Prelims 9C Answers and Worked Solutions | 预考 9C 答案解析与详述

📚 Prelims 9C Answers and Worked Solutions | 预考 9C 答案解析与详述

This article provides full worked answers for the Prelims 9C mathematics paper. The solutions cover the most common question types in algebra, trigonometry, calculus, probability and vectors, with step-by-step reasoning and examiner tips.

本文提供数学预考 9C 试卷的完整答案解析。内容涵盖代数、三角、微积分、概率和向量等最常见题型，给出逐步推理和阅卷提示。

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1. Algebraic Simplification | 代数化简

Question 1 (typical). Simplify (3x² – 2x + 1) – (x² + 4x – 5) and then evaluate at x = -2. Answer: First remove brackets: 3x² – 2x + 1 – x² – 4x + 5 = 2x² – 6x + 6. At x = -2, 2(-2)² – 6(-2) + 6 = 8 + 12 + 6 = 26.

典型题目：化简 (3x² – 2x + 1) – (x² + 4x – 5)，再代入 x = -2。答案：先去括号：3x² – 2x + 1 – x² – 4x + 5 = 2x² – 6x + 6。当 x = -2 时，2(-2)² – 6(-2) + 6 = 8 + 12 + 6 = 26。

Be careful with the minus sign before the second bracket. A common error is to write -x² + 4x – 5 instead of -x² – 4x + 5.

去第二个括号时要注意负号。常见错误是写成 -x² + 4x – 5，而正确结果是 -x² – 4x + 5。

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2. Quadratic Equations | 二次方程

Question 2 (typical). Solve 2x² – 5x – 3 = 0 by factorising. Answer: 2x² – 5x – 3 = (2x + 1)(x – 3) = 0, so x = -1/2 or x = 3.

典型题目：用因式分解解 2x² – 5x – 3 = 0。答案：2x² – 5x – 3 = (2x + 1)(x – 3) = 0，所以 x = -1/2 或 x = 3。

For quadratic equations that do not factorise, use the formula x = (-b ± √(b² – 4ac)) / 2a. Always check the discriminant Δ = b² – 4ac first. If Δ > 0 there are two real roots, if Δ = 0 there is one repeated root, and if Δ < 0 there are no real roots.

对于不能因式分解的二次方程，使用公式 x = (-b ± √(b² – 4ac)) / 2a。先检验判别式 Δ = b² – 4ac。若 Δ > 0，有两个实根；若 Δ = 0，有一个重根；若 Δ < 0，无实根。

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3. Inequalities | 不等式

Question 3 (typical). Solve x² – 4x – 5 > 0. Answer: Factorise to (x – 5)(x + 1) > 0. Critical values are x = -1 and x = 5. Testing intervals gives x < -1 or x > 5.

典型题目：解 x² – 4x – 5 > 0。答案：因式分解为 (x – 5)(x + 1) > 0。临界值为 x = -1 和 x = 5。检验区间得 x < -1 或 x > 5。

Remember to use open intervals for strict inequalities and closed intervals for ≤ or ≥. A sign table helps avoid mistakes when multiple critical values are involved.

注意严格不等式用开区间，≤ 或 ≥ 用闭区间。当涉及多个临界值时，使用符号表可避免错误。

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4. Polynomials and the Factor Theorem | 多项式与因式定理

Question 4 (typical). Show that (x – 2) is a factor of f(x) = x³ – 4x² + x + 6, then factorise f(x) fully. Answer: f(2) = 8 – 16 + 2 + 6 = 0, so (x – 2) is a factor. Division gives f(x) = (x – 2)(x² – 2x – 3) = (x – 2)(x – 3)(x + 1).

典型题目：证明 (x – 2) 是 f(x) = x³ – 4x² + x + 6 的因式，并将 f(x) 完全分解。答案：f(2) = 8 – 16 + 2 + 6 = 0，所以 (x – 2) 是因式。多项式除法得 f(x) = (x – 2)(x² – 2x – 3) = (x – 2)(x – 3)(x + 1)。

The factor theorem states that if f(a) = 0, then (x – a) is a factor. Use synthetic division or long division to reduce the degree of the polynomial.

因式定理指出：若 f(a) = 0，则 (x – a) 是因式。用综合除法或长除法降低多项式次数。

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5. Trigonometric Identities | 三角恒等式

Question 5 (typical). Prove that (1 – cos θ)(1 + cos θ) = sin² θ. Answer: Expand the left side: 1 – cos² θ = sin² θ, using sin² θ + cos² θ = 1.

典型题目：证明 (1 – cos θ)(1 + cos θ) = sin² θ。答案：展开左边：1 – cos² θ = sin² θ，使用了 sin² θ + cos² θ = 1。

Common identities tested include sin² θ + cos² θ = 1, tan θ = sin θ / cos θ, and double-angle formulas such as sin 2θ = 2 sin θ cos θ and cos 2θ = cos² θ – sin² θ.

常考的恒等式包括 sin² θ + cos² θ = 1、tan θ = sin θ / cos θ，以及倍角公式如 sin 2θ = 2 sin θ cos θ 和 cos 2θ = cos² θ – sin² θ。

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6. Trigonometric Equations | 三角方程

Question 6 (typical). Solve 2 sin x = 1 for 0° ≤ x ≤ 360°. Answer: sin x = 1/2, so principal value x = 30°. The second solution is x = 180° – 30° = 150°. Thus x = 30°, 150°.

典型题目：在 0° ≤ x ≤ 360° 内解 2 sin x = 1。答案：sin x = 1/2，主值 x = 30°。第二个解为 x = 180° – 30° = 150°。因此 x = 30°, 150°。

Always give answers in degrees or radians as specified, and check the range. For cosine, the second solution uses 360° – principal value. For tangent, add 180° to the principal value.

务必按题目指定的度数或弧度给出答案并检查范围。对于余弦，第二个解用 360° 减去主值。对于正切，在主值上加 180°。

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7. Differentiation and Stationary Points | 微分与驻点

Question 7 (typical). Differentiate y = 3x⁴ – 5x² + 2x – 7. Answer: dy/dx = 12x³ – 10x + 2.

典型题目：求 y = 3x⁴ – 5x² + 2x – 7 的导数。答案：dy/dx = 12x³ – 10x + 2。

For stationary points, set dy/dx = 0 and solve. Use the second derivative d²y/dx² to classify maximum, minimum or point of inflection.

求驻点时令 dy/dx = 0 并解方程。用二阶导数 d²y/dx² 判断极大值、极小值或拐

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