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IGCSE 附加数学 0606 Paper 1 模拟卷｜80 分・2 小时
练习卷 · IGCSE 附加数学 0606 Paper 1（2 小时・80 分）
IGCSE 附加数学 0606 Paper 1 模拟卷：10 题 80 分、2 小时、可用计算机，涵盖根式与指数、二次函数、余式定理、对数方程、弧长与扇形、R 形三角方程、二项式、排列组合、向量及运动学微积分，附完整答案与评分要点。
卷面资料 年级：IGCSE ｜ 科目：附加数学 ｜ 时限：120 分钟 ｜ 总分：80 分 ｜ 题数：10
甲部 代数、对数与圆弧量度（36 分）
全部题目必须作答，须展示步骤；除另有指明外，答案须准确至三位有效数字。
| # | 题目 | 答案与评分要点 |
|---|---|---|
| 1 | (a) Simplify √50 + √18 − √8, giving your answer in the form k√2. (b) Evaluate 27^(2/3). (c) Show that 6 ÷ (√5 − 1) = 3(√5 + 1) ÷ 2. | (a) √50 = 5√2, √18 = 3√2 and √8 = 2√2, so the sum is 6√2. (b) 27^(1/3) = 3 and 3² = 9, so 27^(2/3) = 9. (c) Multiply top and bottom by (√5 + 1); the denominator becomes 5 − 1 = 4, so 6(√5 + 1) ÷ 4 = 3(√5 + 1) ÷ 2 ≈ 4.854. [2, 2, 2] |
| 2 | (a) Express f(x) = 2x² − 12x + 7 in the form a(x − h)² + k and state the minimum value of f. (b) Solve f(x) = 0, giving your answers in surd form. | (a) f(x) = 2(x − 3)² − 11, so the minimum value is -11 at x = 3. (b) 2(x − 3)² = 11 gives x − 3 = ±√22 ÷ 2, so x = 0.655 or x = 5.345, i.e. x = 3 ± √22 ÷ 2. [3, 4] |
| 3 | p(x) = x³ + ax² + bx − 6, where a and b are constants. (x − 1) is a factor of p(x), and p(x) leaves a remainder of 12 when divided by (x + 2). (a) Find the values of a and b. (b) Factorise p(x) completely. | (a) p(1) = 0 gives a + b = 5; p(−2) = 12 gives 2a − b = 13. Solving: a = 6, b = -1. (b) p(x) = (x − 1)(x² + 7x + 6) = (x − 1)(x + 1)(x + 6). [5, 3] |
| 4 | (a) Solve log₂(x + 3) + log₂(x − 3) = 4. (b) Solve 3^(2x − 1) = 5, giving x correct to 3 significant figures. | (a) The logs combine to log₂(x² − 9) = 4, so x² − 9 = 16, x² = 25 and x = 5; x = −5 is rejected because x − 3 must be positive. (b) 2x − 1 = log₃5, so x = (1 + ln 5 ÷ ln 3) ÷ 2 = 1.232. [4, 3] |
| 5 | A sector OAB of a circle has radius 9 cm and angle AOB = 1.2 radians. Find (a) the length of the arc AB, (b) the area of the sector, (c) the area of the minor segment cut off by the chord AB, correct to 2 decimal places. | (a) Arc = rθ = 9 × 1.2 = 10.8 cm. (b) Area = ½r²θ = 0.5 × 81 × 1.2 = 48.6 cm². (c) Triangle OAB = ½r² sin θ = 0.5 × 81 × sin 1.2 = 37.75 cm², so the segment = 48.6 − 37.75 = 10.85 cm² (the chord is 10.16 cm). [2, 3, 3] |
乙部 三角、二项式、排列组合、向量与微积分（44 分）
全部题目必须作答；数值答案须准确至三位有效数字，角度以弧度表示者另予指明。
| # | 题目 | 答案与评分要点 |
|---|---|---|
| 6 | (a) Express 3 sin θ + 4 cos θ in the form R sin(θ + α), where R and α are positive constants and α is in radians, and state the maximum value of 3 sin θ + 4 cos θ. (b) Hence solve 3 sin θ + 4 cos θ = 2 for 0 ≤ θ ≤ 2π, giving your answers correct to 3 significant figures. | (a) R = √(3² + 4²) = 5 and tan α = 4 ÷ 3, so α = 0.927 rad; 3 sin θ + 4 cos θ = 5 sin(θ + 0.927), whose maximum value is 5. (b) sin(θ + α) = 0.4 gives θ + α = 0.412 or 2.730, so θ = 1.80 or θ = 5.77. [4, 4] |
| 7 | (a) Find the coefficient of x³ in the expansion of (2 + x/3)^6. (b) Find the coefficient of x⁴ in the same expansion. (c) State the constant term. | (a) 6C3 × 2³ × (1/3)³ = 20 × 8 ÷ 27 = 5.926, so the coefficient is 160/27. (b) 6C4 × 2² × (1/3)⁴ = 15 × 4 ÷ 81 = 20/27 ≈ 0.7407. (c) The constant term is 2⁶ = 64. [3, 2, 2] |
| 8 | A committee of 5 is chosen from 6 boys and 5 girls. (a) How many committees contain exactly 3 girls? (b) If a committee of 5 is chosen at random, find the probability that it contains exactly 3 girls. (c) How many committees contain more girls than boys? | (a) 5C3 × 6C2 = 10 × 15 = 150. (b) 11C5 = 462, so the probability is 150 ÷ 462 = 25/77 ≈ 0.325. (c) 150 + (5C4 × 6C1) + 5C5 = 150 + 30 + 1 = 181. [3, 2, 3] |
| 9 | Relative to an origin O, the position vectors of A and B are a = 3i − j + 2k and b = i + 4j − k. Find (a) a·b, (b) the exact values of |a| and |b|, (c) the angle between a and b in degrees correct to 1 decimal place, (d) a unit vector parallel to a − b, in exact form. | (a) a·b = 3 − 4 − 2 = -3. (b) |a| = √14 ≈ 3.742 and |b| = √18 ≈ 4.243. (c) cos θ = −3 ÷ √252 ≈ -0.1890, so θ = 100.9°. (d) a − b = 2i − 5j + 3k with length √38, so the unit vector is (2i − 5j + 3k) ÷ √38. [2, 2, 3, 3] |
| 10 | A particle moves in a straight line so that its displacement s metres from a fixed point O at time t seconds is s = t³ − 6t² + 9t for t ≥ 0. (a) Find an expression for the velocity and the times at which the particle is instantaneously at rest. (b) Find the acceleration at t = 2 s. (c) Find the total distance travelled in the first 4 seconds. | (a) v = ds/dt = 3t² − 12t + 9 = 3(t − 1)(t − 3), so v = 0 at t = 1 s and t = 3 s. (b) a = dv/dt = 6t − 12, so at t = 2 s the acceleration is 0 m s⁻² (the velocity is minimum there). (c) s(0) = 0, s(1) = 4, s(3) = 0 and s(4) = 4, so the total distance is 4 + 4 + 4 = 12 m (the displacement is only 4 m). [3, 2, 6] |
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