The absolute minimum value, of the function
f(x)=|x²-x+1|+[x²-x+1],
where [t] denotes the greatest integer function, in the interval [-1,2], is :

Solution

Maths sir · 4 steps

Call x²-x+1 one quantity g; it never goes negative, so the modulus drops and only [g] needs care.

1. Complete the square. Let g = x² - x + 1 = (x - 1/2)² + 3/4. So g ≥ 3/4 for every x. Since g is always positive, |g| = g.

   f(x) = g + [g], g = (x-1/2)² + 3/4

2. Range on the interval. On [-1, 2] the smallest value of g is 3/4, at x = 1/2. The largest is 3, at both x = -1 and x = 2. So g takes every value in [3/4, 3].
3. Smallest value of f. Both g and [g] never decrease when g increases, so f is smallest where g is smallest. At g = 3/4 we get [3/4] = 0.
4. So. The absolute minimum is f(1/2) = 3/4 + 0 = 3/4, which is option (a).

The idea to keep. When |g| and [g] act on the same expression, first find the range of g. Since g + [g] never decreases as g grows, its minimum sits where g is smallest.

Answer from the NTA final answer key. Solution by PrepWiser.