Question

15£®É躯Êýf£¨x£©=e^x-ax£¬ÆäÖÐeÊÇ×ÔÈ»¶ÔÊýµÄµ×Êý£¬a¡ÊR£®
£¨1£©Èôº¯Êýy=f£¨x£©µÄͼÏóÔÚx=ln2´¦µÄÇÐÏßlµÄÇãб½ÇΪ0£¬ÇóÇÐÏßlµÄ·½³Ì£»
£¨2£©¼Çº¯Êýy=f£¨x£©Í¼ÏóΪÇúÏßC£¬ÉèµãA£¨x₁£¬f£¨x₁£©£©£¬B£¨x₂£¬f£¨x₂£©£©£¨x₁£¼x₂£©ÊÇÇúÏßCÉϲ»Í¬µÄÁ½¶¨µã£¬µãMΪÏ߶ÎABµÄÖе㣬¹ýµãM×÷xÖáµÄ´¹Ïß½»ÇúÏßCÓÚµãN£¬¼ÇÖ±ÏßABµÄбÂÊΪk£®Èôx₁=-x₂£¬ÊÔÎÊ£ºÇúÏßCÔÚµãN´¦µÄÇÐÏßÊÇ·ñƽÐÐÓÚÖ±ÏßAB£¿Çë˵Ã÷ÀíÓÉ£®

Answer 1

·ÖÎö £¨1£©Çóµ¼Êý£¬ÀûÓú¯Êýy=f£¨x£©µÄͼÏóÔÚx=ln2´¦µÄÇÐÏßlµÄÇãб½ÇΪ0£¬Çó³öa£¬¼´¿ÉÇóÇÐÏßlµÄ·½³Ì£»
£¨2£©Éè³öÏ߶ÎABµÄÖеãMµÄ×ø±ê£¬µÃµ½NµÄ×ø±ê£¬ÓÉÁ½µãʽÇó³öABµÄбÂÊ£¬ÔÙÓɵ¼ÊýµÃµ½ÇúÏßC¹ýNµãµÄÇÐÏßµÄбÂÊ£¬ÓÉбÂÊÏàµÈµÃ$\frac{{e}^{{x}_{2}}-{e}^{{x}_{1}}}{2{x}_{2}}$-a=1-a£¬¹¹Ô캯Êý£¬È·¶¨µ¥µ÷ÐÔ£¬¼´¿ÉµÃ³ö½áÂÛ£®

½â´ð ½â£ºf¡ä£¨x£©=e^x-a£®----£¨1·Ö£©
£¨1£©Óɺ¯Êýy=f£¨x£©µÄͼÏóÔÚx=ln2´¦µÄÇÐÏßlµÄÇãб½ÇΪ0£¬
¼´f¡ä£¨ln2£©=tan0=0£¬
Ôòe^ln2-a=0£¬¼´a=2£¬---£¨3·Ö£©
ÓÖf£¨ln2£©=2-2ln2£¬
¹ÊÇÐÏßlµÄ·½³ÌΪy=2-2ln2£»----£¨5·Ö£©
£¨2£©ÓÉÌâÒâÖªx₁=-x₂£¬k=$\frac{{e}^{{x}_{2}}-{e}^{{x}_{1}}}{{x}_{2}-{x}_{1}}$-a=$\frac{{e}^{{x}_{2}}-{e}^{{x}_{1}}}{2{x}_{2}}$-a£¬----£¨8·Ö£©
µãNµÄºá×ø±ê$\frac{{x}_{1}+{x}_{2}}{2}$=0Ϊ£¬
ÇúÏßCÔÚµãN´¦ÇÐÏßбÂÊk¡ä=f¡ä£¨0£©=1-a£¬----£¨10·Ö£©
¼ÙÉèÇúÏßCÔÚµãN´¦µÄÇÐÏßƽÐÐÓÚÖ±ÏßAB£¬
Ôò$\frac{{e}^{{x}_{2}}-{e}^{{x}_{1}}}{2{x}_{2}}$-a=1-a£¬¼´${e}^{{x}_{2}}$-$\frac{1}{{e}^{{x}_{2}}}$-2x₂=0£¬ÆäÖÐx₂£¾0£¬----£¨12·Ö£©
Éèg£¨x£©=e^x-$\frac{1}{{e}^{x}}$-2x£¨x£¾0£©£¬g¡ä£¨x£©=£©=e^x+$\frac{1}{{e}^{x}}$-2¡Ý0£¬
Ôòg£¨x£©ÔÚ£¨0£¬+¡Þ£©Éϵ¥µ÷µÝÔö£¬Ôòg£¨x£©£¾g£¨0£©=0£¬----£¨14·Ö£©
¹Ê${e}^{{x}_{2}}$-$\frac{1}{{e}^{{x}_{2}}}$-2x₂=0²»³ÉÁ¢£¬
Òò´ËÇúÏßCÔÚµãN´¦µÄÇÐÏß²»Æ½ÐÐÓÚÖ±ÏßAB£®----£¨16·Ö£©

µãÆÀ ±¾Ì⿼²éÀûÓõ¼ÊýÇóº¯ÊýµÄÇÐÏß·½³Ì£¬ÑµÁ·ÁËÀûÓù¹Ô캯Êý·¨Ö¤Ã÷ÎÊÌ⣬ÊÇѹÖáÌ⣮