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

$(A)\; \left(\frac{1}{3}\right)^{\frac{1}{3}} \quad (B)\; \frac{1}{2} \quad (C)\; 1 \quad (D)\; 0$

Can you answer this question?

Toolbox:
• $\large\frac{d}{dx}$$(x^n)=nx^{n-1} Step 1: Let y=[x(x-1)+1]^{\Large\frac{1}{3}} Differentiating with respect to x we get \large\frac{dy}{dx}=\frac{1}{3}$$[x(x-1)+1]^{\Large\frac{-2}{3}}\times 2x-1$
$\quad\;\;=\large\frac{2x-1}{3[x(x-1)+1]^{\Large\frac{2}{3}}}$
$\large\frac{dy}{dx}$$=0 at x=\large\frac{1}{2} Step 2: \large\frac{dy}{dx} changes sign from -ve to +ve at x=\large\frac{1}{2} y is minimum at x=\large\frac{1}{2} Value of y at x=0,(0+1)^{\Large\frac{1}{3}} \Rightarrow 1^{\Large\frac{1}{3}}$$=1$
Value of $y$ at $x=1$,
$(0+1)^{\Large\frac{1}{3}}$$=1^{\Large\frac{1}{3}}$
The maximum value of $y$ is 1
Part (C) is the correct answer.
answered Aug 9, 2013
edited Aug 19, 2013

+1 vote