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

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

1 Answer

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 by sreemathi.v
edited Aug 19, 2013 by sharmaaparna1
 

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