考點(diǎn):函數(shù)的零點(diǎn)與方程根的關(guān)系
專題:綜合題,函數(shù)的性質(zhì)及應(yīng)用
分析:函數(shù)y=ax與y=logax關(guān)于y=x對(duì)稱,只需要討論與y=x有解即可,構(gòu)造函數(shù)h(x)=ax-x,只須h(x)的最小值小于等于0,進(jìn)而得到實(shí)數(shù)a的取值范圍.
解答:
解:由于函數(shù)y=a
x與y=log
ax關(guān)于y=x對(duì)稱,只需要討論與y=x有解即可,
令h(x)=a
x-x,則函數(shù)h(x)有零點(diǎn),
當(dāng)a>1時(shí),令h′(x)=a
xlna-1=0,則x=
loga,
當(dāng)0<x<
loga時(shí),h′(x)<0,此時(shí)函數(shù)h(x)為減函數(shù);
當(dāng)x>
loga時(shí),h′(x)>0,此時(shí)函數(shù)h(x)為增函數(shù);
故當(dāng)x=
loga時(shí),函數(shù)h(x)取最小值,
若函數(shù)h(x)有零點(diǎn),則h(
loga)≤0,
即
=log
ae≤
loga,
即e≤
,
即0<lna≤
,
即1<a≤
e,
故實(shí)數(shù)a的取值范圍是(1,
e],
故答案為:(1,
e].
點(diǎn)評(píng):本題考查的知識(shí)點(diǎn)是根的存在性及根的個(gè)數(shù)判斷,反函數(shù),導(dǎo)數(shù)法判斷函數(shù)的單調(diào)性,導(dǎo)數(shù)法求函數(shù)的最值,對(duì)數(shù)的運(yùn)算性質(zhì),是指數(shù)函數(shù),對(duì)數(shù)函數(shù),函數(shù)零點(diǎn),導(dǎo)數(shù)等的綜合應(yīng)用,運(yùn)算量大,綜合性可,轉(zhuǎn)化困難,屬于難題.