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📝 Continuous functions on an interval (17 MCQs)

📖 From Calculus • 2. Limits and Continuity an Introduction • 17 questions available

←Previous Topic:Continuity definition in calculus📝 Continuous functions on an intervalContinuity of polynomial and rational functions:Next Topic→

What is Continuous functions on an interval?

Continuous Functions on an Interval Definition:
A function $f(x)$ is said to be continuous on a closed interval $[a, b]$ if it is continuous at every single point inside the open interval $(a, b)$ , and additionally, the right-hand limit at $a$ equals $f(a)$ while the left-hand limit at $b$ equals $f(b)$ . Intuitively, this means the graph can be drawn over the entire interval without ever lifting your pencil from the paper, with no breaks, jumps, or holes.

Working:
To check continuity on an interval, we first verify that for any interior point $c$ in $(a, b)$ , the condition $\lim_{x \to c} f(x) = f(c)$ holds true. Then, we separately check the two endpoints: we ensure $\lim_{x \to a^+} f(x) = f(a)$ and $\lim_{x \to b^-} f(x) = f(b)$ . If all these conditions are satisfied, the function is continuous on the entire closed interval.

Example:
Show that $f(x) = x^2$ is continuous on the interval $[1, 3]$ .
For any $c$ in $(1, 3)$ , we know $\lim_{x \to c} x^2 = c^2 = f(c)$ .
At the left endpoint: $\lim_{x \to 1^+} x^2 = 1 = f(1)$ .
At the right endpoint: $\lim_{x \to 3^-} x^2 = 9 = f(3)$ .
Since all conditions are met, $f(x) = x^2$ is continuous on $[1, 3]$ .

Reason:
Continuity on an interval is a crucial property because it guarantees the application of powerful theorems like the Intermediate Value Theorem and the Extreme Value Theorem. These theorems allow us to conclude the existence of solutions to equations and the presence of maximum and minimum values, which are essential in optimization and real-world modeling.

Easy

Medium

Hard

📝 All Continuous functions on an interval MCQs

Q1. A function $f$ is continuous on $[0,2]$ and satisfies $f(0)=5$ , $f(2)=-3$ . Which statement must be true?

A.There exists

c \in (0,2)

with

f(c)=0

B.There exists

c \in (0,2)

with

f(c)=1

C.There exists

c \in (0,2)

with

f(c)=4

D.All of the above ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: By the Intermediate Value Theorem (IVT), since 5 and -3 bracket every value between -3 and 5, there must be points where $f$ equals 0, 1, and 4. This tests the full range of the IVT, not just the zero.

Q2. Let $f(x)=\frac{x^2-4}{x-2}$ for $x\neq 2$ . How should $f(2)$ be defined to make $f$ continuous on $[0,3]$ ?

A.0

B.2

C.4 ✅

D.undefined

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Simplifying gives $f(x)=x+2$ for $x\neq2$ . The limit as $x\to2$ is 4. Defining $f(2)=4$ removes the hole. This requires recognizing removable discontinuity and applying continuity definition.

Q3. A student claims: $f(x)=\frac{1}{x}$ is continuous on $[-1,1]$ because it is continuous everywhere except at 0. What is the error?

A.The function is not defined at 0, so interval must exclude 0.

B.The function is not bounded on the interval.

C.Both A and B ✅

D.The student is correct

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Continuity on a closed interval requires continuity at every point in the interval. Since $f$ is not defined at 0, it fails. Also, it is unbounded near 0, so even an extension would fail. The error is both definitional and behavioral.

Q4. The graph of $g$ is a continuous curve on $[-2,3]$ with $g(-2)=1$ and $g(3)=5$ . If the horizontal line $y=3$ intersects the graph exactly twice, what can we conclude?

A.The function is not one-to-one

B.The function has at least one local extremum

C.Both A and B ✅

D.The function must be constant somewhere

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: By IVT, $y=3$ must be attained at least once. Exactly two intersections mean the function goes above and below 3, requiring a turning point (extremum) between them. Also, two distinct points with same output violate one-to-one.

Q5. If $f$ is continuous on $[a,b]$ and $f(a)f(b)<0$ , which of the following is FALSE?

A.There exists

c\in(a,b)

with

f(c)=0

B.The equation

f(x)=0

has at least one solution in

(a,b)

C.The function must change sign at least once

f

must be strictly increasing or decreasing ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The sign change condition guarantees a root by IVT, but it does not imply monotonicity. The function could oscillate wildly while still crossing zero. This tests the limitation of IVT.

Q6. A tank is being filled at a rate that varies continuously. The volume $V(t)$ in liters at time $t$ minutes is continuous on $[0,10]$ , with $V(0)=50$ , $V(10)=200$ . Which is guaranteed?

A.There is a time when volume is exactly 100 L ✅

B.There is a time when fill rate is exactly 15 L/min

C.Both A and B

D.Neither

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: IVT guarantees volume 100 L because 100 is between 50 and 200. However, rate is derivative, not given continuous, and even if continuous, MVT requires differentiability. This distinguishes between value and rate.

Q7. Let $f$ be continuous on $[1,5]$ . If $f(1)=2$ and $f(5)=10$ , and $f$ is one-to-one, what is the minimum number of times the line $y=7$ intersects the graph?

A.0

B.1 ✅

C.2

D.Cannot determine

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: One-to-one continuous function is strictly monotonic. Since 7 lies between 2 and 10, IVT gives exactly one intersection. Strict monotonicity prevents more than one. This combines IVT with monotonicity.

Q8. Which function is continuous on its entire domain but NOT uniformly continuous?

f(x)=\sin x

\mathbb{R}

f(x)=x^2

[0,1]

f(x)=\frac{1}{x}

(0,1]

✅

f(x)=\sqrt{x}

[0,\infty)

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: $1/x$ on (0,1] is continuous but not uniformly continuous because it blows up near 0. Uniform continuity requires a single $\delta$ for all points; near 0, the slope is unbounded, violating it.

Q9. Suppose $f$ is continuous on $[0,1]$ and $f(x)\in\mathbb{Q}$ for all $x\in[0,1]$ . What must be true?

f

is constant ✅

f

is rational-valued only at rational points

f

is discontinuous everywhere

f

is the Dirichlet function

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: A continuous function on a connected interval has an image that is an interval. The only intervals contained in $\mathbb{Q}$ are single points. Therefore, $f$ must be constant. This connects continuity, connectedness, and density of irrationals.

Q10. Given $f(x)=\begin{cases} x\sin(1/x), & x\neq0 \\ 0, & x=0 \end{cases}$ . Which statement is correct?

f

is continuous at 0

f

is differentiable at 0

C.Both A and B ✅

D.Neither

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: $\lim_{x\to0} x\sin(1/x)=0=f(0)$ , so continuous. Also, $\lim_{h\to0} \frac{h\sin(1/h)-0}{h} = \lim \sin(1/h)$ which does not exist, so not differentiable. This tests continuity vs differentiability.

Q11. A function $h$ is continuous on $[-2,2]$ and $h(-2)=4$ , $h(2)=4$ . Which of the following is necessarily true?

A.There exists

c\in(-2,2)

with

h(c)=4

✅

B.There exists

c\in(-2,2)

with

h(c)>4

C.There exists

c\in(-2,2)

with

h(c)<4

D.None of the above

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Since endpoints both equal 4, by IVT, for any value between 4 and itself, only 4 is guaranteed. But actually, the function attains 4 at endpoints; by continuity, it attains 4 somewhere in between trivially. However, if we require interior, it might be constant 4. So option A is guaranteed (endpoints included? But c in open interval—if constant, any c works). So A is correct.

Q12. If $f$ is continuous on $[0,2]$ and $f(0)=1$ , $f(2)=3$ , which equation MUST have a solution in $(0,2)$ ?

f(x)=x

f(x)=x^2

C.Both A and B ✅

D.Neither

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Define $g(x)=f(x)-x$ . $g(0)=1>0$ , $g(2)=1>0$ — no sign change. But $f(x)=x^2$ : define $h(x)=f(x)-x^2$ . $h(0)=1>0$ , $h(2)=3-4=-1<0$ , so by IVT, solution. For $f(x)=x$ , not guaranteed. So only B. Correction: The correct answer should be B only. Let's adjust.

Q13. (Corrected) If $f$ is continuous on $[0,2]$ and $f(0)=1$ , $f(2)=3$ , which equation MUST have a solution in $(0,2)$ ?

f(x)=x

f(x)=x^2

✅

f(x)=x+1

f(x)=x^2+1

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: For $f(x)=x^2$ , define $h(x)=f(x)-x^2$ . $h(0)=1>0$ , $h(2)=3-4=-1<0$ , so by IVT, solution. For $f(x)=x$ , $g(0)=1>0$ , $g(2)=1>0$ , no sign change. So only B is guaranteed.

Q14. The graph of $f$ is continuous on $[0,4]$ and consists of a line from (0,0) to (2,4) and then a line from (2,4) to (4,0). How many solutions does $f(x)=x$ have?

A.0

B.1

C.2 ✅

D.3

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: First segment: $f(x)=2x$ on [0,2]. Solve $2x=x \Rightarrow x=0$ (endpoint). Second segment: line from (2,4) to (4,0): slope -2, equation $f(x)=-2x+8$ . Solve $-2x+8=x \Rightarrow 3x=8 \Rightarrow x=8/3$ . So interior solution at 8/3, and endpoint 0? But open interval? For (0,4), only 8/3, so 1 solution. Let's correct: On [0,2], only x=0 which is endpoint; on (2,4), x=8/3 is interior. So total 1 in (0,4).

Q15. (Corrected) The graph of $f$ is continuous on $[0,4]$ with $f(0)=0, f(2)=4, f(4)=0$ , linear pieces. How many solutions to $f(x)=x$ in (0,4)?

A.0

B.1 ✅

C.2

D.3

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: First piece: from (0,0) to (2,4): $f=2x$ , solve $2x=x$ gives x=0 (not in open). Second piece: from (2,4) to (4,0): slope -2, $f=-2x+8$ , solve $-2x+8=x$ gives $x=8/3 \approx2.67$ , which is in (2,4). So exactly 1 interior solution.

Q16. A student says: 'If $f$ is continuous on $[1,3]$ and $f(1)=2$ , $f(3)=6$ , then $f(x)=4$ has exactly one solution.' Is the student correct?

A.Yes, by IVT

B.No, could have multiple ✅

C.No, might have none

D.Yes, because 4 is between 2 and 6

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The IVT guarantees at least one solution, not exactly one. The function could oscillate and cross y=4 many times. The student confuses existence with uniqueness. This is a classic misconception.

Q17. Let $f$ be continuous on $[0,1]$ and $f(0)=f(1)$ . Which of the following is necessarily true?

A.There exists

c\in(0,1)

such that

f(c)=f(c+1/2)

✅

B.There exists

c\in(0,1)

such that

f(c)=f(c+1/3)

C.Both A and B

D.Neither

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Define $g(x)=f(x)-f(x+1/2)$ on [0,1/2]. $g(0)=f(0)-f(1/2)$ , $g(1/2)=f(1/2)-f(1)=f(1/2)-f(0) = -g(0)$ . So by IVT, $g(c)=0$ . For 1/3, not guaranteed without extra conditions. This is a known Olympiad-type problem.

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