سوال ۳۳ کنکور ارشد مهندسی کامپیوتر ۱۴۰۱

Derivative (FTC + Leibniz rule). Since the integrand is continuous, the Fundamental Theorem of Calculus gives $F'(x) = \frac{\sin^4 x}{1+x^4} \ge 0 \quad\text{for all } x.$

No extremum — $F$ is monotone. The derivative is non-negative everywhere, so $F$ is non-decreasing on all of $\mathbb{R}$. We do have $F'(x)=0$ exactly at $x=k\pi$ (where $\sin x = 0$), but $F'$ does not change sign there — it stays $\ge 0$ on both sides. A point is a relative extremum only if the derivative changes sign, so these are mere stationary points, not extrema. Hence options 3 and 4 (which claim extrema at $k\pi$) are wrong.

Boundedness. Using $0 \le \sin^4 t \le 1$, for $x \ge 0$: $0 \le F(x) \le \int_0^x \frac{dt}{1+t^4} \le \int_0^{\infty} \frac{dt}{1+t^4} = \frac{\pi}{2\sqrt{2}} < \infty.$ The integrand $\dfrac{\sin^4 t}{1+t^4}$ is even in $t$, so $F$ is odd: $F(-x) = -F(x)$. Therefore for $x<0$, $\;-\frac{\pi}{2\sqrt{2}} \le F(x) \le 0$. Combining, $|F(x)| \le \frac{\pi}{2\sqrt{2}}$ for every $x\in\mathbb{R}$: $F$ is bounded.

Conclusion. $F$ is bounded and strictly increasing (its only stationary points are the isolated zeros $k\pi$), so it has no extremum. As $x\to+\infty$, $F(x)\to L=\int_0^{\infty}\frac{\sin^4 t}{1+t^4}\,dt$, a finite supremum that is approached but never attained — so there is not even an absolute maximum. This is exactly option 2.

Pitfall: A zero of $F'$ is a relative extremum only when $F'$ changes sign. Here $F'\ge 0$ everywhere, so the points $k\pi$ are inflection-type stationary points, not maxima or minima.