## SUBJECT 1

Solve in the set of positive real numbers the equation: $[x]+\left[x^{2}\right]=2$, ( where $[x]$ is the integer part of $x$ )

( adapted from GMB )
## SUBJECT 1

![](https://cdn.mathpix.com/cropped/2024_06_07_0faeb188ae535db9d977g-2.jpg?height=74&width=1396&top_left_y=477&top_left_x=244)

![](https://cdn.mathpix.com/cropped/2024_06_07_0faeb188ae535db9d977g-2.jpg?height=63&width=1396&top_left_y=551&top_left_x=244)

$x \in [1, \sqrt{2}) \Rightarrow x^{2} \in [1,2) \Rightarrow 1+1=2 \Rightarrow x \in [1, \sqrt{2})$ is a solution............................................

$x \in [\sqrt{2}, 2) \Rightarrow x^{2} \in [2,4) \Rightarrow [x] + \left\lfloor x^{2} \right\rfloor \geq 1+2=3 (\text{F})$. Therefore, the solution is $x \in [1, \sqrt{2}) \ldots \ldots \ldots .1 p$