Question

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

Answer 1

$f\left(x\right)=(x-a{)}^m(x-s{)}^n$

$f\left(a\right)=0$

$a\left(b\right)=0$

thus by Rolle's theorem $fC$ st.

$f^{\prime }\left(c\right)=0$ $c\in (a,b)$

$f^{\prime }\left(x\right)=\frac{d}{dx}\left(f\right(x\left)\right)=m(x-a{)}^{m-1}(x-b{)}^n+n(x-a{)}^m(x-b{)}^{n-1}$

$f^{\prime }\left(x\right)=0$

$\Rightarrow m.(x-a{)}^{m-1}(x-b{)}^n+n(x-a{)}^m(x-b{)}^{n-1}=0$

$\Rightarrow (x-a{)}^{m-1}.(x-b{)}^{n-1}\left(m\right(x-b)+n(x-a\left)\right)=0$

$\Rightarrow m(x-b)+n(x-a)=0$

$\Rightarrow x=\frac{mb+na}{m+n}$

$a<\frac{mb+na}{m+n}<b$

$\Rightarrow c=\frac{mb+na}{m+n}$