AP Calculus AB Practice Quiz: Removing Discontinuities
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 9 questions to check your progress.
Question 1 of 9
All Questions (9)
A) -1
B) 1
C) 2
D) 5
Correct Answer: A
For the piecewise function to be continuous at the boundary x=2, the value of the expression on the left side must equal the value of the expression on the right side. We set the two expressions equal at x=2: k(2) + 5 = (2)^2 - 1. This simplifies to 2k + 5 = 4 - 1, or 2k + 5 = 3. Solving for k gives 2k = -2, so k = -1.
A) 0
B) 3
C) 6
D) 9
Correct Answer: C
To remove the discontinuity, we must define f(3) to be equal to the limit of f(x) as x approaches 3. We can find the limit by factoring the numerator: lim(x→3) (x-3)(x+3) / (x-3). The (x-3) terms cancel, leaving lim(x→3) (x+3). Evaluating the limit gives 3+3 = 6. Therefore, f(3) should be defined as 6.
A) The limit of g(x) as x approaches c exists.
B) The function g(c) is defined.
C) The limit of g(x) as x approaches c from the left does not equal the limit as x approaches c from the right.
D) The limit of g(x) as x approaches c is infinite.
Correct Answer: A
According to the definition, a discontinuity at a point can be removed if the limit of the function exists at that point. The discontinuity is removed by defining or redefining the function's value at that point to be equal to the value of the limit.
A) 2
B) 3
C) 4
D) 5
Correct Answer: C
For f to be continuous at the boundary x=1, the limit from the left must equal the limit from the right. We set the two pieces of the function equal to each other at x=1: (1)^2 + a(1) = 3(1) + 2. This simplifies to 1 + a = 3 + 2, or 1 + a = 5. Solving for a gives a = 4.
A) -3
B) -1
C) 1
D) 3
Correct Answer: D
For f(x) to have a removable discontinuity at x=2, the limit as x approaches 2 must exist. This requires that (x-2) is a factor of the numerator, which means the numerator must be zero when x=2. We set the numerator equal to zero at x=2: (2)^2 + k(2) - 10 = 0. This simplifies to 4 + 2k - 10 = 0, or 2k - 6 = 0. Solving for k gives 2k = 6, so k = 3.
A) g(x) and h(x) must be equal for all x.
B) The derivative of g(x) at x=a must equal the derivative of h(x) at x=a.
C) The limit of g(x) as x approaches a from the left must equal the value of h(a).
D) The function must simply be defined at x=a.
Correct Answer: C
For a piecewise function to be continuous at a boundary point 'a', three conditions must be met: the limit from the left must exist, the limit from the right must exist, and the function value at 'a' must be equal to these limits. In this case, lim(x→a⁻) g(x) must equal lim(x→a⁺) h(x), which is h(a). Therefore, the value of the expression on one side (the left-hand limit) must equal the value of the function at the boundary.
A) 0
B) 1
C) 2
D) 3
Correct Answer: C
For f to be continuous, it must be continuous at the boundaries x=0 and x=2. At x=0, we set the first two expressions equal: 2(0) + 1 = a(0) + b, which gives b = 1. At x=2, we set the second and third expressions equal: a(2) + b = (2)^2 - 1, which gives 2a + b = 3. Substituting b=1 into the second equation gives 2a + 1 = 3, so 2a = 2, and a = 1. Therefore, a + b = 1 + 1 = 2.
A) 0
B) 1
C) 4
D) The value cannot be determined.
Correct Answer: C
To remove the discontinuity at x=0, f(0) must be defined as the limit of f(x) as x approaches 0. We use the known limit property lim(u→0) sin(u)/u = 1. We can rewrite the expression as lim(x→0) 4 * (sin(4x) / 4x). As x approaches 0, 4x also approaches 0. Therefore, the limit is 4 * 1 = 4. The function must be defined as f(0) = 4 to be continuous.
A) -1
B) 0
C) 1
D) 2
Correct Answer: C
For the function to be continuous at the boundary x=1, the value of the expression defining the function on one side of the boundary must equal the value of the expression on the other side. We set the two expressions equal at x=1: (1)^2 + 1 = 3 - a(1). This simplifies to 2 = 3 - a. Solving for a gives a = 3 - 2, so a = 1.