Question

If the value of c prescribed in Rolle's theorem for the function
f (x) = 2x (x − 3)ⁿ on the interval $[0,2\sqrt{3}]\mathrm{is}\frac{3}{4},$ write the value of n (a positive integer).

Answer 1

We have
$f\left(x\right)=2x{\left(x-3\right)}^n$

Differentiating the given function with respect to x, we get

$$\begin{gathered} f\text{'}\left(x\right)=2\left[xn{\left(x-3\right)}^{n-1}+{\left(x-3\right)}^n\right] \\ \Rightarrow f\text{'}\left(x\right)=2{\left(x-3\right)}^n\left[\frac{xn}{\left(x-3\right)}+1\right] \\ \Rightarrow f\text{'}\left(c\right)=2{\left(c-3\right)}^n\left[\frac{cn}{\left(c-3\right)}+1\right] \end{gathered}$$

Given:
$f\text{'}\left(\frac{3}{4}\right)=0$

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