The maximum value of [x (x - 1) + 1]^1/3, 0 ≤ x ≤ 1 is
(A) (1/3)^1/3   (B) 1/2   (C) 1   (D) 0

Solution:

Maxima and minima are known as the extrema of a function

Maxima and minima are the maximum or the minimum value of a function within the given set of ranges.

Let f (x) = [x (x - 1) + 1]^1/3

Therefore,

On differentiating wrt x, we get

f' (x) = (2x - 1)/[x (x - 1) + 1]^2/3

Now,

f' (x) = 0

⇒ x = 1/2

Then, we evaluate the value of f at critical point at x = 1/2 and at the end points of the interval [0, 1]

i.e., at x = 0 and x = 1.

f (0) = [0 (0 - 1) + 1]^1/3

f (1) = [1(1 - 1) + 1]^1/3

f (1/2) = [1/2(- 1/2) + 1/2]^1/3

= (3/4)^1/3

Hence, we can conclude that the maximum value of f in the interval [0, 1] is 1.

Thus, the correct option is C

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 29

The maximum value of [x (x - 1) + 1]^1/3, 0 ≤ x ≤ 1 is (A) (1/3)^1/3 (B) 1/2 (C) 1 (D) 0

Summary:

The maximum value of [x (x - 1) + 1]^1/3, 0 ≤ x ≤ 1 is 1. Thus, the correct option is C