2x) dx = ½(x − ½ sin 2x) + C = ½ x − ¼ sin 2x + C。
8. Question 8: Vectors | 问题8：向量
Given points A(1, 2, 3) and B(3, 6, 9), find the vector AB→, its magnitude, and the angle between AB→ and the vector i + j + k.
给定点 A(1, 2, 3) 和 B(3, 6, 9)，求向量 AB→、其模长，以及 AB→ 与向量 i + j + k 的夹角。
The vector AB→ is obtained by subtracting coordinates: (3 − 1) i + (6 − 2) j + (9 − 3) k = 2i + 4j + 6k.
向量 AB→ 通过坐标相减得到：(3 − 1) i + (6 − 2) j + (9 − 3) k = 2i + 4j + 6k。
The magnitude |AB→| = √(2² + 4² + 6²) = √(4 + 16 + 36) = √56 = 2√14.
模长 |AB→| = √(2² + 4² + 6²) = √(4 + 16 + 36) = √56 = 2√14。
To find the angle θ between AB→ and v = i + j + k, use the dot product: AB→ · v = 2×1 + 4×1 + 6×1 = 12. |v| = √(1²+1²+1²) = √3. So cos θ = 12 / (|AB→| |v|) = 12 / (2√14 × √3) = 12 / (2√42) = 6/√42. Simplifying gives cos θ = 6/√42 = √(36/42) = √(6/7). Hence θ = arccos(√(6/7)).
为求 AB→ 与 v = i + j + k 的夹角 θ，使用点积：AB→ · v = 2×1 + 4×1 + 6×1 = 12。|v| = √(1²+1²+1²) = √3。因此 cos θ = 12 / (|AB→| |v|) = 12 / (2√14 × √3) = 12 / (2√42) = 6/√42。化简得 cos θ = 6/√42 = √(36/42) = √(6/7)。所以 θ = arccos(√(6/7))。
The angle is left in exact form, as is standard in CIE Pure Mathematics 1.
夹角保留精确值形式，这是 CIE 纯数学1的常规要求。
9. Summary and Exam Tips | 总结与应试技巧
Mastering each topic in pure mathematics requires familiarity with standard techniques such as binomial expansion, calculus rules, trigonometric identities, and vector operations. Always check your answers for algebraic errors and ensure that you write solutions in the required exact form unless otherwise stated.
掌握纯数学的每个专题需要熟悉标准方法，如二项式展开、微积分法则、三角恒等式以及向量运算。务必检查答案是否出现代数错误，并确保除非题目另有说明，解答均以精确形式书写。
When solving past papers, time yourself and review the mark schemes to understand where marks are awarded. This paper from May 2018 exemplifies the balance of routine and slightly challenging problems that characterises the CIE Pure Mathematics 1 exam.
在做历年真题时，给自己计时并对照评分标准，理解得分点。这份2018年5月的试卷典型地体现了常规题与稍具挑战性题目之间的平衡，这正是 CIE 纯数学1考试的特点。
Published by TutorHao | CIE Pure Mathematics 1 Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply