AP Calculus Practice with Real Questions and Key Concept Consolidation | AP微积分真题自测与考点巩固

📚 AP Calculus Practice with Real Questions and Key Concept Consolidation | AP微积分真题自测与考点巩固

Welcome to our intensive AP Calculus practice session, where we combine authentic exam-style questions with targeted concept reviews. This article is designed to help you identify weak areas and consolidate your understanding of key topics in both AB and BC calculus. Each section presents a typical problem, its detailed solution, and the underlying ideas you must master.

欢迎参加我们的AP微积分强化练习，我们将真题风格的题目与针对性概念复习相结合。本文旨在帮助你发现薄弱环节，并巩固AB和BC微积分中关键主题的理解。每一节都提供一道典型题目、详细解答以及你必须掌握的核心思想。

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1. Limits and Continuity | 极限与连续性

Question: Evaluate lim(x→3) (x² - 9)/(x - 3). Is the function continuous at x = 3?

题目：求 lim(x→3) (x² − 9)/(x − 3)。该函数在 x = 3 处是否连续？

3) to obtain the expression x + 3. The limit then equals 3 + 3 = 6. Since the limit exists but f(3) is undefined, the function has a removable discontinuity at x = 3.

解答：直接代入得到 0/0，是不确定式。分子因式分解：x² − 9 = (x − 3)(x + 3)。约去公因子（当 x ≠ 3 时成立），得到 x + 3。因此极限等于 3 + 3 = 6。由于极限存在但 f(3) 无定义，该函数在 x = 3 处有可去间断。

. A removable discontinuity occurs when the limit exists but fails to equal the function value (or the function is not defined there). Algebraic simplification often reveals the true limiting behaviour.

核心考点巩固：函数在点 a 处连续必须满足三个条件：f(a) 有定义，lim(x→a) f(x) 存在，且 lim(x→a) f(x) = f(a)。当极限存在但与函数值不等（或函数未定义）时，出现可去间断。代数化简常常能揭示真正的极限行为。

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2. Derivative Definition and Differentiability | 导数的定义与可导性

Question: Use the limit definition to find f'(4) for f(x) = √x. Determine whether f is differentiable at x = 0.

题目：利用极限定义求 f(x) = √x 在 x = 4 处的导数 f'(4)。判断 f 在 x = 0 处是否可导。

Solution: f'(4) = lim(h→0) [√4 + h − √4] / h = lim(h→0) [√4 + h − 2] / h. Multiply numerator and denominator by the conjugate √4 + h + 2: [ (4 + h) − 4 ] / [ h(√4 + h + 2) ] = h / [ h(√4 + h + 2) ] = 1 / [ √4 + h e limit as h → 0 gives 1 / (2 + 2) = 1/4. At x = 0, the limit definition yields lim(h→0) [√0 + h − 0] / h = lim(h→0) 1/√h, which does not exist (infinite), so f is not differentiable at x = 0.

解答：f'(4) = lim(h→0) [√4 + h − √4] / h = lim(h→0) [√4 + h − 2] / h。分子分母同乘以共轭式 √4 + h + 2：[ (4 + h) − 4 ] / [ h(√4 + h + 2) ] = h / [ h(√4 + h + 2) ] = 1 / [ √4 + h + 2 ]。取 h → 0 的极限得 1 / (2 + 2) = 1/4。在 x = 0 处，极限定义给出 lim(h→0) [√0 + h − 0] / h = lim(h→0) 1/√h，该极限不存在（趋于无穷），因此 f 在 x = 0 处不可导。

Key Concept Consolidation: The derivative f'(a) exists if and only if the two-sided limit of the difference quotient exists and is finite. Differentiability implies continuity, but continuity does not guarantee differentiability (e.g., a sharp corner or vertical tangent). Always check the limit from both sides when a function changes definition.

核心考点巩固：导数 f'(a) 存在当且仅当差商的双侧极限存在且有限。可导必然连续，但连续不一定可导（如尖点或垂直切线）。当函数分段定义时，务必检查双侧极限。

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3. Basic Differentiation Rules | 基本求导法则

Question: Differentiate y = 3x⁴ − 5x² + 2eˣ − 7 ln x with respect to x.

题目：对 y = 3x⁴ − 5x² + 2eˣ − 7 ln x 关于 x 求导。

erm by term. The derivative of 3x⁴ is 12x³. For −5x², we obtain −10x. The derivative of 2eˣ is 2eˣ. The derivative of −7 ln x is −7/x. Therefore, dy/dx = 12x³ − 10x + 2eˣ − 7/x.

解答：逐项应用幂函数求导法则、指数函数求导法则和对数函数求导法则。3x⁴ 的导数为 12x³。−5x² 的导数为 −10x。2eˣ 的导数为 2eˣ。−7 ln x 的导数为 −7/x。因此，dy/dx = 12x³ − 10x + 2eˣ − 7/x。

Key Concept Consolidation: Memorise the fundamental rules: d/dx [xⁿ] = n xⁿ⁻¹, d/dx [eˣ] = eˣ akes like misapplying the power rule or forgetting the chain rule when it is needed later.

核心考点巩固：熟记基本公式：d/dx [xⁿ] = n xⁿ⁻¹，d/dx [eˣ] = eˣ，d/dx [ln x] = 1/x。对于和与常数倍，逐项求导。这些法则是AP微积分的基石。小心常见错误，如幂法则用错，或后续需要链式法则时忘记使用。

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4. Implicit Differentiation | 隐函数微分

y² = 7, find dy/dx in terms of x and y. Then find the slope of the tangent line at (1, 2).

题目：已知曲线 x² + xy + y² = 7，用 x 和 y 表示 dy/dx。然后求点 (1, 2) 处切线的斜率。

y/dx). For y², use the chain rule: 2y·(dy/dx). The right side derivative is 0. Thus 2x + y + x dy/dx + 2y dy/dx = 0. Collect dy/dx terms: (x + 2y) dy/dx = −(2x + y). Hence dy/dx = −(2x + y)/(x + 2y). At (1,2), plug in to get −(2·1 + 2)/(1 + 4) = −4/5.

解答：方程两边关于 x 求导，注意 y 是 x 的函数。对 x² 得 2x。对 xy 使用乘积法则：(1)·y + x·(dy/dx)。对 y² 使用链式法则：2y·(dy/dx)。右边导数为 0。因此 2x + y + x dy/dx + 2y dy/dx = 0。合并 dy/dx 项：(x + 2y) dy/dx = −(2x + y)。故 dy/dx = −(2x + y)/(x + 2y)。在点 (1,2) 处代入得 −(2·1 + 2)/(1 + 4) = −4/5。

Key Concept Consolidation: Implicit differentiation is essential when you cannot easily solve for y. Always append dy/dx whenever you differentiate a function of y. After finding the derivative expression, avoid simplifying until you substitute any given point to reduce errors. Practise checking your result with explicit solutions when possible.

核心考点巩固：当无法方便解出 y 时，隐函数微分必不可少。每当对 y 的函数求导时，务必乘上 dy/dx。求出导数表达式后，若非必要不要过早化简，代入给定点可减少错误。在可能时用显式解检验你的结果。

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5. Related Rates | 相关变化率

Question: A spherical balloon is being inflated at a rate of 10 cm³/s. How fast is the surface area increasing when the radius is 5 cm?

题目：一个球形气球以 10 cm³/s 的速率充气。当半径为 5 cm 时，表面积增加得有多快？

/dt). Plug r = 5: 10 = 4π(25) (dr/dt) → dr/dt = 10 / (100π) = 1/(10π) cm/s. Next, dS/dt = 8πr (dr/dt). Substitute r = 5 and dr/dt: dS/dt = 8π·5·(1/(10π)) = 40π/(10π) = 4 cm²/s.

解答：设 V 为体积，S 为表面积。已知 dV/dt = 10。公式：V = (4/3)πr³，S = 4πr²。两边对 t 求导：dV/dt = 4πr² (dr/dt)。代入 r = 5：10 = 4π(25) (dr/dt) → dr/dt = 10 / (100π) = 1/(10π) cm/s。接着，dS/dt = 8πr (dr/dt)。代入 r = 5 和 dr/dt：dS/dt = 8π·5·(1/(10π)) = 40π/(10π) = 4 cm²/s。

Key Concept Consolidation: Related rates problems require you to identify an equation linking the variables, differentiate with respect to time using the chain rule, and then substitute known values (only after differentiation). Always check units and ensure the sign of the rate matches the context (increasing vs decreasing). A common mistake is plugging in constants before differentiating.

核心考点巩固：相关变化率问题需要你找到联系变量的方程，利用链式法则关于时间求导，然后代入已知数值（只能在求导之后代入）。务必检查单位并确保变化率的正负号与情境相符（增加还是减少）。常见错误是在求导前就代入常数。

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6. Mean Value Theorem & L'Hopital's Rule | 中值定理与洛必达法则

Question: (a) State the Mean Value Theorem. (b) Use L'Hopital's Rule to evaluate lim(x→0) (eˣ − 1 − x) / x².

题目：(a) 叙述中值定理。(b) 利用洛必达法则求 lim(x→0) (eˣ − 1 − x) / x²。

continuous on [a, b] and differentiable on (a, b), there exists a c in (a, b) such that f' − f(a)) / (b − a). (b) As x → 0, both numerator and denominator approach 0, so we apply L'Hopital. Differentiate numerator: eˣ − 1; denominator: 2x. The new limit lim(x→0) (eˣ − 1)/(2x) is still 0/0. Apply L'Hopital again: derivative of numerator is eˣ, denominator is 2. Now limit = 1/2. Thus, lim(x→0) (eˣ − 1 − x) / x² = 1/2.

解答：(a) 若 f 在 [a, b] 上连续，在 (a, b) 内可导，则存在 c ∈ (a, b) 使得 f'(c) = (f(b) − f(a)) / (b − a)。(b) 当 x → 0 时，分子与分母均趋于 0，故使用洛必达法则。分子求导：eˣ − 1；分母求导：2x。新的极限 lim(x→0) (eˣ − 1)/(2x) 仍为 0/0 型。再次使用洛必达法则：分子导数 eˣ，分母导数 2。此时极限为 1/2。因此，lim(x→0) (eˣ − 1 − x) / x² = 1/2。

Key Concept Consolidation: The MVT links average and instantaneous rates of change. L'Hopital' he conditions before differentiating. If repeated application is needed, keep differentiating until an indeterminate form is resolved. For limits that are not initially indeterminate, algebraic manipulation may be required.

核心考点巩固：中值定理将平均变化率与瞬时变化率联系起来。洛必达法则仅适用于 0/0 或 ∞/∞ 型的不确定式。求导前务必验证条件。若需多次使用，则持续求导直到不确定式消失。对于初看并非不确定式的极限，可能需要代数变形。

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7. Curve Sketching with Derivatives | 函数图像与导数应用

critical points, intervals of increase/decrease, inflection points, and intervals of concavity. Sketch a rough graph.

题目：对于 f(x) = x³ − 3x²，求出所有临界点、增减区间、拐点以及凹凸区间。画出草图。

Solution: First derivative: f'(x) = 3x² − 6x = 3x(x − 2). Critical points where f'(x) = 0: x = 0, x = 2. Test sign of f': for x < 0, f' > 0 (increasing); 0 < x < 2, f' < 0 (decreasing); x > 2, f' > 0 (increasing). Second derivative: f''(x) = 6x − 6 = 6(x − 1). Inflection point at x = 1 (f'' changes sign). Concave down for x < 1 (f'' < 0), concave up for x > 1 (f'' > 0). Local max at (0,0), local min at (2, −4), inflection point at (1, −2).

解答：一阶导数：f'(x) = 3x² − 6x = 3x(x − 2)。令 f'(x) = 0 得临界点 x = 0, x =

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