from a handpicked tutor in LIVE 1-to-1 classes

# 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

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

visual curriculum