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


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

$f(x)=2 x(x-3)^{n}$ on the interval $[0,2 \sqrt{3}]$ is $\frac{3}{4}$, write the value of $n$ (a positive integer).


We have

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

Differentiating the given function with respect to x, we get

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

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


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



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

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




$\Rightarrow n=3$

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