Mathematics

Mathematics

Problem

Let f:R→(0,∞)f:\mathbb{R}\rightarrow (0,\infty) be such that f(x)+ex+x2f(x)≤ex+ex2∀x>0, then lim⁡x→1f(x) isf(x)+\frac{e^{x+x^2}}{f(x)}\leq e^x+e^{x^2}\quad \forall x>0,\text{ then }\lim_{x\rightarrow 1} f(x)\text{ is} (a) 1 (b) 1e\frac{1}{e} (c) ee (d) 2e2e
Intuitive explanation

This problem hinges on recognizing that the inequality is actually tight—it becomes an equality at the limit point. The key insight is to treat f(x)f(x) as an unknown that satisfies a constraint, then use the AM-GM inequality backwards: the sum f(x)+ex+x2f(x)f(x) + \frac{e^{x+x^2}}{f(x)} is minimized when these two terms are equal. Since the right side is fixed, equality must hold at x=1x=1, which forces f(1)=e1+1=ef(1) = \sqrt{e^{1+1}} = e.

Intuitive explanation

This problem hinges on recognizing that the inequality is actually tight—it becomes an equality at the limit point. The key insight is to treat f(x)f(x) as an unknown that satisfies a constraint, then use the AM-GM inequality backwards: the sum f(x)+ex+x2f(x)f(x) + \frac{e^{x+x^2}}{f(x)} is minimized when these two terms are equal. Since the right side is fixed, equality must hold at x=1x=1, which forces f(1)=e1+1=ef(1) = \sqrt{e^{1+1}} = e.

Browse the problem library