**Question:**

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

**Solution:**

(a) 1

The given function is $f(x)=x^{3}-3 x$.

This is a polynomial function, which is continuous and derivable in $R$.

Therefore, the function is continuous on $[0, \sqrt{3}]$ and derivable on $(0, \sqrt{3})$.

Differentiating the given function with respect to *x,* we get

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

$\Rightarrow f^{\prime}(c)=3 c^{2}-3$

$\therefore f^{\prime}(c)=0$

$\Rightarrow 3 c^{2}-3=0$

$\Rightarrow c^{2}=1$

$\Rightarrow c=\pm 1$

Thus, $c=1 \in[0, \sqrt{3}]$ for which Rolle's theorem holds.

Hence, the required value of *c* is 1.