2020屆山東高三理科數(shù)學(xué)一輪復(fù)習(xí)課件第二章§23二次函數(shù)與冪函數(shù)

1、高考數(shù)學(xué),山東專用,2.3,二次函數(shù)與冪函數(shù),考點(diǎn)一,二次函數(shù),課標(biāo)卷、其他自主命題省,區(qū)、市,卷題組,1,2017,浙江,5,4,分,若函數(shù),f,x,x,2,ax,b,在區(qū)間,0,1,上的最大值是,M,最小值是,m,則,M,m,A,與,a,有關(guān),且與,b,有關(guān),B,與,a,有關(guān),但與,b,無關(guān),C,與,a,無關(guān),且與,b,無關(guān),D,與,a,無關(guān),但與,b,有關(guān),五年高考,答案,B,本題考查二次函數(shù)在閉區(qū)間上的最值,二次函數(shù)的圖象,考查數(shù)形結(jié)合思想和分類,討論思想,解法一,令,g,x,x,2,ax,則,M,m,g,x,max,g,x,min,故,M,m,與,b,無關(guān),又,a,1,時(shí),g,x,m

2、ax,g,x,min,2,a,2,時(shí),g,x,max,g,x,min,3,故,M,m,與,a,有關(guān),故選,B,解法二,1,當(dāng),1,即,a,2,時(shí),f,x,在,0,1,上為減函數(shù),M,m,f,0),f,1),a,1,2,當(dāng),1,即,2,a,1,時(shí),M,f,0,m,f,從而,M,m,f,0),f,b,a,2,3,當(dāng),0,即,1,a,0,時(shí),M,f,1,m,f,從而,M,m,f,1),f,a,2,a,1,4,當(dāng),0,即,a,0,時(shí),f,x,在,0,1,上為增函數(shù),M,m,f,1),f,0),a,1,2,a,1,2,2,a,2,a,2,a,2,4,a,b,1,4,2,a,1,2,2,a,2,a,1,4

3、,2,a,即有,M,m,M,m,與,a,有關(guān),與,b,無關(guān),故選,B,2,2,1,0,1,1,1,0,4,1,2,1,4,1,2,a,a,a,a,a,a,a,a,a,2,2015,陜西,12,5,分,對(duì)二次函數(shù),f,x,ax,2,bx,c,a,為非零,四位同學(xué)分別給出下列結(jié)論,其中,有且只有一個(gè)結(jié)論是錯(cuò)誤的,則錯(cuò)誤的結(jié)論是,A.-1,是,f,x,的零點(diǎn),B.1,是,f,x,的極值點(diǎn),C.3,是,f,x,的極值,D,點(diǎn),2,8,在曲線,y,f,x,上,整數(shù),答案,A,由已知得,f,x,2,ax,b,則,f,x,只有一個(gè)極值點(diǎn),若,A,B,正確,則有,解得,b,2,a,c,-3,a,則,f,x,a

4、x,2,2,ax,3,a,由于,a,為非零整數(shù),所以,f,1)=-4,a,3,則,C,錯(cuò),而,f,2)=-3,a,8,則,D,也錯(cuò),與題意不符,故,A,B,中有一個(gè)錯(cuò)誤,C,D,都正確,若,A,C,D,正確,則有,由得,代入中并整理得,9,a,2,4,a,0,又,a,為非零整數(shù),則,9,a,2,4,a,為整數(shù),故方程,9,a,2,4,a,0,無整數(shù)解,故,A,錯(cuò),0,2,0,a,b,c,a,b,2,0,4,2,8,4,3,4,a,b,c,a,b,c,ac,b,a,8,3,8,2,3,b,a,c,a,64,9,64,9,若,B,C,D,正確,則有,解得,a,5,b,-10,c,8,則,f,x,5

5、,x,2,10,x,8,此時(shí),f,1)=23,0,符合題意,故選,A,2,0,3,4,2,8,a,b,a,b,c,a,b,c,3,2015,四川,9,5,分,如果函數(shù),f,x,m,2,x,2,n,8,x,1,m,0,n,0,在區(qū)間,上單調(diào)遞減,那么,mn,的最大值為,A.16,B.18,C.25,D,1,2,1,2,2,81,2,答案,B,當(dāng),m,2,時(shí),f,x,n,8,x,1,在區(qū)間,上單調(diào)遞減,則,n,80,n,8,于是,mn,16,則,mn,無,最大值,當(dāng),m,0,2,時(shí),f,x,的圖象開口向下,要使,f,x,在區(qū)間,上單調(diào)遞減,需,即,2,n,m,18,又,n,0,則,mn,m,m,2

6、,9,m,而,g,m,m,2,9,m,在,0,2,上為增函數(shù),m,0,2,時(shí),g,m,g,2)=16,故,m,0,2,時(shí),mn,無最大值,當(dāng),m,2,時(shí),f,x,的圖象開口向上,要使,f,x,在區(qū)間,上單調(diào)遞減,需,2,即,2,m,n,12,而,2,m,n,2,mn,18,當(dāng)且僅當(dāng),即,時(shí),取,此時(shí)滿足,m,2,故,mn,max,18,故選,B,1,2,2,1,2,2,8,2,n,m,1,2,9,2,m,1,2,1,2,1,2,2,8,2,n,m,2,m,n,2,12,2,m,n,m,n,3,6,m,n,4,2019,浙江,16,4,分,已知,a,R,函數(shù),f,x,ax,3,x,若存在,t,R

7、,使得,f,t,2),f,t,則實(shí)數(shù),a,的最大,值是,2,3,答案,4,3,解析,本題考查絕對(duì)值不等式的解法及二次函數(shù)的最值等相關(guān)知識(shí),以三次函數(shù)為背景,對(duì)不,等式化簡變形,考查學(xué)生運(yùn)算求解能力,將不等式有解問題轉(zhuǎn)化為函數(shù)值域,最值,問題,考查學(xué),生的化歸與轉(zhuǎn)化思想、數(shù)形結(jié)合思想,突出考查了數(shù)學(xué)運(yùn)算的核心素養(yǎng),f,t,2),f,t,a,t,2,3,t,2),at,3,t,6,at,2,12,at,8,a,2,3,at,2,6,at,4,a,1,3,at,2,6,at,4,a,1,a,3,t,2,6,t,4,3,t,2,6,t,4=3,t,1,2,1,1,若存在,t,R,使不等式成立,則需,a

8、,0,故,a,3,t,2,6,t,4,a,只需,a,即可,0,a,故,a,的最大值為,2,3,2,3,2,3,1,3,1,3,1,3,2,3,4,3,2,4,3,3,4,3,4,3,疑難突破,能夠?qū)⒃^對(duì)值不等式化繁為簡,將問題簡化為一元二次不等式有解問題,再進(jìn)一,步轉(zhuǎn)化為值域交集非空是求解本題的關(guān)鍵,5,2017,北京文,11,5,分,已知,x,0,y,0,且,x,y,1,則,x,2,y,2,的取值范圍是,答案,1,1,2,解析,解法一,由題意知,y,1,x,y,0,x,0,0,x,1,則,x,2,y,2,x,2,(1,x,2,2,x,2,2,x,1=2,當(dāng),x,時(shí),x,2,y,2,取最小值

9、,當(dāng),x,0,或,x,1,時(shí),x,2,y,2,取最大值,1,x,2,y,2,解法二,由題意可知,點(diǎn),x,y,在線段,AB,上,如圖,x,2,y,2,表示點(diǎn),x,y,與原點(diǎn)的距離的平方,x,2,y,2,的最小值為原點(diǎn)到直線,x,y,1=0,的距離的平方,即,又易知,x,2,y,2,max,1,x,2,y,2,2,1,2,x,1,2,1,2,1,2,1,1,2,2,2,2,1,1,1,1,2,1,1,2,考點(diǎn)二,冪函數(shù),1,2016,課標(biāo)全國文,7,5,分,已知,a,b,c,2,則,A,b,a,c,B,a,b,c,C,b,c,a,D,c,a,b,4,3,2,2,3,3,1,3,5,答案,A,解法一

10、,a,c,2,而函數(shù),y,在,0,上單調(diào)遞增,所以,即,b,a,c,故選,A,解法二,a,1,b,而函數(shù),y,是單調(diào)遞增函數(shù),知,c,2,1,a,b,所以,c,a,b,故,選,A,4,3,2,2,3,4,1,3,5,2,3,5,2,3,x,2,3,3,2,3,4,2,3,5,4,3,2,1,3,6,2,3,3,1,3,9,1,3,x,1,3,5,1,3,6,1,3,9,2,2018,上海,7,5,分,已知,若冪函數(shù),f,x,x,為奇函數(shù),且在,0,上遞減,則,1,1,2,1,1,2,3,2,2,答案,1,規(guī)律方法,冪函數(shù),y,x,R,的性質(zhì)及圖象特征,所有的冪函數(shù)在,0,上都有定義,并且圖象都

11、過點(diǎn),1,1,若,0,則冪函數(shù)的圖象過原點(diǎn),并且在區(qū)間,0,上為增函數(shù),若,0,則冪函數(shù)的圖象在區(qū)間,0,上為減函數(shù),當(dāng),為奇數(shù)時(shí),冪函數(shù)為奇函數(shù),當(dāng),為偶數(shù)時(shí),冪函數(shù)為偶函數(shù),解析,本題主要考查冪函數(shù)的性質(zhì),冪函數(shù),f(x)=x,為奇函數(shù),可取,1,1,3,又,f(x)=x,在,0,上遞減,0,故,1,1,2015,廣東文,21,14,分,設(shè),a,為實(shí)數(shù),函數(shù),f,x,x,a,2,x,a,a,a,1,1,若,f,0,1,求,a,的取值范圍,2,討論,f,x,的單調(diào)性,3,當(dāng),a,2,時(shí),討論,f,x,在區(qū)間,0,內(nèi)的零點(diǎn)個(gè)數(shù),4,x,教師專用題組,解析,1,f,0),a,2,a,a,a,1)

12、=,a,a,當(dāng),a,0,時(shí),f,0)=0,1,對(duì)于任意的,a,0,恒成立,當(dāng),a,0,時(shí),f,0)=2,a,令,2,a,1,解得,0,a,綜上,a,的取值范圍是,2,函數(shù),f,x,的定義域?yàn)槿w實(shí)數(shù),R,由已知得,f,x,則,f,x,當(dāng),x,a,時(shí),f,x,2,x,(2,a,1)=2,x,a,10,所以,f,x,在區(qū)間,a,上單調(diào)遞減,當(dāng),x,a,時(shí),f,x,2,x,(2,a,1)=2,x,a,10,所以,f,x,在區(qū)間,a,上單調(diào)遞增,3,令,h,x,f,x,由,2,得,1,2,1,2,2,2,2,1,2,2,1,x,a,x,a,x,a,x,a,x,x,a,2,2,1,2,2,1,x,a,x

13、,a,x,a,x,a,4,x,h,x,則,h,x,當(dāng),0,x,a,時(shí),h,x,2,x,(2,a,1),2,x,a,1,0,所以,h,x,在區(qū)間,0,a,上單調(diào)遞減,當(dāng),x,a,時(shí),因?yàn)?a,2,所以,x,2,即,0,1,所以,h,x,2,x,a,0,所以,h,x,在區(qū)間,a,上單調(diào)遞增,h,1)=40,h,2,a,2,a,0,若,a,2,則,h,a,a,2,a,-4+2+2=0,2,2,4,2,1,2,0,4,2,1,x,a,x,a,x,a,x,x,a,x,x,a,x,2,2,4,2,2,1,0,4,2,2,1,x,a,x,a,x,x,a,x,a,x,2,4,x,2,4,x,2,4,x,2,4

14、,1,x,2,a,4,a,此時(shí),h,x,在,0,上有唯一一個(gè)零點(diǎn),若,a,2,則,h,a,a,2,a,0,此時(shí),h,x,在區(qū)間,0,a,上和,a,上各有,一個(gè)零點(diǎn),共兩個(gè)零點(diǎn),綜上,當(dāng),a,2,時(shí),f,x,在區(qū)間,0,內(nèi)有一個(gè)零點(diǎn),當(dāng),a,2,時(shí),f,x,在區(qū)間,0,內(nèi)有兩個(gè)零點(diǎn),4,a,3,2,4,a,a,a,2,1,4,a,a,a,4,x,4,x,2,2015,浙江,18,15,分,已知函數(shù),f,x,x,2,ax,b,a,b,R,記,M,a,b,是,f,x,在區(qū)間,1,1,上的最大值,1,證明,當(dāng),a,2,時(shí),M,a,b,2,2,當(dāng),a,b,滿足,M,a,b,2,時(shí),求,a,+,b,的最大

15、值,解析,1,證明,由,f,x,b,得圖象的對(duì)稱軸為直線,x,由,a,2,得,1,故,f,x,在,1,1,上單調(diào),所以,M,a,b,max,f,1)|,f,1),當(dāng),a,2,時(shí),由,f,1),f,1)=2,a,4,得,max,f,1),f,1,2,即,M,a,b,2,當(dāng),a,2,時(shí),由,f,1),f,1)=-2,a,4,得,max,f,1),f,1,2,即,M,a,b,2,綜上,當(dāng),a,2,時(shí),M,a,b,2,2,由,M,a,b,2,得,1,a,b,=,f,1),2,|1,a,b,=,f,1),2,故,a,b,3,a,b,3,由,a,+,b,得,a,+,b,3,當(dāng),a,2,b,-1,時(shí),a,+

16、,b,=3, ,f,x,=,x,2,2,x,1,此時(shí)易知,f,x,在,1,1,上的最大值為,2,即,M,2,-1)=2,2,2,a,x,2,4,a,2,a,2,a,0,0,a,b,ab,a,b,ab,所以,a,+,b,的最大值為,3,考點(diǎn)一,二次函數(shù),三年模擬,A,組,2017,2019,年高考模擬考點(diǎn)基礎(chǔ)題組,1,2018,河南安陽模擬,5,已知函數(shù),f,x,x,2,4,x,a,x,0,1,若,f,x,有最小值,2,則,f,x,的最大值為,A.1,B.0,C.-1,D.2,答案,A,f,x,x,2,4,x,a,x,2,2,a,4,函數(shù),f,x,x,2,4,x,a,在,0,1,上單調(diào)遞增,當(dāng),

17、x,0,時(shí),f,x,取得最小值,當(dāng),x,1,時(shí),f,x,取得最大值,f,0),a,-2,f,1)=3,a,3-2=1,故選,A,2,2018,山東鄒城第一中學(xué)期中,10,定義運(yùn)算,ad,bc,若函數(shù),f,x,在,m,上單,調(diào)遞減,則實(shí)數(shù),m,的取值范圍是,A.(,5,B.(,5,C.-5,D.(-5,a,b,c,d,1,2,3,3,x,x,x,答案,A,f,x,x,1,x,3)-(-2,3,x,x,2,10,x,3,所以,m,m,5,故選,A,10,2,3,2018,湖北荊州模擬,8,二次函數(shù),f,x,滿足,f,x,2),f,x,2,且,f,0)=3,f,2)=1,若在,0,m,上有最大值,3

18、,最小值,1,則,m,的取值范圍是,A.(0,B.2,C.(0,2,D.2,4,答案,D,二次函數(shù),f,x,滿足,f,2,x,f,2,x,其圖象的對(duì)稱軸是,x,2,又,f,0)=3,f,4)=3,又,f,2,f,0,f,x,的圖象開口向上,f,0)=3,f,2)=1,f,4)=3,f,x,在,0,m,上的最大值為,3,最小值為,1,由二次函數(shù)的性質(zhì)知,2,m,4,故選,D,4,2017,天津紅橋期中,14,若函數(shù),f,x,ax,2,2,x,3,在區(qū)間,4,上是單調(diào)遞增的,則實(shí)數(shù),a,的取值范,圍是,答案,1,0,4,解析,當(dāng),a,0,時(shí),f,x,2,x,3,在,4,上單調(diào)遞增,滿足題意,當(dāng),a

19、,0,時(shí),要使函數(shù),f,x,ax,2,2,x,3,在區(qū)間,4,上單調(diào)遞增,則實(shí)數(shù),a,滿足,解得,a,0,綜上,a,0,0,1,4,a,a,1,4,1,4,5,2017,江西九江七校聯(lián)考,19,已知二次函數(shù),f,x,圖象的對(duì)稱軸為直線,x,-2,f,x,的圖象截,x,軸所,得的線段長為,2,且滿足,f,0)=1,1,求,f,x,的解析式,2,若,f,k,對(duì),x,1,1,恒成立,求實(shí)數(shù),k,的取值范圍,3,1,2,x,解析,1,由題意可設(shè),f,x,a,x,2,x,2,a,0,由,f,0)=1,得,a,1,f,x,x,2,x,2,x,2,4,x,1,2,當(dāng),x,1,1,時(shí),令,t,則,t,由,1,

20、知,f,x,的圖象開口向上,對(duì)稱軸為直線,x,-2,f,t,在,t,上單調(diào)遞增,f,t,min,f,實(shí)數(shù),k,的取值范圍是,3,3,3,3,1,2,x,1,2,2,1,2,2,1,2,13,4,13,4,1,2018,陜西西安檢測,3,函數(shù),y,的圖象大致是,2,3,x,考點(diǎn)二,冪函數(shù),答案,C,y,其定義域?yàn)?x,R,排除,A,B,又,0,1,圖象在第一象限為上凸的,排除,D,故選,C,2,3,x,2,3,x,2,3,2,2018,湖北宜昌月考,4,函數(shù),f,x,m,2,m,1,是冪函數(shù),且在,x,0,上是減函數(shù),則實(shí)數(shù),m,的值為,A.2,B.3,C.4,D.5,2,2,3,m,m,x,答

21、案,A,由題意知,m,2,m,1=1,解得,m,2,或,m,-1,當(dāng),m,2,時(shí),m,2,2,m,3=-3,f,x,x,3,符合題意,當(dāng),m,-1,時(shí),m,2,2,m,3=0,f,x,x,0,不合題意,綜上,m,2,3,2018,河南洛陽二模,7,已知點(diǎn),在冪函數(shù),f,x,a,1,x,b,的圖象上,則函數(shù),f,x,是,A,奇函數(shù),B,偶函數(shù),C,定義域內(nèi)的減函數(shù),D,定義域內(nèi)的增函數(shù),1,2,a,答案,A,點(diǎn),在冪函數(shù),f,x,a,1,x,b,的圖象上,a,1=1,解得,a,2,且,2,b,b,-1,f,x,x,1,函數(shù),f,x,在定義域,0,0,上是奇函數(shù),且在每一個(gè)區(qū)間內(nèi)是減函數(shù),故選,A

22、,1,2,a,1,2,4,2018,山東濰坊三模,8,已知,a,b,c,lo,則,a,b,c,的大小關(guān)系是,A,a,b,c,B,b,a,c,C,c,a,b,D,a,c,b,2,3,2,3,2,3,3,4,3,4,g,2,3,答案,A,因?yàn)?y,是定義域上的增函數(shù),所以,1,即,0,a,b,1,又,c,lo,lo,1,所以,a,b,c,故選,A,2,3,x,2,3,2,3,2,3,3,4,3,4,g,2,3,3,4,g,3,4,5,2018,寧夏銀川月考,10,已知冪函數(shù),f,x,的圖象過點(diǎn),2,則,f,4,的值為,2,答案,2,解析,設(shè)冪函數(shù),f,x,x,a,f,x,的圖象過點(diǎn),2,2,a,解

23、得,a,f,4),2,2,2,1,2,1,2,4,B,組,2017,2019,年高考模擬專題綜合題組,時(shí)間,45,分鐘,分值,70,分,一、選擇題,每小題,5,分,共,35,分,1,2018,北京東城月考,6,已知冪函數(shù),f,x,n,2,2,n,2,n,Z,的圖象關(guān)于,y,軸對(duì)稱,且在,0,上是減函數(shù),則,n,的值為,A.-3,B.1,C.2,D.1,或,2,2,3,n,n,x,答案,B,由于,f,x,為冪函數(shù),所以,n,2,2,n,2=1,解得,n,1,或,n,-3,經(jīng)檢驗(yàn),只有,n,1,符合題意,故選,B,2,2019,山東聊城第一中學(xué)模擬,4,已知函數(shù),f,x,是偶函數(shù),當(dāng),x,0,時(shí),

24、f,x,則在,2,0,上,下列函,數(shù)中與,f,x,的單調(diào)性相同的是,A,y,x,2,1,B,y,x,1,C,y,e,x,D,y,1,3,x,3,2,1,0,1,0,x,x,x,x,答案,C,由已知得,f,x,在,2,0,上單調(diào)遞減,由此可知只有選項(xiàng),C,符合題意,故選,C,3,2018,湖北武漢模擬,10,冪函數(shù),y,x,當(dāng),取不同的正數(shù)時(shí),在區(qū)間,0,1,上它們的圖象是一組美,麗的曲線,如圖,設(shè)點(diǎn),A,1,0,B,0,1,連接,AB,線段,AB,恰好被其中的兩個(gè)冪函數(shù),y,x,a,y,x,b,的圖象,三等分,即有,BM,MN,NA,則,a,A.0,B.1,C,D.2,1,b,1,2,答案,A

25、,因?yàn)?BM,MN,NA,點(diǎn),A,1,0,B,0,1,所以,M,N,分別代入,y,x,a,y,x,b,得,a,lo,b,lo,a,lo,0,故選,A,1,2,3,3,2,1,3,3,1,3,g,2,3,2,3,g,1,3,1,b,1,3,g,2,3,2,3,1,1,log,3,4,2018,湖北襄樊調(diào)研,11,設(shè),a,b,是關(guān)于,x,的一元二次方程,x,2,2,mx,m,6=0,的兩個(gè)實(shí)根,則,a,1,2,b,1,2,的最小值是,A.,B.18,C.8,D.-6,49,4,答案,C,方程,x,2,2,mx,m,6=0,的兩個(gè)根為,a,b,且,4,m,2,m,6,0,m,3,或,m,2,令,y,

26、a,1,2,b,1,2,a,b,2,2,ab,2,a,b,2=4,m,2,6,m,10=4,由二次函數(shù)的性質(zhì)知,當(dāng),m,3,時(shí),函數(shù),y,4,m,2,6,m,10,取得最小值,最小值為,8,故,a,1,2,b,1,2,的最小,值是,8,故選,C,2,6,a,b,m,ab,m,2,3,4,m,49,4,5,2018,山東德州期中,1,已知,f,x,ax,2,b,a,x,c,b,其中,a,b,c,若,a,b,c,0,x,1,x,2,為,f,x,的兩個(gè)零,點(diǎn),則,x,1,x,2,的取值范圍為,A,B.(2,2,C.(1,2,D.(1,2,3,2,3,2,3,3,答案,A,a,b,c,a,b,c,0,

27、a,0,c,0,b,a,c,0,由根與系數(shù)的關(guān)系可得,x,1,x,2,2,x,1,x,2,1,4,x,1,x,2,4,4,a,b,c,a,0,x,1,x,2,0,x,1,x,2,0,2,0,1,0,2,412,x,1,x,2,故選,A,c,a,a,b,a,2,a,c,a,c,a,c,b,a,2,a,c,a,2,c,a,2,1,2,x,x,2,1,2,x,x,2,2,c,a,2,1,c,a,2,c,a,4,c,a,2,2,c,a,a,b,a,c,b,a,c,a,2,c,a,c,a,1,2,9,4,2,2,c,a,3,2,3,2,6,2017,山東菏澤一模,10,設(shè),min,m,n,表示,m,n,

28、二者中較小的一個(gè),已知函數(shù),f,x,x,2,8,x,14,g,x,min,x,0,若,x,1,5,a,a,4,x,2,0,使得,f,x,1,g,x,2,成立,則,a,的最大,值為,A.-4,B.-3,C.-2,D.0,2,2,1,log,4,2,x,x,答案,C,當(dāng),log,2,4,x,時(shí),解得,x,1,當(dāng),0,x,1,時(shí),log,2,4,x,當(dāng),x,1,時(shí),log,2,4,x,g,x,min,x,0),當(dāng),0,x,1,時(shí),g,x,的值域?yàn)?2,當(dāng),x,1,時(shí),g,x,的值域?yàn)?0,2,g,x,的值域?yàn)?2,f,x,x,2,8,x,14,x,4,2,2,其圖象的對(duì)稱軸為直線,x,-4,f,x,

29、在,5,-4,上為減函數(shù),在,4,a,上為增函數(shù),f,5)=-1,f,a,a,2,8,a,14,當(dāng),4,a,3,時(shí),函數(shù),f,x,的值域?yàn)?2,-1,當(dāng),a,3,時(shí),函數(shù),f,x,的值域?yàn)?2,a,2,8,a,14,2,1,2,x,2,1,2,x,2,1,2,x,2,2,1,log,4,2,x,x,2,2,log,4,0,1,1,1,2,x,x,x,x,x,1,5,a,a,4,x,2,0,使得,f,x,1,g,x,2,成立,a,2,8,a,14,2,解得,6,a,2,3,a,2,綜上所述,a,的取值范圍為,4,-2,a,的最大值為,2,故選,C,思路分析,根據(jù)新定義求出函數(shù),g,x,的解析式,

30、再求出函數(shù),g,x,的值域,進(jìn)而求出,f,x,的值域,由,x,1,5,a,a,4,x,2,0,使得,f,x,1,g,x,2,成立,知,f,x,的值域是,g,x,的值域的子集,由此求,出實(shí)數(shù),a,的取值范圍,進(jìn)而得出,a,的最大值,7,2018,山東平度第一中學(xué)模擬,12,已知函數(shù),f,x,g,x,x,2,2,x,設(shè),b,為實(shí)數(shù),若存,在實(shí)數(shù),a,使得,f,a,g,b,2,成立,則實(shí)數(shù),b,的取值范圍為,A.(-1,3,B.-1,3,C.(,1,3,D.(,1,3,2,1,1,ln,2,1,x,x,x,x,x,3,4,答案,B,當(dāng),a,1,時(shí),f,a,a,1,2,當(dāng)且僅當(dāng),a,-2,時(shí)取等號(hào),f

31、,a,的取值范圍是,當(dāng),a,1,時(shí),f,a,ln,a,2,f,a,的取值范圍是,0,綜上,f,a,的取值范圍是,要存在實(shí)數(shù),a,使得,f,a,g,b,2,成立,則函數(shù),g,b,b,2,2,b,解得,1,b,3,2,1,a,a,2,1,1,a,a,1,1,1,2,1,a,a,1,1,a,1,0,4,1,4,3,4,9,4,二、填空題,共,5,分,8,2018,山東菏澤第一中學(xué)第一次月考,16,已知函數(shù),f,x,x,g,x,x,2,2,ax,4,若對(duì)任意,x,1,0,1,存在,x,2,1,2,使,f,x,1,g,x,2,則實(shí)數(shù),a,的最小值是,1,1,x,答案,9,4,解析,由題意得,原不等式可轉(zhuǎn)化為,f,x,min,g,x,min,顯然,f,x,在區(qū)間,0,1,上是單調(diào)遞增函數(shù),所,以,f,x,min,f,0)=-1,當(dāng),a,1,時(shí),g,x,min,g,1)=5-2,a,1,解得,a,3,與,a,1,矛盾,舍去,當(dāng),a,2,時(shí),g,x,min,g,2,8-4,a,1,解得,a,符合題意,當(dāng),1,a,2,時(shí),g,x,min,g,a,4,a,2