Functions Question 18
Question 18 - 31 January - Shift 2
The absolute minimum value, of the function $f(x)=|x^{2}-x+1|+[x^{2}-x+1]$, where $[t]$ denotes the greatest integer function, in the interval $[-1,2]$, is :
(1) $\frac{3}{4}$
(2) $\frac{3}{2}$
(3) $\frac{1}{4}$
(4) $\frac{5}{4}$
Show Answer
Answer: (1)
Solution:
Formula: Properties of Greatest Integer Function, Operations on functions, Maximum and Minimum value
$ f(x)=|x^{2}-x+1|+\lfloor x^{2}-x+1\rfloor ; x \in\lfloor-1,2\rfloor $
Let $g(x)=x^{2}-x+1$
$=(x-\frac{1}{2})^{2}+\frac{3}{4}$
$\because|x^{2}-x+1|$ and $[x^{2}-x+2]$
Both have minimum value at $x=1 / 2$
$\Rightarrow$ Minimum $f(x)=\frac{3}{4}+0=\frac{3}{4}$
