Medium MCQ +4 / -1 PYQ · JEE Mains 2023

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 :

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

Solution

$\mathrm{f}(\mathrm{x})=\left|\mathrm{x}^{2}-\mathrm{x}+1\right|+\left[\mathrm{x}^{2}-\mathrm{x}+1\right] ; \mathrm{x} \in[-1,2]$ <br/><br/>Let $g(x)=x^{2}-x+1$ <br/><br/>$=\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4}$ <br/><br/>$$ \because\left|\mathrm{x}^{2}-\mathrm{x}+1\right| \text { and }\left[\mathrm{x}^{2}-\mathrm{x}+1\right] $$ <br/><br/>Both have minimum value at $\mathrm{x}=1 / 2$ <br/><br/>$\Rightarrow$ Minimum $\mathrm{f}(\mathrm{x})=\frac{3}{4}+0$ <br/><br/>$=\frac{3}{4}$

About this question

Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations

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