Question

1£®ÒÑÖªº¯Êýf£¨x£©=$\left\{\begin{array}{l}{-{x}^{3}+a{x}^{2}+bx£¨x¡Ü1£©}\\{c£¨{e}^{x-1}-1£©£¨x¡Ý1£©}\end{array}\right.$£¬ÔÚx=0£¬x=$\frac{2}{3}$´¦´æÔÚ¼«Öµ
£¨1£©ÇóʵÊýa£¬bµÄÖµ£»
£¨2£©º¯Êýy=f£¨x£©µÄͼÏóÉÏ´æÔÚÁ½µãA£¬B£¬Ê¹µÃ¡÷AOBÊÇÒÔ×ø±êÔ­µãOΪֱ½Ç¶¥µãµÄÖ±½ÇÈý½ÇÐΣ¬ÇÒб±ßABµÄÖеãÔÚyÖáÉÏ£¬ÇóʵÊýcµÄÈ¡Öµ·¶Î§£»
£¨3£©µ±c=eʱ£¬ÌÖÂÛ¹ØÓÚxµÄ¹ý³Ìf£¨x£©=kx£¨k¡ÊR£©µÄʵ¸ù¸öÊý£®

Answer 1

·ÖÎö £¨1£©µ±x£¼1ʱ£¬ÏȶԺ¯Êýf£¨x£©½øÐÐÇóµ¼£¬ÓÉÌâÒâÖªx=0£¬x=$\frac{2}{3}$ÊÇ·½³Ìf¡ä£¨x£©=0µÄÁ½Êµ¸ù£¬ÓÉΤ´ï¶¨Àí¿ÉÇó³öa£¬bµÄÖµ£®
£¨2£©¸ù¾Ý·Ö¶Îº¯Êý£¬·ÖÀàÌÖÂÛ£¬ÀûÓÃ$\overrightarrow{OA}•\overrightarrow{OB}$=0£¬½áºÏº¯Êý˼Ïë¼´¿ÉÇóʵÊýcµÄÈ¡Öµ·¶Î§£®
£¨3£©½«·½³Ìת»¯Îªº¯Êýy=kÓëy=f£¨x£©£¬½«·½³Ì¸ùµÄÎÊÌâת»¯Îªº¯ÊýͼÏó½»µãÎÊÌâ½â¾ö£®

½â´ð ½â£º£¨1£©µ±x£¼1ʱ£¬f¡ä£¨x£©=-3x²+2ax+b£®
Óɼ«ÖµµãµÄ±ØÒªÌõ¼þ¿ÉÖªx=0£¬x=$\frac{2}{3}$ÊÇ·½³Ìf¡ä£¨x£©=0µÄÁ½¸ù£¬
Ôò0+$\frac{2}{3}$=$\frac{2a}{3}$£¬0¡Á$\frac{2}{3}$=-$\frac{b}{3}$£¬½âµÃa=1£¬b=0£®
£¨2£©ÓÉ£¨1£©Öª£¬f£¨x£©=$\left\{\begin{array}{l}{-{x}^{3}+{x}^{2}£¨x£¼1£©}\\{c£¨{e}^{x-1}-1£©£¨x¡Ý1£©}\end{array}\right.$£¬
¸ù¾ÝÌõ¼þµÃA£¬BµÄºá×ø±ê»¥ÎªÏà·´Êý£¬²»·ÁÉèA£¨-t£¬t³+t²£©£¬B£¨t£¬f£¨t£©£©£¬£¨t£¾0£©£®
Èôt£¼1£¬Ôòf£¨t£©=-t³+t²£¬
ÓÉÌâÒâ$\overrightarrow{OA}•\overrightarrow{OB}$=0£¬¼´-t²+£¨t³+t²£©£¨-t³+t²£©=0£¬´Ëʱt=0£¬²»ºÏÌâÒ⣬ÉáÈ¥£»
Èôt¡Ý1£¬Ôòf£¨t£©=c£¨e^t-1-1£©£®
ÓÉÓÚABµÄÖеãÔÚyÖáÉÏ£¬ÇÒ¡ÏAOBÊÇÖ±½Ç£¬ËùÒÔBµã²»¿ÉÄÜÔÚxÖáÉÏ£¬¼´t¡Ù1£®
ͬÀíÓÉ$\overrightarrow{OA}•\overrightarrow{OB}$=0£¬¼´-t²+£¨t³+t²£©•c£¨e^t-1-1£©=0£¬
¡àc=$\frac{1}{£¨{e}^{t-1}-1£©£¨t+1£©}$£®
ÓÉÓÚº¯Êýg£¨t£©=$\frac{1}{£¨{e}^{t-1}-1£©£¨t+1£©}$£¨t£¾1£©µÄÖµÓòÊÇ£¨0£¬+¡Þ£©£¬
¡àʵÊýcµÄÈ¡Öµ·¶Î§ÊÇ£¨0£¬+¡Þ£©
£¨3£©µ±c=eʱ£¬f£¨x£©=$\left\{\begin{array}{l}{-{x}^{3}+{x}^{2}£¨x£¼1£©}\\{\frac{e}{x}£¨{e}^{x-1}-1£©£¨x¡Ý1£©}\end{array}\right.$£®
µ±x¡Ý1ʱ£¬f¡ä£¨x£©=$\frac{e}{{x}^{2}}$£¾0£¬´Ëʱº¯ÊýÔÚ[1£¬+¡Þ£©ÉÏÊÇÔöº¯Êý£¬
Èçͼ£¬ÓÖµ±x=$\frac{1}{2}$ʱ£¬f£¨x£©È¡µÃ¼«´óÖµ$\frac{1}{4}$£¬
ÓÉͼÏóÖªµ±k¡Ê£¨0£¬$\frac{1}{4}$£©Ê±£¬º¯Êýy=kÓëy=f£¨x£©ÓÐ3¸ö²»Í¬µÄ½»µã£¬¼´·½³ÌÓÐ3¸öʵ¸ù£®
¹ÊʵÊýkµÄÈ¡Öµ·¶Î§Îª£¨0£¬$\frac{1}{4}$£©£®

µãÆÀ ±¾ÌâÖ÷Òª¿¼²éº¯ÊýµÄµ¥µ÷ÐÔ¡¢¼«ÖµµãÓëÆäµ¼º¯ÊýÖ®¼äµÄ¹Øϵ£¬ÒÔ¼°Ñо¿·½³Ì¸ùµÄ¸öÊýÎÊÌ⣬´ËÀàÎÊÌâÊ×Ñ¡µÄ·½·¨ÊÇͼÏ󷨼´¹¹Ô캯ÊýÀûÓú¯ÊýͼÏó½âÌ⣬Æä´ÎÊÇÖ±½ÓÇó³öËùÓеĸù£®