Track / Stage 5 / 97 of 400 #697 of 1964

Problem 697

AIME late

Algebra Difficulty 5.2 Find the answer

$\begin{array}{l} A=\left\{x \mid \log _{2}(x-1)<1\right\}, \\ B=\{x|| x-a \mid<2\} . \end{array}$

If $A \cap B \neq \varnothing$ , then the range of real number $a$ is

Your answer

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solutions — 2

Solution 1

- 1. $(-1,5)$ .

From $\log _{2}(x-1)<1 \Rightarrow A=(1,3)$ , $|x-a|<2 \Rightarrow B=(a-2, a+2)$ .
If $A \cap B=\varnothing$ , then $a+2 \leqslant 1$ or $a-2 \geqslant 3$ $\Rightarrow a \leqslant-1$ or $a \geqslant 5$ .
Therefore, when $A \cap B \neq \varnothing$ , the range of values for $a$ is $(-1,5)$ .

Solution 2

-1. $(-1,5)$ .
From $\log _{2}(x-1)<1 \Rightarrow A=(1,3)$ ,
$|x-a|<2 \Rightarrow B=(a-2, a+2) \text {. }$

If $A \cap B=\varnothing$ , then
$a+2 \leqslant 1$ or $a-2 \geqslant 3$
$\Rightarrow a \leqslant-1$ or $a \geqslant 5$ .
Therefore, when $A \cap B \neq \varnothing$ , the range of values for $a$ is $(-1,5)$ .

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