指數(shù)函數(shù)與對(duì)數(shù)函數(shù)的交點(diǎn)個(gè)數(shù)問(wèn)題

指數(shù)函數(shù)yax與對(duì)數(shù)函數(shù)ylogax交點(diǎn)個(gè)數(shù)問(wèn)題安徽省渦陽(yáng)縣第三中學(xué)胡維大論題：指數(shù)函數(shù)yax(a0且a1)與對(duì)數(shù)函數(shù)ylogax(a0且a1)交點(diǎn)個(gè)數(shù)問(wèn)題．分a1及0a1兩種情況進(jìn)行議論．〔一〕：當(dāng)a1時(shí)，過(guò)原點(diǎn)O(0,0)作yax的切線l，設(shè)切點(diǎn)為P(x0,y0)∵yaxlna∴klyxx0ax0lna又∵kly0ax0∴ax0ax0lnax0x0x0∴x01logaelna進(jìn)而klalogaelnaelna1當(dāng)kl1，即elna1，亦即aeeee時(shí)，P在yx上，∴x0y0這樣就有ax0x0，∴l(xiāng)ogax0x0∴P(x0,y0)是yax與ylogax的公共點(diǎn).當(dāng)kl1,即elna1，亦即aee時(shí)，yax與yx相離,yax與ylogax沒(méi)有公共點(diǎn).當(dāng)kl1,即elna1，亦即1aee時(shí)，yax與yx有兩個(gè)公共點(diǎn)M(x1,y1),N(x2,y2),同理可知M(x1,y1),N(x2,y2)均是yax與ylogax的公共點(diǎn).引理：當(dāng)a1時(shí)，yax與ylogax不可能有不在yx上的公共點(diǎn).證明：用反證法.假定yax與ylogax有公共點(diǎn)Q(s,t),s,t0,st,當(dāng)st時(shí),ast①,logast②由②得ats③∵yax單增,又∵st∴asat由此式聯(lián)合①③可知ts,與st矛盾.同應(yīng)當(dāng)ts時(shí)亦矛盾.進(jìn)而假定不真.所以,引理得證.由上可知：當(dāng)1aee時(shí),yax與ylogax有兩個(gè)公共點(diǎn),當(dāng)a當(dāng)a

ee

時(shí),時(shí),

yax與ylogax有唯一公共點(diǎn),yax與ylogax沒(méi)有公共點(diǎn).〔二〕：當(dāng)0a1時(shí),作函數(shù)f(x)axlogax,limf(x)limf(x)易知x0,x不妨設(shè)aem(m0),那么f(x)emx1lnx,memxm2xf(x)mxemx過(guò)原點(diǎn)作yemx的切線,那么切線的斜率kelnemme當(dāng)mem2,即me時(shí),f(x)0恒建立.進(jìn)而f(x)單增,∴f(x)有唯一的零點(diǎn).當(dāng)mem2,即me時(shí),不妨設(shè)yemx與ym2x交于兩點(diǎn)M(x1,y1),N(x2,y2),(x1x2)那么當(dāng)x(0,x1)時(shí),f(x)0,當(dāng)x(x1,x2)時(shí),f(x)0,當(dāng)x(x2,)時(shí),f(x)0∴f(x)的單增區(qū)間為(0,x1),(x2,),單減區(qū)間為(x1,x2)f(x)在xx1處取得極大值,且f(x1)0,f(x)在xx2處取得極小值,且f(x2)0,再由零點(diǎn)存在定理可知,f(x)有三個(gè)零點(diǎn),分別在區(qū)間(0,x1),(x1,x2),(x2,)之內(nèi).對(duì)上述f(x1)0及f(x2)0的結(jié)論,可證明如下：設(shè)yemx〔me〕與yx的交點(diǎn)為(x3,x3)1me1∵me∴m11∴eee即當(dāng)x1yemx的函數(shù)值小于yx的函數(shù)值，數(shù)形聯(lián)合可知x1ee∵yemx〔me〕與yx的交點(diǎn)為(x3,x3)∴x3emx3進(jìn)而lnx3mx3于是f(x3)emx31lnx3x31mx30mmemx3m2x31m2x3又∵f(x3)x3mx3emx3mx3emx31m2x32(1mx)(1mx)mx2emx333mx32emx33∵m0,x30，∴1mx30，2mx30mx3e又∵x31，∴l(xiāng)nx31，∴1mx31lnx30e∴f(x3)0，又∵當(dāng)x(0,x1)時(shí),f(x)0,當(dāng)x(x1,x2)時(shí),f(x)0,當(dāng)x(x2,)時(shí),f(x)0∴x3(x1,x2)又∵f(x)在區(qū)間(x1,x2)單減及f(x3)0可知f(x1)0且f(x2)0由上可知：當(dāng)me即eea1時(shí)，yax與ylogax有唯一公共點(diǎn),且此公共點(diǎn)在yx上，當(dāng)me即0aee時(shí)，yax與ylogax有三個(gè)公共點(diǎn),且有兩個(gè)不在直線yx上，但對(duì)于yx對(duì)稱，而第三個(gè)公共點(diǎn)在直線yx上.綜合上述,我們能夠獲得如下結(jié)論：當(dāng)1aee時(shí),yax與ylogax有兩個(gè)公共點(diǎn),且兩個(gè)公共點(diǎn)均在直線yx上.當(dāng)aee時(shí),yax與ylogax有唯一公共點(diǎn),且該公共點(diǎn)在直線yx上.當(dāng)a當(dāng)e

ee時(shí),yax與ylogax沒(méi)有公共點(diǎn).ea1時(shí)，yax與ylogax有唯一