Home / AP Calculus BC : 1.12 Confirming Continuity over an Interval- Exam Style questions with Answer- MCQ

AP Calculus BC : 1.12 Confirming Continuity over an Interval- Exam Style questions with Answer- MCQ

AP Calculus BC-Questions and Answers - All Topics
Question

Let \( f \) be the function given by

\( f(x) = \frac{|x^2 – 2|(x + 0.4)}{(x^2 – 2)(x + 0.4)} \)

On which of the following open intervals is \( f \) continuous?

A) \( (-2, -1) \)

B) \( (-1, 0) \)

C) \( (0, 1) \)

D) \( (1, 2) \)

▶️ Answer/Explanation
Solution
1. Identify points of discontinuity:
The denominator \((x^2 – 2)(x + 0.4)\) is zero when:
  • \( x^2 – 2 = 0 \implies x = \pm \sqrt{2} \approx \pm 1.414 \)
  • \( x + 0.4 = 0 \implies x = -0.4 \)
2. Check continuity in each interval:
  • A) \( (-2, -1) \): Contains \( x = -1.414 \) (discontinuity). ❌
  • B) \( (-1, 0) \): Contains \( x = -0.4 \) (discontinuity). ❌
  • C) \( (0, 1) \): No discontinuities. ✅
  • D) \( (1, 2) \): Contains \( x = 1.414 \) (discontinuity). ❌
✅ Answer: C) \( (0, 1) \)
Question

The graph of the function \( f \) is shown below:

Graph of function f

On which of the following intervals is \( f \) continuous?

A) \( (0, 1) \)

B) \( (1, 2) \)

C) \( (2, 3) \)

D) \( (3, 4) \)

▶️ Answer/Explanation
Solution
1. Identify continuity requirements:
A function is continuous on an interval if there are no breaks, jumps, or holes in its graph within that interval.
2. Analyze the graph:
  • A) \( (0, 1) \): The graph shows a jump discontinuity at \( x = 1 \). ❌
  • B) \( (1, 2) \): The graph has a hole or jump at \( x = 2 \). ❌
  • C) \( (2, 3) \): The graph breaks at \( x = 3 \). ❌
  • D) \( (3, 4) \): The graph is unbroken and smooth. ✅
✅ Answer: D) \( (3, 4) \)
Question

The function \( f \) is continuous on the interval \( -2 < x < 5 \) and is not continuous on the interval \( -2 \leq x \leq 5 \). Which of the following could not be an expression for \( f(x) \)?

A) \( \frac{(x+2)}{(x-5)} \)

B) \( \frac{(x-5)}{(x+2)} \)

C) \( (x+2)(x−5) \)

D) \( \frac{1}{(x+2)(x-5)} \)

▶️ Answer/Explanation
Solution
1. Understand the condition:
\( f \) must be continuous on \( (-2, 5) \) but discontinuous at \( x = -2 \) and \( x = 5 \).
2. Analyze each option:
  • A) \( \frac{x+2}{x-5} \): Discontinuous at \( x = 5 \) (denominator zero). Continuous on \( (-2, 5) \). ✔️ Possible.
  • B) \( \frac{x-5}{x+2} \): Discontinuous at \( x = -2 \) (denominator zero). Continuous on \( (-2, 5) \). ✔️ Possible.
  • C) \( (x+2)(x-5) \): A polynomial, continuous everywhere (including \( x = -2 \) and \( x = 5 \)). Violates the given condition. ❌ Not possible.
  • D) \( \frac{1}{(x+2)(x-5)} \): Discontinuous at \( x = -2 \) and \( x = 5 \). Continuous on \( (-2, 5) \). ✔️ Possible.
3. Conclusion:
Only option C is continuous on \( -2 \leq x \leq 5 \), contradicting the question’s requirement.
✅ Answer: C) \( (x+2)(x−5) \)
Question

Let \( f \) be the function defined by:

\( f(x) = \begin{cases} x^2 & \text{for } x < 2 \\ 5 & \text{for } x = 2 \\ x + 3 & \text{for } x > 2 \end{cases} \)

For what values of \( x \) is \( f \) NOT continuous?

A) 0 only

B) 1 only

C) 2 only

D) 0 and 2 only

E) 0, 1, and 2

▶️ Answer/Explanation
Solution
Check continuity at \( x = 0 \), \( x = 1 \), and \( x = 2 \).
At \( x = 0 \): \( f(0) = 0^2 = 0 \), limit \( \lim_{x \to 0^-} x^2 = 0 \), so continuous.
At \( x = 1 \): \( f(1) = 1^2 = 1 \), limit \( \lim_{x \to 1^-} x^2 = 1 \), so continuous.
At \( x = 2 \): \( f(2) = 5 \), left limit \( \lim_{x \to 2^-} x^2 = 4 \), right limit \( \lim_{x \to 2^+} (x + 3) = 5 \).
Since \( 4 \neq 5 \), the left and right limits differ, so \( f \) is not continuous at \( x = 2 \).
✅ Answer: C)
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