Question

3£®ÒÑÖªº¯Êýf£¨x£©ÊǶ¨ÒåÔÚDÉϵĺ¯Êý£¬Èô´æÔÚÇø¼ä[m£¬n]⊆D£¬Ê¹º¯Êýf£¨x£©ÔÚ[m£¬n]ÉϵÄÖµÓòǡΪ[km£¬kn]£¬Ôò³Æº¯Êýf£¨x£©ÊÇkÐͺ¯Êý£®¸ø³öÏÂÁÐ˵·¨£º
¢Ùº¯Êýf£¨x£©=$\frac{3x-1}{x}$²»¿ÉÄÜÊÇkÐͺ¯Êý£»
¢ÚÈôº¯Êýy=-$\frac{1}{2}$x²+xÊÇ3Ðͺ¯Êý£¬Ôòm=-4£¬n=0£»
¢ÛÉ躯Êýf£¨x£©=x³+2x²+x£¨x¡Ü0£©ÊÇkÐͺ¯Êý£¬ÔòkµÄ×îСֵΪ$\frac{4}{9}$£»
¢ÜÈôº¯Êýy=$\frac{£¨{a}^{2}+a£©x-1}{{a}^{2}x}$£¨a¡Ù0£©ÊÇ1Ðͺ¯Êý£¬Ôòn-mµÄ×î´óֵΪ$\frac{2\sqrt{3}}{3}$£®
ÏÂÁÐÑ¡ÏîÕýÈ·µÄÊÇ£¨¡¡¡¡£©

• A£® ¢Ù¢Û
• B£® ¢Ú¢Û
• C£® ¢Ù¢Ü
• D£® ¢Ú¢Ü

Answer 1

·ÖÎö ¸ù¾ÝÌâÄ¿ÖеÄж¨Ò壬½áºÏº¯ÊýÓë·½³ÌµÄ֪ʶ£¬ÖðÒ»Åж¨ÃüÌâ¢Ù¢Ú¢Û¢ÜÊÇ·ñÕýÈ·£¬´Ó¶øÈ·¶¨ÕýÈ·µÄ´ð°¸£®

½â´ð ½â£º¶ÔÓÚ¢Ù£¬f£¨x£©µÄ¶¨ÒåÓòÊÇ{x|x¡Ù0}£¬¼ÙÉèf£¨x£©ÊÇkÐͺ¯Êý£¬Ôò·½³Ì$\frac{3x-1}{x}$=kxÓÐÏàÒìÁ½Êµ¸ù£¬
¼´kx²-3x+1=0£¨k¡Ù0£©ÓÐÏàÒìÁ½Êµ¸ù£¬¡à¡÷=3²-4k£¾0£¬µÃk$£¼\frac{9}{4}$£®
¡à¼ÙÉè³ÉÁ¢£¬º¯Êýf£¨x£©=$\frac{3x-1}{x}$ÊÇkÐͺ¯Êý£¬¹Ê¢Ù´íÎó£»
¶ÔÓÚ¢Ú£¬y=-$\frac{1}{2}$x²+xÊÇ3Ðͺ¯Êý£¬¼´-$\frac{1}{2}$x²+x=3x£¬½âµÃx=0£¬»òx=-4£¬¡àm=-4£¬n=0£¬¹Ê¢ÚÕýÈ·£»
¶ÔÓÚ¢Û£¬f£¨x£©=x³+2x²+x£¨x¡Ü0£©ÊÇkÐͺ¯Êý£¬Ôòx³+2x²+x=kxÓжþ²»µÈ¸ºÊµÊý¸ù£¬¼´x²+2x+£¨1-k£©=0Óжþ²»µÈ¸ºÊµÊý¸ù£¬
¡à$\left\{\begin{array}{l}{1-k£¾0}\\{4-4£¨1-k£©£¾0}\end{array}\right.$£¬½âµÃ0£¼k£¼1£¬¹Ê¢Û´íÎó£»
¶ÔÓڢܣ¬y=$\frac{£¨{a}^{2}+a£©x-1}{{a}^{2}x}$£¨a¡Ù0£©ÊÇ1Ðͺ¯Êý£¬¼´£¨a²+a£©x-1=a²x²£¬¡àa²x²-£¨a²+a£©x+1=0£¬
¡à·½³ÌµÄÁ½¸ùÖ®²îx₁-x₂=$\sqrt{£¨{x}_{1}-{x}_{2}£©^{2}-4{x}_{1}{x}_{2}}$=$\sqrt{£¨\frac{a+1}{a}£©^{2}-4•\frac{1}{{a}^{2}}}$=$\sqrt{1+\frac{2}{a}+\frac{1}{{a}^{2}}-\frac{4}{{a}^{2}}}$=$\sqrt{1+\frac{2}{a}-\frac{3}{{a}^{2}}}$¡Ü$\frac{2\sqrt{3}}{3}$£¬¼´n-mµÄ×î´óֵΪ$\frac{2\sqrt{3}}{3}$£¬¹Ê¢ÜÕýÈ·£®
×ÛÉÏ£¬ÕýÈ·µÄÃüÌâÊǢڢܣ®
¹ÊÑ¡£ºD£®

µãÆÀ ±¾ÌâÊÇж¨ÒåÌ⣬¿¼²éÁËÃüÌâµÄÕæ¼ÙÅжÏÓëÓ¦Ó㬿¼²éÁËÔÚж¨ÒåϺ¯ÊýµÄ¶¨ÒåÓò¡¢ÖµÓòÎÊÌâÒÔ¼°½â·½³ÌµÄÎÊÌ⣬ÊÇÖеµÌâÒ²ÊÇÒ×´íÌ⣮