Question

6£®Èç¹û¶¨ÒåÔÚRÉϵĺ¯Êýf£¨x£©Âú×㣺¶ÔÓÚÈÎÒâx₁¡Ùx₂£¬¶¼ÓÐx₁f£¨x₁£©+x₂f£¨x₂£©¡Ýx₁f£¨x₂£©+x₂f£¨x₁£©£¬Ôò³Æf£¨x£©Îª¡°Hº¯Êý¡±£®¸ø³öÏÂÁк¯Êý£º
¢Ùy=-x³+x+l£»
¢Úy=3x-2£¨sinx-cosx£©£»
¢Ûy=l-ex£»
¢Üf£¨x£©=$\left\{\begin{array}{l}{lnx£¨x¡Ý1£©}\\{0£¨x£¼1£©}\end{array}\right.$£¬
ÆäÖС°Hº¯Êý¡±µÄ¸öÊýÓУ¨¡¡¡¡£©

• A£® 3¸ö
• B£® 2¸ö
• C£® 1¸ö
• D£® 0¸ö

Answer 1

·ÖÎö ¸ù¾ÝÌâÒ⣬½«x₁f£¨x₁£©+x₂f£¨x₂£©¡Ýx₁f£¨x₂£©+x₂f£¨x₁£©±äÐοɵÃ[f£¨x₁£©-f£¨x₂£©]£¨x₁-x₂£©¡Ý0£¬½ø¶ø·ÖÎö¿ÉµÃÈôº¯Êýf£¨x£©Îª¡°Hº¯Êý¡±£¬Ôòº¯Êýf£¨x£©ÎªÔöº¯Êý»ò³£Êýº¯Êý£»¾Ý´ËÒÀ´Î·ÖÎöËù¸øº¯ÊýµÄµ¥µ÷ÐÔ£¬×ۺϿɵô𰸣®

½â´ð ½â£º¸ù¾ÝÌâÒ⣬¶ÔÓÚx₁f£¨x₁£©+x₂f£¨x₂£©¡Ýx₁f£¨x₂£©+x₂f£¨x₁£©£¬
ÔòÓÐf£¨x₁£©£¨x₁-x₂£©-f£¨x₂£©£¨x₁-x₂£©¡Ý0£¬
¼´[f£¨x₁£©-f£¨x₂£©]£¨x₁-x₂£©¡Ý0£¬
·ÖÎö¿ÉµÃ£ºÈôº¯Êýf£¨x£©Îª¡°Hº¯Êý¡±£¬Ôòº¯Êýf£¨x£©ÎªÔöº¯Êý»ò³£Êýº¯Êý£»
¶ÔÓÚ¢Ù¡¢y=-x³+x+l£¬ÓÐy¡ä=-3x²+l£¬²»ÊÇÔöº¯ÊýÒ²²»Êdz£Êýº¯Êý£¬ÔòÆä²»ÊÇ¡°Hº¯Êý¡±£¬
¶ÔÓÚ¢Ú¡¢y=3x-2£¨sinx-cosx£©£»ÓÐy¡ä=3-2£¨sinx+cosx£©=3-2$\sqrt{2}$sin£¨x+$\frac{¦Ð}{4}$£©£¬ÓÐy¡ä¡Ý0£¬
y=3x-2£¨sinx-cosx£©ÎªÔöº¯Êý£¬ÔòÆäÊÇ¡°Hº¯Êý¡±£¬
¶ÔÓÚ¢Û¡¢y=l-ex=-ex+1£¬ÊǼõº¯Êý£¬ÔòÆä²»ÊÇ¡°Hº¯Êý¡±£¬
¶ÔÓڢܡ¢f£¨x£©=$\left\{\begin{array}{l}{lnx£¨x¡Ý1£©}\\{0£¨x£¼1£©}\end{array}\right.$£¬µ±x£¼1ʱÊdz£Êýº¯Êý£¬µ±x¡Ý1ʱÊÇÔöº¯Êý£¬ÔòÆäÊÇ¡°Hº¯Êý¡±£¬
¹Ê¡°Hº¯Êý¡±ÓÐ2¸ö£¬
¹ÊÑ¡£ºB£®

µãÆÀ ±¾Ì⿼²éº¯Êýµ¥µ÷ÐÔµÄÅж¨ÓëÓ¦Ó㬹ؼüÊÇÒÀ¾Ýx₁f£¨x₁£©+x₂f£¨x₂£©¡Ýx₁f£¨x₂£©+x₂f£¨x₁£©£¬Åжϳöº¯ÊýµÄµ¥µ÷ÐÔ£®