Track / Stage 4 / 54 of 340 #314 of 2000

Problem 314

AMC 12 late, AIME early

Algebra Difficulty 4.6 Find the answer

4. The even function $f(x)$ defined on $\mathbf{R}$ satisfies $f(x+1)=-f(x)$ ,
and is increasing in the interval $[-1,0]$ , then ().
(A) $f(3)<f(\sqrt{3})<f(2)$
(B) $f(2)<f(3)<f(\sqrt{3})$
(C) $f(3)<f(2)<f(\sqrt{3})$
(D) $f(2)<f(\sqrt{3})<f(3)$

Your answer

Official solution

4. A.

From the problem, we know
$\begin{array}{l} f(x)=-f(x+1)=f(x+2) . \\ \text { Then } f(3)=f(1)=f(-1), \\ f(2)=f(0), f(\sqrt{3})=f(\sqrt{3}-2) . \end{array}$

And $-1<\sqrt{3}-2<0$ , $f(x)$ is increasing in the interval $[-1,0]$
so, $f(3)<f(\sqrt{3})<f(2)$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.