AP FRQ Practice

AP-Calc-BC-FRQ #1 · 9 pts

Calculus BC 积分 / 面积 / 体积综合应用题

Let R be the region in the first quadrant bounded by the graphs of y = ln(x + 1) and y = 2 - x, and the y-axis, as shown in the figure below.

Figure Context: The two curves intersect at some point x = a where 0 < a < 2. Verify: ln(a+1) = 2 - a ⇒ a ≈ 1.278 (numerical only — do not solve algebraically).

(a) [2 pts] Find the area of region R. Show the integral expression and evaluate numerically.

Sub (a):

(b) [3 pts] Region R is the base of a solid. For this solid, each cross-section perpendicular to the x-axis is a semicircle with diameter extending between the two curves. Find the volume of this solid.

Sub (b):

(c) [4 pts] Write, but do NOT evaluate, an integral expression for the volume of the solid generated when R is rotated about the horizontal line y = 3. Then, using the method of cylindrical shells, also set up a volume integral for rotation about the y-axis.

Sub (c):

✅ 参考解答 & 分步计分 Rubric（点击展开）

(a) Area — 2 points:

1 pt Correct integrand & limits: A = ∫₀ᵃ [ (2 − x) − ln(x + 1) ] dx where a ≈ 1.278

1 pt Numerical answer: ≈ 0.445 (accept 0.44 – 0.45)

(b) Volume, semicircular cross-section — 3 points:

1 pt Diameter = (2 − x) − ln(x+1), ⇒ radius = ½·[difference]

1 pt Area(semicircle) = ½πr² = ⅛π·[(2−x)−ln(x+1)]² ⇒ V = ⅛π ∫₀ᵃ [(2−x)−ln(x+1)]² dx

1 pt Numerical evaluation: ≈ 0.058π ≈ 0.182 (accept ±0.01)

(c) Washer + Shells — 4 points:

2 pts Rotation about y = 3 (washer): V = π ∫₀ᵃ [ (3 − ln(x+1))² − (3 − (2 − x))² ] dx = π ∫₀ᵃ [ (3 − ln(x+1))² − (1 + x)² ] dx. (Limits 1; integrand 1)

2 pts Shells about y-axis: V = 2π ∫₀ᵃ x·[(2 − x) − ln(x+1)] dx. (2π & x factor 1; height factor 1)

⚠️ 注意：要求"do NOT evaluate"，写出正确积分式即可得全分；数值积分错误但积分式正确不扣分。

FRQ #1 总分（0-9）：

AP-Calc-BC-FRQ #2 · 9 pts

Calculus BC 微分方程 / 斜率场 / 级数

Consider the differential equation dy/dx = (1 + y) · cos(πx).

(a) [3 pts] On the axes provided, sketch a slope field for the given differential equation at the six points indicated: (0, 0), (0, 1), (1, 0), (1, 1), (0.5, 0), (0.5, 1). Then describe all points (x, y) in the xy-plane where the solution curves have horizontal tangent lines.

Sub (a):

(b) [3 pts] Find the particular solution y = f(x) to the differential equation with the initial condition f(0) = 1. State the domain of this solution.

Sub (b):

(c) [3 pts] The function g satisfies the same differential equation with g(0) = 0. Write the first four non-zero terms of the Maclaurin series for g(x). Then use this polynomial to approximate g(0.5).

Sub (c):

✅ 参考解答 & 分步计分 Rubric（点击展开）

(a) Slope field + horizontal tangents — 3 points:

1 pt At least 4 of 6 slopes correct (verify: (0,0)=1, (0,1)=2, (1,0)=-1, (1,1)=-2, (0.5,0)=0, (0.5,1)=0)

1 pt All 6 slopes drawn with appropriate orientation — positive/negative/zero

1 pt dy/dx = 0 ⇒ (1+y)·cos(πx)=0 ⇒ y = −1 OR x = ½ + k, k ∈ ℤ. Describe as: "y = −1 at all x, or x = ½ + any integer".

(b) Particular solution — 3 points:

1 pt Separation: dy/(1+y) = cos(πx) dx, integration attempt

1 pt Correct antiderivatives: ln|1+y| = (1/π) sin(πx) + C; apply IC: ln(2) = C ⇒ y = 2e^((1/π)sin(πx)) − 1

1 pt Domain: (−∞, ∞) — (1+y) never zero because 2e^(bounded) − 1 ≥ 2e^(−1/π) − 1 > 0, so no singularity.

(c) Maclaurin series — 3 points:

1 pt g(0)=0, g'(0)=1 ⇒ first term x; g''=π(1+y)(−sin πx) + y'·cos(πx)... = 0 at 0; g'''(0)= −π²; fourth derivative = 0; fifth non-zero. Accept: x − (π²/6)x³ + (π⁴/120)x⁵ − ...

1 pt First four non-zero: x − (π²/6)x³ + (π⁴/120)x⁵ − (π⁶/5040)x⁷ (or equivalent coefficients)

1 pt Approx g(0.5) ≈ 0.5 − (π²/6)(0.125) = 0.5 − (9.8696/6)(0.125) ≈ 0.5 − 0.2056 ≈ 0.294 (accept 0.28 – 0.31; adding further terms ≈ 0.302)

FRQ #2 总分（0-9）：

AP-Stat-FRQ #1 · 4 pts (Investigative Task)

Statistics 假设检验 / 置信区间

A cereal manufacturer claims that its boxes contain a mean of 500 grams of cereal, with a population standard deviation of 8 grams. A consumer advocacy group suspects the actual mean is less than advertised, and purchases a random sample of 36 boxes. The sample mean weight is 497.2 grams.

(a) [1 pt] Construct and interpret a 95% confidence interval for the true population mean weight of cereal in the boxes.

Sub (a):

(b) [1 pt] Based only on your confidence interval from (a), is there convincing evidence the true mean is less than 500 grams? Justify your answer.

Sub (b):

(c) [2 pts] Perform a significance test at the α = 0.05 level to investigate the group's suspicion that the mean is less than 500 grams. Include the hypotheses, test statistic, p-value, and conclusion in context.

Sub (c):

✅ 参考解答 & 分步计分 Rubric（点击展开）

(a) 95% CI — 1 point (Essentially Correct):

✓ Conditions: (1) random sample stated; (2) n=36 ≥ 30 ⇒ CLT applies, sampling dist of x̄ approx Normal; (3) σ=8 known ⇒ one-sample z-interval for μ

✓ Calculation: x̄ ± z*·σ/√n = 497.2 ± 1.96·(8/6) = 497.2 ± 2.613 ⇒ (494.59, 499.81) grams

✓ Interpretation: "We are 95% confident that the interval from 494.59 g to 499.81 g captures the true mean weight of cereal in all such boxes." (Context required)

(b) CI-based decision — 1 point:

✓ Yes, there is convincing evidence at α = 0.05. Reason: the entire 95% CI lies below 500 grams — the hypothesized value 500 is not in the interval. ⇒ We reject H₀: μ = 500 in favor of μ < 500 (two-sided 5% ⇔ one-sided 2.5% logic still supports; link clearly to interval).

(c) Significance test — 2 points:

1 pt H₀: μ = 500 g; Hₐ: μ < 500 g (define μ = true mean weight). Test stat: z = (x̄ − μ₀)/(σ/√n) = (497.2 − 500)/(8/6) = −2.1 (correct formula & value)

1 pt p-value = P(Z < −2.1) ≈ 0.0179. Since 0.0179 < 0.05, we reject H₀. There is convincing evidence at the α = 0.05 level that the true mean cereal weight is less than 500 g. (context required for credit)

FRQ #1 总分（0-4）：

AP-Stat-FRQ #2 · 4 pts

Statistics 回归分析 / 概率分布

Researchers collected data on 50 high school seniors to investigate the relationship between hours of study per week (x, in hours) and college entrance exam score (y, on a 400–1600 scale). Computer output for a least-squares regression is shown below.

Predictor	Coef	SE Coef	T	P
Constant	892.50	38.20	23.37	0.000
Hours (x)	14.80	3.10	4.77	0.000

s = 82.3 R-sq = 32.1% R-sq(adj) = 30.7%

(a) [1 pt] Write the equation of the least-squares regression line. Define any variables you use. Interpret the slope of the regression line in context.

Sub (a):

(b) [1 pt] One student in the sample studied 20 hours per week and scored 1250. Calculate and interpret the residual for this student.

Sub (b):

(c) [2 pts] Assume exam scores follow a normal distribution and the residual standard deviation s = 82.3 is a reasonable estimate of σ. For a student who studies 15 hours per week: (i) predict the exam score; (ii) find the probability that this student scores 1050 or higher. Show your work.

Sub (c):

✅ 参考解答 & 分步计分 Rubric（点击展开）

(a) LSRL equation + slope interpretation — 1 point:

✓ Equation: ŷ = 892.50 + 14.80x, where ŷ = predicted entrance exam score, and x = weekly hours of study. (MUST use ŷ or "predicted y" — writing y loses credit.)

✓ Slope: For each additional hour of weekly study, the predicted entrance exam score increases by approximately 14.8 points, on average. (Context: "predicted," "on average", score units required.)

(b) Residual — 1 point:

✓ Predicted: ŷ = 892.50 + 14.80·(20) = 892.50 + 296.0 = 1188.5

✓ Residual = y − ŷ = 1250 − 1188.5 = +61.5. Interpretation: This student scored 61.5 points higher than the score predicted by the regression model for a student studying 20 hours/week.

(c) Prediction + probability — 2 points:

1 pt (i) ŷ = 892.50 + 14.80·15 = 892.50 + 222 = 1114.5 predicted score.

1 pt (ii) Y | x = 15 ~ Normal(μ = 1114.5, σ = 82.3). z = (1050 − 1114.5)/82.3 ≈ −0.7837. P(Y ≥ 1050) = P(Z ≥ −0.78) ≈ 0.782, or more precisely ≈ 0.783. (Correct z, correct upper tail direction, correct answer ≈ 0.78 to 0.785; wrong direction or value ≈ 0.217 no credit.)

FRQ #2 总分（0-4）：

AP-PhysC-Mech-FRQ #1 · 15 pts

Physics C 牛顿定律 / 能量 / 动量综合

A block of mass m₁ = 3.0 kg is placed on a rough incline plane angled at θ = 30° above the horizontal. It is connected by a light string over a massless, frictionless pulley to a hanging block of mass m₂ = 2.0 kg. The coefficient of kinetic friction between m₁ and the incline is μₖ = 0.25. The system is released from rest.

Use g = 10 m/s² for simplicity; show all work including free-body diagrams described in words.

(a) [5 pts] On the figures below, draw and label all forces acting on each block. Then calculate the magnitude of the acceleration of the two-block system and the tension in the string.

Sub (a):

(b) [5 pts] Starting from rest, m₂ falls a distance h = 1.5 m. Using energy methods, calculate the speed of the blocks at this point. Then calculate the work done by friction on m₁ during this displacement.

Sub (b):

(c) [5 pts] When m₂ has fallen 1.5 m, the string snaps. Immediately after the string breaks, m₂ continues falling and strikes the floor 0.3 s later, but m₁ slides farther up the ramp (still with friction) before coming to rest. Find the extra distance along the incline that m₁ travels after the string breaks. (You may use speed from part (b); if you couldn't solve it, assume v = 2.0 m/s for partial credit.)

Sub (c):

✅ 参考解答 & 分步计分 Rubric（点击展开）

(a) FBD + Newton's 2nd Law — 5 points:

2 pts FBDs: m₁ has (m₁g↓, N⊥plane, fₖ ↓plane, T ↑plane). m₂ has (m₂g↓, T↑). Correct directions & labels; no extras.

2 pts Equations: m₂g − T = m₂a; T − m₁g·sinθ − μₖ·m₁g·cosθ = m₁a. Add: m₂g − m₁g(sinθ + μₖcosθ) = (m₁+m₂)a.

1 pt Plug in g=10: LHS = 20 − 30(0.5 + 0.25·(√3/2)) = 20 − 30(0.5 + 0.2165) = 20 − 21.495 ≈ −1.495 N. Wait: system accelerates down incline? Actually m₁ down-plane = 15 N + friction ≈ +6.5 N ⇒ T must be 21.5 N, vs m₂g=20 N ⇒ m₁ slides down the plane, m₂ rises. ⇒ Flip signs: a ≈ −0.30 m/s² (meaning system accelerates with m₁ moving down). Accept a ≈ 0.30 m/s² with stated direction; then T ≈ 20 − 2(0.3) ≈ 19.4 N. Rubric: correct algebra & direction; arithmetic sign error minor.

(b) Energy + work by friction — 5 points:

1 pt Correct sign ΔU: m₂ rises h if m₁ goes down, else falls. (From part (a), m₁ moves down, so m₂ rises h: ΔU_grav = +m₂gh − m₁g·h·sinθ = +30 − 22.5 = +7.5 J.)

2 pts W_nonconservative = −fₖ·d along plane. d_along = h/sinθ = 1.5/0.5 = 3.0 m. fₖ = 0.25·30·(√3/2) ≈ 6.495 N. ⇒ W_f = −6.495·3 ≈ −19.49 J.

2 pts W_net = ΔKE ⇒ −7.5 − 19.49 = ½·5·v² ⇒ Wait actually conservation: Initial energy = final: 0 = ΔU + W_f + ½(m₁+m₂)v² ⇒ v ≈ √(2·|11.99|/5) ≈ 2.19 m/s. If using the a from (a) · d along = 0.30·3 ⇒ v²=1.8 ⇒ v≈1.34 m/s, accept based on consistent signs.

(c) Extra distance after string breaks — 5 points:

2 pts Correct deceleration: after string breaks, net force on m₁ along plane = −m₁g·sinθ − μₖm₁g·cosθ (both oppose up-plane motion) ⇒ a' = −g(sinθ + μₖ cosθ) ≈ −10·(0.5 + 0.2165) ≈ −7.165 m/s².

2 pts Kinematics: 0 = v² + 2a'd ⇒ d = v² / (2g(sinθ+μₖcosθ)). With v ≈ 2.0 m/s: d ≈ 4.0 / (14.33) ≈ 0.28 m. With v ≈ 2.19: d ≈ 0.33 m. Accept 0.27 – 0.35 m.

1 pt Clear reasoning: use v² = u² + 2as OR energy ½mv² = (m₁g sinθ + fₖ)·d with consistent variables.

FRQ #1 总分（0-15）：

AP-PhysC-Mech-FRQ #2 · 15 pts

Physics C 转动 / 平动综合（Rolling + Conservation）

A uniform solid sphere of mass M = 0.80 kg and radius R = 0.05 m is released from rest at the top of an incline of height H = 2.0 m. The sphere rolls without slipping to the bottom. (Moment of inertia of a solid sphere about its center: I = ⅖MR².)

(a) [4 pts] Using energy conservation, derive an expression for the translational speed v_cm of the sphere at the bottom of the incline in terms of g, H, and any necessary constants. Then compute its numerical value. Use g = 10 m/s².

Sub (a):

(b) [4 pts] At the bottom, the sphere rolls onto a horizontal, frictionless track and collides elastically with a block of mass m = 1.2 kg that is initially at rest. Because the track is frictionless, the sphere continues to rotate with the same angular velocity as before the collision, but its translational motion changes. Find the translational velocity of both the sphere and the block immediately after the collision (linear elastic collision; ignore rotation for the collision calculation, as no friction ⇒ no torque ⇒ ω unchanged).

Sub (b):

(c) [3 pts] Immediately after the collision, what is the magnitude and direction of the total angular momentum of the sphere about the collision point (the point on the track where the collision took place)? Answer symbolically first, then evaluate numerically. Is angular momentum conserved about this point during the collision? Briefly justify.

Sub (c):

(d) [4 pts] After leaving the frictionless track, the sphere moves onto a long carpet where the coefficient of kinetic friction is μₖ = 0.30. The sphere is now translating backward (relative to positive) but spinning as before, so friction acts until pure rolling is re-established. Find the time and the distance the sphere travels on the carpet before it again rolls without slipping, and find its final center-of-mass velocity.

Sub (d):

✅ 参考解答 & 分步计分 Rubric（点击展开）

(a) Rolling without slipping, energy conservation — 4 points:

1 pt Correct energies: MgH = ½Mv_cm² + ½Iω² (potential → translational + rotational KE; no slipping ⇒ ω = v/R)

1 pt Substitute I = ⅖MR²: MgH = ½Mv² + ½(⅖MR²)(v²/R²) = ½Mv² + ⅕Mv² = (7/10)Mv². Cancel M.

1 pt Algebra: v = √(10gH/7) (symbolic result)

1 pt Numerics: v = √(10·10·2 / 7) = √(200/7) ≈ √28.571 ≈ 5.345 m/s (accept 5.3 – 5.4 m/s)

(b) Elastic translational collision — 4 points:

1 pt Correctly apply linear elastic equations: (1) momentum Mv₀ = Mv₁ + mv₂, (2) v₂ − v₁ = v₀ (relative speed preserved).

1 pt Correct formulas: v₁ = (M−m)/(M+m)·v₀, v₂ = 2M/(M+m)·v₀.

1 pt v₁ = (−0.4/2.0)·5.345 ≈ −1.07 m/s (sphere rebounds, negative ⇒ backward toward incline).

1 pt v₂ = (1.6/2.0)·5.345 ≈ +4.28 m/s (block forward). Units required.

(c) Angular momentum — 3 points:

1 pt L_total = L_spin + L_orbital = Iω₀ + M·v₁·R (perpendicular, since point on ground directly below ⇒ r=R). ω₀ = v₀/R ⇒ L = ⅖MR²·(v₀/R) + Mv₁R = MR(⅖v₀ + v₁) = MR(0.4·5.345 − 1.069) = MR(2.138 − 1.069) = MR·1.069 ≈ 0.8·0.05·1.069 ≈ 0.0428 kg·m²/s.

1 pt Direction: into page (right-hand rule: spin forward rolling + after slight correction from rebounding sign; accept consistent direction reasoning).

1 pt Angular momentum IS conserved about the collision point because the only impulsive force (normal/contact between block & sphere) acts at or through this point, so its torque is zero (lever arm zero). Weight and normal also cancel/equal lever.

(d) Re-establishing pure rolling on carpet — 4 points:

1 pt a = +μₖg = 3.0 m/s² (forward), α = τ/I = (−μₖMg·R)/(⅖MR²) = −(5/2)(μₖg/R) = −2.5·(3/0.05) = −150 rad/s². (Opposes spin.)

1 pt Pure rolling condition: v_cm = ω·R. v(t) = v₁ + at; ω(t) = ω₀ + αt; ω₀ = v₀/R ≈ 106.9 rad/s.

1 pt Solve: v₁ + at = (ω₀ + αt)·R ⇒ −1.069 + 3t = 5.345 − 7.5t ⇒ 10.5t = 6.414 ⇒ t ≈ 0.611 s (0.60 – 0.62 acceptable). v_final = −1.069 + 3·0.611 ≈ +0.764 m/s (forward direction).

1 pt Distance: d = v₁·t + ½at² ≈ −1.069·0.611 + ½·3·(0.611)² ≈ −0.653 + 0.560 ≈ −0.093 m (about 9.3 cm backward displacement before forward rolling sets in). OR 0.08 – 0.11 m accept.

FRQ #2 总分（0-15）：

AP FRQ Practice v1

📐 AP Calculus BC · Free Response Questions

📊 AP Statistics · Free Response Questions

⚙️ AP Physics C: Mechanics · Free Response Questions

📋 答题历史记录（AP 全科目）