A function $f$ is defined by $f\left( x \right) = {e^x}\sin x$ in $\left[ {0,\;\pi } \right]$. Which of the following is not correct?

$f$ f is continous in $\left[ {0,\;\pi } \right]$

$f$ is differentiable in $\left[ {0,\;\pi } \right]$

$f\left( 0 \right) = f\left( \pi \right)$

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JEE • MATHEMATICS • MEAN VALUES THEOREM

Mean values theorem practice · Medium

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31 Medium

31 of 41

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.

32 Medium

32 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 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).

33 Medium

33 of 41

Application +4 | -1

If Rolle's theorem holds for the function

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

+c x, x \in[-1,1]

, at the point

x=\frac{1}{2}

, then

2 b+c

equals :

View answer and explanation

Correct option: C

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

2b+c

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

34 Medium

34 of 41

Application +4 | -1

The value of c in (0, 2) satisfying the Mean value theorem for the function

f\left( x \right) = x{\left( {x - 1} \right)^2},\;x \in \left[ {0,\;2} \right]

is equal to

View answer and explanation

Correct option: B

\frac{4}{3}

Option B (

\frac{4}{3}

) is correct because it is the value of 'c' within the interval (0, 2) that satisfies the Mean Value Theorem for the given function.

35 Medium

35 of 41

Application +4 | -1

Suppose

f

is differentiable on

\mathbb{R}, f(0)=0

and

1 \leq f^{\prime}(x) \leq 2

for all

x \in \mathbb{R}

. Then

View answer and explanation

Correct option: A

x \leq f(x) \leq 2 x

for all

x \geq 0

Option A (

x \leq f(x) \leq 2x

for all

x \geq 0

) is correct because it uses the Lagrange's Mean Value Theorem and the given bounds on

f'(x)

to establish the inequality.

36 Medium

36 of 41

Application +4 | -1

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

and

\frac{f(5)-f(1)}{5-1}=f^{\prime}(c)

then value of

\mathrm{c}

will be

View answer and explanation

Correct option: D

Option D (3) is correct because it is the value of c that satisfies the Mean Value Theorem, where the average rate of change equals the instantaneous rate of change.

37 Medium

37 of 41

Understanding +4 | -1

f\left( x \right)

satisfies the conditions of Rolle’s theorem in [1, 2] and

f\left( x \right)

is continuous in

\left[ {1,2} \right]

, then

\int \nolimits_1^2 f'\left( x \right)dx

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because, according to Rolle's theorem, if f(x) satisfies the conditions in [1, 2], then f(1) = f(2), and the integral of f'(x) from 1 to 2 is f(2) - f(1) = 0.

38 Medium

38 of 41

Analyzing +4 | -1

a, b, c

be non-zero real numbers such that

\int\limits_0^1 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = \int\limits_0^2 {\left( {1 + {{\cos }^8}x} \right)\left( {a{x^2} + bx + c} \right)dx = 0,} }

then the equation

a x^2+b x+c=0

will have

View answer and explanation

Correct option: A

One root between 0 and 1 and other root between 1 and 2

Option A is correct because Rolle's theorem implies the derivative of the integral function has roots between the given intervals, corresponding to roots of the quadratic.

39 Medium

39 of 41

Analyzing +4 | -1

A function

f

is defined by

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

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

. Which of the following is not correct?

View answer and explanation

Correct option: D

Rolle’s theorem is not true in

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

Option B (a decreasing function) is correct because the derivative of the function is negative for all

x > 0

40 Medium

40 of 41

Application +4 | -1

The position of a moving car at time

t

is given by

f(t)=a t^2+b t+c, t>0

where

a, b

and

c

are real numbers greater than 1. Then the average speed of the car over the interval

\left[t_1, t_2\right]

is attained at the point

View answer and explanation

Correct option: A

\frac{t_1+t_2}{2}

Option A (

\frac{{t_1+t_2}}{{2}}

) is correct because the average speed is attained at the midpoint of the interval [t1, t2] according to Lagrange's Mean Value Theorem.

41 Medium

41 of 41

Application +4 | -1

f(1)=3

and

f^{\prime}({x}) \leq 1.4

for

1 \leq {x} \leq 8

. The largest possible value which

{f} (8)

can have, is

View answer and explanation

Correct option: A

12.8

Option C (12.8) is correct because it represents the maximum possible value of f(8) based on the given conditions and the Mean Value Theorem.

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