If fx is differentiable in -a- b- and for c - a- b- a2 - b2 f-c - kfa - fb- then k is equal to

f( x ) is differentaible in [ a,b] and c∈(a,b) ⇒ #160;f( x ) also continuous in [ a,b] ∴ By LMVT f'( c )=f( b ) f( a ) ...

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Standard XII

Mathematics

Definition of Function

If fx is diff...

Question

If f(x) is differentiable in [a, b] and for c $ϵ$ $(a, b), (a^{2} - b^{2})$ $f^{'} (c) = k (f (a) - f (b))$ , then k is equal to

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Solution

$f (x)$ is differentaible in $[a, b]$ and $c \in (a, b)$
$\Rightarrow f (x)$ also continuous in $[a, b]$
$∴$ By LMVT $f^{'} (c) = \frac{f (b) - f (a)}{b - a}$ and $c \in (a, b)$
$∴ (a^{2} - b^{2}) f^{'} (c) = k (f (a) - f (b))$
$\Rightarrow k = (a + b) .$
Hence, solved.

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Similar questions

When M.V.T 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. Assertion :Statement-1 : If

a, b

R

and

a < b

, then there is atleast one real number

c \in (a, b)

such that

\frac{c}{a + b} = \frac{b^{2} + a^{2}}{4 c^{2}}

.
Reason: Statement-2 : If

f (x)

is continuous in

[a, b]

and derivable in

(a, b) & f^{'} (c) = 0

for atleast one

c \in (a, b)

, then it necessarily implies that

f (a) = f (b)

LMVT says that if y = f(x) be a given function which is ;

a.Continuous in [a,b]

b. Differentiable in (a,b)

Then, f'(c) = $\frac{f (b) - f (a)}{b - a}$ for some c $ϵ$ (a,b)

Find the value of c for the function f(x) = $- x^{2}$ +4x-5 and the interval [-1,1]

___

Q. Assertion :If

f

is differentiable on an open interval

(a, b)

such that

| f^{'} (x) | \leq M

for all

x ϵ (a, b)

, then

| f (x) - f (y) | \leq M | x - y |

for all

x, y ϵ (a, b)

. Reason: If

f (x)

is a continuous function defined on

[a, b]

such that it is differentiable on

(a, b)

, then there exists

c ϵ (a, b)

such that

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

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If f(x) is differentiable in [a, b] and for c ϵ (a,b),(a2−b2) f′(c)=k(f(a)−f(b)), then k is equal to

f(x) is differentaible in [a,b] and c∈(a,b)⇒f(x) also continuous in [a,b]∴ By LMVT f′(c)=f(b)−f(a)b−a and c∈(a,b)∴(a2−b2)f′(c)=k(f(a)−f(b))⇒k=(a+b).Hence, solved.

If f(x) is differentiable in [a, b] and for c $ϵ$ $(a, b), (a^{2} - b^{2})$ $f^{'} (c) = k (f (a) - f (b))$ , then k is equal to

$f (x)$ is differentaible in $[a, b]$ and $c \in (a, b)$
$\Rightarrow f (x)$ also continuous in $[a, b]$
$∴$ By LMVT $f^{'} (c) = \frac{f (b) - f (a)}{b - a}$ and $c \in (a, b)$
$∴ (a^{2} - b^{2}) f^{'} (c) = k (f (a) - f (b))$
$\Rightarrow k = (a + b) .$
Hence, solved.