A function fis defined by $f(x)=2+(x-1)^{2 / 3}$ on $[0,2]$. Which of the following is not correct?

$f$ is not derivable in $(0,2)$

If $f\left( x \right)$ is a function given by $f\left( x \right) = \left| {\begin{array}{*{20}{c}} {\sin x}&{\sin a}&{\sin b}\\ {\cos x}&{\cos a}&{\cos b}\\ {\tan x}&{\tan a}&{\tan b} \end{array}} \right|,{\rm{where\;}}0 < a < b < \frac{\pi }{2}$ Then the equation $f'\left( x \right) = 0$

Has at most one root in $\left( {a,b} \right)$

Has exactly one root in $\left( {a,b} \right)$

Has no root in $\left( {a,b} \right)$

Back to MEAN VALUES THEOREM

Your Progress

0 / 41 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • MEAN VALUES THEOREM

Mean values theorem practice · Easy

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →

Showing 1-30 of 41 questions

1 Easy

1 of 41

Analyzing +4 | -1

The Rolle’s theorem is applicable in the interval

- 1 \le x \le 1

for the function

View answer and explanation

Correct option: B

f\left( x \right) = {x^2}

Option B (

f(x) = x^2

) is correct because it satisfies all conditions of Rolle's theorem: continuity, differentiability, and

f(-1) = f(1)

2 Easy

2 of 41

Analyzing +4 | -1

f(x)

satisfies the conditions for Rolle's theorem on

[3,5]

, then

\int\limits_3^5 {f\left( x \right)} \,dx

equals

View answer and explanation

Correct option: D

-4 / 3

Option D (-4/3) is correct because integrating the function f(x) that satisfies Rolle's theorem conditions on [3,5] gives -4/3.

3 Easy

3 of 41

Analyzing +4 | -1

If the functions

f(x)

and

g(x)

are continuous on

[a, b]

and differentiable on

(a, b),

then in the interval

(a, b),

the equation

\left|\begin{array}{ll}f^{\prime}(x) & f(a) \\ g^{\prime}(x) & g(a)\end{array}\right|=\frac{1}{a-b}\left|\begin{array}{ll}f(a) & f(b) \\ g(a) & g(b)\end{array}\right|

View answer and explanation

Correct option: A

Has at least one root

Option A is correct because Lagrange's Mean Value Theorem applied to a function defined by the determinant guarantees at least one root for the given equation in the interval (a, b).

4 Easy

4 of 41

Analyzing +4 | -1

0 < a < b < \frac{\pi }{2}\,\,{\text{and}}\,\,f\left( {a,b} \right) = \frac{{\tan b - \tan a}}{{b - a}},

then

View answer and explanation

Correct option: B

f(a, b)>1

Option B (f(a, b)>1) is correct because by applying Lagrange's mean value theorem to f(x) = tan x, we find that f(a, b) = sec²(c) > 1.

5 Easy

5 of 41

Application +4 | -1

If the function

f(x)=x^3-6 k x^2+5 x

satisfies the conditions of Lagrange's mean value theorem for the interval

[1,2]

and the tangent to the curve

y=f(x)

x=\frac{7}{4}

is parallel to the chord that joins the points of intersection of the curve with the ordinates

\mathrm{x}=1

and

\mathrm{x}=2

. Then the value of

k

View answer and explanation

Correct option: B

\frac{35}{48}

The correct answer is B (

\frac{35}{48}

) because it satisfies the conditions of Lagrange's mean value theorem and the given tangent condition.

6 Easy

6 of 41

Application +4 | -1

Let a function

f

be differentiable on R. If

f(1)=-2

and

f(x) \geq 2, \forall x \in[1,6]

View answer and explanation

Correct option: B

f(6) \geq 8

Option B (

f(6) \geq 8

) is correct because by applying the Mean Value Theorem and using the given condition

f'(x) \geq 2

, we can deduce that

f(6) \geq 8

7 Easy

7 of 41

Analyzing +4 | -1

Let

f

be twice differentiable function satisfying

f(1)=1, f(2)=4, f(3)=9,

then

View answer and explanation

Correct option: C

There exists at least one

x \in(1,3)

such that

f^{\prime \prime}(x)=2

Option C is correct because by defining a function g(x) = f(x) - x² and applying Rolle's theorem twice, we can show that there exists at least one x in (1,3) such that f''(x) = 2.

8 Easy

8 of 41

Analyzing +4 | -1

Let

f(x)

be twice differentiable continuous function defined on

[a, b]

such that

f^{\prime}(x)>0

and

f^{\prime \prime}(x)<0

for all

x \in(a, b).

c \in(a, b),

then

\frac{f(c)-f(a)}{f(b)-f(c)}

is greater than

View answer and explanation

Correct option: D

\frac{c-a}{b-c}

Option D is correct because by applying Lagrange's mean value theorem and using the given conditions on the first and second derivatives, we can derive the inequality relating the function values.

9 Easy

9 of 41

Understanding +4 | -1

f^{\prime \prime}(x)<0

for all

x \in(a, b),

then

f^{\prime}(x)=0

View answer and explanation

Correct option: B

At most once in

(a, b)

Option B (At most once in

(a, b)

) is correct because if

f'(x) = 0

at two distinct points, Rolle's theorem would imply

f''(x) = 0

at some point, contradicting

f''(x) < 0

10 Easy

10 of 41

Analyzing +4 | -1

For

f(x)=\ln \left(\frac{x^2+\alpha}{7 x}\right)

, Roll's theorem is applicable on

[3,4]

with

c \in(3,4)

such that

f^{\prime}(c)=0.

The value of

f^{\prime \prime}(c)

View answer and explanation

Correct option: A

\frac{1}{12}

Option A is correct because by applying Rolle's theorem to the given function and solving for the constant α, we can find the value of c and then evaluate f''(c).

11 Easy

11 of 41

Application +4 | -1

If the function

f\left( x \right) = {x^3} - 6\;{x^2} + ax + b

defined on [1,3], satisfies the Rolle’s theorem for

c = \frac{{2\sqrt 3 + 1}}{{\sqrt 3 }},

, then

View answer and explanation

Correct option: C

a = 11,b \in R

Option C (

a = 11,b \in R

) is correct because Rolle's theorem requires f(1) = f(3), which leads to a = 11, and b can be any real number.

12 Easy

12 of 41

Application +4 | -1

2 a+3 b+6 c=0,

then the equation

a x^2+b x+c=0

has in the interval

(0,1)

View answer and explanation

Correct option: A

At least one root

The correct answer is A (At least one root) because applying Rolle's theorem to the function

f(x)

guarantees the existence of at least one root in the interval (0, 1).

13 Easy

13 of 41

Application +4 | -1

Let

f

be continuous on [1, 5] and differentiable in

\left( {1,5} \right)

. If

f\left( 1 \right) = - 3

and

f'\left( x \right) \ge 9

for all

x \in \left( {1,\;5} \right)

, then

View answer and explanation

Correct option: A

f\left( 5 \right) \ge 33

Option A (

f\left( 5 \right) \ge 33

) is correct because it represents the minimum possible value of f(5) based on the Mean Value Theorem and the given conditions.

14 Easy

14 of 41

Analyzing +4 | -1

f\left( x \right)

satisfies the condition for Rolle’s theorem is [3, 5], then

\int \nolimits_3^5 \,f\left( x \right)\;dx

equals

View answer and explanation

Correct option: D

\frac{{ - 4}}{3}

Option D (

\frac{{ - 4}}{3}

) is correct because it is the definite integral of a function that satisfies Rolle's theorem over the given interval.

15 Easy

15 of 41

Analyzing +4 | -1

A function fis defined by

f(x)=2+(x-1)^{2 / 3}

[0,2]

. Which of the following is not correct?

View answer and explanation

Correct option: D

Rolle's theorem is applicable on

[0,2]

Option D (Rolle's theorem is applicable on [0,2]) is not correct because f(x) is not differentiable at x=1, which is a requirement for Rolle's theorem.

Explore Other Chapters in MATHEMATICS

View All →

SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

16 Easy

16 of 41

Analyzing +4 | -1

Between any two real roots of the equation

e^x \sin x-1=0,

the equation

e^x \cos x+1=0

has

View answer and explanation

Correct option: A

At least one root

Option A is correct because Rolle's theorem applied to a transformed function guarantees at least one root for the given equation between any two roots of the original equation.

17 Easy

17 of 41

Application +4 | -1

Let the function,

f:[ - 7,0] \to R

be continuous on [-7, 0] and differentiable on (-7, 0). If

f( - 7) = - 3

and

f'(x) \le 2

, for all

x \in ( - 7,0)

then for all such functions

f,f\left( { - 1} \right) + f\left( 0 \right)

lies in the interval

View answer and explanation

Correct option: D

( - \infty ,\mathrm{ 20}]

Option D (

( - \infty ,\mathrm{ 20}])

is correct because it represents the interval in which

f( - 1) + f(0)

lies, derived using the Mean Value Theorem and the given conditions.

18 Easy

18 of 41

Application +4 | -1

Let

f(x)=e^x, x \in[0,1],

then a number c of the Lagrange's mean value theorem is

View answer and explanation

Correct option: A

\log _e(e-1)

The correct answer is A (

\log _e(e-1)

) because applying Lagrange's Mean Value Theorem to the function

f(x) = e^x

leads to

c = \log_e(e-1)

19 Easy

19 of 41

Application +4 | -1

The function

f(x)=x(x+3) e^{-x / 2}

satisfies all the conditions of Rolle's theorem in

[-3,0].

The value of

c

View answer and explanation

Correct option: C

-2

The correct answer is C (-2) because applying Rolle's theorem requires finding a value

c

where the derivative is zero, which leads to

c = -2

20 Easy

20 of 41

Analyzing +4 | -1

[0,1]

Lagrange's mean value theorem is not applicable to

View answer and explanation

Correct option: A

f(x)= \begin{cases}\frac{1}{2}-x, & x<\frac{1}{2} \\ \left(\frac{1}{2}-x\right)^2, & x \geq \frac{1}{2}\end{cases}

Option A is correct because the function is not differentiable at x=1/2, violating a condition for Lagrange's Mean Value Theorem.

21 Easy

21 of 41

Analyzing +4 | -1

Let

f:[0,4] \in R

be a continuous function such that

|f(x)| \leq 2

for all

x \in[0,4]

and

\int\limits_0^4 {f\left( t \right)dt = 2.}

Then, for all

x \in[0,4]

the value of

\int\limits_0^x {f\left( t \right)dt}

lies in the interval

View answer and explanation

Correct option: A

[-6+2 x, 10-2 x]

Option A is correct because by applying Lagrange's mean value theorem and using the given conditions, we can derive the interval in which the integral lies.

22 Easy

22 of 41

Understanding +4 | -1

The value of

c

in Lagrange's theorem for the function

f(x)=|x|

in the interval

[-1,1],

View answer and explanation

Correct option: D

Non-existent in the interval

Option D (Non-existent in the interval) is correct because

f(x) = |x|

is not differentiable at

x=0

in the interval

(-1, 1)

, and Lagrange's theorem requires differentiability on the open interval.

23 Easy

23 of 41

Application +4 | -1

Let

f(x)=3 x^4-4 x^2+5

. Then a value of

c

in Lagrange's mean value theorem for

f(x)

on the interval

[-1,1]

View answer and explanation

Correct option: C

\sqrt{\frac{2}{3}}

Option C (

\sqrt{\frac{2}{3}}

) is correct because it is a value within the interval

[-1, 1]

that satisfies Lagrange's Mean Value Theorem for the given function.

24 Easy

24 of 41

Understanding +4 | -1

Rolle's theorem is not applicable to the function

f(x)=|x|

for

-2 \leq x \leq 2

because

View answer and explanation

Correct option: B

f

is not derivable at

x=0

Option B (

f

is not derivable at

x=0

) is correct because Rolle's theorem requires the function to be differentiable on the open interval, which is not true for

f(x) = |x|

x=0

25 Easy

25 of 41

Application +4 | -1

2 a+3 b+6 c=0,(a, b, c \in R)

then the quadratic equation

a x^{2}+b x+c=0

has

View answer and explanation

Correct option: A

at least one root in

[0,1]

Option A (at least one root in

[0,1]

) is correct because Rolle's theorem can be applied to

f(x)

to show the existence of a root in the interval.

26 Easy

26 of 41

Analyzing +4 | -1

c

is a point at which Rolle’s Theorem holds for the function,

f\left( x \right) = {\log _e}\left( {\frac{{{x^2} + \alpha }}{{7x}}} \right)

in the interval

\left[ {3,\,4} \right]

where

\alpha \, \in \,R,

then

f''(c)

is equal to

View answer and explanation

Correct option: C

\frac{1}{{12}}

Option C (

\frac{1}{{12}}

) is correct because after finding

\alpha

using Rolle's Theorem, we find

c

and then evaluate

f''(c)

27 Easy

27 of 41

Application +4 | -1

The value of c prescribed by Lagrange's mean value theorem, when

f(x)=\sqrt{x^2-4}, a=2

and

b=3,

View answer and explanation

Correct option: B

\sqrt{5}

Option B (

\sqrt{{5}}

) is correct because it is the value of c that satisfies Lagrange's Mean Value Theorem for the given function and interval.

28 Easy

28 of 41

Understanding +4 | -1

a

and

b

are two distinct real roots of the polynomial

f\left( x \right)

such that

a < b

, then there exists a real number

c

lying between

a

and

b

, such that

View answer and explanation

Correct option: B

f'\left( c \right) = 0

Option B (

f'\left( c \right) = 0

) is correct because Rolle's Theorem states that between any two roots of a polynomial

f(x)

, there is a root of

f'(x)

29 Easy

29 of 41

Analyzing +4 | -1

Let

f(x)=a x^5+b x^4+c x^3+d x^2+e x,

where

a, b, c,

d, e \in R

and

f(x)=0

has a positive root

\alpha.

Then,

View answer and explanation

Correct option: D

All of the above

Option D is correct because Rolle's theorem and properties of polynomial roots ensure that all statements about the derivatives of f(x) are true.

30 Easy

30 of 41

Analyzing +4 | -1

f\left( x \right)

is a function given by

f\left( x \right) = \left| {\begin{array}{cccccccccccccccccccc} {\sin x}&{\sin a}&{\sin b}\\ {\cos x}&{\cos a}&{\cos b}\\ {\tan x}&{\tan a}&{\tan b} \end{array}} \right|,\mathrm{where\;}0 < a < b < \frac{\pi }{2}

Then the equation

f'\left( x \right) = 0

View answer and explanation

Correct option: A

Has at least one root in

\left( {a,b} \right)

Option A (Has at least one root in (a,b)) is correct because Rolle's theorem applies, as f(a) = f(b) = 0, implying the existence of at least one root for f'(x) in (a,b).

Revision notes and key takeaways

Review MEAN VALUES THEOREM in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 41 single-correct-answer MCQs on MEAN VALUES THEOREM.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

Showing 1-30 of 41 questions

Continue Your Practice

📚 Switch Set (EASY)

Chapter Sets

ALL 86 MCQs EASY 41 MCQs HARD 4 MCQs MEDIUM 41 MCQs

0 /41

0✓ 0✗

0% Accuracy