Question

# If $$f(x) = (x - a)^m (x - b)^n m, n$$ are positive integers, satisfies the conditions of Rolle's theorem on $$[a, b]$$, then find '$$c$$'.

Solution

## $$f(x)=(x-a)^{m}(x-s)^{n}$$$$f(a)=0$$$$a(b)=0$$thus by Rolle's theorem $$f\ C$$ st.$$f'(c)=0$$                   $$c\in(a,b)$$$$f'(x)=\dfrac{d}{dx}(f(x))=m(x-a)^{m-1}(x-b)^{n}+n(x-a)^{m}(x-b)^{n-1}$$$$f'(x)=0$$$$\Rightarrow m.(x-a)^{m-1}(x-b)^{n}+n(x-a)^{m}(x-b)^{n-1}=0$$$$\Rightarrow (x-a)^{m-1}.(x-b)^{n-1}(m(x-b)+n(x-a))=0$$$$\Rightarrow m(x-b)+n(x-a)=0$$$$\Rightarrow x=\dfrac{mb+na}{m+n}$$$$a<\dfrac{mb+na}{m+n}<b$$$$\Rightarrow c=\dfrac{mb+na}{m+n}$$Maths

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