Mixed Deck — All AP CALC BC Topics Flashcards

What is lim(x→3) (x² − 9)/(x − 3)?

Answer: 6

Factor the numerator as (x − 3)(x + 3) and cancel (x − 3). The limit is then lim(x→3) (x + 3) = 6.

A function f is continuous on [1, 4] with f(1) = −3 and f(4) = 5. Which statement must be true?

Answer: f(c) = 0 for at least one c in (1, 4)

0 lies between f(1) = −3 and f(4) = 5, so the Intermediate Value Theorem guarantees at least one c in (1, 4) with f(c) = 0.

A particle moves along a line with position s(t) = t³ − 9t² + 15t for t ≥ 0. At which time is the particle speeding up?

Answer: t = 2

v(t) = 3(t − 1)(t − 5) and a(t) = 6t − 18. At t = 2, v = −9 and a = −6 have the same sign, so the speed is increasing.

Which of the following series converges?

Answer: Σ_{n=2}^∞ 1/(n(ln n)²)

By the integral test, ∫₂^∞ dx/(x(ln x)²) = 1/ln 2 is finite, so the series converges. The other three series diverge.

What is lim(x→0) sin(5x)/(2x)?

Answer: 5/2

Rewrite as (5/2)·[sin(5x)/(5x)]. Since sin(u)/u → 1 as u → 0, the limit is 5/2.

A farmer has 200 feet of fencing to enclose a rectangular pen along a straight river. No fence is needed along the river, so only three sides are fenced. What is the maximum possible area of the pen?

Answer: 5,000 ft²

With width x and length y along the river, 2x + y = 200, so A = x(200 − 2x). A′ = 200 − 4x = 0 gives x = 50, y = 100, and A = 5,000 ft².

If y = ln(cos(3x)), what is dy/dx?

Answer: −3 tan(3x)

Apply the chain rule twice: d/dx ln(u) = u'/u with u = cos(3x), and u' = −3 sin(3x). This gives −3 sin(3x)/cos(3x) = −3 tan(3x).

If F(x) = ∫ from 0 to x² of √(1 + t³) dt, what is F′(x)?

Answer: 2x·√(1 + x⁶)

By the Fundamental Theorem of Calculus with the chain rule, F′(x) = √(1 + (x²)³) · d/dx(x²) = 2x·√(1 + x⁶).

What is lim(x→0) (1 − cos x)/x²?

Answer: 1/2

Multiply by (1 + cos x)/(1 + cos x) to get sin²x / [x²(1 + cos x)]. This approaches 1·(1/2) = 1/2.

What is lim(x→2) |x − 2|/(x − 2)?

Answer: Does not exist

From the right the expression equals 1, and from the left it equals −1. The one-sided limits differ, so the two-sided limit does not exist.

Let f(x) = x^2 for x ≤ 1 and f(x) = 2x − 1 for x > 1. Which statement is true at x = 1?

Answer: f is differentiable at x = 1 and f'(1) = 2

Both pieces equal 1 at x = 1, so f is continuous. The left derivative is 2x = 2 and the right derivative is 2, so f'(1) = 2.

In the Maclaurin series for f(x) = e^(−x²), what is the coefficient of x⁴?

Answer: 1/2

Substitute u = −x² into eᵘ = 1 + u + u²/2 + …. The u²/2 term is (−x²)²/2 = x⁴/2, so the coefficient is 1/2.

What is the length of the curve y = x^(3/2) from x = 0 to x = 4?

Answer: (8/27)(10√10 − 1)

Here y′ = (3/2)x^(1/2), so the integrand is √(1 + 9x/4). Integrating gives (8/27)(1 + 9x/4)^(3/2), and evaluating from 0 to 4 gives (8/27)(10√10 − 1).

Air is pumped into a spherical balloon at a rate of 100 cm³/s. How fast is the radius increasing at the instant the radius is 5 cm? (V = (4/3)πr³)

Answer: 1/π cm/s

Differentiating gives dV/dt = 4πr²(dr/dt). With r = 5, 100 = 100π(dr/dt), so dr/dt = 1/π cm/s.

What is the area of the region enclosed by the curves x = y² and x = y + 2?

Answer: 9/2

The curves intersect where y² = y + 2, at y = −1 and y = 2, and x = y + 2 is the right-hand curve. The area is ∫₋₁² (y + 2 − y²) dy = 9/2.

Which expression equals ∫ 1/((x − 1)(x + 2)) dx?

Answer: (1/3)·ln|(x − 1)/(x + 2)| + C

Partial fractions give 1/((x−1)(x+2)) = (1/3)/(x−1) − (1/3)/(x+2). Integrating gives (1/3)ln|x−1| − (1/3)ln|x+2| + C.

What is the relative maximum value of f(x) = x·e^(−x)?

Answer: 1/e

f′(x) = e^(−x)(1 − x) = 0 at x = 1, and f″(1) = e^(−1)(1 − 2) < 0, so x = 1 is a maximum. The value is f(1) = 1/e.

The Maclaurin series for arctan x is Σ_{n=0}^∞ (−1)ⁿ x^(2n+1)/(2n+1) for |x| ≤ 1. What is the sum of Σ_{n=0}^∞ (−1)ⁿ / (3ⁿ(2n + 1))?

Answer: π√3/6

Setting x = 1/√3 gives x^(2n+1) = 3^(−n)/√3, so arctan(1/√3) = π/6 equals (1/√3) times the given sum. The sum is therefore √3 · π/6 = π√3/6.

What is lim(x→2⁻) 1/(x − 2)?

Answer: −∞

For x slightly less than 2, x − 2 is a small negative number, so 1/(x − 2) is a large negative number. The limit is −∞.

Euler's method with two steps of size 0.5 is used to approximate y(1) for the differential equation dy/dx = x + y with y(0) = 1. What is the approximation?

Answer: 2.5

At (0, 1) the slope is 1, so y(0.5) ≈ 1 + 0.5(1) = 1.5. At (0.5, 1.5) the slope is 2, so y(1) ≈ 1.5 + 0.5(2) = 2.5.