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1、課下層級(jí)訓(xùn)練(十二)對(duì)數(shù)與對(duì)數(shù)函數(shù)A級(jí)基礎(chǔ)強(qiáng)化訓(xùn)練1若函數(shù)yf(x)是函數(shù)yax(a0,且a1)的反函數(shù),且f(2)1,則f(x)()Alog2xBClogx D2x2【答案】A由題意知f(x)logax(a0,且a1),f(2)1,loga21,a2.f(x)log2x.2(2019山東煙臺(tái)月考)函數(shù)f(x)xa滿足f(2)4,那么函數(shù)g(x)|loga(x1)|的圖象大致為()A BC D【答案】C方法一函數(shù)g(x)|loga(x1)|的定義域?yàn)椋簒|x1,從而排除D;由g(x)|loga(x1)| 0,排除B;x0時(shí),g(x)0,排除A方法二由f(2)4,即2a4,得a2.先作出ylog
2、2x的圖象,再將此函數(shù)圖象向左平移1個(gè)單位,得函數(shù)ylog2(x1)的圖象,最后將此函數(shù)圖象x軸上方部分不變,下方部分關(guān)于x軸對(duì)稱進(jìn)行翻折,即得g(x)|loga(x1)|的圖象3(2019山西晉中月考)已知a2,blog2,clog,則()Aabc BacbCcba Dcab【答案】D02201,blog2log210,cloglog23log221,caB4(2019福建龍巖月考)已知函數(shù)f(x)ln x,g(x)lg x,h(x)log3x,直線ya(a0)與這三個(gè)函數(shù)的交點(diǎn)的橫坐標(biāo)分別是x1,x2,x3,則x1,x2,x3的大小關(guān)系是()Ax2x3x1 Bx1x3x2Cx1x2x3 D
3、x3x2x1【答案】A分別作出三個(gè)函數(shù)的大致圖象,如圖所示,由圖可知,x2x3x1.5(2019山東濟(jì)南月考)已知log23a,log35b,則lg 6()A BC D【答案】Dlog23a,log35b,a,b,解得lg 2,lg 3.lg 6lg 2lg 3.6若函數(shù)f(x)lg(x22ax1a)在區(qū)間(,1上遞減,則a的取值范圍為()A1,2) B1,2C1,) D2,)【答案】A令函數(shù)g(x)x22ax1a(xa)21aa2,對(duì)稱軸為xa,要使函數(shù)在(,1上遞減,則有即解得1a0,a1)的圖象過定點(diǎn)A,若點(diǎn)A也在函數(shù)f(x)2xb的圖象上,則f(log23)_.【答案】1由題意得A(2
4、,0),因此f(2)4b0,b4,從而f(log23)341.8設(shè)f(x)loga(1x)loga(3x)(a0,a1),且f(1)2.(1)求a的值及f(x)的定義域;(2)求f(x)在區(qū)間上的最大值【答案】解(1)f(1)2,loga42(a0,a1),a2.由得x(1,3),函數(shù)f(x)的定義域?yàn)?1,3)(2)f(x)log2(1x)log2(3x)log2(1x)(3x)log2(x1)24,當(dāng)x(1,1時(shí),f(x)是增函數(shù);當(dāng)x(1,3)時(shí),f(x)是減函數(shù),故函數(shù)f(x)在上的最大值是f(1)log242.9已知函數(shù)f(x)是定義在R上的偶函數(shù),f(0)0,當(dāng)x0時(shí),f(x)lo
5、gx.(1)求函數(shù)f(x)的解析式;(2)解不等式f(x21)2.【答案】解(1)當(dāng)x0,則f(x)log(x)因?yàn)楹瘮?shù)f(x)是偶函數(shù),所以f(x)f(x)所以函數(shù)f(x)的解析式為(2)因?yàn)閒(4)log42,f(x)是偶函數(shù),所以不等式f(x21)2可化為f(|x21|)f(4)又因?yàn)楹瘮?shù)f(x)在(0,)上是減函數(shù),所以|x21|4,解得x,即不等式的解集為x|xkg(x)恒成立,求實(shí)數(shù)k的取值范圍【答案】解(1)h(x)(42log2x)log2x2(log2x1)22,因?yàn)閤1,4,所以log2x0,2,故函數(shù)h(x)的值域?yàn)?,2(2)由f(x2)f()kg(x),得(34log2x)(3log2x)klog2x,令tlog2x,因?yàn)閤1,4,所以tlog2x0,2,所以(34t)(3t)kt對(duì)一切t0,2恒成立,當(dāng)t0時(shí),kR;當(dāng)t(0,2時(shí),k恒成立,即k4t15,因?yàn)?t 12,當(dāng)且僅當(dāng)4t,即t時(shí)取等號(hào),所以4t15的最小值為3,綜上,k(,3) 5