If the value of c prescribed in Rolle's theorem for the function

Solution:

We have

$f(x)=2 x(x-3)^{n}$

Differentiating the given function with respect to x, we get

$f^{\prime}(x)=2\left[x n(x-3)^{n-1}+(x-3)^{n}\right]$

$\Rightarrow f^{\prime}(x)=2(x-3)^{n}\left[\frac{x n}{(x-3)}+1\right]$

 

$\Rightarrow f^{\prime}(c)=2(c-3)^{n}\left[\frac{c n}{(c-3)}+1\right]$

Given:

$f^{\prime}\left(\frac{3}{4}\right)=0$

$\therefore 2\left(\frac{-9}{4}\right)^{n}\left[\frac{\frac{3}{4} n}{\left(\frac{-9}{4}\right)}+1\right]=0$

$\Rightarrow 2\left(\frac{-9}{4}\right)^{n}\left[\frac{-n}{3}+1\right]=0$

$\Rightarrow\left[\frac{-n}{3}+1\right]=0$

$\Rightarrow-n+3=0$

 

$\Rightarrow n=3$