Browse Questions

The value of c in Rolle's theorem for the function $f(x)=x^3-3x$ in the interval $[0,\sqrt 3]$ is

$\begin{array}{1 1}(A)\;1 & (B)\;-1\\(C)\;\frac{3}{2} & (D)\;\frac{1}{3}\end{array}$

Toolbox:
• Rolle's Theorem :Let $f$ be a real valued function defined once close interval $[a,b]$,such that
• $\quad(i)$ it is continuous on the closed interval $[a,b]$.
• $\quad(ii)$ it is differentiable in the open interval $(a,b)$.
• $\quad(iii)$ $f(a)=f(b)$,then there exists a real numbers $c\in (a,b)$ such that $f'(c)=0$
Step 1:
$f(x)=x^3-3x\quad[0,\sqrt 3]$
$f'(x)=3x^2-3$
$f'(c)=3c^2-3$
Since $c$ lies between the given interval $[0,\sqrt 3]$ for $f'(c)=0$
Step 2:
(i.e)$3c^2-3=0$
$3(c^2-1)=0$
$\Rightarrow c=\pm 1$
But only $+1$ lies in the interval $[0,\sqrt 3]$
Hence the correct option is $A$
edited Jul 5, 2013