AP Calculus AB & BC: Free-Response Questions Answers and Solutions | AP微积分AB与BC大题答案及解析

📚 AP Calculus AB & BC: Free-Response Questions Answers and Solutions | AP微积分AB与BC大题答案及解析

AP Calculus AB and BC free-response questions (FRQs) demand more than just correct answers—they require clear reasoning, proper notation, and a solid understanding of fundamental theorems. This guide walks you through representative FRQ types, complete with step-by-step solutions and detailed explanations to help you master the section.

AP微积分AB与BC的自由响应题（FRQ）不仅要求正确答案，更需要清晰的推理过程、规范的符号以及扎实的基本定理理解。本文将带你梳理典型大题类型，配以逐步解答与详细解析，助你彻底掌握该部分。

1. Understanding the FRQ Structure | 理解大题结构

Both AP Calculus AB and BC exams contain 6 free-response questions to be completed in 90 minutes. Part A (2 questions) allows graphing calculators; Part B (4 questions) does not. AB topics include limits, derivatives, integrals, and their applications. BC extends to parametric equations, polar coordinates, vector-valued functions, and infinite series.

AP微积分AB和BC考试均包含6道自由响应题，限时90分钟。Part A（2题）允许使用图形计算器；Part B（4题）不允许。AB内容涵盖极限、导数、积分及其应用；BC则延伸至参数方程、极坐标、向量值函数以及无穷级数。

2. AB & BC Shared: Function Analysis from Derivative Graph | AB与BC共享：基于导数图形的函数分析

Sample Problem: Let f be a function defined on [0, 6] such that its derivative f'(x) is given by f'(x) = (x – 1)(x – 4) sin(x). It is known that f(2) = 3.

题目示例：设函数 f 定义在 [0, 6] 上，其导数 f'(x) = (x – 1)(x – 4) sin(x)。已知 f(2) = 3。

(a) Find the intervals on which f is increasing or decreasing. Justify your answer.

(a) 求 f 的递增和递减区间。请说明理由。

Solution: f is increasing where f'(x) > 0. f'(x) = (x – 1)(x – 4) sin(x). On [0, 6], sin(x) > 0 for x in (0, π) ≈ (0, 3.14) and sin(x) < 0 for x in (π, 6). Sign analysis: x - 1 > 0 for x > 1; x – 4 > 0 for x > 4. Combining sign changes, f'(x) > 0 on (1, π) and (4, 6); f'(x) < 0 on [0, 1) and (π, 4). Hence, f increases on (1, π) and (4, 6); decreases on [0, 1) and (π, 4).

解答：f 递增需要 f'(x) > 0。f'(x) = (x – 1)(x – 4) sin(x)。在 [0, 6] 上，sin(x) 在 (0, π) ≈ (0, 3.14) 为正，在 (π, 6) 为负。符号分析：x – 1 > 0 当 x > 1；x – 4 > 0 当 x > 4。综合符号变化，f'(x) > 0 在 (1, π) 和 (4, 6)；f'(x) < 0 在 [0, 1) 和 (π, 4)。因此，f 在 (1, π) 和 (4, 6) 递增，在 [0, 1) 和 (π, 4) 递减。

(b) Find all relative maxima and minima of f. Justify using the First Derivative Test.

(b) 求 f 的所有相对极大值和极小值。用一阶导数检验说明。

Solution: Critical points where f'(x) = 0 or undefined: x = 1, π, 4. f’ changes from negative to positive at x = 1 => relative min at x = 1. f’ changes from positive to negative at x = π => relative max at x = π. f’ changes from negative to positive at x = 4 => relative min at x = 4. To find f-values, use the given f(2) = 3 and integrate f’ numerically or approximate if needed. (In an exam, you might be asked to evaluate using the Fundamental Theorem with given values.)

解答：临界点满足 f'(x)=0 或不存在：x = 1, π, 4。在 x=1 处 f’ 由负变正 => 相对极小值；在 x=π 处 f’ 由正变负 => 相对极大值；在 x=4 处 f’ 由负变正 => 相对极小值。要求函数值，可利用 f(2)=3 并对 f’ 积分求得近似值（考试中常结合给定值使用微积分基本定理）。

Solution: Inflection points occur where f” changes sign. f”(x) = (2x – 5) sin(x) + (x² – 5x + 4) cos(x). (Obtained from product rule.) Sign chart shows f” changes sign at x ≈ 0.8, 2.9, 5.5 (approximate). These are inflection points.

解答：拐点出现在 f” 变号处。f”(x) = (2x – 5) sin(x) + (x² – 5x + 4) cos(x)（由乘积法则得到）。符号表显示 f” 在 x ≈ 0.8, 2.9, 5.5 变号，即为拐点。

3. Particle Motion & Total Distance | 粒子运动与总路程

Sample Problem: A particle moves along a line with velocity v(t) = t² – 4t + 3 meters per second for 0 ≤ t ≤ 5 seconds. The initial position is s(0) = 2.

题目示例：一个粒子沿直线运动，速度 v(t) = t² – 4t + 3 米/秒，0 ≤ t ≤ 5 秒。初始位置 s(0) = 2。

(a) Find the displacement of the particle during the 5 seconds.

(a) 求这5秒内粒子的位移。

Solution: Displacement = ∫₀⁵ v(t) dt = ∫₀⁵ (t² – 4t + 3) dt = [t³/3 – 2t² + 3t]₀⁵ = (125/3 – 50 + 15) – 0 = 125/3 – 35 = (125 – 105)/3 = 20/3 ≈ 6.667 meters.

解答：位移 = ∫₀⁵ v(t) dt = ∫₀⁵ (t² – 4t + 3) dt = [t³/3 – 2t² + 3t]₀⁵ = (125/3 – 50 + 15) – 0 = 20/3 ≈ 6.667 米。

(b) Find the total distance traveled.

(b) 求总路程。

Solution: Total distance = ∫ |v(t)| dt. Find where v(t) = 0: t² – 4t + 3 = 0 => (t-1)(t-3)=0 => t=1, 3. Sign of v: v>0 on [0,1) and (3,5]; v<0 on (1,3). Distance = ∫₀¹ (t²-4t+3) dt + ∫₁³ -(t²-4t+3) dt + ∫₃⁵ (t²-4t+3) dt. Compute: first part = [t³/3 - 2t² + 3t]₀¹ = (1/3 - 2 + 3) = 4/3. Second = -[t³/3 - 2t² + 3t]₁³ = -[(9 - 18 + 9) - (1/3 - 2 + 3)] = -[0 - (4/3)] = 4/3. Third = [t³/3 - 2t² + 3t]₃⁵ = (125/3 - 50 + 15) - 0 = 20/3. Total distance = 4/3 + 4/3 + 20/3 = 28/3 ≈ 9.333 meters.

解答：总路程 = ∫|v(t)| dt。求 v(t)=0：t=1, 3。v 的符号：v>0 在 [0,1) 和 (3,5]；v<0 在 (1,3)。路程 = ∫₀¹ v dt + ∫₁³ -v dt + ∫₃⁵ v dt。计算：第一部分 = 4/3；第二部分 = 4/3；第三部分 = 20/3。总路程 = 28/3 ≈ 9.333 米。

Solution: Moving left when v(t) < 0, which occurs for t in (1, 3).

解答：当 v(t)<0 时向左运动，即 t∈(1, 3)。

4. Area Between Curves and Volume of Revolution | 曲线间的面积与旋转体体积

Sample Problem: Let R be the region enclosed by y = 4x – x² and y = 2x in the first quadrant.

题目示例：设 R 是由 y = 4x – x² 与 y = 2x 在第一象限围成的区域。

(a) Find the area of R.

(a) 求 R 的面积。

Solution: Intersection: 4x – x² = 2x => x² – 2x = 0 => x=0, 2. Between x=0 and 2, upper curve y = 4x – x², lower y = 2x. Area = ∫₀² [(4x – x²) – 2x] dx = ∫₀² (2x – x²) dx = [x² – x³/3]₀² = (4 – 8/3) = 4/3 square units.

解答：交点：4x – x² = 2x ⇒ x=0, 2。在 [0,2] 上，上曲线为 y=4x-x²，下曲线为 y=2x。面积 = ∫₀² [(4x-x²) – 2x] dx = ∫₀² (2x – x²) dx = [x² – x³/3]₀² = 4/3 平方单位。

(b) Find the volume generated when R is revolved about the x-axis. Set up the integral and evaluate.

(b) 求 R 绕 x 轴旋转所得立体的体积。写出积分并计算。

Solution: Washer method: Volume = π ∫₀² [(4x – x²)² – (2x)²] dx. Simplify: (16x² – 8x³ + x⁴) – 4x² = 12x² – 8x³ + x⁴. Integrate: π [12x³/3 – 8x⁴/4 + x⁵/5]₀² = π [4x³ – 2x⁴ + x⁵/5]₀² = π (32 – 32 + 32/5) = (32π)/5 ≈ 20.106 cubic units.

解答：垫圈法：体积 = π ∫₀² [(4x-x²)² – (2x)²] dx。化简得 12x² – 8x³ + x⁴。积分得 π [4x³ – 2x⁴ + x⁵/5]₀² = (32π)/5 立方单位。

5. Differential Equations & Slope Fields | 微分方程与斜率场

Sample Problem: Consider dy/dx = (x² + 1)/y with y(0) = 1.

题目示例：考虑 dy/dx = (x² + 1)/y，且 y(0) = 1。

(a) Find the particular solution y = f(x) using separation of variables.

(a) 用分离变量法求特解 y = f(x)。

Solution: Separate: y dy = (x² + 1) dx. Integrate both sides: ∫y dy = ∫(x² + 1) dx ⇒ (1/2)y² = (1/3)x³ + x + C. Apply y(0)=1: (1/2)(1)² = C ⇒ C = 1/2. So (1/2)y² = (1/3)x³ + x + 1/2. Multiply by 2: y² = (2/3)x³ + 2x + 1. Since y(0)=1 >0, take positive root: y = √[(2/3)x³ + 2x + 1].

解答：分离变量：y dy = (x² + 1) dx。两边积分：(1/2)y² = (1/3)x³ + x + C。代入 y(0)=1 得 C=1/2。所以 y² = (2/3)x³ + 2x + 1，取正根得 y = √[(2/3)x³ + 2x + 1]。

(b) Sketch a slope field at the nine points (0,0), (±1,0), (±1,±1), (0,±1).

(b) 在九个点 (0,0), (±1,0), (±1,±1), (0,±1) 处画斜率场。

Solution: Slopes at (0,0): dy/dx = (0+1)/0 undefined (vertical segment). At (±1,0) undefined. At (0,1): slope = 1/1 = 1; (0,-1): -1; (1,1): (1+1)/1=2; (-1,1): 2; (1,-1): -2; (-1,-1): -2. Draw short line segments with indicated slopes.

解答：斜率计算：(0,0) 处无定义（垂直线段）；(±1,0) 无定义；(0,1) 斜率为 1；(0,-1) 为 -1；(1,1) 为 2；(-1,1) 为 2；(1,-1) 为 -2；(-1,-1) 为 -2。画出相应短线。

6. BC Exclusive: Power Series Radius & Interval of Convergence | BC专属：幂级数的收敛半径与区间

Sample Problem: Find the radius and interval of convergence for Σₙ₌₁^∞ [(x – 2)ⁿ / (n · 3ⁿ)].

题目示例：求级数 Σₙ₌₁^∞ [(x – 2)ⁿ / (n·3ⁿ)] 的收敛半径与区间。

Solution: Use Ratio Test. Let aₙ = |x-2|ⁿ / (n · 3ⁿ). Then |aₙ₊₁ / aₙ| = |x-2|^(n+1)/((n+1)·3^(n+1)) * (n·3ⁿ)/|x-2|ⁿ = |x-2|/(3) * (n/(n+1)). Taking limit as n→∞ gives |x-2|/3. For convergence, require |x-2|/3 < 1 ⇒ |x-2| < 3. Radius R = 3. Test endpoints: x = -1 and x = 5. At x = -1, series becomes Σ (-3)ⁿ/(n·3ⁿ) = Σ (-1)ⁿ/n, which converges conditionally (alternating harmonic). At x = 5, series = Σ 3ⁿ/(n·3ⁿ) = Σ 1/n, diverges (harmonic). Interval of convergence: [-1, 5).

解答：使用比值法。aₙ = |x-2|ⁿ / (n·3ⁿ)，比值极限为 |x-2|/3 < 1 得 |x-2| < 3，半径 R=3。检查端点：x = -1 得 Σ (-1)ⁿ/n 条件收敛（交错调和级数）；x = 5 得 Σ 1/n 发散。收敛区间为 [-1, 5)。

7. BC Exclusive: Parametric Equations & Polar Curves | BC专属：参数方程与极坐标曲线

Sample Problem: A particle moves such that x(t) = t³ – 3t, y(t) = t² – t. Find the particle’s speed at t = 2, and the equation of the line tangent to the path at t = 2.

题目示例：质点运动满足 x(t) = t³ – 3t，y(t) = t² – t。求 t=2 时的速率，以及 t=2 处路径的切线方程。

Solution: Velocity vector v(t) = (x'(t), y'(t)) = (3t² – 3, 2t – 1). At t=2, v = (3·4 – 3, 4 – 1) = (9, 3). Speed = √(9² + 3²) = √(81+9) = √90 = 3√10. Point at t=2: (x, y) = (2³ – 6, 4 – 2) = (2, 2). Tangent slope = dy/dx = y'(t)/x'(t) = (2t-1)/(3t²-3) = 3/9 = 1/3. Tangent line: y – 2 = (1/3)(x – 2).

解答：速度向量 v(t) = (3t²-3, 2t-1)。t=2 时 v=(9,3)，速率 = √(9²+3²) = 3√10。位置 (2,2)。切线斜率 dy/dx = (2t-1)/(3t²-3) = 3/9 = 1/3。切线方程：y – 2 = (1/3)(x – 2)。

Polar sample: Find the area of one petal of r = 2 cos(3θ).

极坐标示例：求 r = 2 cos(3θ) 一个花瓣的面积。

Solution: One petal traced for θ from -π/6 to π/6. Area = ½ ∫_{-π/6}^{π/6} [2 cos(3θ)]² dθ = 2 ∫_{-π/6}^{π/6} cos²(3θ) dθ. Using symmetry, = 4 ∫₀^{π/6} (1+cos(6θ))/2 dθ = 2 ∫₀^{π/6} (1+cos(6θ)) dθ = 2[θ + (1/6)sin(6θ)]₀^{π/6} = 2[π/6 + 0] = π/3.

解答：一个花瓣对应 θ 从 -π/6 到 π/6。面积 = ½ ∫ (2 cos(3θ))² dθ。利用对称性和半角公式得面积 = π/3。

8. Interpreting Tabular/Graphical FRQs | 图表解释类大题

FRQs often present a table of values for f, f’, or f”. You may need to approximate a derivative or integral using difference quotients and Riemann sums. For instance, using a midpoint sum to approximate ∫₀¹⁰ f(x) dx with given four subintervals.

大题中常给出 f、f’ 或 f

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AP Calculus AB & BC: Free-Response Questions Answers and Solutions | AP微积分AB与BC大题答案及解析