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📝 Intermediate value theorem IVT (18 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 18 questions available

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What is Intermediate value theorem IVT?

Definition:
The Intermediate Value Theorem states that if $f$ is continuous on a closed interval $[a, b]$ and $N$ is any number between $f(a)$ and $f(b)$ , then there exists at least one $c \in (a, b)$ such that $f(c) = N$ . This guarantees that a continuous function takes on every intermediate value between its endpoints, ensuring no gaps in its range over the interval.

Example:
Show $f(x) = x^3 - x - 1$ has a root in $(1,2)$ .
Solution: $f(1) = -1$ , $f(2) = 5$ ; since $0$ is between $-1$ and $5$ , there is $c$ with $f(c)=0$ .

Reason:
IVT is a powerful existence theorem used to prove the existence of solutions to equations, find roots numerically, and justify that continuous functions have no gaps, which is critical in optimization and root-finding algorithms.

Easy

Medium

Hard

📝 All Intermediate value theorem IVT MCQs

Q1. Suppose $f$ is continuous on $[a,b]$ with $f(a)=2$ and $f(b)=5$ . Which of the following statements must be true?

f(c)=3

for some

c\in(a,b)

✅

f(c)=6

for some

c\in(a,b)

f

is constant on

[a,b]

f

attains its maximum at

a

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: By the Intermediate Value Theorem, any value between 2 and 5, such as 3, must be taken by $f$ at some point in the interval. Values outside that range, a constant function, or a maximum at the left endpoint are not guaranteed.

Q2. If $f$ is continuous on $[0,3]$ with $f(0)=-1$ and $f(3)=2$ , define $g(x)=f(x)-1$ . In which interval must a root of $g$ lie?

(0,1)

(1,2)

(2,3)

✅

(0,3)

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: $g(0)=-2$ and $g(3)=1$ have opposite signs, so by the IVT there is at least one $c$ with $g(c)=0$ in $[0,3]$ . Because the endpoints are non‑zero, the root cannot be at 0 or 3, guaranteeing a root somewhere inside the interval.

Q3. A continuous function $h$ on $[a,b]$ satisfies $h(a)=4$ and $h(b)=-4$ . Which statement is a valid deduction?

A.There exists

c

with

h(c)=0

and another

d

with

h(d)=2

. ✅

B.There exists exactly one

c

with

h(c)=0

h

must be decreasing on

[a,b]

D.No value between

-4

and

4

is guaranteed.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The IVT ensures that every value between $-4$ and $4$ is attained. Hence there is at least one point where $h$ equals 0 and at least one where it equals 2. Uniqueness or monotonicity are not implied, making only the first statement certain.

Q4. Which of the following highlights a key difference between the Intermediate Value Theorem and the Mean Value Theorem?

A.IVT concerns function values; MVT concerns derivative values. ✅

B.IVT requires differentiability; MVT requires continuity.

C.Both theorems guarantee a point where the function equals its average value.

D.IVT applies only to polynomials, while MVT applies to all functions.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The IVT guarantees that a continuous function takes every intermediate value, while the MVT guarantees a point where the derivative equals the average rate of change. Differentiability is required for MVT, not for IVT, making option A the correct distinction.

Q5. For the polynomial $p(x)=x^{5}-x+3$ on $[1,2]$ , which set of $k$ values guarantees at least one solution to $p(x)=k$ ?

k\in[3,33]

✅

k\in[1,2]

k\in[0,10]

k\in[5,15]

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Since $p(1)=3$ and $p(2)=33$ and $p$ is continuous, the IVT tells us that every number between 3 and 33 is taken by $p$ somewhere in $[1,2]$ . Thus any $k$ in that interval yields a solution.

Q6. Let $f$ be continuous on $[a,b]$ with $f(a)=1$ and $f(b)=4$ . Let $g$ be continuous on $[a,b]$ with $g(a)=2$ and $g(b)=3$ . Which statement must be true about the equation $f(x)=g(x)$ ?

A.There is at least one solution in

[a,b]

. ✅

B.There is exactly one solution in

[a,b]

C.No solution exists in

[a,b]

D.There are at least two solutions in

[a,b]

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Define $h(x)=f(x)-g(x)$ . Then $h$ is continuous, $h(a)= -1$ and $h(b)=1$ ; opposite signs guarantee, by the IVT, a point where $h(c)=0$ , i.e., $f(c)=g(c)$ . Uniqueness cannot be assured.

Q7. Which interval must contain a root of $f(x)=x^{3}-x-2$ on $[1,2]$ ?

[1,1.5]

[1.5,2]

(1,2)

✅

[0,3]

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Evaluating, $f(1)=-2$ and $f(2)=4$ have opposite signs, so the IVT ensures a zero lies between them, i.e., in the open interval $(1,2)$ .

Q8. In the bisection method, how does the Intermediate Value Theorem justify each iteration?

A.It guarantees the midpoint is a root.

B.It ensures the chosen subinterval still contains a root because the function values at its ends have opposite signs. ✅

C.It proves the function becomes linear after each step.

D.It shows the error decreases exponentially.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: At each step the method evaluates the function at the midpoint; whichever half retains opposite signs at its endpoints must contain a root by the IVT, allowing the algorithm to continue narrowing the interval.

Q9. Which of the following statements about the IVT is false when the function is not monotonic?

A.The IVT still guarantees a root for any intermediate value.

B.The IVT may give multiple points for the same value.

C.The IVT does not require monotonicity.

D.The IVT fails if the function oscillates infinitely often. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The IVT does not depend on monotonicity; it only needs continuity. Even if a function oscillates, every value between the endpoint values is still attained, possibly many times. Hence the statement that the IVT fails under such oscillation is false.

Q10. What does the Intermediate Value Theorem state?

A.If a function is differentiable on

[a,b]

, it attains every value between

f(a)

and

f(b)

B.If a function is continuous on

[a,b]

, it attains every value between

f(a)

and

f(b)

. ✅

C.If a function is continuous on

(a,b)

, it must have a maximum and minimum.

D.If a function is integrable on

[a,b]

, it attains its average value.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The theorem asserts that continuity on a closed interval guarantees the function takes every intermediate value between its endpoint values. Differentiability, extrema, or integrability are not part of the statement.

Q11. If $f(a)=7$ and $f(b)=-3$ , which of the following values must be assumed by $f$ on $[a,b]$ ?

0

✅

5

-5

10

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Because 0 lies between -3 and 7, the IVT ensures that a continuous function on $[a,b]$ must equal 0 at some point in the interval.

Q12. Suppose $f$ is continuous on $[2,5]$ with $f(2)=-2$ and $f(5)=6$ . Let $q(x)=f(x)^{2}-4$ . Which interval must contain a root of $q$ ?

(2,5)

[2,5]

✅

(3,4)

[3,4]

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: At $x=2$ , $q(2)=(-2)^{2}-4=0$ ; at $x=5$ , $q(5)=36-4=32>0$ . Since $q$ is continuous, a zero certainly exists in the closed interval $[2,5]$ ; it may occur at the endpoint, so $[2,5]$ is the guaranteed interval.

Q13. Is the following statement always true? \If $|f(a)-k|<0.001$ for a continuous $f$ on $[a, b]$ , then there exists $c$ with $|f(c)-k|<0.001$ .\

A.Always true

B.Sometimes false ✅

C.ALWAYS FALSE

D.Cannot be determined without more information

💡 Difficulty: medium | ✅ Correct: B

Q14. Why can't the Intermediate Value Theorem be applied to a discontinuous function?

A.Discontinuities may cause the function to skip values between the endpoints. ✅

B.Discontinuous functions are not defined at the endpoints.

C.The theorem requires differentiability, which discontinuous functions lack.

D.It can be applied; continuity is not essential.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The IVT relies on the function's graph being unbroken; a jump or removable discontinuity can prevent the function from attaining some intermediate values, violating the theorem's premise.

Q15. For a continuous $f$ on $[0,1]$ with $f(0)=0$ and $f(1)=1$ , which theorem guarantees a point $c$ such that $f(c)=c$ ?

A.Intermediate Value Theorem

B.Fixed Point Theorem

C.Mean Value Theorem ✅

D.Rolle's Theorem

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Define $g(x)=f(x)-x$ . Then $g$ is continuous, $g(0)=0$ and $g(1)=0$ ; applying the IVT (or directly noting the sign change) shows a point where $g(c)=0$ , i.e., $f(c)=c$ .

Q16. Let $f$ be continuous on $[a,b]$ with $f(a)=2$ and $f(b)=8$ . Define $h(x)=f(x)^{3}-27$ . How many solutions can $h(x)=0$ have in $(a,b)$ ?

A.At least one ✅

B.Exactly one

C.At most one

D.Possibly none

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Compute $h(a)=2^{3}-27=-19$ and $h(b)=8^{3}-27=485$ ; the signs differ, so by the IVT there is at least one root of $h$ in $(a,b)$ . The theorem does not limit the number of roots.

Q17. A temperature sensor records 15 °C at midnight and 25 °C at noon, varying continuously. Which statement is guaranteed by the IVT?

A.The temperature will be exactly 20 °C at noon.

B.The temperature will exceed 30 °C before noon.

C.The temperature will be 20 °C at some time between midnight and noon. ✅

D.The temperature will be constant for an hour.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Since 20 °C lies between the recorded values 15 °C and 25 °C, continuity ensures the temperature must attain 20 °C at some moment in the interval, making option C certain.

Q18. Which condition is NOT required for the Intermediate Value Theorem to hold?

A.The function is continuous on the closed interval.

B.The interval

[a,b]

is closed.

C.The function is differentiable on

[a,b]

. ✅

D.The values

f(a)

and

f(b)

are defined.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Differentiability is unnecessary; the IVT only needs continuity on a closed interval. Hence the statement about differentiability is the one not required.

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