The maximum value of `[x(x-1)+1]^(1/3) ` in 0≤ x ≤ 1 is: ,

Options

A. 0
B. `1/2`
C. 1
D. `(1/3)^(1/3)`

Answer

B

Explanation

`y=(x(x−1)+1)^(1/3) =(x^ 2 −1x+1)^(1/3)`

`(dx)/(dy) = 31​ (x^2 −x+1)^(−2/3) (2x−1)`

`(dx)/(dy)​ =0⇒x=1/2`

`1/3​ (x^2 −x+1) ^(−2/3) =0`

Hence critical point is x=0.5

`So f(0.5)=(1+0.5(−0.5)) 31​ =( 3/4​ )^(1/3)`
​
Check at the endpointsie.x=0 and x=1

f(0)=1 f(1)=1
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