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Question
The maximum value of `[x(x −1) +1]^(1/3)` , 0 ≤ x ≤ 1 is ______.
Options
`(1/3)^(1/3)`
`1/2`
1
0
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Solution
The maximum value of `[x(x −1) +1]^(1/3)` , 0 ≤ x ≤ 1 is 1.
Explanation:
Let, `y = [x (x – 1) + 1]^(1/3)`
Differentiating both sides with respect to x,
`dy/dx = 1/3 [x (x - 1) + 1]^(-2/3) d/dx [x(x - 1) + 1]`
`= 1/3 [x (x - 1) + 1]^(-2/3) × (2x - 1)`
`= (2x - 1)/(3 [x (x - 1) + 1]^(2/3))`
For highest and lowest value, `dy/dx = 0 => 2x - 1 = 0 => x = 1/2`
For highest and lowest value, `dy/dx = 0 => 2x - 1 = 0 => x = 1/2`
At `x= 0, f(0) = 1^(1/3) = 1`
At `x= 1, f(1) = 1^(1/3) = 1`
x `= 1/2 at, f(1/2) = [1/2 (-1/2) xx 1]^(1/3) = (3/4)^(1/3)`
`dy/dx , at x = 1/2` sign is changing from -ve to +ve
∴ y is minimum at x = `1/2`.
maximum value = 1
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