Question

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 1

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.