Let f x is differentiable in -2-5- If f 2-1-5- f 5-1-2- then there exists a number c- 2- c -5 for which f - c is equal toA- 1-10B- 5C- 101D- -1-10

The correct option is A 110 nbsp;f'(c) = f(5) f(2)5 2 = 12 153 nbsp;=310× 3 = 110 ...

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Standard XII

Mathematics

LaGrange's Mean Value theorem

Let f x is di...

Question

Let $f (x)$ is differentiable in $[2, 5]$ . If $f (2) = \frac{1}{5}, f (5) = \frac{1}{2}$ , then there exists a number $c, 2 < c < 5$ for which $f^{'} (c)$ is equal to

\frac{1}{5}

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- \frac{1}{10}

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\frac{1}{10}

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\frac{3}{10}

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Solution

The correct option is C $\frac{1}{10}$
$f^{'} (c) = \frac{f (5) - f (2)}{5 - 2} = \frac{\frac{1}{2} - \frac{1}{5}}{3} = \frac{3}{10 \times 3} = \frac{1}{10}$

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Similar questions

If f (x) is differentiable in the interval [2, 5], where $f (2) = \frac{1}{5}$ and $f (5) = \frac{1}{2}$ , then there exists a number c, 2 < c < 5 for which f ' (c) is equal to

Q. If

f (x)

is differentiable in the interval

[2, 5]

, where

f (2) = \frac{1}{5}

and

f (5) = \frac{1}{2}

, then there exists a number

c

2 < c < 5

for which

f^{'} (c) =

Q. If f(x) is differentiable in the interval [2, 5] and

f (2) = \frac{1}{5}; f (5) = \frac{1}{2}

. Then there exist a number ‘c’, 2 < c < 5 for which f’(c) is equal to

Q. Let

y = f (x) and y = g (x)

be two differentiable function in

[0, 2]

such that

f (0) = 3, f (2) = 5, g (0) = 1 and g (2) = 2.

If there exists atleast one

c ϵ (0, 2)

such that

f^{'} (c) = k g^{'} (c)

, then

k

must be

If f (x) is differentiable in the interval [2, 5], where $f (2) = \frac{1}{5}$ and $f (5) = \frac{1}{2}$ , then there exists a number c, 2 < c < 5 for which f ' (c) is equal to

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Let f(x) is differentiable in [2,5]. If f(2)=15, f(5)=12, then there exists a number c, 2<c<5 for which f′(c) is equal to

The correct option is C 110 f′(c)=f(5)−f(2)5−2 =12−153 =310×3=110

Let $f (x)$ is differentiable in $[2, 5]$ . If $f (2) = \frac{1}{5}, f (5) = \frac{1}{2}$ , then there exists a number $c, 2 < c < 5$ for which $f^{'} (c)$ is equal to

The correct option is C $\frac{1}{10}$
$f^{'} (c) = \frac{f (5) - f (2)}{5 - 2} = \frac{\frac{1}{2} - \frac{1}{5}}{3} = \frac{3}{10 \times 3} = \frac{1}{10}$