Home / AP Calculus BC : 1.16 Working with the Intermediate Value  Theorem (IVT)- Exam Style questions with Answer- MCQ

AP Calculus BC : 1.16 Working with the Intermediate Value  Theorem (IVT)- Exam Style questions with Answer- MCQ

AP Calculus BC-Questions and Answers - All Topics
Question

Let \( f \) be a function such that \( f(1) = -2 \) and \( f(5) = 7 \). Which of the following conditions ensures that \( f(c) = 0 \) for some value \( c \) in the open interval \( (1, 5) \)?

A) \( \int_1^5 f(x) \, dx \) exists.

B) \( f \) is increasing on the closed interval \( [1, 5] \).

C) \( f \) is continuous on the closed interval \( [1, 5] \).

D) \( f \) is defined for all values of \( x \) in the closed interval \( [1, 5] \).

▶️ Answer/Explanation
Solution
Given \( f(1) = -2 \) and \( f(5) = 7 \), with values of opposite signs, the Intermediate Value Theorem ensures \( f(c) = 0 \) for some \( c \) in \( (1, 5) \) if \( f \) is continuous on \( [1, 5] \).
A) The existence of \( \int_1^5 f(x) \, dx \) doesn’t guarantee a zero crossing.
B) Increasing \( f \) means \( f(x) \) grows from -2 to 7, but continuity is needed for the theorem.
C) Continuity on \( [1, 5] \) is the key condition.
D) Being defined everywhere doesn’t ensure a zero if \( f \) is discontinuous.
✅ Answer: C)
Question

Let \( f \) be the function given by \( f(x) = 2x + \tan\left(\frac{x}{3}\right) – 15 \). The Intermediate Value Theorem applied to \( f \) on the closed interval \( [10, 15] \) guarantees a solution in \( [10, 15] \) to which of the following equations?

A) \( f(x) = -15 \)

B) \( f(x) = 0 \)

C) \( f(x) = 5 \)

D) \( f(x) = 15 \)

▶️ Answer/Explanation
Solution
The Intermediate Value Theorem requires \( f \) to be continuous on \( [10, 15] \). Since \( f(x) = 2x + \tan\left(\frac{x}{3}\right) – 15 \) is a sum of continuous functions, it applies.
Evaluate at the endpoints: \( f(10) = 2(10) + \tan\left(\frac{10}{3}\right) – 15 \approx 20 + 1.53 – 15 = 6.53 \).
\( f(15) = 2(15) + \tan\left(\frac{15}{3}\right) – 15 = 30 + \tan(5) – 15 \approx 15 – 3.38 = 11.62 \).
The theorem guarantees \( f(x) = k \) has a solution if \( k \) is between \( 6.53 \) and \( 11.62 \).
Check options: A) \( f(x) = -15 \) (not in range), B) \( f(x) = 0 \) (not in range), C) \( f(x) = 5 \) (in range), D) \( f(x) = 15 \) (not in range).
✅ Answer: C)
Question

Let \( f \) be a function such that \( f(3) < 4 < f(5) \). Which of the following statements provides sufficient additional information to conclude that there is a value \( x = c \) in the interval \( [3, 5] \) such that \( f(c) = 4 \)?

A) \( f \) is defined for all \( x \).

B) \( f \) is increasing for all \( x \).

C) \( f \) is continuous for all \( x \).

D) There is a value \( x = c \) in the interval \( [3, 5] \) such that \( \lim_{x \to c} f(x) = 4 \).

▶️ Answer/Explanation
Solution
Given \( f(3) < 4 < f(5) \), the Intermediate Value Theorem ensures \( f(c) = 4 \) for some \( c \) in \( [3, 5] \) if \( f \) is continuous on \( [3, 5] \).
A) Being defined everywhere doesn’t guarantee continuity.
B) Increasing \( f \) doesn’t ensure it hits 4 without continuity.
C) Continuity on \( [3, 5] \) is sufficient.
D) A limit equaling 4 at some \( c \) doesn’t imply \( f(c) = 4 \) unless \( f \) is continuous there.
✅ Answer: C)
Question

Let \( f \) be a function of \( x \). Which of the following statements, if true, would guarantee that there is a number \( c \) in the interval \( [-5, 4] \) such that \( f(c) = 12 \)?

A) \( f \) is increasing on the interval \( [-5, 4] \), where \( f(-5) = 0 \) and \( f(4) = 20 \).

B) \( f \) is increasing on the interval \( [-5, 4] \), where \( f(-5) = 15 \) and \( f(4) = 30 \).

C) \( f \) is continuous on the interval \( [-5, 4] \), where \( f(-5) = 0 \) and \( f(4) = 20 \).

D) \( f \) is continuous on the interval \( [-5, 4] \), where \( f(-5) = 15 \) and \( f(4) = 30 \).

▶️ Answer/Explanation
Solution
To guarantee \( f(c) = 12 \) for some \( c \) in \( [-5, 4] \), use the Intermediate Value Theorem, which requires continuity and \( f \) values bracketing 12.
A) \( f \) increasing, \( f(-5) = 0 \), \( f(4) = 20 \): 12 is in range, but increasing doesn’t ensure continuity.
B) \( f \) increasing, \( f(-5) = 15 \), \( f(4) = 30 \): 12 is not in range.
C) \( f \) continuous, \( f(-5) = 0 \), \( f(4) = 20 \): 12 is in range, continuity guarantees a solution.
D) \( f \) continuous, \( f(-5) = 15 \), \( f(4) = 30 \): 12 is not in range.
✅ Answer: C)
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