AP Calculus mock exam Free full practice test

Written by STEM-Tutor's tutors · Updated October 2026 · 50 questions and 4 free-response questions · about 160 minutes

A full-length practice test built from our exam-style questions on every unit of AP Calculus. Each answer is marked as you go, and every question has a worked solution to open afterwards.

On this page

Before you start
Section I
Section II
After the mock

Before you start

Set a timer for about 160 minutes and work without notes. Questions marked No calculator should be done by hand, as in Part A of the multiple-choice section. You can type fractions, such as 3/4.
There are no hints in the mock. Answer every question before you open its worked solution: opening it first counts the question as missed.
Write the free-response questions on paper, then mark them yourself with the scoring guide.

Section I: 50 exam-style questions

Q1Exam-styleNo calculator

Find lim_x→0 sin(5x)/x.

Hint

Make the bottom match the angle inside the sine.

Worked solution

Write sin(5x)/x = 5 × sin(5x)/(5x).
As x → 0, 5x → 0, so sin(5x)/(5x) → 1.
The limit is 5.

Q2Exam-styleNo calculator

f(x) = x² + k for x < 2, and f(x) = 3x − 1 for x ≥ 2. What value of k makes f continuous at x = 2?

Hint

The two pieces must meet at x = 2.

Worked solution

From the left: 2² + k = 4 + k.
From the right, and f(2): 3(2) − 1 = 5.
Set them equal: 4 + k = 5, so k = 1.

Q3Exam-styleNo calculator

Find lim_x→4 (√x − 2)/(x − 4). You may give a fraction.

Hint

Multiply the top and bottom by the conjugate, √x + 2.

Worked solution

Substitution gives 0/0.
(√x − 2)(√x + 2) = x − 4, so the fraction becomes 1/(√x + 2).
At x = 4: 1/(2 + 2) = 1/4.

Q4Exam-styleNo calculator

f is continuous on [1, 3], with f(1) = −2 and f(3) = 5. Which must be true?

Hint

Which theorem guarantees a value between f(1) and f(3)?

Worked solution

f is continuous, and 0 lies between f(1) = −2 and f(3) = 5.
By the intermediate value theorem, f(c) = 0 for some c in (1, 3).
The other options need more information. The answer is A.

Q5ChallengeNo calculator

Find lim_x→0 (1 − cos x)/x².

Hint

Multiply the top and bottom by 1 + cos x, or use L'Hôpital's rule twice.

Worked solution

Multiply by (1 + cos x)/(1 + cos x): the top becomes 1 − cos²x = sin²x.
So the expression is (sin x / x)² × 1/(1 + cos x).
As x → 0 this tends to 1² × 1/2 = 1/2.

Q6Exam-styleCalculator

f(x) = x²e^x. Find f'(1), to 3 decimal places.

Hint

Use the product rule with u = x² and v = e^x.

Worked solution

f'(x) = 2x e^x + x²e^x.
f'(1) = 2e + e = 3e.
3e = 8.155.

Q7Exam-styleNo calculator

Find the y-intercept of the tangent line to y = x³ − 2x at x = 1.

Hint

You need the point and the slope at x = 1.

Worked solution

Point: y(1) = 1 − 2 = −1, so (1, −1).
Slope: y' = 3x² − 2, so y'(1) = 1.
Tangent: y + 1 = 1(x − 1), so y = x − 2. The y-intercept is −2.

Q8Exam-styleNo calculator

f(x) = (x + 1)/(x − 1). Find f'(2).

Hint

Use the quotient rule.

Worked solution

f'(x) = [(1)(x − 1) − (x + 1)(1)]/(x − 1)² = −2/(x − 1)².
f'(2) = −2/1 = −2.

Q9Exam-styleNo calculator

For f(x) = |x| at x = 0, which is true?

Hint

Look at the slope from each side.

Worked solution

The graph is unbroken at 0, so f is continuous.
The slope is −1 from the left and +1 from the right: a corner, so no derivative.
The answer is B. Differentiable always implies continuous, but not the other way round.

Q10Exam-styleNo calculator

Find d/dx (sin x cos x) at x = 0.

Hint

Product rule, or rewrite as ½ sin 2x.

Worked solution

d/dx(sin x cos x) = cos²x − sin²x = cos 2x.
At x = 0: cos 0 = 1.

Q11Exam-styleNo calculator

For the circle x² + y² = 25, find dy/dx at (3, 4).

Hint

Differentiate both sides with respect to x, treating y as a function of x.

Worked solution

2x + 2y(dy/dx) = 0, so dy/dx = −x/y.
At (3, 4): dy/dx = −3/4 = −0.75.

Q12Exam-styleNo calculator

f(x) = x³ + x, so f(1) = 2. Find (f⁻¹)'(2).

Hint

Use (f⁻¹)'(b) = 1/f'(a) where f(a) = b.

Worked solution

f(1) = 2, so a = 1.
f'(x) = 3x² + 1, so f'(1) = 4.
(f⁻¹)'(2) = 1/4.

Q13Exam-styleNo calculator

Find d/dx arctan(2x) at x = 0.

Hint

Chain rule with the arctan derivative.

Worked solution

d/dx arctan(2x) = 2/(1 + 4x²).
At x = 0: 2.

Q14Exam-styleNo calculator

y = (x² + 1)³. Find dy/dx at x = 1.

Hint

Chain rule: 3(inside)² × (derivative of inside).

Worked solution

dy/dx = 3(x² + 1)² × 2x.
At x = 1: 3 × 4 × 2 = 24.

Q15ChallengeNo calculator

The curve x² + xy + y² = 7 passes through (1, 2). Find dy/dx there.

Hint

Differentiate implicitly. The xy term needs the product rule.

Worked solution

2x + (y + x dy/dx) + 2y dy/dx = 0.
dy/dx = −(2x + y)/(x + 2y).
At (1, 2): −4/5 = −0.8.

Q16Exam-styleNo calculator

A 10 m ladder leans against a wall. The bottom slides away at 1 m/s. How fast is the top sliding down when the bottom is 6 m from the wall?

Hint

x² + y² = 100. Differentiate with respect to t.

Worked solution

When x = 6, y = 8.
2x dx/dt + 2y dy/dt = 0, so dy/dt = −x(dx/dt)/y = −6/8.
dy/dt = −0.75 m/s: the top slides down at 0.75 m/s.

Q17Exam-styleNo calculator

Use a linear approximation of f(x) = √x at x = 4 to estimate √4.1.

Hint

Tangent line at x = 4: f(4) = 2 and f'(4) = 1/(2√4).

Worked solution

f'(x) = 1/(2√x), so f'(4) = 1/4.
L(x) = 2 + (1/4)(x − 4).
L(4.1) = 2 + 0.025 = 2.025. (The true value is 2.0248.)

Q18Exam-styleNo calculator

Find lim_x→0 (e^x − 1 − x)/x².

Hint

It is 0/0. Apply L'Hôpital's rule, perhaps twice.

Worked solution

0/0, so differentiate top and bottom: (e^x − 1)/(2x), still 0/0.
Again: e^x/2.
At x = 0: 1/2.

Q19Exam-styleNo calculator

A particle has velocity v(t) = t² − 4t + 3. Find its acceleration at t = 1.

Hint

Acceleration is the derivative of velocity.

Worked solution

a(t) = 2t − 4.
a(1) = −2.

Q20Exam-styleNo calculator

For the particle with v(t) = t² − 4t + 3, at t = 0.5 the particle is:

Hint

Compare the signs of v(0.5) and a(0.5).

Worked solution

v(0.5) = 0.25 − 2 + 3 = 1.25 > 0.
a(0.5) = 1 − 4 = −3 < 0.
Opposite signs mean it is slowing down: B.

Q21Exam-styleNo calculator

Find the absolute maximum value of f(x) = x³ − 3x on [0, 2].

Hint

Check the critical points inside the interval and both endpoints.

Worked solution

f'(x) = 3x² − 3 = 0 gives x = 1 (in the interval).
f(0) = 0, f(1) = −2, f(2) = 2.
The absolute maximum is 2, at x = 2.

Q22Exam-styleNo calculator

Find the x-coordinate of the point of inflection of f(x) = x³ − 6x² + 4.

Hint

Find where f'' changes sign.

Worked solution

f''(x) = 6x − 12.
f''(x) = 0 at x = 2, and it changes sign there.
x = 2.

Q23Exam-styleNo calculator

A rectangle has a perimeter of 40 m. What is its largest possible area?

Hint

Write the area in terms of one side length.

Worked solution

Sides x and 20 − x, so A = x(20 − x) = 20x − x².
A'(x) = 20 − 2x = 0 at x = 10.
A = 10 × 10 = 100 m², a square.

Q24Exam-styleNo calculator

f'(x) changes from positive to negative at x = c. Then f has:

Hint

Think about what f does before and after c.

Worked solution

f' > 0 before c: f increases. f' < 0 after c: f decreases.
So f has a local maximum at c. The answer is B.

Q25ChallengeNo calculator

Squares of side x are cut from the corners of a 12 cm by 12 cm sheet to make an open box. What is the largest possible volume?

Hint

V(x) = x(12 − 2x)² for 0 < x < 6.

Worked solution

V'(x) = (12 − 2x)² − 4x(12 − 2x) = (12 − 2x)(12 − 6x).
V'(x) = 0 at x = 2 (x = 6 gives zero volume).
V(2) = 2 × 8² = 128 cm³.

Q26Exam-styleNo calculator

Evaluate ∫₁^e (1/x) dx.

Hint

The antiderivative of 1/x is ln x.

Worked solution

[ln x] from 1 to e = ln e − ln 1.
= 1 − 0 = 1.

Q27Exam-styleNo calculator

Evaluate ∫₀¹ 2x(x² + 1)³ dx.

Hint

Let u = x² + 1. Change the limits too.

Worked solution

u = x² + 1, du = 2x dx; x = 0 gives u = 1, x = 1 gives u = 2.
∫₁² u³ du = [u⁴/4] from 1 to 2.
(16 − 1)/4 = 15/4 = 3.75.

Q28Exam-styleNo calculator

f(0) = 1, f(2) = 3 and f(4) = 7. Use the trapezoidal rule with these two subintervals to estimate ∫₀⁴ f(x) dx.

Hint

Average the two ends of each strip, times its width.

Worked solution

Strip 1: 2 × (1 + 3)/2 = 4.
Strip 2: 2 × (3 + 7)/2 = 10.
Total = 14.

Q29Exam-styleNo calculator

F(x) = ∫₂^x² √t dt. Find F'(2).

Hint

Fundamental theorem with the chain rule: f(upper limit) × derivative of the upper limit.

Worked solution

F'(x) = √(x²) × 2x.
F'(2) = 2 × 4 = 8.

Q30Exam-styleNo calculator

Find the average value of f(x) = x² on [0, 3].

Hint

Average value = (1/(b − a)) ∫_a^b f(x) dx.

Worked solution

∫₀³ x² dx = 9.
Average = 9/3 = 3.

Q31Exam-styleCalculator

Solve dy/dx = xy with y(0) = 2. Find y(1), to 3 decimal places.

Hint

Separate the variables: dy/y = x dx.

Worked solution

ln|y| = x²/2 + C, so y = Ae^x²/2.
y(0) = 2 gives A = 2.
y(1) = 2e^0.5 = 3.297.

Q32Exam-styleCalculator

A population satisfies dP/dt = 0.05P with P(0) = 1000. Find P(10), to the nearest whole number.

Hint

This is exponential growth.

Worked solution

P(t) = 1000e^0.05t.
P(10) = 1000e^0.5.
≈ 1649.

Q33Exam-styleBC only

Use Euler's method with step 0.5 for dy/dx = x + y, y(0) = 1, to estimate y(1).

Hint

Two steps, each: new y = y + 0.5 × (x + y).

Worked solution

Step 1: y(0.5) ≈ 1 + 0.5(0 + 1) = 1.5.
Step 2: y(1) ≈ 1.5 + 0.5(0.5 + 1.5) = 2.5.

Q34Exam-styleNo calculator

dy/dt = ky with k < 0 models:

Hint

The solution is y = y₀e^kt. What happens when k is negative?

Worked solution

y = y₀e^kt with k < 0 shrinks by the same factor each unit of time.
That is exponential decay: B.

Q35ChallengeCalculator

Solve dy/dx = (y − 1)/x² with y(1) = 3. Find y(2), to 3 decimal places.

Hint

Separate: dy/(y − 1) = dx/x².

Worked solution

ln|y − 1| = −1/x + C. At (1, 3): ln 2 = −1 + C, so C = 1 + ln 2.
y − 1 = 2e^{1 − 1/x}.
y(2) = 1 + 2e^0.5 = 4.297.

Q36Exam-styleCalculator

The region under y = √x from x = 0 to x = 4 is rotated about the x-axis. Find the volume, as a decimal.

Hint

Disc method: V = π∫ y² dx.

Worked solution

V = π∫₀⁴ x dx.
= π × 8 = 8π.
≈ 25.13.

Q37Exam-styleNo calculator

A particle has velocity v(t) = t − 2 for 0 ≤ t ≤ 4. Find the total distance it travels.

Hint

v changes sign at t = 2.

Worked solution

From 0 to 2 it moves back: area 2.
From 2 to 4 it moves forward: area 2.
Total distance = 4 (the displacement is 0).

Q38Exam-styleNo calculator

Find the area enclosed by y = 4 − x² and the x-axis. Give a decimal.

Hint

The curve crosses the x-axis at x = ±2.

Worked solution

∫₋₂² (4 − x²) dx.
= [4x − x³/3] = (8 − 8/3) − (−8 + 8/3) = 32/3.
≈ 10.667.

Q39Exam-styleNo calculator

The base of a solid is the region under y = √x from x = 0 to x = 4. Cross-sections perpendicular to the x-axis are squares. Find the volume.

Hint

Each square has side √x, so area x.

Worked solution

Area of a cross-section: (√x)² = x.
V = ∫₀⁴ x dx = 8.

Q40ChallengeCalculator

The region between y = x and y = x² is rotated about the x-axis. Find the volume, to 4 decimal places.

Hint

Washer method: outer radius x, inner radius x².

Worked solution

V = π∫₀¹ (x² − x⁴) dx.
= π(1/3 − 1/5) = 2π/15.
≈ 0.4189.

Q41Exam-styleNo calculator

x = 3cos t, y = 3sin t. Find dy/dx at t = π/4.

Hint

dy/dx = (3cos t)/(−3sin t).

Worked solution

dy/dx = −cos t/sin t = −cot t.
At π/4: −1.

Q42Exam-styleNo calculator

Find the arc length of x = t, y = (2/3)t^3/2 for 0 ≤ t ≤ 3. Give a decimal.

Hint

dx/dt = 1 and dy/dt = t^1/2.

Worked solution

L = ∫₀³ √(1 + t) dt.
= (2/3)[(1 + t)^3/2] from 0 to 3 = (2/3)(8 − 1) = 14/3.
≈ 4.667.

Q43Exam-styleNo calculator

x = t², y = t³. Find d²y/dx² at t = 1.

Hint

d²y/dx² = [d/dt(dy/dx)] ÷ (dx/dt).

Worked solution

dy/dx = 3t/2, so d/dt(dy/dx) = 3/2.
Divide by dx/dt = 2t = 2.
3/4 = 0.75.

Q44Exam-styleCalculator

Find the area enclosed by the cardioid r = 1 + cos θ, as a decimal.

Hint

A = ½∫₀^2π (1 + cos θ)² dθ.

Worked solution

Expand: 1 + 2cos θ + cos²θ. Over a full turn, ∫cos θ = 0 and ∫cos²θ = π.
A = ½(2π + 0 + π) = 3π/2.
≈ 4.712.

Q45ChallengeCalculator

Find the slope dy/dx of the polar curve r = θ at θ = π/2. Give a decimal.

Hint

dy/dx = (r' sinθ + r cosθ)/(r' cosθ − r sinθ).

Worked solution

r = θ and r' = 1. At π/2: sin = 1 and cos = 0.
dy/dx = (1 + 0)/(0 − π/2) = −2/π.
≈ −0.637.

Q46Exam-styleNo calculator

Find Σ_n=0^∞ 3(−1/2)ⁿ.

Hint

Geometric with a = 3 and r = −1/2.

Worked solution

|r| < 1, so the sum is 3/(1 + 1/2).
= 2.

Q47Exam-styleNo calculator

Apply the ratio test to Σ n/2ⁿ. The series:

Hint

Find lim |a_n+1/a_n|.

Worked solution

a_n+1/a_n = [(n + 1)/2ⁿ⁺¹] × [2ⁿ/n] = (n + 1)/(2n).
This tends to 1/2 < 1, so the series converges: B.

Q48Exam-styleNo calculator

Find the radius of convergence of Σ xⁿ/3ⁿ.

Hint

It is geometric with ratio x/3.

Worked solution

It converges when |x/3| < 1.
That is |x| < 3, so R = 3.

Q49Exam-styleNo calculator

Use the degree-2 Maclaurin polynomial for cos x to estimate cos(0.2).

Hint

cos x ≈ 1 − x²/2.

Worked solution

1 − 0.2²/2 = 1 − 0.02.
= 0.98. (The true value is 0.98007.)

Q50ChallengeNo calculator

The first 3 terms of Σ (−1)ⁿ⁺¹/n² are used to estimate its sum. What is the alternating series error bound?

Hint

The error is at most the first omitted term.

Worked solution

The first omitted term is the 4th: 1/4².
Error ≤ 1/16 = 0.0625.

Section II: 4 free-response questions

Free-response question 1: Limits and continuity

No-calculator free-response style · 6 points · about 15 minutes

Let g(x) = (x² − 4)/(x − 2) for x ≠ 2, and g(2) = k.

(a) Find lim_x→2 g(x). Show your work.
(b) For what value of k is g continuous at x = 2? Justify your answer using the definition of continuity.
(c) Let h(x) = (2x² + 3)/(x² − 1). Find lim_x→∞ h(x) and state what it tells you about the graph of h.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

For x ≠ 2, g(x) = (x − 2)(x + 2)/(x − 2) = x + 2, so the limit is 4.

1 point for: Factorises or otherwise simplifies to x + 2
1 point for: Limit = 4

Part (b)

g is continuous at 2 when g(2) = lim_x→2 g(x). The limit is 4, so k = 4. Then g(2) exists, the limit exists, and they are equal.

1 point for: k = 4
1 point for: Justification naming all three conditions, or g(2) = the limit

Part (c)

Dividing by x², the limit is 2. So y = 2 is a horizontal asymptote of h.

1 point for: Limit = 2
1 point for: Horizontal asymptote y = 2

Your mark: 0 of 6

Free-response question 2: Contextual applications of differentiation

No-calculator free-response style · 6 points · about 15 minutes

A particle moves along the x-axis with position x(t) = t³ − 6t² + 9t + 1 for 0 ≤ t ≤ 5.

(a) Find the times when the particle is at rest.
(b) Is the particle speeding up or slowing down at t = 4? Give a reason.
(c) Find the total distance travelled from t = 0 to t = 3.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

v(t) = 3t² − 12t + 9 = 3(t − 1)(t − 3) = 0, so t = 1 and t = 3.

1 point for: v(t) correct
1 point for: t = 1 and t = 3

Part (b)

v(4) = 9 > 0 and a(4) = 6(4) − 12 = 12 > 0. Same sign, so it is speeding up.

1 point for: Values or signs of v(4) and a(4)
1 point for: Speeding up, with the same-sign reason

Part (c)

x(0) = 1, x(1) = 5, x(3) = 1. It turns at t = 1, so the distance is 4 + 4 = 8.

1 point for: Splits at the turning point t = 1
1 point for: Total distance 8

Your mark: 0 of 6

Free-response question 3: Differential equations

No-calculator free-response style · 6 points · about 15 minutes

Consider the differential equation dy/dx = (y − 1)/x², with y(1) = 3.

(a) Write the equation of the tangent line to the solution at x = 1, and use it to estimate y(1.1).
(b) Find the particular solution y = f(x).

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

Slope at (1, 3) = 2. y ≈ 3 + 2(x − 1), so y(1.1) ≈ 3.2.

1 point for: Slope 2
1 point for: Estimate 3.2

Part (b)

dy/(y − 1) = dx/x²; ln|y − 1| = −1/x + C; C = 1 + ln 2; y = 1 + 2e^{1 − 1/x}.

1 point for: Separates the variables
1 point for: Correct antiderivatives
1 point for: Uses the initial condition
1 point for: Solves for y

Your mark: 0 of 6

Free-response question 4: Infinite sequences and series (BC)

No-calculator free-response style · 6 points · about 15 minutes

Let f(x) = e^2x.

(a) Write the first four nonzero terms of the Maclaurin series for f.
(b) Write the general term.
(c) Find the radius of convergence. Show your reasoning.
(d) Use the first four terms to approximate e^0.2.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

Substitute 2x into e^u: 1 + 2x + 2x² + (4/3)x³ + …

1 point for: Uses the e^u series
1 point for: All four terms correct

Part (b)

(2x)ⁿ/n! = 2ⁿxⁿ/n!.

1 point for: Correct general term

Part (c)

Ratio: |2x|/(n + 1) → 0 < 1 for every x, so the radius is infinite.

1 point for: Ratio test set up
1 point for: Radius is infinite

Part (d)

x = 0.1: 1 + 0.2 + 0.02 + 0.001333 ≈ 1.22133. (True value 1.22140.)

1 point for: Uses x = 0.1 and gets about 1.2213

Your mark: 0 of 6

After the mock

Note every question you missed and its unit, then go back to that unit's practice page: all AP Calculus units. Then try the real thing with official AP Calculus past papers.