函數(shù)單調(diào)性

1、1.下列函數(shù)中在上為增函數(shù)的是()(0)???A.B.21yx??22yxx??C.D.11xy??1xyx??【答案】A【解析】∵的對(duì)稱軸為x=0且開口向下21yx??∴為其單調(diào)遞增區(qū)間.(0)???2.若R則M的取值范圍為…()24(aMaa???0)a???A.B.(4][4)????????(4]????C.D.[44][4)???【答案】A【解析】∵當(dāng)a0時(shí)當(dāng)a1.∴x1.4.已知函數(shù)f(x)=log則f(x)的值域?yàn)椤?)

2、123(32)xx???A.B.(22)(2)????C.D.()?????[0)???【答案】C【解析】∵∴時(shí)取”=“).令則30x??132(03xxx???1323xtx????0t??又∵真數(shù)大于0∴t0.∴y=log的值域?yàn)镽選C.2(0)tt?5.函數(shù)y=ln的單調(diào)遞增區(qū)間是.11xx??【答案】(11)【解析】根據(jù)題意需即函數(shù)的定義域?yàn)?11)原函數(shù)的遞增區(qū)間即為函數(shù)101xx????在(11)上的遞增區(qū)間由于u′′.故函

3、數(shù)u(x)=1()1xuxx???1()()1xxx???202(1)x???的遞增區(qū)間為(11)即為原函數(shù)的遞增區(qū)間.11xx??圍是()A.(01)B.C.D.2(0)3?32[)83?3[1)8?【答案】C【解析】本題考查對(duì)函數(shù)單調(diào)性概念的理解程度注意函數(shù)在兩個(gè)區(qū)間上如果分別單調(diào)并不能簡(jiǎn)單地說(shuō)函數(shù)在并區(qū)間上單調(diào)故由題意知需滿足:.132001(32)161aaaaa??????????????????3283a???7.函數(shù)

4、在上為增函數(shù)則a的取值范圍是.xyxa??(2)????【答案】2a?【解析】1xayxaxa??????依題意得函數(shù)的單調(diào)增區(qū)間為、(a)要使在上為增函數(shù)()a??????xyxa??(2)????只需即2.2a????a?8.已知函數(shù)f(x)=在上是增函數(shù)則a的取值范圍是.(23)43(1)(1)xaxaxax?????????(???)??【答案】12a??【解析】若函數(shù)f(x)=在上是增函數(shù)則(23)43(1)(1)xaxa

5、xax?????????()?????解得故.2301(23)43aaaaa???????????????3212aaa??????????????12a??9.已知函數(shù)的定義域?yàn)槿魧?duì)任意N都有則實(shí)數(shù)c的取值()cfxxx??(0)????x?()(3)fxf??范圍是.【答案】[612]【解析】若則f(x)在(0上遞增不合題意0c??)??若c的圖象如圖所示則0()cfxxx????(2)(3)(4)(3)ffff???????解得