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If f(x) and g(x) are continuous and diff...

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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`(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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