Question

The minimum value of |a + bω + cω²|, where a,b and c are all not equal integers and ω(≠ 1) is a cube root of unity, is

(a) √3

(b) 1/2

(c) 1

(d) 0

Answer 1

The correct option is (c) 1

Explanation:

Since, a,b,c are all integers but not all simultaneously equal.

⇒ lf a = b then a ≠ c and, b ≠ c 

Because, difference of integers = integer. 

⇒ (b - c)² ≥ 1{as minimum difference of two consecutive integers is (+ 1)} also (c - a)² ≥ I and we have taken a = b ⇒ (a - b)² = 0