Home / AP Calculus BC : 1.13 Removing  Discontinuities- Exam Style questions with Answer- MCQ

AP Calculus BC : 1.13 Removing  Discontinuities- Exam Style questions with Answer- MCQ

Question

\(f\left( x \right) = \begin{cases} e^{bx} & \text{for } \, x \leq 3 \\ \frac{2}{3}x + b & \text{for } \, x > 3 \end{cases}\)

Let f be the function defined above. For what values of b is f continuous at x=3?

A) 0.394 only

B) 0.274 only

C) −4.500 and 0.394

D) −1.998 and 0.274

▶️Answer/Explanation
Markscheme

Answer: D

Explanation:

For continuity at x=3, the left-hand limit (x→3⁻) must equal the right-hand limit (x→3⁺) and both must equal f(3).

1. Left-hand limit (x≤3): \( \lim_{x→3^-} f(x) = e^{3b} \)

2. Right-hand limit (x>3): \( \lim_{x→3^+} f(x) = \frac{2}{3}(3) + b = 2 + b \)

3. Set them equal: \( e^{3b} = 2 + b \)

4. Solving this equation gives two solutions: b ≈ -1.998 and b ≈ 0.274

Therefore, the correct answer is D) −1.998 and 0.274.

Question

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

\( g(x) = \begin{cases} \frac{x^2 – 4}{4x + 8} & \text{for } x \neq -2 \\ k & \text{for } x = -2 \end{cases} \)

For what value of \( k \) is \( g \) continuous at \( x = -2 \)?

A) -2

B) -1

C) -12

D) 0

▶️ Answer/Explanation
Solution
For \( g \) to be continuous at \( x = -2 \), \( g(-2) = k \) must equal \( \lim_{x \to -2} g(x) \).
Compute the limit: \( \lim_{x \to -2} \frac{x^2 – 4}{4x + 8} = \lim_{x \to -2} \frac{(x-2)(x+2)}{4(x+2)} \).
Cancel \( (x+2) \): \( \lim_{x \to -2} \frac{x-2}{4} = \frac{-2-2}{4} = -1 \).
Thus, \( k = -1 \).
✅ Answer: B)
Question

\(f(x) = \begin{cases} 2c + c \sin\left(\frac{\pi}{2}x\right) & \text{for } x < 3 \\ 7 & \text{for } x = 3 \\ 2c + 5x & \text{for } x > 3 \end{cases}\)

Let f be the function defined above. For what value of c, if any, is f continuous at x=3?

A) −4

B) 7

C) −15

D) There is no such c.

▶️Answer/Explanation
Markscheme

Answer: D

Explanation:

For continuity at x=3, the following must hold:

1. Left-hand limit (x→3⁻): \( \lim_{x→3^-} f(x) = 2c + c \sin\left(\frac{3\pi}{2}\right) = 2c – c = c \)

2. Right-hand limit (x→3⁺): \( \lim_{x→3^+} f(x) = 2c + 15 \)

3. Function value: \( f(3) = 7 \)

For continuity, all three must be equal:

\( c = 2c + 15 = 7 \)

This leads to two equations:

1. \( c = 7 \)

2. \( 2c + 15 = 7 \) ⇒ \( c = -4 \)

Since c cannot be both 7 and -4 simultaneously, there is no value of c that makes f continuous at x=3.

Question

Which of the following are point(s) of discontinuity of the function \(f(x)=\frac{3x+1}{2x^{3}-8x^{2}-64x}\)?

I. 0

II. 4

III. 8

(A) I only

(B) II only

(C) I and II only

(D) I and III only

▶️Answer/Explanation
Markscheme

Answer: D

Explanation:

1. Find where the denominator equals zero:

\(2x^{3}-8x^{2}-64x = 0\)

Factor: \(2x(x^{2}-4x-32) = 0\)

Solutions: \(x = 0\) or \(x = \frac{4 \pm \sqrt{16+128}}{2} = \frac{4 \pm 12}{2}\)

Thus: \(x = 0\), \(x = 8\), or \(x = -4\)

2. The function is discontinuous at these points (I: 0 and III: 8)

3. At x = 4 (II), the function is continuous since the denominator is not zero

4. Therefore, the correct answer is (D) I and III only

Rational functions are continuous everywhere except where the denominator is 0.

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