29) The maximum value of [ x ( x-1)+ 1 ] ^{1/3 } , 0\leq x \leq 1 
 (A) \left ( \frac{1}{3} \right ) ^{1/3}\: \: (B) 1 /2\: \: (C) 1\: \: (D) 0

Answers (1)

Given function is
f(x) = [ x ( x-1)+ 1 ] ^{1/3 }
f^{'}(x) = \frac{1}{3}.[(x-1)+x].\frac{1}{[x(x-1)+1]^\frac{2}{3}} = \frac{2x-1}{3[x(x-1)+1]^\frac{2}{3}}
f^{'}(x) = 0\\ \frac{2x-1}{3[x(x-1)+1]^\frac{2}{3}} = 0\\ x =\frac{1}{2}
Hence, x = 1/2 is the critical point s0 we need to check the value at x = 1/2 and at the end points of given range i.e. at x = 1 and x = 0
f(\frac{1}{2}) = [ \frac{1}{2} ( \frac{1}{2}-1)+ 1 ] ^{1/3 } = \left ( \frac{3}{4} \right )^\frac{1}{3}
f(0) = [ 0 ( 0-1)+ 1 ] ^{1/3 } = \left ( 1 \right )^\frac{1}{3} = 1
f(1) = [ 1 ( 1-1)+ 1 ] ^{1/3 } = \left ( 1 \right )^\frac{1}{3} = 1
Hence, by this we can say that maximum value of given function is 1 at x = 0 and x = 1

option c is correct 

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