AP Calculus AB practice questions and exam guide

∫ sin(2x) dx =

1. cos(2x) + C
2. -cos(2x)/2 + C
3. -cos(2x) + C
4. cos(2x)/2 + C

Answer: B — -cos(2x)/2 + C

B) Use u-sub or recognize: derivative of -cos(2x)/2 = sin(2x). C) Forgets the factor of 1/2. A/D) Sign errors.

Evaluate lim_{x→0} sin(5x)/x.

1. 0
2. 1
3. 5
4. Does not exist

Answer: C — 5

Use lim_{u→0} sin(u)/u = 1 with u = 5x: lim sin(5x)/x = 5 · lim sin(5x)/(5x) = 5 · 1 = 5. (B) is the unscaled fundamental limit.

A logistic equation dP/dt = kP(1 - P/M) has carrying capacity:

1. k
2. P
3. M
4. 1

Answer: C — M

C) M is the carrying capacity — population approaches M as t→∞. A) Growth rate. B) Current population. D) Just a scaling factor.

The area between y = x² and y = 2x (from x=0 to x=2) is:

1. 4/3
2. 2
3. 8/3
4. 4

Answer: A — 4/3

A) Top function is 2x. ∫₀² (2x - x²) dx = [x² - x³/3]₀² = 4 - 8/3 = 4/3. B/C) Computes one curve only. D) Wrong simplification.

d/dx[(2x + 1)⁵] =

1. 5(2x+1)⁴
2. 10(2x+1)⁴
3. 10x(2x+1)⁴
4. (2x+1)⁵

Answer: B — 10(2x+1)⁴

B) Chain rule: 5(2x+1)⁴ · 2 = 10(2x+1)⁴. A) Forgets to multiply by the inner derivative (2). C) Mixes power-rule with multiplying by x. D) Forgets to differentiate.

The Intermediate Value Theorem requires that f be:

1. Differentiable on (a, b)
2. Continuous on [a, b]
3. Strictly increasing
4. A polynomial

Answer: B — Continuous on [a, b]

B) IVT requires continuity on a closed interval [a, b]. A) Differentiability would be needed for MVT, not IVT. C) Not a requirement. D) Polynomials work but so do many others.

The derivative of f(x) = √x at x = 4 is:

1. 1/4
2. 2
3. 1/2
4. 4

Answer: A — 1/4

A) f'(x) = 1/(2√x). f'(4) = 1/(2·2) = 1/4. B) Forgets the derivative entirely (gives √4). C) Forgets to evaluate at x=4. D) Returns function value, not derivative.

d/dx [ln(x² + 1)] equals:

1. 2x / (x² + 1)
2. 1 / (x² + 1)
3. 2x · ln(x² + 1)
4. 1 / (2x)

Answer: A — 2x / (x² + 1)

Chain rule: d/dx[ln u] = u'/u. With u = x² + 1, u' = 2x. So the derivative is 2x/(x²+1).

Solve dy/dx = 2y with y(0) = 3:

1. y = 3e^(2x)
2. y = 3 + 2x
3. y = e^(2x) + 2
4. y = 3e^x

Answer: A — y = 3e^(2x)

A) Separable: dy/y = 2 dx → ln|y| = 2x + C → y = Ce^(2x). y(0)=3 → C=3. B) Linear, wrong. C) Wrong form. D) Forgets the 2 in exponent.

Compute ∫₀^(π/2) cos(x) dx.

1. 0
2. 1
3. π/2
4. 2

Answer: B — 1

Antiderivative of cos x is sin x. [sin x]₀^(π/2) = sin(π/2) − sin(0) = 1 − 0 = 1.

The derivative of arctan(x) is:

1. 1 / (1 + x²)
2. 1 / √(1 − x²)
3. − 1 / (1 + x²)
4. sec²(x)

Answer: A — 1 / (1 + x²)

d/dx[arctan x] = 1/(1+x²). (B) is the derivative of arcsin x.

lim(x→0) (e^x - 1)/x by L'Hôpital's rule equals:

1. 0
2. Cannot apply L'Hôpital
3. e
4. 1

Answer: D — 1

D) Form is 0/0; derivatives give e^x/1 → e^0/1 = 1. A) That's a wrong limit. C) Misreads. B) L'Hôpital applies here.

What does lim_{x→a} f(x) = L mean?

As x approaches a from either side, f(x) gets arbitrarily close to L (without necessarily equaling L).

Find volume rotating y = x² on [0, 2] around x-axis.

V = π ∫₀² x⁴ dx = π · 32/5 = 32π/5.

Use log diff: d/dx[xˣ].

y = xˣ → ln y = x ln x → y'/y = ln x + 1 → y' = xˣ(ln x + 1).

What does f'' = 0 imply?

Possible inflection point — must check sign of f'' on both sides for actual change in concavity.

Logistic differential equation?

dP/dt = kP(1 − P/M), where M is the carrying capacity.

Evaluate d/dx ∫_0^(x²) cos t dt.

Chain rule: cos(x²) · 2x.