The maximum value of [x(x − 1) + 1]1/3, 0≤ x ≤ 1 is :(a) 0(b) 1/

1.	The maximum value of [x(x − 1) + 1]1/3, 0≤ x ≤ 1 is :(a) 0(b) 1/2(c) 1(d) \(\sqrt[3]{\frac{1}{3}}\)
Answer» Option : (c) Let f(x) = [x(x − 1) + 1]^1/3, 0 ≤ x ≤ 1 f'(x) = \(\frac{2x-1}{3(x^2-x+1)^{\frac{2}{3}}}\) Let f'(x) = 0 ⇒ x = \(\frac{1}{2}\)∈ [0,1] f(0) = 1, f(\(\frac{1}{2}\)) = \((\frac{3}{4})^\frac{1}{3}\) and f(1) = 1 ∴ Maximum value of f(x) is 1.

The maximum value of [x(x − 1) + 1]1/3, 0≤ x ≤ 1 is :(a) 0(b) 1/2(c) 1(d) \(\sqrt[3]{\frac{1}{3}}\)

Answer»

Option : (c)

Let f(x) = [x(x − 1) + 1]^1/3, 0 ≤ x ≤ 1

f'(x) = \(\frac{2x-1}{3(x^2-x+1)^{\frac{2}{3}}}\)

Let f'(x) = 0

⇒ x = \(\frac{1}{2}\)∈ [0,1]

f(0) = 1, f(\(\frac{1}{2}\))

= \((\frac{3}{4})^\frac{1}{3}\) and f(1) = 1

∴ Maximum value of f(x) is 1.

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The maximum value of [x(x − 1) + 1]1/3, 0≤ x ≤ 1 is :(a) 0(b) 1/2(c) 1(d) \(\sqrt[3]{\frac{1}{3}}\)

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