Hint: Differentiate the given function twice and find whether f(x) is increasing by substituting [ - 1,1] and differentiate the function once and substitute ( - 1,1] to find whether the function has roots.

Complete step-by-step answer:

I. To check a function either it is increasing or decreasing we have to double differentiate the function and check the function in their domain either it is increasing or decreasing in this question our function is $f(x) = {x^3} - 1 $

So let's find the first derivative $f'(x) = 3{x^2} $

Now the second derivative is $f''(x) = 6x$, we check the function for $x \in [ - 1,1]$

here $f''(x) \in [ - 6,6]$ $\therefore$ the function f(x) is increasing.

II. To find the root of $f'(x)$ we have to equate $f'(x) = 0$.

$ \Rightarrow 3{x^2} = 0{\text{ }} \Rightarrow x = 0{\text{ }} $

there is one root of $f'(x)$ in ( - 1,1].

$ \therefore$ Statement I is correct and II is incorrect

Correct answer is option A.

Note: - To check whether a function either is increasing or decreasing we have to differentiate the function when the first derivative is always positive in the given domain then it is strictly increasing.