Question

Let $nϵN$. If the value of $c$ prescribed in Rolle's theorem for the function $f\left(x\right)=2x{(x-3)}^n$ on $[0,3]$ is $\frac{3}{4}$, then $n$ is equal to

• A. 1
• B. 3
• C. 5
• D. 7

Answer 1

Answer: (B)

3

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

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

$=2(x-3{)}^n[\frac{nx}{x-3}+1]$

Since, Rolle's theorem holds for $c=\frac{3}{4}$

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

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

$\Rightarrow n=3$