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CIE A Level 數學 9709 Paper 3（Pure Mathematics 3）模擬卷｜75 分・1 小時 50 分
練習卷 · CIE A Level 數學 9709 Paper 3（Pure Mathematics 3）
CIE A Level 數學 9709 Paper 3（Pure Mathematics 3）模擬卷：11 題共 75 分、1 小時 50 分，涵蓋對數與指數方程、部分分式、微分與積分、微分方程、牛頓法、三角方程、向量與複數，每題附完整答案與評分要點，題目版與答案版 PDF 可即時下載，方便考生即時對答案。
卷面資料 年級：CIE A Level ｜ 科目：數學 ｜ 時限：110 分鐘 ｜ 總分：75 分 ｜ 題數：11
甲部：對數、指數方程與部分分式
全部題目均須作答。不得使用計算機。
| # | 題目 | 答案與評分要點 |
|---|---|---|
| 1 | Without using a calculator, simplify: (a) log248 − log23, (b) log510 + log52.5, (c) 2log36 − log34. | (a) log2(48 ÷ 3) = log216 = 4 [2] (b) log5(10 × 2.5) = log525 = 2 [2] (c) log3(6² ÷ 4) = log39 = 2 [2] |
| 2 | Solve 4x − 9(2x) + 8 = 0. | Let y = 2x: y² − 9y + 8 = 0 → (y − 1)(y − 8) = 0, so y = 1 or y = 8 [3] 2x = 1 → x = 0; 2x = 8 → x = 3 [2] Check: x = 0 gives 1 − 9 + 8 = 0 ✓; x = 3 gives 64 − 72 + 8 = 0 ✓ [1] |
| 3 | Express (7x + 8)/((x + 1)(x + 2)) in partial fractions. | (7x + 8)/((x + 1)(x + 2)) ≡ A/(x + 1) + B/(x + 2) [1] 7x + 8 ≡ A(x + 2) + B(x + 1) [1] x = −1: 1 = A; x = −2: −6 = −B so B = 6 [3] Hence 1/(x + 1) + 6/(x + 2); coefficients check: A + B = 7, 2A + B = 8 ✓ [2] |
| 4 | f(x) = x³ − 4x² + 5x − 6. (a) Find the remainder when f(x) is divided by (x − 2). (b) Given that (x − 3) is a factor, factorise f(x) completely and explain why the quadratic factor has no real roots. | (a) Remainder = f(2) = 8 − 16 + 10 − 6 = −4 [2] (b) x³ − 4x² + 5x − 6 = (x − 3)(x² − x + 2) [3] For x² − x + 2, discriminant = 1 − 8 = −7 < 0, so no real roots [1] |
乙部：微分、積分、微分方程與數值方法
全部題目均須作答。須寫出運算步驟。
| # | 題目 | 答案與評分要點 |
|---|---|---|
| 5 | f(x) = x³ + 3x² − 9x + 5. Find the coordinates of the stationary points and determine their nature. | f′(x) = 3x² + 6x − 9 = 3(x + 3)(x − 1) = 0 → x = −3 or x = 1 [3] f(−3) = −27 + 27 + 27 + 5 = 32; f(1) = 1 + 3 − 9 + 5 = 0 [2] f″(x) = 6x + 6: f″(−3) = −12 < 0 → maximum at (−3, 32); f″(1) = 12 > 0 → minimum at (1, 0) [3] |
| 6 | Evaluate ∫14 (2x + 3/√x) dx, giving an exact value. | ∫(2x + 3x−1/2)dx = x² + 6√x [3] At x = 4: 16 + 12 = 28; at x = 1: 1 + 6 = 7 [2] Value = 28 − 7 = 21 [2] |
| 7 | A curve satisfies dy/dx = 2y/x and passes through (1, 3). Find y in terms of x. | Separate: ∫(1/y)dy = ∫(2/x)dx → ln y = 2 ln x + c [3] y = Ax² [1] x = 1, y = 3 → A = 3, so y = 3x² [2] Check: dy/dx = 6x and 2y/x = 6x²/x = 6x ✓ [1] |
| 8 | f(x) = x³ − 2x − 5. (a) Show that f(x) = 0 has a root between x = 2 and x = 3. (b) Use the Newton-Raphson method with x0 = 2 to find x1 and x2, giving x2 correct to 4 decimal places. | (a) f(2) = −1 and f(3) = 16; sign change and f is continuous, so a root lies in (2, 3) [2] f′(x) = 3x² − 2 [1] x1 = 2 − (−1)/10 = 2.1 [2] f(2.1) = 0.061, f′(2.1) = 11.23 → x2 = 2.1 − 0.061/11.23 = 2.0946 (4 d.p.) [3] |
丙部：三角學、向量與複數
全部題目均須作答。角度以度為單位。
| # | 題目 | 答案與評分要點 |
|---|---|---|
| 9 | Solve 2sin²θ + 3cosθ = 3 for 0° ≤ θ ≤ 360°. | 2(1 − cos²θ) + 3cosθ − 3 = 0 → 2cos²θ − 3cosθ + 1 = 0 [2] (2cosθ − 1)(cosθ − 1) = 0 → cosθ = 1/2 or cosθ = 1 [2] cosθ = 1/2 → θ = 60°, 300° [2] cosθ = 1 → θ = 0°, 360° [1] |
| 10 | Vectors a = 2i − j + 3k and b = −i + 4j − 2k. (a) Find a·b. (b) Find the angle between a and b. | (a) a·b = (2)(−1) + (−1)(4) + (3)(−2) = −2 − 4 − 6 = −12 [2] |a| = √14, |b| = √21 [2] cosθ = −12/√294 = −0.6999 → θ = 134.4° [2] |
| 11 | z = (4 − 2i)/(1 + i). Find z in the form a + bi, and state |z| and arg z (in degrees). | Multiply by (1 − i)/(1 − i): (4 − 2i)(1 − i)/2 = (4 − 4i − 2i + 2i²)/2 [2] = (2 − 6i)/2 = 1 − 3i [2] |z| = √(1 + 9) = √10 ≈ 3.1623 [2] arg z = arctan(−3/1) ≈ −71.6° [1] |
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