Sweep the Floor and the Ceiling

$\lceil{x}\rceil$ and $\lfloor{x}\rfloor$ are integers, so adding 0.5 and then doing a floor function changes nothing.

So we have $\lceil{x}\rceil+\lfloor{x}\rfloor=12$

If $x$ contained a decimal part, the floor and the ceiling functions would return two consecutive integers, the sum of which would have to be an odd number. 12 is even, so $x$ is an integer. Floor and ceiling functions do not change the value of an integer, therefore.

$x+x=12$

$x=6$

Relevant wiki: Floor and Ceiling Functions - Problem Solving

$\left\lfloor{\left \lceil{x}\right \rceil\ + \left \lfloor{x}\right \rfloor + 0.5}\right\rfloor$

$\implies$ $f(x)= \lfloor {\lceil{x} \rceil} + \lfloor{x}\rfloor \rfloor$ [As fractions does'nt make any change in floor function]

Now $f(x)=12$ .

So $\lfloor {\lceil{x} \rceil} + \lfloor{x}\rfloor \rfloor=12$ $\implies$ $13> \lceil {x} \rceil + \lfloor{x}\rfloor \geq 12$ .

Now both $\lceil x\rceil$ and $\lfloor x \rfloor$ both are integers by their definetion.

So \lceil {x} \rceil + \lfloor{x}\rfloor = 12

And now see that $\lfloor{x}\rfloor = \lceil{x}\rceil$ for even integers as we know that both when added will give $2x-1$ which is always odd.

$\implies$ $\lfloor{x}\rfloor=\lceil{x}\rceil = 6$ .

Hence , We have $6>x\geq 5$ [For Ceiling] and or $7> x\geq 6$ [For Boxed] .

Combining both we have $x= 6$ as our only choice