IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

neeraj_agarwal_1990

If f(x) is a continuous function in [a,b] and differentiable in (a,b) ,then there exists a 'c' (c

(a,b)) ,such that f'(c)/f(c) =?
[ans: 1/(a-c) - 1/(b-c)]

taruntanuj007

by LMVT f'(c) = f(b) - f(a) / b-a

and by IVT (intermedaite value theorum) f(c) =[ f(b) + f(a) ]/2

now try and solve it nudg me if u cant!!!

neeraj_agarwal_1990

Whats this ivt

taruntanuj007

Neeraj pick up any book of calculus u'll find it there or read goiit stuff

iitkgp_bipin

Let a function g be defined as :
g(x) = (x-a)(x-b)f(x)

Since f(x) is continuous in [a,b] and differentiable in (a,b) , hence g(x) is also continuous in [a,b] and differentiable in (a,b).

Now g(a) = 0 and g(b) = 0

Hence Rolle's theorem is applicable to g(x) in (a,b).
There exists at least one c in (a,b) for which g'(c) = 0

g'(x) = (x-a)f(x) + (x-b)f(x) + (x-a)(x-b)f'(x)
g'(c) = (c-a)f(c) + (c-b)f(c) + (c-a)(c-b)f'(c) = 0

Divide the whole equation be (c-a)(c-b)f(c) :
1/(c-b) + 1/(c-a) + f'(c)/f(c) = 0

f'(c)/f(c) = 1/(a-c) + 1/(b-c)

neeraj_agarwal_1990

Thnx sir ....
but how do we know the definition of g(x)??

suns_uday

since f(a)=f(b) is not mentioned rolles theorem is not applicable

so, according to remaining 2 theorems lmvt and ivt we can find f'(c)and f(c)

and we can easily find f'(c)/f(c)

elessar_iitkgp

Hats off Bipin!!

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