The maximum value of ${[x(x - 1) + 1]^{1/3}},0 \le x \le 1$ is

(A) ${\left( {\cfrac{1}{3}} \right)^{1/3}}$

(B) $\cfrac{1}{2}$

(C) $1$

(D) $0$

Option C is correct

Let $f(x) = {[x(x - 1) + 1]^{1/3}} = {({x^2} - x + 1)^{1/3}},0 \le x \le 1$

$\Rightarrow f'(x) = \cfrac{1}{3}{({x^2} - x + 1)^{\cfrac{1}{3} - 1}}(2x - 1) = \cfrac{1}{3}{({x^2} - x + 1)^{\cfrac{{ - 2}}{3}}}(2x - 1)$

For critical points, $f'(x) = 0 \Rightarrow 2x - 1 = 0 = x = \cfrac{1}{2} \in [0,1]$

For maximum or minimum value of $f$, we evaluate $f(0),f(1)$ and $f(1/2)$ .

Here, $f(0) = {(0 - 0 + 1)^{1/3}} = 1,f(1) = {(1(0) + 1)^{1/3}} = 1$
and $f\left( {\cfrac{1}{2}} \right) = {\left( {\cfrac{1}{4} - \cfrac{1}{2} + 1} \right)^{1/3}} = {\left( {\cfrac{3}{4}} \right)^{1/3}}$

Therefore the maximum value of $f(x)$ is $1$.