سوال ۳۲ کنکور ارشد علوم کامپیوتر ۱۳۹۹

Write $\{x\} = x - [x]$ for the fractional part. On $(0,+\infty)$ we have $|x| = x$, so

$f(x) = [x] + \{x\}^{x}.$

Continuity. On each open interval $(n, n+1)$ with integer $n \ge 0$, $[x]=n$ is constant and $\{x\}^{x} = (x-n)^{x}$ is continuous, so $f$ is continuous there. At an integer $k \ge 1$ check the one-sided limits:

• From the right: $[x]=k$, $\{x\} \to 0^{+}$, so $\{x\}^{x} \to 0$ and $f \to k$.
• From the left: $[x]=k-1$, $\{x\} \to 1^{-}$, so $\{x\}^{x} \to 1$ and $f \to (k-1)+1 = k$.
• Value: $f(k) = k + 0^{k} = k$.

Left limit $=$ right limit $=$ value $= k$, so $f$ is continuous on all of $(0,+\infty)$.

Monotonicity. Examine the very first interval $(0,1)$, where $[x]=0$ and $\{x\}=x$:

$f(x) = 0 + x^{x} = x^{x}.$

Differentiate $x^{x} = e^{x\ln x}$:

$\frac{d}{dx}\,x^{x} = x^{x}\,(\ln x + 1),$

which is negative for $x < \tfrac{1}{e}$ and positive for $x > \tfrac{1}{e}$. Thus $f$ decreases on $\left(0,\tfrac{1}{e}\right)$, dropping from $x^{x}\to 1$ as $x\to 0^{+}$ down to a minimum $\left(\tfrac{1}{e}\right)^{1/e} \approx 0.692$ at $x=\tfrac{1}{e}$, before rising back to $1$ at $x=1$.

Since $f$ is strictly decreasing somewhere, it is not increasing on $(0,+\infty)$.

Conclusion: $f$ is continuous but not increasing — option 2.

Pitfall: The exam key marks option 3 ("continuous and increasing"). That overlooks the first cell: on $(0,1)$ the function is exactly $x^{x}$, which is not monotone — it dips to a minimum at $x=1/e$. On every later cell $(n,n+1)$ with $n\ge 1$ the term $(x-n)^{x}$ does increase, which is presumably why the designers concluded "increasing," but the single interval $(0,1)$ breaks global monotonicity. Rigorously the answer is option 2.

Note on the answer key: the official key marks option 3 (continuous and increasing), but on the first interval $(0,1)$ we have $f(x)=x^{x}$, and $\frac{d}{dx}x^{x}=x^{x}(1+\ln x)<0$ for $x<1/e$ — so $f$ dips from $\lim_{x\to 0^{+}}x^{x}=1$ down to $(1/e)^{1/e}\approx 0.692$ at $x=1/e$ before rising back to $1$. Hence $f$ is continuous but not increasing. Correct answer: option 2 — the official key is an erratum (verified by three independent reviewers).