Question

12£®£¨1£©Èô²»µÈʽ$sin£¨2x+\frac{¦Ð}{3}£©-\frac{1}{a}£¾0$¶Ô$x¡Ê[\frac{¦Ð}{6}£¬\frac{¦Ð}{2}]$µÄËùÓÐʵÊýx¶¼³ÉÁ¢£¬ÇóaµÄÈ¡Öµ·¶Î§£»
£¨2£©Èô²»µÈʽx²-2ax+2a+1£¾0¶Ô0¡Üx¡Ü1µÄËùÓÐʵÊýx¶¼³ÉÁ¢£¬ÇóaµÄÈ¡Öµ·¶Î§£»
£¨3£©Éèa£¾0ÇÒa¡Ù1£¬f£¨x£©=x²-a^x£¬¶Ôx¡Ê£¨-1£¬1£©£¬¾ùÓÐ$f£¨x£©£¼\frac{1}{2}$£¬ÇóaµÄ·¶Î§£®
£¨4£©Íê³ÉÌî¿Õ

	ÓÃͼÏóÓïÑÔ±íÊö	Óú¯Êý×îÖµ±íÊö
ÔÚ£¨a£¬b£©ÄÚ£¬Èô¶ÔÈÎÒâµÄxÓÐf£¨x£©£¾g£¨x£©³ÉÁ¢	¢Ù	¢Ú
ÔÚ£¨a£¬b£©ÄÚ£¬Èô´æÔÚx0£¬Ê¹f£¨x£©£¾g£¨x£©³ÉÁ¢	¢Û	¢Ü

Answer 1

·ÖÎö £¨1£©ÓÉ$x¡Ê[\frac{¦Ð}{6}£¬\frac{¦Ð}{2}]$£¬¿ÉµÃ$£¨2x+\frac{¦Ð}{3}£©$¡Ê$[\frac{2¦Ð}{3}£¬\frac{4¦Ð}{3}]$£¬ÓÚÊÇ$sin£¨2x+\frac{¦Ð}{3}£©$¡Ê$[-\frac{\sqrt{3}}{2}£¬1]$£¬Òò´Ë$\frac{1}{a}$£¼$-\frac{\sqrt{3}}{2}$£¬½â³ö¼´¿ÉµÃ³ö£®
£¨2£©²»µÈʽx²-2ax+2a+1£¾0»¯Îªa£¨2-2x£©+x²+1£¾0£¬ÓÉÓÚ¶Ô0¡Üx¡Ü1µÄËùÓÐʵÊýx¶¼³ÉÁ¢£¬ÀûÓÃÒ»´Îº¯ÊýµÄµ¥µ÷ÐÔ¼´¿ÉµÃ³ö£®
£¨3£©²»µÈʽ$f£¨x£©£¼\frac{1}{2}$£¬»¯Îªu£¨x£©=x²-$\frac{1}{2}$£¼a^x£®»­³öͼÏ󣬼´¿ÉµÃ³ö£®
£¨4£©·Ö±ð»­³öº¯Êýf£¨x£©Óëg£¨x£©µÄͼÏ󣬼´¿ÉµÃ³ö£®

½â´ð ½â£º£¨1£©¡ß$x¡Ê[\frac{¦Ð}{6}£¬\frac{¦Ð}{2}]$£¬¡à$£¨2x+\frac{¦Ð}{3}£©$¡Ê$[\frac{2¦Ð}{3}£¬\frac{4¦Ð}{3}]$£¬
¡à$sin£¨2x+\frac{¦Ð}{3}£©$¡Ê$[-\frac{\sqrt{3}}{2}£¬1]$£¬
¡à$\frac{1}{a}$£¼$-\frac{\sqrt{3}}{2}$£¬
½âµÃ$-\frac{2\sqrt{3}}{3}$£¼a£¼0£®
¡àaµÄÈ¡Öµ·¶Î§ÊÇ$£¨-\frac{2\sqrt{3}}{3}£¬0£©$£®
£¨2£©²»µÈʽx²-2ax+2a+1£¾0»¯Îªa£¨2-2x£©+x²+1£¾0£¬
¡ß¶Ô0¡Üx¡Ü1µÄËùÓÐʵÊýx¶¼³ÉÁ¢£¬
¡à$\left\{\begin{array}{l}{2a+1£¾0}\\{2£¾0}\end{array}\right.$£¬½âµÃ$a£¾-\frac{1}{2}$£®
£¨3£©²»µÈʽ$f£¨x£©£¼\frac{1}{2}$£¬»¯Îªu£¨x£©=x²-$\frac{1}{2}$£¼a^x£®»­³öͼÏ󣺵±1£¼aʱ£¬¿ÉµÃ£º£¨-1£©²-$\frac{1}{2}$£¼a^-1£¬½âµÃ1£¼a£¼2£»Í¬Àí¿ÉµÃ£ºµ±0£¼a£¼1ʱ£¬${1}^{2}-\frac{1}{2}$£¼a¹£¬½âµÃ$\frac{1}{2}£¼a£¼1$£®
×ÛÉϿɵÃaµÄÈ¡Öµ·¶Î§ÊÇ£º$£¨\frac{1}{2}£¬1£©$¡È£¨1£¬2£©£®
£¨4£©¢ÙÈçͼËùʾ£¬ÔÚÇø¼ä£¨a£¬b£©ÄÚ£¬º¯Êýy=f£¨x£©µÄͼÏóÔÚº¯Êýy=g£¨x£©µÄͼÏóµÄÉÏ·½£»
¢Úº¯Êý[f£¨x£©-g£¨x£©]_min£¾0£®
¢ÛÈçͼËùʾ£¬´æÔÚx₀¡Ê£¨a£¬b£©£¬Ê¹µÃº¯Êýy=f£¨x£©µÄͼÏóÔÚx=x₀µÄµãÔÚº¯Êýy=g£¨x£©µÄͼÏóÖÐx=x₀µãµÄÉÏ·½£»
¢Ü´æÔÚx₀¡Ê£¨a£¬b£©£¬f£¨x₀£©-g£¨x₀£©£¾0£®

µãÆÀ ±¾Ì⿼²éÁ˺¯ÊýµÄͼÏóÓëÐÔÖÊ£¬¿¼²éÁËÊýÐνáºÏ˼Ïë·½·¨¡¢ÍÆÀíÄÜÁ¦Óë¼ÆËãÄÜÁ¦£¬ÊôÓÚÖеµÌ⣮