Track / Stage 5 / 319 of 400 #919 of 1964

Problem 919

AIME late

Number theory Difficulty 5.8 Prove it

2. Given real numbers $x, y$ satisfy $x^{2}+2 \cos y=1$ . Then the range of $x-\cos y$ is $\qquad$ .

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Official solution

2. $[-1, \sqrt{3}+1]$ .

Since $x^{2}=1-2 \cos y \in[-1,3]$ , we have $x \in[-\sqrt{3}, \sqrt{3}]$ .
From $\cos y=\frac{1-x^{2}}{2}$ , we know
$x-\cos y=\frac{1}{2}(x+1)^{2}-1 \text {. }$

Therefore, when $x=-1$ , $x-\cos y$ has a minimum value of -1, at which point, $y$ can be $\frac{\pi}{2}$ ;

When $x=\sqrt{3}$ , $x-\cos y$ has a maximum value of $\sqrt{3}+1$ , at which point, $y$ can be $\pi$ .

Since the range of $\frac{1}{2}(x+1)^{2}-1$ is $[-1, \sqrt{3}+1]$ , we know that the range of $x-\cos y$ is $[-1, \sqrt{3}+1]$ .

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