SAT Math Level 2 Tough Questions Explained | SAT II 数学考前疑难习题精讲

📚 SAT Math Level 2 Tough Questions Explained | SAT II 数学考前疑难习题精讲

As you approach the SAT Subject Test in Mathematics Level 2, practicing challenging problems is essential to secure a high score. This article handpicks several difficult questions from key topics such as functions, trigonometry, complex numbers, matrices, probability, and more. Each problem is followed by a clear, step-by-step solution to deepen your understanding. Let's master the toughest material together.

当您准备SAT II 数学 Level 2 考试时，练习具有挑战性的题目是取得高分的关键。本文精选了函数、三角学、复数、矩阵、概率等核心专题中的若干难题，并为每道题提供清晰分步解答，帮助您深化理解。让我们一同攻克最难的考点。

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1. Composite Functions & Domain | 复合函数与定义域

Given f(x) = 1/(x-2) and g(x) = √x+3, find the domain of (f ∘ g)(x).

问题：已知 f(x) = 1/(x-2) 且 g(x) = √x+3，求 (f ∘ g)(x) 的定义域。

Step 1: Write the composite function. (f ∘ g)(x) = f(g(x)) = 1/(√x+3 - 2).

步骤1：写出复合函数。(f ∘ g)(x) = f(g(x)) = 1/(√x+3 - 2)。

Step 2: The square root requires x+3 ≥ 0 → x ≥ -3.

步骤2：平方根要求 x+3 ≥ 0 → x ≥ -3。

Step 3: Denominator cannot be zero: √x+3 - 2 ≠ 0 → √x+3 ≠ 2 → squaring gives x+3 ≠ 4 → x ≠ 1.

步骤3：分母不能为零：√x+3 - 2 ≠ 0 → √x+3 ≠ 2 → 平方得 x+3 ≠ 4 → x ≠ 1。

Step 4: Combine conditions. The domain is x ≥ -3 and x ≠ 1, i.e., [-3, 1) ∪ (1, ∞).

步骤4：合并条件。定义域为 x ≥ -3 且 x ≠ 1，即 [-3, 1) ∪ (1, ∞)。

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2. Solving Trigonometric Equations | 解三角方程

Solve 2sin²θ - sinθ - 1 = 0 for θ in the interval [0, 2π).

问题：在区间 [0, 2π) 上解方程 2sin²θ - sinθ - 1 = 0。

Step 1: Treat as a quadratic in sinθ. Factor: (2sinθ + 1)(sinθ - 1) = 0.

步骤1：将方程视为关于 sinθ 的二次式。因式分解得 (2sinθ + 1)(sinθ - 1) = 0。

Step 2: Set each factor to zero: 2sinθ + 1 = 0 → sinθ = -1/2; sinθ - 1 = 0 → sinθ = 1.

步骤2：令各因式为零：2sinθ + 1 = 0 → sinθ = -1/2；sinθ - 1 = 0 → sinθ = 1。

/2. sinθ = -1/2 in [0,2π) gives θ = 7π/6 and θ = 11π/6 (reference angle π/6 in QIII, QIV).

步骤3：求角度。sinθ = 1 → θ = π/2。sinθ = -1/2 在 [0,2π) 内得 θ = 7π/6 和 θ = 11π/6（参考角 π/6，分别位于第三、四象限）。

Step 4: Solution set: {π/2, 7π/6, 11π/6}.

步骤4：解集为 {π/2, 7π/6, 11π/6}。

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3. Polar Form of Complex Numbers | 复数的极坐标形式

Express z = -√3 + i in polar form r(cosθ + i sinθ) with 0 ≤ θ < 2π.

问题：将复数 z = -√3 + i 表示为极坐标形式 r(cosθ + i sinθ)，其中 0 ≤ θ < 2π。

Step 1: Compute modulus r = √( (√3)² + 1² ) = √3+1 = 2.

步骤1：计算模 r = √( (√3)² + 1² ) = √3+1 = 2。

Step 2: The argument θ satisfies tanθ = 1/(-√3) = -1/√3. The complex number lies in Quadrant II (negative real, positive imaginary).

步骤2：辐角 θ 满足 tanθ = 1/(-√3) = -1/√3。该复数位于第二象限（实部负、虚部正）。

Step 3: Reference angle is arctan(1/√3) = π/6. Thus θ = π - π/6 = 5π/6.

步骤3：参考角为 arctan(1/√3) = π/6。因此 θ = π - π/6 = 5π/6。

Step 4: Polar form: z = 2(cos 5π/6 + i sin 5π/6).

步骤4：极坐标形式为 z = 2(cos 5π/6 + i sin 5π/6)。

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4. Matrix Multiplication & Determinant | 矩阵乘法与行列式

Given matrices A = [2 -1; 0 3] and B = [1 4; -2 0], compute AB and its determinant.

问题：给定矩阵 A = [2 -1; 0 3] 和 B = [1 4; -2 0]，计算 AB 及其行列式。

Step 1: Write the matrices clearly.

步骤1：清晰写出矩阵。

A =

2 -1 0 3

B =

1 4 -2 0

+ (-1)×0 = 8+0 = 8. Entry (2,1): 0×1 + 3×(-2) = -6. Entry (2,2): 0×4 + 3×0 = 0. So AB = [4 8; -6 0].

步骤2：计算 AB。元素 (1,1)：2×1 + (-1)×(-2) = 2+2 = 4。元素 (1,2)：2×4 + (-1)×0 = 8+0 = 8。元素 (2,1)：0×1 + 3×(-2) = -6。元素 (2,2)：0×4 + 3×0 = 0。因此 AB = [4 8; -6 0]。

Step 3: Determinant of AB = 4×0 - 8×(-6) = 0 + 48 = 48.

步骤3：AB 的行列式 = 4×0 - 8×(-6) = 0 + 48 = 48。

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5. Probability with Combinations | 组合概率问题

A committee of 3 is to be selected from 5 men and 4 women. Find the probability that the committee contains at most one man.

问题：从5名男性和4名女性中选取3人组成委员会。求委员会中最多有1名男性的概率。

Step 1: Total possible committees = C(9,3) = 84.

步骤1：总可能组合数 = C(9,3) = 84。

Step 2: "At most one man" means 0 men, 3 women OR 1 man, 2 women.

步骤2：“最多1名男性”意味着0男3女 或 1男2女。

Step 3: 0 men, 3 women: choose all 3 from 4 women → C(4,3) = 4.

步骤3：0男3女：从4女中选3人 → C(4,3) = 4。

: 1 man, 2 women: choose 1 man from 5 (C(5,1)=5) and 2 women from 4 (C(4,2)=6) → 5×6 = 30.

步骤4：1男2女：从5男选1人 (C(5,1)=5)，从4女选2人 (C(4,2)=6) → 5×6 = 30。

Step 5: Favorable outcomes = 4 + 30 = 34. Probability = 34/84 = 17/42.

步骤5：有利结果 = 4 + 30 = 34。概率 = 34/84 = 17/42。

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6. Exponential Equation with Substitution | 换元法解指数方程

Solve 2e^2x = e^x + 6 for real x.

问题：解实数 x 满足 2e^2x = e^x + 6。

Step 1: Let u = e^x. Then e^2x = u². The equation becomes 2u² = u + 6 → 2u² - u - 6 = 0.

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