已知 \(0 \leqslant {a_1} \leqslant 1\) ，定义 \({a_{n + 1}} = \left\{ {\begin{array}{*{20}{l}} {2{a_n},0 \leqslant {a_n} < \frac{1}{2}} \\ {2{a_n} - 1,{a_n} \geqslant \frac{1}{2}} \end{array}} \right.\).如果 \({a_2} = {a_3}\) ，则 \({a_2} = \)___；如果 \({a_1} < {a_3}\) ，则 \({a_1}\) 的取值范围是___.

（1） \(\because 0 \leqslant {a_1} \leqslant 1\) ，定义 \({a_{n + 1}} = \left\{ {\begin{array}{*{20}{l}} {2{a_n},0 \leqslant {a_n} < \frac{1}{2}} \\\ {2{a_n} - 1,{a_n} \geqslant \frac{1}{2}} \end{array}} \right.\).

若 \(0 \leqslant {a_2} < \frac{1}{2}\) ，则 \({a_3} = 2{a_2} = {a_2}\) ，解得 \({a_2} = 0\)；

若 \({a_2} \geqslant \frac{1}{2}\) ，则 \({a_3} = 2{a_2} - 1 = {a_2}\) ，解得 \({a_2} = 1\) .

综上所述， \({a_2} = 0\) 或 \(1\) ；

（2）①当 \(0 \leqslant {a_1} < \frac{1}{2}\) 时， \({a_2} = 2{a_1}\) .

（i）若 \(0 \leqslant {a_2} < \frac{1}{2}\) ，即 \(0 \leqslant 2{a_1} < \frac{1}{2} \Rightarrow 0 \leqslant {a_1} < \frac{1}{4}\) ， \({a_3} = 2{a_2} = 4{a_1}\) ，

\(\because {a_1} < {a_3} = 4{a_1}\) ， \(\therefore {a_1} > 0\) ，此时， \(0 < {a_1} < \frac{1}{4}\) ；

（ii）若 \({a_2} \geqslant \frac{1}{2}\) ，即 \(2{a_1} \geqslant \frac{1}{2}\) ，得 \(\frac{1}{4} \leqslant {a_1} < \frac{1}{2}\) ， \({a_3} = 2{a_2} - 1 = 4{a_1} - 1\) ，

\(\because {a_1} < {a_3} = 4{a_1} - 1\) ， \(\therefore {a_1} > \frac{1}{3}\) ，此时， \(\frac{1}{3} < {a_1} < \frac{1}{2}\) ；

②当 \(\frac{1}{2} \leqslant {a_1} \leqslant 1\) 时， \({a_2} = 2{a_1} - 1\) .

（i）若 \(0 \leqslant {a_2} < \frac{1}{2}\) ，即 \(0 \leqslant 2{a_1} - 1 < \frac{1}{2} \Rightarrow \frac{1}{2} \leqslant {a_1} < \frac{3}{4}\) ， \({a_3} = 2{a_2} = 4{a_1} - 2\) ，

 \(\because {a_1} < {a_3} = 4{a_2} - 2\) ， \(\therefore {a_1} > \frac{2}{3}\) ，此时， \(\frac{2}{3} < {a_1} < \frac{3}{4}\) ；

（ii）若 \({a_2} \geqslant \frac{1}{2}\) ，即 \(2{a_1} - 1 \geqslant \frac{1}{2} \Rightarrow \frac{3}{4} \leqslant {a_1} \leqslant 1\) ， \({a_3} = 2{a_2} - 1 = 4{a_1} - 3\) ，

\(\because {a_1} < {a_3} = 4{a_1} - 3\) ， \(\therefore {a_1} > 1\) ，此时， \({a_1}\) 不存在.

综上所述， \({a_1}\) 的取值范围是 \(\left( {0,\frac{1}{4}} \right) \cup \left( {\frac{1}{3},\frac{1}{2}} \right) \cup \left( {\frac{2}{3},\frac{3}{4}} \right)\) .

故答案为： \(0\) 或 \(1\) ； \(\left( {0,\frac{1}{4}} \right) \cup \left( {\frac{1}{3},\frac{1}{2}} \right) \cup \left( {\frac{2}{3},\frac{3}{4}} \right)\) .

进入考试题库