AP Calculus AB Practice Questions and Answers

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Planning to take the AP Calculus AB exam soon? Do not worry! We are here to help. This article is packed with everything you need to ace the exam. 

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AP Calculus AB is known to be one of the more difficult AP courses, with a reputation for its challenging curriculum and historically lower pass rates. It requires students to understand calculus concepts deeply and have rigorous problem-solving skills. 

This difficulty is why students must use AP Calculus AB practice questions and answers to help them navigate their exam preparation.

In this article, we’ll break down what you can expect in the AP Calculus AB exam, provide a few practice questions you can use to prepare, and offer key tips to help streamline your study sessions. 

AP Calculus AB Exam Overview

AP Calculus AB is designed to provide high school students with a comprehensive understanding of fundamental calculus principles, equipping them with the skills and knowledge required for college-level courses.

The AP Calculus AB exam is divided into two main sections: the Multiple Choice (MC) section and the Free Response (FRQ) section. The MC section comprises 45 questions in total. Students are allotted 60 minutes to complete the first 30 questions without the use of a calculator and an additional 45 minutes to complete the remaining 15 questions, where a graphing calculator is allowed.

The FRQ section, on the other hand, consists of 6 questions in total, further divided into two parts: A and B. Students are given 1 hour and 45 minutes to complete this section, with the first 75 minutes dedicated to Part A and the final 30 minutes for Part B.

AP Calculus AB Practice Questions and Answers

The best way to prepare for this exam is to complete as many practice questions as possible. Here are a few AP Calculus AB Practice questions and answers you can use to help you prepare for the exam.

Question 1:

=

A) e

B) 1

C) e^h

D) e⁴

E) 4^e

Explanation

Let f(x) = e^x. The given limit may be written as follows: as x ----> h

Lim (e⁴ e^h - e⁴) / h = lim (e^{4 + h} - e⁴) / h = limit [ f(4+h) – f (4) ] / h

which is the definition of the first derivative of f(x) = e^x at x = 4. Hence as x ---> h

lim (e⁴ e^h - e⁴) / h = e ⁴

The answer is D.

Question 2

Curve C is described by the equation 0.25x² + y² = 9. Determine the y coordinates of the points on curve C whose tangent lines have a slope equal to 1.

A) -3 sqrt (5) / 5, 3 sqrt (5) / 5

B) - sqrt(35) / 2 , sqrt(35) / 2

C) -3, 3

D) - sqrt(2) / 2 , sqrt(2) / 2

E) -3 sqrt(2) , 3 sqrt(2)

Explanation

Let us calculate the first derivative. Differentiate both sides of the given equation

0.25 (2x) + 2 y y ' = 0

y ' = - 0.5 x / (2 y)

We now solve the given equation 0.25 x² + y² = 9 for x

x = + or - sqrt [ (9 - y²) / 0.25 ]

Substitute x in y ' = - 0.5 x / (2 y) by + or - sqrt [ (9 - y²) / 0.25 ]

y ' = - 0.5 (+ or - sqrt [ (9 - y²) / 0.25 ] / (2 y)

The slope of the tangent is equal to 1. Hence

- 0.5 (+ or - sqrt [ (9 - y²) / 0.25 ] )/ (2 y) = 1

Solve the above for y. Two solutions

y = 3 sqrt(5) / 5 , y = - 3 sqrt(5) / 5

The answer is A.

Question 3

Find the solution to the differential equation dy/dx = cos(x) / y², where y(π/2) = 0.

A) y = (3 sin(x) - 3)

B) y = sin(x) - 1

C) y = (3 sin(x) - 3)^1/3

D) y = (3 sin(x) - 3)³

E) y = (3 sin(x) - 3)^-1/3

Explanation

The variable in the given differential equation may be separated as follows

y² dy = cos(x) dx

Integrate both sides

∫ y² dy = ∫ cos(x) dx

(1/3) y³ = sin(x) + C , constant of integration

We now use the condition y(π/2) = 0 to find the constant C

(1/3) y³(π/2) = sin(π/2) + C

0 = 1 + C

C = - 1

Substitute C by -1 in (1/3) y³ = sin(x) + C and solve for y

(1/3) y³ = sin(x) - 1

y³ = 3(sin(x) - 1)

y = (3 sin(x) - 3)^1/3

The answer is C.

Question 4

 A) cos⁵(x) + C

B) -(1/5)sin⁵(x) + C

C) sin⁵(x) + C

D) -(1/5)cos⁵(x) + C

E) -5cos⁵(x) + C

Explanation

Let u = cos x and therefore du/dx = - sin x. 

We now substitute cos x by u and sin x by -du/dx in the given integral. Hence

∫cos⁴(x)sin(x) dx = ∫ u⁴ (-du/dx) dx

= - ∫ u⁴du

= (-1/5)u⁵ + C , C constant of integration

= (-1/5)cos⁵(x) + C

The answer is D.

Question 5

A) 2sin(4x² + 1)

B) 2sin(x² + 1)

C) sin(x² + 1)

D) 2 sin(4x² + 1) - 2 sin(3² + 1)

E) 2 sin(4x²)

Explanation

Let u = 2x and therefore du/dx = 2 or dx = du / 2. Hence the given integral becomes

d/dx ∫ ₃^2xsin (t² + 1) dt = 2 d/du ∫ ₃^usin (t² + 1) dt

using the fundamental theorem of calculus, we obtain

= 2 sin (u² + 1)

Substitute u by 2x

= 2 sin (4x² + 1)

The answer is A.

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Question 6

A) 100

B) 108

C) 110

D) 112

E) 114

Explanation

We first analyze the signs of the expressions 4 - x and 2 - 2x between the limits of integration 0 and 10. 4 - x changes sign at x = 4 and 2 - 2x changes sign at x = 1.

for x between 0 and 4: 4 - x is positive and hence |4 - x| = 4 - x

for x between 4 and 10: 4 - x is negative and hence |4 - x| = -(4 - x)

for x between 0 and 1: 2 - 2x is positive and hence |2 - 2x| = 2 - 2x

for x between 1 and 10: 2 - 2x is negative and hence |2 - 2x| = -(2 - 2x)

We now rewrite the given integral as a sum of two integrals as follows.

∫ ₀¹⁰ (|4 - x|+|2 - 2x|) dx =

∫ ₀¹⁰ (|4 - x|) dx + ∫ ₀¹⁰ (|2 - 2x|) dx

We now calculate each of the individual integrals above as follows.

∫ ₀¹⁰ (|4 - x|) dx = ∫ ₀⁴ (4 - x) dx + ∫ ₄¹⁰ -(4 - x) dx = 8 + 18 = 26

and

∫ ₀¹⁰ (|2 - 2x|) dx = ∫ ₀¹ (2 - 2x) dx + ∫ ₁¹⁰ -(2 - 2x) dx = 1 + 81 = 82

We now have

∫ ₀¹⁰ (|4 - x|) dx + ∫ ₀¹⁰ (|2 - 2x|) dx = 26 + 82 = 108

The answer is B.

Question 7

Evaluate the integral:

A) (5 + x^3/4)¹⁰

B) (x^3/4)¹⁰

C) (1/10) (5 + x^3/4)¹⁰

D) (1/10) (5 + x^3/4)¹⁰ / x^1/4

E) (2/15) (5 + x^3/4)¹⁰

Explanation

Let u = 5 + x^3/4 and therefore du/dx = (3/4) 1/x^1/4 and substitute in the given integral

∫ (5 + x^3/4)⁹ / (x^1/4) dx = ∫ [ ( u⁹ ) / (x^1/4) ] (4/3) x^1/4 du

= (4/3) ∫ u⁹ du

= (4/3) (1/10) u¹⁰

= (2/15) (5 + x^3/4)¹⁰

The answer is E.

Question 8

Given that function h is defined by:

find h'(x).

A) (3x² / (x⁶ + 2x³ + 2) + 2)

B) 4 (arctan (x³ + 1) + 2x)³ (3x² / (x⁶ + 2x³ + 2) )

C) 4 (arctan(x³ + 1) + 2x)³

D) 4 (3x² / (x⁶ + 2x³ + 2) + 2)

E) (1/4)(arctan(x³ + 1) + 2x)³

Explanation

Let u = arctan(x³ + 1) + 2x. Hence function h can be written as

h(x) = u⁴ h '(x) = 4 u³ u'

We now let v = arctan(x³ + 1) and calculate u '

u ' = (v ')( 1 / (1 + v²) )

= (3x²) / (1 + (x³ + 1)²)

= (3x²) / (x⁶ + 2x³ + 2)

Hence

h '(x) = 4 (arctan(x³ + 1) + 2x)³ (3x²) / (x⁶ + 2x³ + 2)

The answer is B.

Question 9 

The set of all points (ln(t - 2), 3t), where t is a real number greater than 2, is the graph of

A) y = ln(x/3 - 2)

B) y = 3x

C) x = ln(y - 2)

D) y = 3(e^x + 2)

E) y = ln(x)

Explanation

The given parametric equations may be written as

x(t) = ln (t - 2) and y(t) = 3t

Solve y(t) = 3t for t

t = y / 3

Substitute t by y / 3 in x(t) = ln (t - 2)

x = ln(y / 3 - 2)

Solve for y

y/3 - 2 = e^x y = 3 ( e^x + 2 )

The answer is D.

Question 10 

Let P(x) = 2 x³ + K x + 1. Find K if the remainder of the division of P(x) by x - 2 is equal to 10.

A) -7/2

B) 2/7

C) 7/2

D) -2/7

E) K cannot be determined

Explanation

The Remainder theorem states that the division of P(x) by x - 2 is equal to P(2). Hence

P(2) = 2 (2)³ + K (2) + 1 = 10

Solve for K

K = - 7/2

The answer is A.

Tips to Prepare for The AP Calculus AB Exam

AP courses are rigorous and challenging. After taking AP courses for a whole academic year, the next thing to face is the exams. It is normal for students to worry about passing their AP exams, especially AP Calculus AB. However, individual students can boost their scores when they self-study for the exam.

Below are some tips to prepare for the AP Calculus AB Exam:

• Solve many AP Calculus AB practice questions
• Memorize important formulas
• Know how to use a calculator well
• Practice showing all your work

Remember that success in this exam is as a result of the combination of these tips. 

FAQs: AP Calculus AB Questions

AP Calculus AB exams come with a lot of questions from the students. Here are some frequently asked questions about this AP course:

1. Is AP Calc AB Really Hard?

Students consider AP Calculus AB to be moderately difficult.

2. Is It Hard to Get a 5 on AP Calculus AB?

It is not hard to get a 5 on AP Calculus AB. In 2023, 22% of exam takers that took AP Calculus AB got a 5. For you to get a 5, you will need to be committed to studying the course material and preparing well for the exam. 

3. How Long is the AP Calculus AB Exam?

The AP Calculus AB exam is three hours long.

4. How Long Should I Study for the AP Calculus AB Exam?

There is no definite time to study for the AP Calculus AB exam. However, students who study for two to three months get enough time to prepare for the exam.

5. Is AP Calculus AB Worth It?

The AP Calculus AB is worth the time and energy because of its implication on your academics. It also arms you with the academic skills needed to succeed in college.

6. What Happens If You Fail AP Calculus AB?

Don’t worry if you fail AP Calculus AB at the first attempt. You can always do the exam again to get a better score.

Final Thoughts

Even though AP Calculus AB is not part of the hardest AP classes to pass, it can look challenging and hard. However, the exam is simple enough if you know the AP Calculus AB Practice Questions and Answers. Hence students should take advantage of the available resources and be confident in their abilities.

Good luck!