Question

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

• A. \((\frac{1}{3})^{\frac{1}{3}}\)
• B. \(\frac{1}{2}\)
• C. 1
• D. 0

Answer 1

The Correct Option is C

Let \(ƒ(x)=[x(x-1)+1]\frac{1}{3}.\)

\(∴ƒ'(x)=\frac{2x-1}{3[x(x-1)+1]\frac{2}{3}}\)

Now\(,ƒ'(x)=0⇒x=\frac{1}{2}\)

Then,we evaluate the value of f at critical point \(\frac{1}{2}\) and at the end points of the

interval \([0,1]\){i.e.,at \(x=0 and x=1\)}.

\(ƒ(0)=[0(0-1)+1]\frac{1}{3}=1\)

\(ƒ(1)=[1(1-1)+1]\frac{1}{3}=1\)

\(ƒ(\frac{1}{2})=[\frac{1}{2}(-\frac{1}{2})+1]^{\frac{1}{3}}=(\frac{3}{4})^{\frac{1}{3}}\)

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

The correct answer is C.