AP Prep — Calculus BC Practice

Question 1 of 10

MCQU7Topic 7.5Medium No calc

Let y = f(x) be the particular solution to the differential equation dy/dx = x + 2y with f(0) = 1. Euler's method, starting at x = 0 with a step size of Δx = 0.5, is used to approximate f(1). What is the resulting approximation?

A4.25

C9.5

D3.25

Explanation

Euler's method builds each new y-value from the slope at the point you are LEAVING: yₙ₊₁ = yₙ + (Δx)·(dy/dx evaluated at (xₙ, yₙ)).
Step 1 (from (0, 1)): dy/dx = 0 + 2(1) = 2, so y ≈ 1 + (0.5)(2) = 2 at x = 0.5.
Step 2 (from (0.5, 2)): dy/dx = 0.5 + 2(2) = 4.5, so y ≈ 2 + (0.5)(4.5) = 4.25 at x = 1.
The value 9.5 comes from adding the slope itself at each step instead of the slope times Δx — Euler's increment is Δy = (slope)(Δx), and dropping the 0.5 factor doubles every increment. The value 3.25 comes from evaluating the second slope at the ORIGINAL y-value, 0.5 + 2(1) = 2.5, giving 2 + (0.5)(2.5) = 3.25; each step must use the most recently updated y, not y₀. The value 2 is only the intermediate value at x = 0.5, not the value at x = 1.

Question 2 of 10

MCQU1Topic 1.15Medium No calc

What is lim(x→∞) (√(4x² + 12x + 1) − 2x)?

CThe limit is ∞.

Explanation

This is the indeterminate form ∞ − ∞, so rationalize by multiplying by the conjugate (√(4x² + 12x + 1) + 2x) over itself. The numerator becomes (4x² + 12x + 1) − 4x² = 12x + 1, giving (12x + 1)/(√(4x² + 12x + 1) + 2x). Dividing numerator and denominator by x (with x > 0, so x = √(x²)) yields (12 + 1/x)/(√(4 + 12/x + 1/x²) + 2) → 12/(2 + 2) = 3. Answering 0 comes from replacing √(4x² + 12x + 1) with √(4x²) = 2x and cancelling, which illegally discards the 12x term that survives in the ∞ − ∞ form. Answering 6 uses only the 2x in the conjugate denominator (12/2) instead of the full limit 2 + 2 = 4.

Question 3 of 10

MCQU2Topic 2.2Medium No calc

Let L = lim(h→0) [ (2+h)⁵·ln(2+h) − 32·ln 2 ] / h. The value of L is the derivative of a function at a point. What is the value of L?

A40

B16 ln 2 + 16

C80 ln 2

D80 ln 2 + 16

Explanation

The limit has the form lim(h→0) [f(2+h) − f(2)]/h with f(x) = x⁵·ln x, since f(2) = 2⁵·ln 2 = 32 ln 2. So L = f′(2). By the product rule, f′(x) = 5x⁴·ln x + x⁵·(1/x) = 5x⁴ ln x + x⁴. At x = 2: 5·16·ln 2 + 16 = 80 ln 2 + 16. The option 80 ln 2 comes from differentiating only the power factor and treating ln x as a constant — the product rule requires the second term x⁵·(1/x). The option 16 ln 2 + 16 keeps both product-rule terms but drops the factor 5 that the power rule produces on x⁵. The option 40 is the product-rule-as-f′g′ error: 5x⁴·(1/x) = 5x³ = 40 at x = 2.

Question 4 of 10

MCQU9Topic 9.3Medium No calc Diagram

The parametric curve x(t) = 3t², y(t) = 2t³ is traced for 0 ≤ t ≤ 2. What is the length of this curve?

A10√5

B10√5 − 2

C28

D4(5√5 − 1)

Explanation

Arc length of a parametric curve is L = ∫ₐᵇ √((dx/dt)² + (dy/dt)²) dt. Here dx/dt = 6t and dy/dt = 6t², so the integrand is √(36t² + 36t⁴) = 6t√(1 + t²) for t ≥ 0. Substituting u = 1 + t² gives du = 2t dt, so 6t dt = 3 du and the integral becomes ∫ 3√u du = 2u^(3/2) = 2(1 + t²)^(3/2). Evaluating from 0 to 2 gives 2(5^(3/2)) − 2(1) = 10√5 − 2. Forgetting to divide by the inner coefficient in the u-substitution (using 6·(2/3)u^(3/2) = 4u^(3/2)) produces 4(5√5 − 1), exactly double the correct value. The value 28 comes from treating √(36t² + 36t⁴) as 6t + 6t², i.e. assuming the square root of a sum equals the sum of the square roots — it never does. The value 10√5 comes from evaluating the antiderivative only at the upper limit and omitting the lower limit.

Question 5 of 10

MCQU5Topic 5.1Medium Calc

Let f(x) = e^2x on the closed interval [0, ln 2]. The Mean Value Theorem guarantees a value c in (0, ln 2) with f′(c) equal to the average rate of change of f on the interval. To three decimal places, c ≈

A0.347

B0.386

C0.733

D0.203

Explanation

f is continuous and differentiable everywhere, so the MVT applies. The average rate of change is (f(ln 2) − f(0))/(ln 2 − 0) = (4 − 1)/ln 2 = 3/ln 2 ≈ 4.3281. Since f′(x) = 2e^(2x), set 2e^(2c) = 3/ln 2, so e^(2c) = 3/(2 ln 2) ≈ 2.1640 and c = ½·ln(2.1640) ≈ 0.386. A common error is to use f(ln 2) − f(0) = 3 as the required slope, forgetting to divide by the interval length b − a; that gives 2e^(2c) = 3 and c = ½ ln(1.5) ≈ 0.203. Another error solves e^(2c) = 3/ln 2 (dropping the factor 2 from the chain rule), giving ≈ 0.733, and the midpoint (ln 2)/2 ≈ 0.347 is not what the MVT produces for a non-quadratic function.

Question 6 of 10

MCQU8Topic 8.4Easy No calc Diagram

What is the area of the region enclosed by the parabola y = x² and the line y = 3x + 10?

A125/2

B343/6

C203/2

D275/6

Explanation

Set x² = 3x + 10, so x² − 3x − 10 = (x − 5)(x + 2) = 0 and the curves meet at x = −2 and x = 5. On (−2, 5) the line lies above the parabola, so Area = ∫₋₂⁵ (3x + 10 − x²) dx = [3x²/2 + 10x − x³/3]₋₂⁵ = (75/2 + 50 − 125/3) − (6 − 20 + 8/3) = 343/6. Using 0 as the left limit instead of the negative intersection gives 275/6; integrating only the line, forgetting to subtract the parabola, gives 203/2; mishandling the sign of x³/3 at x = −2 (using 125/3 − 8/3 instead of 125/3 + 8/3) gives 125/2.

Question 7 of 10

MCQU10Topic 10.2Medium No calc

What is the sum of the infinite series ∑(n=2 to ∞) 5·(2/5)ⁿ ?

A4/7

B25/3

C10/3

D4/3

Explanation

A geometric series converges when |r| < 1, and the sum of a convergent geometric series equals (first term)/(1 − r). Here r = 2/5, so |r| < 1 and the series converges. The indexing starts at n = 2, so the first term is not 5 but 5·(2/5)² = 5·(4/25) = 4/5. The sum is therefore (4/5)/(1 − 2/5) = (4/5)/(3/5) = 4/3. The value 25/3 comes from applying a/(1 − r) with a = 5, which is the first term only if the sum began at n = 0; that series has two extra terms (5 and 2) that are not present here. Likewise 10/3 uses the n = 1 first term 2. The value 4/7 comes from dividing by 1 + r instead of 1 − r.

Question 8 of 10

MCQU6Topic 6.5Medium No calc Word Diagram

The graph of a continuous function f on the interval −2 ≤ t ≤ 4 consists of two line segments: one from the point (−2, 0) to the point (1, 3), and a second from (1, 3) to (4, −3). Define g(x) = ∫₋₂ˣ f(t) dt. What is the value of g(4)?

C2.25

D4.5

Explanation

g(4) is the NET signed area under f from t = −2 to t = 4: area above the t-axis counts positive, area below counts negative.

First segment, from (−2, 0) to (1, 3): a triangle above the axis with base 3 and height 3, giving area (1/2)(3)(3) = 4.5.

Second segment, from (1, 3) to (4, −3): this line has slope −2 and crosses the t-axis at t = 2.5. From t = 1 to t = 2.5 it forms a triangle above the axis with base 1.5 and height 3, area (1/2)(1.5)(3) = 2.25. From t = 2.5 to t = 4 it forms a triangle below the axis with base 1.5 and height 3, contributing −2.25.

Net: 4.5 + 2.25 − 2.25 = 4.5.

The value 9 is the TOTAL area (4.5 + 2.25 + 2.25), which is ∫|f| — that ignores the sign of the region below the axis. The value 2.25 uses only the second segment's positive piece, and 0 comes from noticing the second segment's two pieces cancel but forgetting the first segment entirely.

Question 9 of 10

MCQU3Topic 3.1Medium No calc

Let f(x) = e^cos(4x). What is f′(π/8)?

B−4

D−1

Explanation

Chain rule: for f(x) = e^(u(x)), f′(x) = e^(u(x))·u′(x). Here u(x) = cos(4x), so u′(x) = −4sin(4x) (the inner factor 4 comes from a second application of the chain rule). Thus f′(x) = −4sin(4x)·e^(cos(4x)). At x = π/8, 4x = π/2, so sin(4x) = 1 and cos(4x) = 0, giving f′(π/8) = −4·1·e⁰ = −4. The value −1 is what you get if you differentiate cos(4x) as −sin(4x) and drop the inner factor 4. The value 0 comes from misusing the power rule on eᵘ (writing u·e^(u−1)·u′), which vanishes because cos(π/2) = 0; the base e is a constant, so the power rule does not apply.

Question 10 of 10

MCQU4Topic 4.2Medium No calc Word

A particle moves along a straight line with velocity v(t) = (t² − 5t + 4)e^−t/3 meters per second for t ≥ 0. What is the acceleration of the particle at t = 2 seconds?

Ae^−2/3 ≈ 0.513 m/s²

B−(5/3)e^−2/3 ≈ −0.856 m/s²

C(1/3)e^−2/3 ≈ 0.171 m/s²

D−(1/3)e^−2/3 ≈ −0.171 m/s²

Explanation

Acceleration is a(t) = v′(t), which needs the product rule together with the chain rule. With f = t² − 5t + 4 and g = e^(−t/3): f′ = 2t − 5 and g′ = e^(−t/3)·(−1/3). So v′ = (2t − 5)e^(−t/3) − (1/3)(t² − 5t + 4)e^(−t/3) = e^(−t/3)[2t − 5 − (t² − 5t + 4)/3]. At t = 2 the bracket is (−1) − (−2)/3 = −1/3, giving a(2) = −(1/3)e^(−2/3). Multiplying only the two derivatives, f′·g′ = (2t − 5)·(−1/3)e^(−t/3), gives (−1)(−1/3)e^(−2/3) = +(1/3)e^(−2/3) — the product rule is f′g + fg′, not f′g′, and here the mistake even flips the sign. Writing g′ = −e^(−t/3) drops the chain-rule inner factor −1/3 and leaves a bracket of −1 + 2 = 1. Adding rather than subtracting the second product-rule term gives a bracket of −1 − 2/3 = −5/3.

Free Response 1 · Section II

FRQRate and AccumulationU4 Calc

An algae bioreactor is monitored for 24 hours. During the interval 0 ≤ t ≤ 24, where t is measured in hours, new algae grows into the culture at a rate modeled by

G(t) = 9t·e^−t/6 kilograms per hour,

and algae is drawn off by a continuous harvester at a rate modeled by

H(t) = 1.5·ln(1 + t) kilograms per hour.

At time t = 0 the reactor holds 40 kilograms of algae. Let M(t) be the mass of algae, in kilograms, in the reactor at time t hours.

(a) Find the total mass of algae produced by growth during the first 6 hours. Show the integral you evaluate, give the exact value, and give the value rounded to three decimal places, with units.

(b) Find M′(4). Using appropriate units, interpret the meaning of M′(4) in the context of this problem.

(d) For 0 < t < 24, find the time t at which the mass of algae in the reactor is a maximum. Justify your answer.

Free response is self-scored — work it out, then reveal the model answer and scoring checklist to compare.

Model answer

Setup: M′(t) = G(t) − H(t) = 9t·e^−t/6 − 1.5·ln(1 + t), with M(0) = 40 kg.

(a) Total growth = ∫₀⁶ 9t·e^−t/6 dt. Integration by parts with u = t, dv = e^−t/6 dt ⇒ du = dt, v = −6e^−t/6: ∫ 9t·e^−t/6 dt = 9[ −6t·e^−t/6 + ∫ 6e^−t/6 dt ] = 9[ −6t·e^−t/6 − 36e^−t/6 ] + C. Evaluating from 0 to 6: 9[(−36e⁻¹ − 36e⁻¹) − (0 − 36)] = 9[36 − 72e⁻¹] = 324 − 648e⁻¹. Exact value: 324 − 648/e kilograms ≈ 85.614 kilograms of algae are produced by growth during the first 6 hours.

(b) M′(4) = G(4) − H(4) = 36e^−2/3 − 1.5·ln 5 ≈ 18.483 − 2.414 = 16.069. M′(4) ≈ 16.069 kilograms per hour. At time t = 4 hours the mass of algae in the reactor is increasing at a rate of about 16.069 kilograms per hour.

(c) Average rate of change of M on [0, 12] = (M(12) − M(0))/12 = (1/12)·∫₀¹² [9t·e^−t/6 − 1.5·ln(1 + t)] dt. ∫₀¹² 9t·e^−t/6 dt = 9[−6t·e^−t/6 − 36e^−t/6]₀¹² = 9[36 − 108e⁻²] = 324 − 972e⁻² ≈ 192.454. ∫₀¹² 1.5·ln(1 + t) dt = 1.5[(1 + t)ln(1 + t) − (1 + t)]₀¹² = 1.5[(13 ln 13 − 13) + 1] = (39/2)·ln 13 − 18 ≈ 32.017. So M(12) − M(0) ≈ 192.454 − 32.017 = 160.438 kg, and the average rate of change is 160.438/12 ≈ 13.370 kilograms per hour.

(d) M has a critical point where M′(t) = 0, i.e. where 9t·e^−t/6 = 1.5·ln(1 + t). At t = 0 both sides are 0, but for small t > 0, G(t) ≈ 9t exceeds H(t) ≈ 1.5t, so M′ > 0 just to the right of 0. Solving 9t·e^−t/6 − 1.5·ln(1 + t) = 0 on (0, 24) with a calculator gives the only positive solution t ≈ 22.542. Sign of M′: M′(12) = 108e⁻² − 1.5 ln 13 ≈ 14.616 − 3.847 = 10.769 > 0, and M′(24) = 216e⁻⁴ − 1.5 ln 25 ≈ 3.956 − 4.828 = −0.872 < 0. Since M′ changes from positive to negative at t ≈ 22.542, M attains an absolute maximum there on 0 < t < 24. The mass of algae is greatest at t ≈ 22.542 hours.

Scoring · 6 points

(a) 1 pt: correct integral setup ∫₀⁶ 9t·e^−t/6 dt for total growth.

(a) 1 pt: correct antiderivative by parts and value 324 − 648e⁻¹ ≈ 85.614 kilograms with units.

(b) 1 pt: M′(4) = G(4) − H(4) ≈ 16.069 with correct units (kilograms per hour).

(b) 1 pt: interpretation stating the algae mass is increasing at that rate at t = 4 hours.

(c) 1 pt: average rate of change set up as (1/12)∫₀¹² [G(t) − H(t)] dt (or (M(12) − M(0))/12) and evaluated to ≈ 13.370 kg/hr.

(d) 2 pts: t ≈ 22.542 obtained from G(t) = H(t), with a sign-change justification for a maximum (award 1 pt for the value alone).

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