中文（繁體）
中文（简体）
English
剑桥 A-Level 数学 9709 Paper 1 纯数模拟卷（10 题・75 分・1 小时 50 分钟）
TUTORZONE 原创模拟卷 · 剑桥 A-Level 数学 9709 Paper 1（纯数 1）
本卷照剑桥 A-Level 数学 9709 Paper 1（纯数 1）的卷制编写：全卷 10 题，合共 75 分，时限 1 小时 50 分钟，占整个 A-Level 30%。
卷制速览
| 项目 | 内容 |
|---|---|
| 试卷 | Paper 1 Pure Mathematics 1 |
| 题数 | 10 题（25 部分） |
| 分数 | 75 分 |
| 时限 | 1 小时 50 分钟 |
| 占比 | A-Level 总分 30% |
| 计算机 | 不准使用计算机 |
题目
全卷 10 题、25 部分，答案须写出步骤；除特别指明外，非准确值取 3 位有效数字。
1. （6 分）二次方程与配方法
(a) Solve the equation 2x² − 7x + 3 = 0. （3 分）
(b) Express 2x² − 12x + 7 in the form a(x + b)² + c, and state the minimum value of 2x² − 12x + 7. （3 分）
2. （7 分）二项式定理
(a) Find the first three terms, in ascending powers of x, of the expansion of (1 + 2x)⁸. （3 分）
(b) Find the coefficient of x³ in the expansion. （2 分）
(c) Hence estimate the value of (1.02)⁸, giving your answer to 4 decimal places. （2 分）
3. （7 分）微分与驻点
(a) A curve has equation y = x³ − 6x² + 9x + 2. Find dy/dx. （2 分）
(b) Find the coordinates of the stationary points of the curve. （3 分）
(c) Determine the nature of each stationary point. （2 分）
4. （8 分）圆的方程
(a) The circle with equation x² + y² − 6x + 10y + 18 = 0 has centre C and radius r. Find the coordinates of C and the value of r. （4 分）
(b) The point P has coordinates (10, −5). Find the shortest distance from P to the circle. （4 分）
5. （8 分）三角方程
(a) Show that the equation 3sin²θ − 2cosθ − 2 = 0 may be written as 3cos²θ + 2cosθ − 1 = 0. （3 分）
(b) Hence solve 3cos²θ + 2cosθ − 1 = 0 for cosθ. （2 分）
(c) Hence solve 3sin²θ − 2cosθ − 2 = 0 for 0° ≤ θ ≤ 360°. （3 分）
6. （8 分）定积分与面积
(a) The curve y = x² + 2 crosses the lines x = 1 and x = 3. Find the area of the region bounded by the curve, the x-axis and these two lines. （4 分）
(b) Find the area of the region enclosed between y = x² + 2 and y = 3x + 2 for 0 ≤ x ≤ 3. （4 分）
7. （8 分）等比数列
(a) A geometric progression has first term 3 and common ratio 2/3. Find the sum to infinity. （3 分）
(b) Find the sum of the first 5 terms, giving your answer as an exact fraction. （3 分）
(c) Find the smallest value of n for which the sum of the first n terms exceeds 8.9. （2 分）
8. （8 分）余式定理与因式定理
(a) The polynomial f(x) = 2x³ + ax² + bx − 6 has factors (x − 1) and (x + 2). Find the values of a and b. （5 分）
(b) Hence factorise f(x) completely. （3 分）
9. （8 分）链式法则与相关变化率
(a) Given that y = (3x − 1)⁶, find dy/dx and hence the value of dy/dx at x = 1. （4 分）
(b) The radius of a circle increases at 0.5 cm per second. Find the rate of increase of the area when the radius is 6 cm, giving your answer in exact form. （4 分）
10. （7 分）向量
(a) The points A and B have coordinates (2, −1) and (5, 3). Find the vector AB and its magnitude. （3 分）
(b) Find a unit vector in the direction of AB. （2 分）
(c) The vector (k, 8) is parallel to AB. Find k. （2 分）
答案与评分要点
以下为每部分参考答案与评分要点，M 为方法分、A 为答案分、B 为独立分。
第 1 题 · 6 分
(a) x = 1/2 or x = 3
M1 factorises correctly as (2x − 1)(x − 3), or uses the formula with a = 2, b = −7, c = 3; A1 both roots correct.
(b) 2(x − 3)² − 11; minimum value −11 (at x = 3)
M1 takes out the factor 2 and completes the square; A1 a = 2, b = −3, c = −11; A1 minimum value −11 stated (allow “minimum at x = 3”).
第 2 题 · 7 分
(a) 1 + 16x + 112x²
M1 uses the binomial coefficients ⁱC₁ and ⁱC₂; A1 16x; A1 +112x².
(b) 448
M1 ⁱC₃ × 2³ = 56 × 8; A1 448.
(c) 1.1712 (true value 1.1717)
M1 substitutes x = 0.01 in the three terms; A1 1.1712.
第 3 题 · 7 分
(a) 3x² − 12x + 9
M1 differentiates term by term; A1 3x² − 12x + 9.
(b) (1, 6) and (3, 2)
M1 sets dy/dx = 0 and factorises 3(x − 1)(x − 3); A1 x = 1, 3; A1 both y-values correct.
(c) (1, 6) is a maximum; (3, 2) is a minimum
M1 finds d²y/dx² = 6x − 12; A1 correct classification of both points.
第 4 题 · 8 分
(a) C(3, −5), r = 4
M1 completes the square in x and in y; A1 centre (3, −5); M1 r² = 9 + 25 − 18; A1 r = 4.
(b) 3 units
M1 distance CP = 7; M1 subtracts the radius; A1 3.
第 5 题 · 8 分
(a) 3(1 − cos²θ) − 2cosθ − 2 = 0, then collect terms
M1 uses sin²θ = 1 − cos²θ; A1 expands; A1 reaches the printed form with all signs correct.
(b) cosθ = 1/3 or cosθ = −1
M1 factorises (3cosθ − 1)(cosθ + 1); A1 both values.
(c) θ = 70.5°, 180°, 289.5°
B1 θ = 180°; M1 θ = 70.5°; A1 289.5° (all to 1 d.p., no extras in range).
第 6 题 · 8 分
(a) 38/3 ≈ 12.67 square units
M1 integrates to x³/3 + 2x; M1 substitutes 3 and 1; A1 15 − 7/3; A1 38/3.
(b) 9/2 = 4.5 square units
M1 forms the difference 3x − x²; M1 integrates; A1 27/2 − 9; A1 9/2.
第 7 题 · 8 分
(a) S∞ = 9
M1 uses a/(1 − r) with r = 2/3; A1 3 ÷ (1/3) = 9; A1 states the condition that r is between −1 and 1.
(b) 211/27 ≈ 7.815
M1 uses Sₙ = a(1 − rⁿ)/(1 − r); M1 substitutes n = 5 and r = 2/3; A1 211/27.
(c) n = 12 (S₁₁ ≈ 8.896, S₁₂ ≈ 8.931)
M1 solves or tests the inequality; A1 n = 12.
第 8 题 · 8 分
(a) a = 5, b = −1
M1 f(1) = 0 gives a + b = 4; M1 f(−2) = 0 gives 4a − 2b = 22; M1 solves the pair; A1 a = 5; A1 b = −1.
(b) f(x) = (x − 1)(x + 2)(2x + 3)
M1 divides out one factor; A1 reaches (x − 1)(x + 2)(2x + 3); A1 all three factors correct.
第 9 题 · 8 分
(a) dy/dx = 18(3x − 1)⁵; at x = 1 the value is 576
M1 uses the chain rule; A1 18(3x − 1)⁵; M1 substitutes x = 1; A1 576.
(b) 6π ≈ 18.85 cm² per second
M1 dA/dr = 2πr; M1 multiplies by dr/dt; A1 6π; A1 cm² per second with correct units.
第 10 题 · 7 分
(a) AB = (3, 4); magnitude 5
M1 subtracts coordinates; A1 (3, 4); A1 5.
(b) (3/5, 4/5)
M1 divides AB by 5; A1 (3/5, 4/5).
(c) k = 6
M1 sets up k/3 = 8/4; A1 k = 6.
常见失分位
- 配方法忘记先抽出系数 2，直接写成 (x − 6)² − 29 之类的错误形式，最小值亦因此写错。
- 三角方程只写出 70.5° 一个解，漏掉 180° 与 289.5°，或写出区间以外的解。
- 面积题没有先比较两条曲线的高低，把积分次序倒转，得出负面积。
- 链式法则漏乘内层导数 3，答案变成 3(3x − 1)⁵ 而不是 18(3x − 1)⁵。
- 相关变化率只写出 12π 而没有乘 dr/dt，或漏写单位 cm² per second。
时间分配
| 题目 | 分数 | 建议时间 |
|---|---|---|
| 第 1–2 题 | 13 分 | 约 19 分钟 |
| 第 3–4 题 | 15 分 | 约 22 分钟 |
| 第 5–6 题 | 16 分 | 约 23 分钟 |
| 第 7–8 题 | 16 分 | 约 23 分钟 |
| 第 9–10 题 | 15 分 | 约 22 分钟 |
下载与打印
题目卷与答案卷均为 A4 版本，可直接打印：
