AP Calculus AB/BC Practice Exam
20 multiple-choice questions covering every major topic from limits through series. College Board difficulty level.
20
Questions
45
Minutes
AB + BC
Scope
4 pts
Per Question
📚 Concept Review & Key Formulas
01
Limits & Continuity
▾
lim[x→a] f(x) = L ⟺ both one-sided limits equal L
Squeeze Theorem: g(x) ≤ f(x) ≤ h(x) and lim g = lim h = L ⟹ lim f = L
lim[x→0] (sin x)/x = 1 , lim[x→0] (1−cos x)/x = 0
lim[x→±∞] (1 + 1/n)ⁿ = e
Squeeze Theorem: g(x) ≤ f(x) ≤ h(x) and lim g = lim h = L ⟹ lim f = L
lim[x→0] (sin x)/x = 1 , lim[x→0] (1−cos x)/x = 0
lim[x→±∞] (1 + 1/n)ⁿ = e
✦ Removable discontinuity: hole — factor & cancel
✦ Jump discontinuity: one-sided limits differ
✦ Infinite discontinuity: vertical asymptote
✦ A function is continuous at a iff: f(a) exists, lim f(x) exists, and they are equal
✦ Jump discontinuity: one-sided limits differ
✦ Infinite discontinuity: vertical asymptote
✦ A function is continuous at a iff: f(a) exists, lim f(x) exists, and they are equal
Example
lim[x→0] (sin 3x)/(2x) = (3/2)·lim[u→0] (sin u)/u = 3/2
02
Derivatives — Rules & Applications
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Power: d/dx[xⁿ] = nxⁿ⁻¹
Chain: d/dx[f(g(x))] = f'(g(x))·g'(x)
Product: (uv)' = u'v + uv'
Quotient: (u/v)' = (u'v − uv')/v²
d/dx[eˣ] = eˣ , d/dx[ln x] = 1/x
d/dx[sin x] = cos x , d/dx[cos x] = −sin x
d/dx[tan x] = sec²x , d/dx[arctan x] = 1/(1+x²)
d/dx[arcsin x] = 1/√(1−x²)
Chain: d/dx[f(g(x))] = f'(g(x))·g'(x)
Product: (uv)' = u'v + uv'
Quotient: (u/v)' = (u'v − uv')/v²
d/dx[eˣ] = eˣ , d/dx[ln x] = 1/x
d/dx[sin x] = cos x , d/dx[cos x] = −sin x
d/dx[tan x] = sec²x , d/dx[arctan x] = 1/(1+x²)
d/dx[arcsin x] = 1/√(1−x²)
✦ MVT: f'(c) = [f(b)−f(a)]/(b−a) for some c in (a,b)
✦ f is increasing ↔ f' > 0 ; concave up ↔ f'' > 0
✦ Inflection point: f'' changes sign
✦ L'Hôpital: 0/0 or ∞/∞ → differentiate top & bottom
✦ f is increasing ↔ f' > 0 ; concave up ↔ f'' > 0
✦ Inflection point: f'' changes sign
✦ L'Hôpital: 0/0 or ∞/∞ → differentiate top & bottom
Example
d/dx[sin(x²)] = cos(x²)·2x = 2x cos(x²)
03
Integrals — Techniques & FTC
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FTC I: d/dx[∫[a to x] f(t)dt] = f(x)
FTC II: ∫[a to b] f(x)dx = F(b) − F(a)
∫xⁿdx = xⁿ⁺¹/(n+1) + C (n ≠ −1)
∫eˣdx = eˣ + C , ∫(1/x)dx = ln|x| + C
∫sin x dx = −cos x + C , ∫cos x dx = sin x + C
∫sec²x dx = tan x + C
Integration by Parts: ∫u dv = uv − ∫v du
FTC II: ∫[a to b] f(x)dx = F(b) − F(a)
∫xⁿdx = xⁿ⁺¹/(n+1) + C (n ≠ −1)
∫eˣdx = eˣ + C , ∫(1/x)dx = ln|x| + C
∫sin x dx = −cos x + C , ∫cos x dx = sin x + C
∫sec²x dx = tan x + C
Integration by Parts: ∫u dv = uv − ∫v du
✦ u-substitution: replace inner function with u
✦ Average value: (1/(b−a))·∫[a to b] f(x)dx
✦ Net displacement vs. total distance: ∫v(t)dt vs. ∫|v(t)|dt
✦ Area between curves: ∫[a to b] [f(x)−g(x)]dx
✦ Average value: (1/(b−a))·∫[a to b] f(x)dx
✦ Net displacement vs. total distance: ∫v(t)dt vs. ∫|v(t)|dt
✦ Area between curves: ∫[a to b] [f(x)−g(x)]dx
Example
∫₀¹ 2x·e^(x²) dx : let u=x², du=2x dx → ∫₀¹ eᵘ du = e¹−e⁰ = e − 1
04
BC Topics — Series & Parametric
▾
Geometric: Σarⁿ = a/(1−r) , |r|<1
Taylor: f(x) = Σ f⁽ⁿ⁾(a)/n! · (x−a)ⁿ
Maclaurin: eˣ = Σxⁿ/n! , sin x = Σ(−1)ⁿx²ⁿ⁺¹/(2n+1)!
cos x = Σ(−1)ⁿx²ⁿ/(2n)! , 1/(1−x) = Σxⁿ
Ratio Test: L = lim|aₙ₊₁/aₙ| ; L<1 converges
Parametric: dy/dx = (dy/dt)/(dx/dt) , Arc Length = ∫√[(dx/dt)²+(dy/dt)²]dt
Taylor: f(x) = Σ f⁽ⁿ⁾(a)/n! · (x−a)ⁿ
Maclaurin: eˣ = Σxⁿ/n! , sin x = Σ(−1)ⁿx²ⁿ⁺¹/(2n+1)!
cos x = Σ(−1)ⁿx²ⁿ/(2n)! , 1/(1−x) = Σxⁿ
Ratio Test: L = lim|aₙ₊₁/aₙ| ; L<1 converges
Parametric: dy/dx = (dy/dt)/(dx/dt) , Arc Length = ∫√[(dx/dt)²+(dy/dt)²]dt
✦ p-series: Σ1/nᵖ converges iff p>1
✦ Alternating Series Est.: |error| ≤ |first omitted term|
✦ Polar area: (1/2)∫r² dθ
✦ Radius of convergence from Ratio Test
✦ Alternating Series Est.: |error| ≤ |first omitted term|
✦ Polar area: (1/2)∫r² dθ
✦ Radius of convergence from Ratio Test
Example
Σ(n=0 to ∞) (1/2)ⁿ = 1/(1−1/2) = 2 (geometric, a=1, r=1/2)
📝 Practice Questions
Q1
Limits
What is the value of the following limit?
lim[x→3] (x² − 9) / (x − 3)
✦ Solution
Factor the numerator: x² − 9 = (x−3)(x+3)(x²−9)/(x−3) = (x−3)(x+3)/(x−3) = x+3 for x≠3
Therefore: lim[x→3] (x+3) = 3+3 = 6
Answer: (C) 6
Q2
Limits — Trig
Evaluate:
lim[x→0] sin(5x) / (3x)
✦ Solution
Use the standard limit lim[u→0] sin(u)/u = 1:sin(5x)/(3x) = (5/3)·[sin(5x)/(5x)]
As x→0, 5x→0, so sin(5x)/(5x)→1
Result: 5/3 · 1 = 5/3
Answer: (A) 5/3
Q3
Derivatives — Chain Rule
If f(x) = e^(cos x), then f'(x) =
✦ Solution
Chain rule: d/dx[e^(u)] = eᵘ · u', where u = cos x, u' = −sin xf'(x) = e^(cos x) · (−sin x) = −sin x · e^(cos x)
Answer: (B) −sin x · e^(cos x)
Q4
Implicit Differentiation
Given x² + y² = 25, find dy/dx.
✦ Solution
Differentiate both sides with respect to x:2x + 2y·(dy/dx) = 0
2y·(dy/dx) = −2x
dy/dx = −x/y
Answer: (C) −x/y
Q5
Related Rates
A spherical balloon is being inflated so that its volume increases at 10 cm³/s. How fast is the radius increasing when the radius is 5 cm? (V = (4/3)πr³)
✦ Solution
Differentiate V = (4/3)πr³ with respect to t:dV/dt = 4πr² · (dr/dt)
Substitute dV/dt = 10 and r = 5:
10 = 4π(25) · (dr/dt)
dr/dt = 10/(100π) = 1/(10π) cm/s
Answer: (A) 1/(10π) cm/s
Q6
Mean Value Theorem
Let f(x) = x³ − 3x on [0, 2]. By the Mean Value Theorem, there exists c in (0, 2) such that f'(c) equals:
✦ Solution
f(0) = 0, f(2) = 8 − 6 = 2MVT slope = [f(2)−f(0)]/(2−0) = (2−0)/2 = 1
So f'(c) must equal 1. (Verify: f'(x)=3x²−3, set =1 → x²=4/3 → x=2/√3 ≈ 1.15 ∈ (0,2) ✓)
Answer: (B) 1
Q7
Definite Integral — u-sub
Evaluate:
∫₀² x · e^(x²) dx
✦ Solution
Let u = x², then du = 2x dx, so x dx = du/2When x=0: u=0; when x=2: u=4
∫₀² x·e^(x²)dx = ∫₀⁴ eᵘ·(du/2) = (1/2)[eᵘ]₀⁴ = (1/2)(e⁴−1)
Answer: (B) (e⁴ − 1)/2
Q8
FTC Part I
Let F(x) = ∫₁ˣ (t² + 1) dt. Then F'(x) =
✦ Solution
By the Fundamental Theorem of Calculus Part I:F'(x) = d/dx ∫₁ˣ (t²+1)dt = x² + 1
Simply substitute x for the variable of integration t.
Answer: (B) x² + 1
Q9
Area Between Curves
The area enclosed between y = x² and y = x (from x = 0 to x = 1) is:
✦ Solution
On [0,1]: y=x is above y=x² (check x=1/2: 1/2 > 1/4 ✓)Area = ∫₀¹ (x − x²)dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6
Answer: (A) 1/6
Q10
L'Hôpital's Rule
Evaluate using L'Hôpital's Rule:
lim[x→0] (eˣ − 1 − x) / x²
✦ Solution
Form 0/0 at x=0. Apply L'Hôpital once:lim (eˣ − 1)/(2x) → still 0/0
Apply L'Hôpital a second time:
lim eˣ/2 = e⁰/2 = 1/2
Answer: (C) 1/2
Q11
Concavity & Inflection
The function f(x) = x⁴ − 4x³ has an inflection point at x =
✦ Solution
f'(x) = 4x³ − 12x²f''(x) = 12x² − 24x = 12x(x−2)
f''(x) = 0 at x = 0 and x = 2
Check sign changes: f''<0 on (0,2), f''>0 outside → inflection at both x=0 and x=2
Answer: (C) x = 0 and x = 2
Q12
Integration by Parts
Evaluate:
∫ x · eˣ dx
✦ Solution
Integration by Parts: ∫u dv = uv − ∫v duLet u = x → du = dx
Let dv = eˣ dx → v = eˣ
∫x·eˣ dx = x·eˣ − ∫eˣ dx = x·eˣ − eˣ + C = eˣ(x−1) + C
Note: (B) and (C) are the same expression. Answer: (B)/(C) eˣ(x−1) + C
Q13
Differential Equations
The general solution of dy/dx = 2xy is:
✦ Solution
Separate variables: dy/y = 2x dx∫dy/y = ∫2x dx → ln|y| = x² + K
|y| = e^(x²+K) = e^K · e^(x²) → y = Ce^(x²)
Answer: (C) y = Ce^(x²)
Q14
Volume of Revolution (Disk)
The volume of the solid formed by revolving y = √x around the x-axis from x = 0 to x = 4 is:
✦ Solution
Disk method: V = π∫[a to b] [f(x)]² dxV = π∫₀⁴ (√x)² dx = π∫₀⁴ x dx = π[x²/2]₀⁴ = π(16/2) = 8π
Answer: (A) 8π
Q15
Geometric Series
What is the sum of the infinite series?
Σ (n=0 to ∞) 3·(2/3)ⁿ
✦ Solution
Geometric series formula: S = a/(1−r), where a = first term, r = common ratioHere a = 3·(2/3)⁰ = 3, r = 2/3, |r| < 1 so series converges
S = 3/(1 − 2/3) = 3/(1/3) = 9
Answer: (B) 9
Q16
Taylor / Maclaurin Series
The first three nonzero terms of the Maclaurin series for f(x) = sin x are:
✦ Solution
Maclaurin series for sin x:sin x = x − x³/3! + x⁵/5! − ··· = x − x³/6 + x⁵/120 − ···
Note: (A) and (D) are the series for cos x.
Answer: (B) x − x³/6 + x⁵/120
Q17
Ratio Test
The series Σ (n=1 to ∞) xⁿ/n! converges for:
✦ Solution
Apply Ratio Test: L = lim|aₙ₊₁/aₙ||aₙ₊₁/aₙ| = |x|ⁿ⁺¹/(n+1)! · n!/|x|ⁿ = |x|/(n+1)
As n→∞: L = |x|·lim[1/(n+1)] = |x|·0 = 0 < 1
L = 0 < 1 for all x, so the series converges for all real x. (This is the series for eˣ.)
Answer: (C) All real numbers x
Q18
Parametric Curves
If x(t) = t² and y(t) = t³ − t, find dy/dx at t = 1.
✦ Solution
For parametric: dy/dx = (dy/dt)/(dx/dt)dx/dt = 2t , dy/dt = 3t² − 1
dy/dx = (3t²−1)/(2t)
At t = 1: dy/dx = (3·1−1)/(2·1) = 2/2 = 1
Answer: (A) 1
Q19
Polar Area
The area enclosed by the polar curve r = 2cos θ for 0 ≤ θ ≤ π is:
✦ Solution
Polar area formula: A = (1/2)∫[α to β] r² dθA = (1/2)∫₀^π (2cos θ)² dθ = (1/2)∫₀^π 4cos²θ dθ = 2∫₀^π cos²θ dθ
Use identity cos²θ = (1+cos 2θ)/2:
= 2∫₀^π (1+cos 2θ)/2 dθ = ∫₀^π (1+cos 2θ)dθ = [θ + sin2θ/2]₀^π = π
Answer: (A) π
Q20
Improper Integral
Determine whether the following integral converges, and if so, find its value:
∫₁^∞ (1/x²) dx
✦ Solution
Evaluate as a limit:∫₁^∞ x⁻² dx = lim[b→∞] ∫₁ᵇ x⁻² dx = lim[b→∞] [−x⁻¹]₁ᵇ
= lim[b→∞] (−1/b + 1/1) = 0 + 1 = 1
The integral converges to 1. (Compare: ∫1/x dx = ln x → diverges; p-integral with p=2>1 converges.)
Answer: (C) 1
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📋 Answer Key & Explanations
Q1CFactor & cancel
Q2Asin(u)/u limit
Q3BChain rule
Q4CImplicit diff.
Q5ARelated rates
Q6BMVT
Q7Bu-substitution
Q8BFTC Part I
Q9AArea between curves
Q10CL'Hôpital ×2
Q11Cf'' sign change
Q12BIntegration by parts
Q13CSeparable ODE
Q14ADisk method
Q15BGeometric series
Q16BMaclaurin sin x
Q17CRatio test → 0
Q18AParametric dy/dx
Q19APolar area
Q20CImproper integral