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

 

Option: 1

\frac{1}{4}


Option: 2

\frac{3}{2}


Option: 3

\frac{5}{4}


Option: 4

\frac{3}{4}


Answers (1)

best_answer

\mathrm{f(x)=\left|x^2-x+1\right|+\left[x^2-x+1\right]}
\mathrm{\mathrm{x} \in[-1,2] Here \quad \mathrm{x}^2-\mathrm{x}+1>0, \forall \mathrm{x} \in \mathrm{R}}
Minimum value of \mathrm{x}^2-\mathrm{x}+1 occurs at \mathrm{a=\frac{1}{2} \in[-1,2]}
 \mathrm{So,\operatorname{Min} f(x)=f\left(\frac{1}{2}\right)}
\mathrm{ =\frac{3}{4}+\left[\frac{3}{4}\right]=\frac{3}{4} }

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Riya

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