If f(x) and g(x) are continuous and differentiable functions, then prove that there exists c in [a,b] such that (f'(c))/(f(a)-f(c))+(g'(c))/(g(b)-g(c))=1.

If f(x) and g(x) are continuous and differentiable functions, then pro

1.	If f(x) and g(x) are continuous and differentiable functions, then prove that there exists c in [a,b] such that (f'(c))/(f(a)-f(c))+(g'(c))/(g(b)-g(c))=1.
Answer» Solution :`(F'(c))/(f(a)-f(c))+(g'(c))/(g(b)-g(c))=1` Put c = x `rArr""(f'(x))/(f(x)-f(a))+(g'(x))/(g(x)-g(b))=-1` `rArr""(d)/(dx)log_(E)(f(x)-f(a))+(d)/(dx)log_(e)(g(x)-g(b))=-1` `rArr""d[(log_(e)(g(x)-f(a))(g(x)-g(b))]=-dx` `rArr""(log_(e)(f(x)-f(a))(g(x)-g(b))]=-x+K` `rArr""(f(x)-f(a))(g(x)-g(b))=e^(-x+k)` `rArr""(f(x)-f(a))(g(x)-g(b))e^(x)-e^(k)=0` Now let `H(x)=(f(x)-f(a))(g(x)-g(b))e^(x)-e^(k)` `h(a)=h(b)=-e^(k)` Then by ROLLE's Theorem, there exists atleast one `c in (a,b)` such that `h'(c)=0` `rArr""(f'(c))/(f(a)-f(c))+(g'(c))/(g(b)-g(c))=1`

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If f(x) and g(x) are continuous and differentiable functions, then prove that there exists c in [a,b] such that (f'(c))/(f(a)-f(c))+(g'(c))/(g(b)-g(c))=1.

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