Home / AP Calculus BC : 1.10 Exploring Types of 3 Discontinuities- Exam Style questions with Answer- MCQ

AP Calculus BC : 1.10 Exploring Types of 3 Discontinuities- Exam Style questions with Answer- MCQ

AP Calculus BC-Questions and Answers - All Topics
Question
Graph of function f

The graph of a function \( f \) is shown in the figure above. At what value of \( x \) does \( f \) have a jump discontinuity?

A) \( x = 1 \)

B) \( x = 3 \)

C) \( x = 4 \)

D) \( x = 5 \)

▶️ Answer/Explanation
Solution
1. Jump Discontinuity Identification:
  • A jump discontinuity occurs where the function has finite but unequal left-hand and right-hand limits.
  • In the graph, this is visible at \( x = 1 \), where the function “jumps” from one value to another.
2. Graph Analysis:
  • At \( x = 1 \), the graph shows an abrupt vertical gap between two points.
  • No similar gaps are present at \( x = 3 \), \( x = 4 \), or \( x = 5 \).
3. Conclusion:
The function \( f \) has a jump discontinuity at \( x = 1 \).
✅ Answer: A) \( x = 1 \)
Question

A)

Graph A

B)

Graph B

C)

Graph C

D)

Graph D

The function f has a removable discontinuity at x = 3. Which of the following could be the graph of f?

A) Graph A

B) Graph B

C) Graph C

D) Graph D

▶️ Answer/Explanation
Solution
1. Removable Discontinuity Identification:
  • A removable discontinuity occurs when the left-hand and right-hand limits of the function exist and are equal (limx→3⁻ f(x) = limx→3⁺ f(x)), but the function is either undefined or has a different value at x = 3.
  • This is typically represented on a graph by a hole at x = 3, where the function approaches the same y-value from both sides but is not defined at that point.
✅ Answer: A) Graph A
Question

The function \( f(x) \) is defined as:

\( f(x) = \begin{cases} \frac{2x^2 – 5x – 3}{x – 3} & \text{if } x \neq 3 \\ 9 & \text{if } x = 3 \end{cases} \)

Which of the following statements is true?

A) \( f \) is continuous at \( x = 3 \).

B) \( f \) has a removable discontinuity at \( x = 3 \).

C) \( f \) has a jump discontinuity at \( x = 3 \).

D) \( f \) has a discontinuity due to a vertical asymptote at \( x = 3 \).

▶️ Answer/Explanation
Solution
1. Evaluate the Limit at \( x = 3 \):
  • For \( x \neq 3 \), the function is \( f(x) = \frac{2x^2 – 5x – 3}{x – 3} \).
  • Simplify the expression: \( 2x^2 – 5x – 3 = (x – 3)(2x + 1) \), so \( f(x) = \frac{(x – 3)(2x + 1)}{x – 3} = 2x + 1 \) (for \( x \neq 3 \)).
  • Compute the limit: \( \lim_{x \to 3} f(x) = \lim_{x \to 3} (2x + 1) = 2(3) + 1 = 7 \).
2. Compare with the Function Value at \( x = 3 \):
  • At \( x = 3 \), the function is defined as \( f(3) = 9 \).
  • Since \( \lim_{x \to 3} f(x) = 7 \) but \( f(3) = 9 \), the limit exists but does not equal the function value at \( x = 3 \).
3. Determine the Type of Discontinuity:
  • A removable discontinuity occurs when the limit exists but does not equal the function value. Here, the limit is 7, and \( f(3) = 9 \), so there is a removable discontinuity at \( x = 3 \).
  • The function is not continuous at \( x = 3 \) (rules out option A).
  • There is no jump discontinuity, as the left and right limits are equal (rules out option C).
  • There is no vertical asymptote, as the limit is finite (rules out option D).
4. Conclusion:
The function \( f \) has a removable discontinuity at \( x = 3 \).
✅ Answer: B) \( f \) has a removable discontinuity at \( x = 3 \).
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