JEE MAIN - Mathematics (2023 - 31st January Evening Shift - No. 10)
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 :
$f(x)=\left|x^{2}-x+1\right|+\left[x^{2}-x+1\right]$,
where $[t]$ denotes the greatest integer function, in the interval $[-1,2]$, is :
$\frac{3}{4}$
$\frac{3}{2}$
$\frac{1}{4}$
$\frac{5}{4}$
Explanation
$\mathrm{f}(\mathrm{x})=\left|\mathrm{x}^{2}-\mathrm{x}+1\right|+\left[\mathrm{x}^{2}-\mathrm{x}+1\right] ; \mathrm{x} \in[-1,2]$
Let $g(x)=x^{2}-x+1$
$$ =\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4} $$
$$ \because\left|\mathrm{x}^{2}-\mathrm{x}+1\right| \text { and }\left[\mathrm{x}^{2}-\mathrm{x}+1\right] $$
Both have minimum value at $\mathrm{x}=1 / 2$
$\Rightarrow$ Minimum $\mathrm{f}(\mathrm{x})=\frac{3}{4}+0$
$=\frac{3}{4}$
Let $g(x)=x^{2}-x+1$
$$ =\left(x-\frac{1}{2}\right)^{2}+\frac{3}{4} $$
$$ \because\left|\mathrm{x}^{2}-\mathrm{x}+1\right| \text { and }\left[\mathrm{x}^{2}-\mathrm{x}+1\right] $$
Both have minimum value at $\mathrm{x}=1 / 2$
$\Rightarrow$ Minimum $\mathrm{f}(\mathrm{x})=\frac{3}{4}+0$
$=\frac{3}{4}$
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