Question

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

• A. $\frac{3}{2}$
• B. $\frac{1}{4}$
• C. $\frac{5}{4}$
• D. $\frac{3}{4}$

Hint:

The greatest integer function \( \left\lfloor t \right\rfloor \) returns the largest integer less than or equal to \( t \). Always check where the function reaches its minimum value.

Answer 1

The Correct Option is D

$f(x)=\left|x^2-x+1\right|+\left[x^2-x+1\right];x\in [-1,2]$
Let $g(x)=x^2-x+1$
$={\left(x-\frac{1}{2}\right)}^2+\frac{3}{4}$
$\because \left|x^2-x+1\right|\text{and}\left[x^2-x+2\right]$
Both have minimum value at $x=1/2$
$\Rightarrow$ Minimum $f(x)=\frac{3}{4}+0$
$=\frac{3}{4}$
So, the correct option is (D) : $\frac{3}{4}$

Answer 2

The given function is: \[ f(x) = |x^2 - x + 1| + \left\lfloor x^2 - x + 1 \right\rfloor \quad \text{where} \, x \in [-1, 2]. \] Step 1: Let \( g(x) = x^2 - x + 1 \). Thus, the function becomes: \[ f(x) = |g(x)| + \left\lfloor g(x) \right\rfloor. \] 
Step 2: We need to find the values of \( x \) in the interval \( [-1, 2] \) that minimize \( f(x) \). 
Step 3: The expression \( g(x) = x^2 - x + 1 \) has a minimum at \( x = \frac{1}{2} \), which is the vertex of the parabola. 
Step 4: At \( x = \frac{1}{2} \), we have: \[ g\left( \frac{1}{2} \right) = \left( \frac{1}{2} \right)^2 - \frac{1}{2} + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4}. \] Step 5: Both \( |g(x)| \) and \( \left\lfloor g(x) \right\rfloor \) reach their minimum values at \( x = \frac{1}{2} \), where \( |g(x)| = \frac{3}{4} \) and \( \left\lfloor g(x) \right\rfloor = 0 \). 
Step 6: Therefore, the minimum value of \( f(x) \) is: \[ f\left( \frac{1}{2} \right) = \frac{3}{4} + 0 = \frac{3}{4}. \]