When M-V-T is verified for f x on - a - b - then there exists c such thatA- c -a- b s-t f-c-fa-fb-b-aB- c -a- b s-t- f-c-fa-fb-b-aC- c -a- b s-t- f-c-fa-fb-b-aD- c -a- b- s-t- f-c-fa-fb-b-a

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Byju's Answer

Standard XII

Mathematics

Cartesian Product

When M.V.T is...

Question

When M.V.T is verified for f(x) on [a, b] then there exists c such that

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Solution

The correct option is B

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Similar questions

When Rolle's Theorem is verified for f(x) on [a, b] then there exists c such that

According to LMVT, if a function f(x) is continuous on [a, b] and differentiable on the interval (a, b) then which of the following option should be correct for some value c from the interval (a,b)?( c can take any value from the interval (a,b) )

Q. Find '

c

', so that

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

, if

f (x) = e^{x}

a = 0, b = 1

Q. The value of

c

prescribed by Lagrange's mean value theorem (where

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

for some

c

[a, b]

), when

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

and

b = 3

, is

Q. Assertion :If both functions

f (t)

and

g (t)

are continuous on the closed interval

[a, b]

, differentiable on the open interval

(a, b)

and

g^{'} (t)

is not zero on that open interval, then there exists some

c

(a, b)

such that

\frac{f^{'} (c)}{g^{'} (c)} = \frac{f (b) - f (a)}{g (b) - g (a)}

Reason: If

f (t)

and

g (t)

are continuous and differntiable in

[a, b]

, then there exists some

c

(a, b)

such that

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

and

g^{'} (c) = \frac{g (b) - g (a)}{b - a}

from Lagrange's mean value theorm.

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