The maximum value of


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

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

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

(C) 1

(D) 0


Let $f(x)=[x(x-1)+1]^{\frac{1}{3}}$.

$\therefore f^{\prime}(x)=\frac{2 x-1}{3[x(x-1)+1]^{\frac{2}{3}}}$

Now, $f^{\prime}(x)=0 \Rightarrow x=\frac{1}{2}$

Then, we evaluate the value of $f$ at critical point $x=\frac{1}{2}$ and at the end points of the interval $[0,1]\{$ i.e., at $x=0$ and $x=1\}$.




Hence, we can conclude that the maximum value of f in the interval [0, 1] is 1.

The correct answer is C.

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