سوال ۲۷ کنکور ارشد مهندسی کامپیوتر ۱۴۰۵

Let $[x]$ be the greatest-integer (floor) function, and write the fractional part $\{x\} = x - [x]$, so $0 \le \{x\} < 1$.

Then $f(x) = [x] + (x - [x]) = [x] + \{x\} = x,$ because the integer part plus the fractional part reconstruct $x$ exactly.

Why the individual jumps cancel. Taken separately, both $[x]$ and $\{x\}$ have a jump at every integer $n$, but in opposite directions:

The $+1$ jump of $[x]$ is exactly cancelled by the $-1$ jump of $\{x\}$, so the sum is continuous at every integer. On the non-integers both terms are already continuous.

Hence $f(x) = x$ is continuous on all of $\mathbb{R}$, and its set of discontinuity points is empty: $\varnothing$. This is option 1.

Exam trick: $[x] + (x-[x]) = x$ is an identity. Even if the fractional part were raised to a positive integer power $k$ (or the superscript is merely a footnote marker for the definition of $[x]$), $\{x\}^{k} \to 1$ as $x \to n^{-}$ still cancels the floor's unit jump, so the answer stays $\varnothing$.