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 $3(a+2 c)=4(b+3 d)$, 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 $f'\left( {\frac{\pi }{4}} \right) \ne 0$

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JEE • MATHEMATICS • MEAN VALUES THEOREM

Mean values theorem practice · Mixed difficulty

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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 31-60 of 86 questions

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31 Easy

31 of 86

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.

32 Easy

32 of 86

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.

33 Hard

33 of 86

Application +4 | -1

If the function

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

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

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

, then

View answer and explanation

Correct option: D

a = 11

Option D (

a = 11

) is correct because it satisfies Rolle's theorem conditions, where the derivative at a point within the interval [1, 3] is zero, allowing us to solve for 'a'.

34 Medium

34 of 86

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.

35 Medium

35 of 86

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

36 Medium

36 of 86

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.

37 Medium

37 of 86

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.

38 Medium

38 of 86

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.

39 Medium

39 of 86

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].

40 Easy

40 of 86

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.

41 Medium

41 of 86

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)

42 Medium

42 of 86

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.

43 Medium

43 of 86

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.

44 Medium

44 of 86

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).

45 Easy

45 of 86

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.

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46 Easy

46 of 86

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)

47 Medium

47 of 86

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.

48 Medium

48 of 86

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).

49 Hard

49 of 86

Analyzing +4 | -1

In [0, 1] Lagranges Mean value theorem is NOT applicable to

View answer and explanation

Correct option: A

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{1}{2} - x,\;\;\;\;\;x < \frac{1}{2}}\\ {{{\left( {\frac{1}{2} - x} \right)}^2},\;\;\;x \ge \frac{1}{2}} \end{array}} \right.

The correct answer is A because Lagrange's Mean Value Theorem requires the function to be differentiable, and option A is not differentiable at x = 1/2.

50 Medium

50 of 86

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).

51 Easy

51 of 86

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

52 Medium

52 of 86

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.

53 Medium

53 of 86

Application +4 | -1

27 a+9 b+3 c+d=0,

then the equation

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

has at least one real root lying between

View answer and explanation

Correct option: C

0 and 3

Option C (0 and 3) is correct because Rolle's theorem guarantees a root between 0 and 3 for the given equation, based on the provided condition.

54 Easy

54 of 86

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.

55 Easy

55 of 86

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.

56 Easy

56 of 86

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.

57 Easy

57 of 86

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.

58 Easy

58 of 86

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

59 Easy

59 of 86

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.

60 Medium

60 of 86

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 function

f\left( x \right) = {a_0}{x^n} + {a_1}{x^{n - 1}} + {a_2}{x^{n - 2}} + \ldots + {a_n}

has in (0,1)

View answer and explanation

Correct option: A

At least one zero

Option A is correct because by constructing a function and applying Rolle's theorem, we can show that the given polynomial has at least one zero in (0,1).

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