If $3(a+2 c)=4(b+3 d)$, then the equation $a x^3+b x^2+c x+d=0$ will have

No real solution equation $p(x)=b$ has

Atleast one real root in $(0,1)$

Atleast one real root in $(0,2)$

If $f\left( x \right) = \sin x/{e^x}$ in $\left[ {0,\pi } \right]$, then $f\left( x \right)$

Satisfies Rolle’s Theorem and $c = \frac{\pi }{4}$, So that ${f'^{\left( {\frac{\pi }{4}} \right)}} = 4$

Does not satisfy Rolle’s Theorem but $f'\left( {\frac{\pi }{4}} \right) > 0$

Satisfies Lagrange’s Mean Value Theorem but $f'\left( {\frac{\pi }{4}} \right) \ne 0$

Back to MEAN VALUES THEOREM

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JEE • MATHEMATICS • MEAN VALUES THEOREM

Mean values theorem practice · Medium

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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.

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Showing 1-30 of 41 questions

1 Medium

1 of 41

Application +4 | -1

f(x)

and

g(x)

are differentiable functions in

[0,1]

such that

f(0)=2, f(1)=6, g(0)=0, g(1)=

2 , then there exists

0 < \mathrm{c} < 1

such that

View answer and explanation

Correct option: C

f^{\prime}(c)=2 g^{\prime}(c)

Option C (

f^{\prime}(c)=2 g^{\prime}(c)

) is correct because it is derived by applying Rolle's theorem to the function

F(x) = f(x) - 2g(x)

2 Medium

2 of 41

Analyzing +4 | -1

Assertion (A): If Rolle's theorem be applied in

f(x)

, then L.M.V. theorem is also applicable for

f(x)

.
Reason (R): Both Rolle's theorem & L.M.V. theorem can not be applied in

f(x)=|\sin | x ||

\left[-\frac{\pi}{3}, \frac{\pi}{3}\right]

The codes are given below, select the correct codes.

View answer and explanation

Correct option: B

Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).

Option B (Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).) is correct because while both the assertion and the reason are true statements, the reason does not directly explain the assertion.

3 Medium

3 of 41

Analyzing +4 | -1

Let

f\left( x \right) = \sqrt {x - 1} + \sqrt {x + 24 - 10\sqrt {x - 1} } ,\;1 \le x \le 26

be real valued function, then

f'\left( x \right)

for

1 < x < 26

View answer and explanation

Correct option: A

Option A is correct because the function simplifies to a constant value of 5 within the given interval, making its derivative zero.

4 Medium

4 of 41

Application +4 | -1

The value of c, in the Lagrange’s Mean value theorem

\frac{{f\left( b \right) - f\left( a \right)}}{{b - a}} = f'\left( c \right)

for the function

f\left( x \right) = x\left( {x - 1} \right)\left( {x - 2} \right)

in the interval [0, 1/2], is

View answer and explanation

Correct option: B

1 - \frac{{\sqrt {21} }}{6}

Option B (

1 - \frac{{\sqrt {21} }}{6}

) is correct because it satisfies Lagrange's Mean Value Theorem within the given interval [0, 1/2].

5 Medium

5 of 41

Analyzing +4 | -1

The number of real values of

k

such that the equation

x^3-3 x-k=0

has two distinct real roots in the interval

(0,1)

View answer and explanation

Correct option: A

Option A (0) is correct because Rolle's theorem cannot be satisfied within the interval (0,1) for the given function, thus there are no such real values of k.

6 Medium

6 of 41

Application +4 | -1

If the function

f\left( x \right) = a{x^3} + b{x^2} + 11x - 6

satisfies the condition of Rolle’s theorem in [1, 3] and

f'\left( {2 + \frac{1}{{\sqrt 3 }}} \right) = 0

, then the values of

a,b

are respectively

View answer and explanation

Correct option: C

1, - 6

Option C (

1, - 6

) is correct because these values of a and b satisfy the conditions of Rolle's theorem and the given derivative condition.

7 Medium

7 of 41

Application +4 | -1

The function

f(x)=(x-3)^2

satisfies all the conditions of mean value theorem in

\{3,4\}

. A point on

y=(x-3)^2

, where the tangent is parallel to the chord joining

(3,0)

and

(4,1)

View answer and explanation

Correct option: B

\left(\frac{7}{2}, \frac{1}{4}\right)

Option B

\left(\frac{7}{2}, \frac{1}{4}\right)

is correct because it satisfies the Mean Value Theorem, where the tangent's slope equals the chord's slope.

8 Medium

8 of 41

Application +4 | -1

f(x)=x^{3}-6 x^{2}+a x+b

is defined on

[1,3]

satisfies 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 the value of 'b' does not affect this condition, so b can be any real number.

9 Medium

9 of 41

Application +4 | -1

The value of

c

in Lagrange’s theorem for the function

f\left( x \right) = {\log _e}\sin x

in the interval

\left[ {\pi /6,5\;\pi /6} \right]

View answer and explanation

Correct option: B

\frac{\pi }{2}

Option B (

\frac{\pi}{2}

) is correct because it satisfies the condition of Lagrange's Mean Value Theorem within the given interval.

10 Medium

10 of 41

Analyzing +4 | -1

\frac{{{C_0}}}{1}\,\, + \,\,\frac{{{C_1}}}{2}\,\, + \,\,\frac{{{C_2}}}{3} = 0

, where

{C_0},\,\,{C_1},\,\,{C_2}

are all real, then the quadratic equation

{C_2}\,{x^2}\,\, + \,\,{C_1}\,x\,\, + \,\,{C_0} = 0

has

View answer and explanation

Correct option: A

At least one root in

\left( {\mathrm{0,1}} \right)

The correct answer is A (At least one root in (0,1)) because Rolle's theorem guarantees a root within the interval (0,1) given the conditions.

11 Medium

11 of 41

Understanding +4 | -1

Let

A = R \times R,R

the real number system and

R = \left\{ {\left( {\left( {x,y} \right),\left( {a,b} \right) \in A \times A} \right)} \right\}

either

\left( {\left( {x,y} \right),\left( {a,b} \right)} \right) \in R

and

\left( {\left( {a,b} \right),\left( {p,q} \right)} \right) \in R?

View answer and explanation

Correct option: A

\left( {\left( {x,y} \right),\left( {p,q} \right)} \right) \in R

Option A (((x,y),(p,q))

\in

R) is correct because the relation R is transitive, meaning if (x,y) is related to (a,b) and (a,b) is related to (p,q), then (x,y) is related to (p,q).

12 Medium

12 of 41

Analyzing +4 | -1

f\left( x \right)

and

g\left( x \right)

are differentiable functions for

0 \le x \le 1

such that

f\left( 0 \right) = 2,g\left( 0 \right) = 0,f\left( 1 \right) = 6,g\left( 1 \right) = 2,

then in the interval (0,1)

View answer and explanation

Correct option: B

f'\left( x \right) = 2\;g'\left( x \right)

for at least one

x

Option B (

f'\left( x \right) = 2\;g'\left( x \right)

for at least one

x

) is correct because it follows from Rolle's theorem applied to the function

\phi(x) = f(x) - 2g(x)

13 Medium

13 of 41

Understanding +4 | -1

f\left( x \right)

satisfies the requirements of Rolle’s Theorem in

\left[ {\mathrm{1, 2}} \right]

and

f\left( x \right)

is continuous in

\left[ {\mathrm{1, 2}} \right]

, then

\int \limits_1^2 \,\,f'\left( x \right)\,\,dx

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because Rolle's Theorem implies f(1) = f(2), and the integral of f'(x) from 1 to 2 is simply f(2) - f(1), which equals zero.

14 Medium

14 of 41

Analyzing +4 | -1

a+b+c=0,

then the equation

3 a x^2+2 b x+c=0

has, in the interval

(0,1).

View answer and explanation

Correct option: A

At least one root

Option A (At least one root) is correct because applying Rolle's theorem to the polynomial f(x) = ax³ + bx² + cx shows that its derivative has at least one root in (0,1).

15 Medium

15 of 41

Application +4 | -1

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

satisfies the relation

f(x)=f(0)+x f^{\prime}(0)+\frac{x^2}{2} f^{\prime \prime}(\theta x)

then as

x \rightarrow 1

, the value of

\theta

View answer and explanation

Correct option: D

\frac{9}{25}

Option D (9/25) is correct because it is the value of θ that satisfies the given relation as x approaches 1, found by using Taylor's theorem.

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16 Medium

16 of 41

Analyzing +4 | -1

f\left( x \right)

satisfies the conditions for Rolle’s theorem in [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 (-4/3) is correct because it is the definite integral of a function that satisfies Rolle's theorem over the given interval, calculated using a specific example.

17 Medium

17 of 41

Application +4 | -1

If the Mean value theorem is

f\left( b \right) - f\left( a \right) = \left( {b - a} \right)f'\left( c \right)

Then, for the function

{x^2} - 2x + 3\,in\,\left[ {1,\frac{3}{2}} \right]

, the value of

c

View answer and explanation

Correct option: B

\frac{5}{4}

Option B (

\frac{5}{4}

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

18 Medium

18 of 41

Application +4 | -1

Rolle's theorem holds for the function

f(x)=x^3+

m x^2+n x

on the interval

[1,2]

and the value of

c

is 4/3. Then

View answer and explanation

Correct option: B

m=-5, n=8

Option B (

m=-5, n=8

) is correct because it satisfies both the Rolle's theorem condition

f(1)=f(2)

and the condition

f^{\prime}(c)=0

c=4/3

19 Medium

19 of 41

Application +4 | -1

The value of

c

for which the conclusion of Lagrange's theorem holds for the function

f(x)=\sqrt{a^2-x^2}, a>1

on the interval

[1, a]

View answer and explanation

Correct option: C

\sqrt{\frac{a(a+1)}{2}}

Option C (

\sqrt{\frac{{a(a+1)}}{{2}}}

) is the correct value of c, obtained by applying Lagrange's Mean Value Theorem to the given function and interval.

20 Medium

20 of 41

Application +4 | -1

n

is a positive integer. If the value of c prescribed in Rolle’s theorem for the function

f\left( x \right) = 2x{\left( {x - 3} \right)^n}

on the interval [0,3] is 3/4, then the value of

n

View answer and explanation

Correct option: C

Option C (3) is correct because it is the value of n that satisfies Rolle's theorem for the given function and interval, found by setting the derivative equal to zero at c=3/4.

21 Medium

21 of 41

Analyzing +4 | -1

Let

f\left( x \right) = \left( {x - 4} \right)\left( {x - 5} \right)\left( {x - 6} \right)\left( {x - 7} \right)

then

View answer and explanation

Correct option: B

Three roots of

f'\left( x \right) = 0

lie in

\left( {4,\;5} \right) \cup \left( {5,\;6} \right) \cup \left( {6,\;7} \right)

Option B (Three roots of

f'(x) = 0

lie in

(4, 5) \cup (5, 6) \cup (6, 7)

) is correct because, according to Rolle's theorem, there exists a root of the derivative between each pair of consecutive roots of the original function.

22 Medium

22 of 41

Analyzing +4 | -1

Let

f:[0,4] \rightarrow R

, be a differentiable function. Then, there exists real numbers

a, b \in(0,4)

such that,

(f(4))^2-(f(0))^2=K f^1(a) f(b)

Where

{K}

, is

View answer and explanation

Correct option: B

Option A (6) is correct because solving the trigonometric equation yields 6 solutions in the interval [0, 2π].

23 Medium

23 of 41

Application +4 | -1

Suppose

f

and

g

are differentiable functions on

[0,1]

and

f(0)=2, g(0)=0, f(1)=6, g(1)=2

. Then, there exists

c \in(0,1)

such that

f^{\prime}(c)=k g^{\prime}(c)

where the value of

k

View answer and explanation

Correct option: A

Option A (2) is correct because by constructing the function

H(x) = f(x) - 2g(x)

and applying Rolle's theorem, we find that

f'(c) = 2g'(c)

for some

c

(0,1)

24 Medium

24 of 41

Application +4 | -1

Let

f

be differentiable for all

x

. If

f\left( 1 \right) = - 2

and

f'\left( x \right) \ge 2

for all

x \in \left[ {1,\;6} \right],

, then

View answer and explanation

Correct option: D

f\left( 6 \right) \ge 8

Option D (

f(6) \ge 8

) is correct because by Lagrange's mean value theorem, the minimum value of

f(6)

is 8 given the conditions.

25 Medium

25 of 41

Analyzing +4 | -1

Rolle's theorem is not applicable to one of the following functions:

View answer and explanation

Correct option: C

f(x)=\tan x

[0, \pi]

Option C (

f(x)=\tan x

[0, \pi]

) is correct because tan(x) is not continuous on the interval

[0, \pi]

due to the asymptote at

x = \pi/2

, thus violating a condition of Rolle's theorem.

26 Medium

26 of 41

Application +4 | -1

The equation

x{\log _e}x = 3 - x

has, in the interval (1,3)

View answer and explanation

Correct option: C

At least one root

The correct answer is At least one root because Rolle's theorem guarantees the existence of at least one root within the interval (1,3).

27 Medium

27 of 41

Application +4 | -1

If the Rolle's theorem holds for the function

f(x)=2 x^{3}+a x^{2}+b x

in the interval

[-1,1]

for the point

\mathrm{c}=\frac{1}{2}

, then the value of

2 \mathrm{a}+\mathrm{b}

is:

View answer and explanation

Correct option: B

-1

Option B (-1) is correct because the value of

2a+b

is -1 after applying Rolle's theorem conditions to the given function.

28 Medium

28 of 41

Analyzing +4 | -1

3(a+2 c)=4(b+3 d)

, then the equation

a x^3+b x^2+c x+d=0

will have

View answer and explanation

Correct option: B

Atleast one real root in

(-1,0)

Option B (Atleast one real root in (-1,0)) is correct because by constructing a function f(x) and applying Rolle's theorem, we can show that there exists at least one root of f'(x) = 0 in the interval (-1,0).

29 Medium

29 of 41

Analyzing +4 | -1

\frac{a_0}{n+1}+\frac{a_1}{n}+\frac{a_2}{n-1}+\ldots+\frac{a_{n-1}}{2}+a_n=0

, then the equation

a_0 x^n+a_1 x^{n-1}+\ldots+a_{n-1} x+a_n=0

has, in the interval

(0,1)

View answer and explanation

Correct option: B

Atleast one root

Option B, Atleast one root, is correct because Rolle's theorem guarantees the existence of at least one root in the interval (0, 1).

30 Medium

30 of 41

Analyzing +4 | -1

f\left( x \right) = \sin x/{e^x}

\left[ {0,\pi } \right]

, then

f\left( x \right)

View answer and explanation

Correct option: C

Satisfies Rolle’s Theorem but

f'\left( {\frac{\pi }{4}} \right) = 0

Option C is correct because the function satisfies Rolle's Theorem as f(0) = f(pi) = 0, and the derivative f'(pi/4) equals 0.

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