高考數(shù)學(xué)理科人教通用版核心講練大一輪復(fù)習(xí)練考題預(yù)測(cè)全過(guò)關(guān)2-2函數(shù)的單調(diào)性與最值

溫馨提示：此套題為Word版，請(qǐng)按住Ctrl,滑動(dòng)鼠標(biāo)滾軸，調(diào)節(jié)合適的觀看比例，答案解析附后。關(guān)閉Word文檔返回原板塊。練考題預(yù)測(cè)·全過(guò)關(guān)1.(2019·全國(guó)卷Ⅲ)設(shè)f(x)是定義域?yàn)镽的偶函數(shù),且在(0,+∞)上單調(diào)遞減,則 ()log314>f(log314>f(C.f(2-32)>f(D.f(2-23)>f(【解析】選C.依據(jù)題意,函數(shù)f(x)為偶函數(shù)且函數(shù)f(x)在(0,+∞)上單調(diào)遞減,則函數(shù)f(x)在(∞,0)上單調(diào)遞增;因?yàn)閒log314=f(log34)=f(又因?yàn)?<2-32<2-2所以f(2-32)>f(22.(2019·全國(guó)卷Ⅱ)設(shè)函數(shù)f(x)的定義域?yàn)镽,滿足f(x+1)=2f(x),且當(dāng)x∈(0,1]時(shí),f(x)=x(x1).若對(duì)任意x∈(∞,m],都有f(x)≥89,則m的取值范圍是A.-∞,94 C.-∞,52 【解析】選B.如圖,令f(x)=89,結(jié)合圖象可得f(x1)=49,則f(x2)=當(dāng)x∈(0,1]時(shí),f(x)=x(x1)=29,解得x=13或23,當(dāng)f(x)=89時(shí),x=73或83,即若f(x)≥89,對(duì)任意x3.定義在R上的函數(shù)f(x)=x3+m,與函數(shù)g(x)=f(x)kx在[1,1]上具有相同的單調(diào)性,則k的取值范圍是 ()A.(∞,0] B.(∞,3]C.[3,+∞) D.[0,+∞)【解析】′(x)=3x2≤0在[1,1]上恒成立,故f(x)在[1,1]上單調(diào)遞減,結(jié)合題意知,g(x)=x3+mkx在[1,1]上單調(diào)遞減,故g′(x)=3x2k≤0在[1,1]上恒成立,故k≥3x2在[1,1]上恒成立,故k≥0.4.已知f(x)=12xm,g(x)=ln(x2+1),若?x1∈[2,1],?x2∈[0,1],使得f(x1)≥g(x2),則實(shí)數(shù)mA.(∞,4] B.(∞,2]C.(∞,2ln2] D.(∞,4ln2]【解析】選B.要使命題成立需滿足f(x1)min≥g(x2)min,函數(shù)g(x)=ln(x2+1)在[0,1]上是增函數(shù),所以g(x2)min=g(0)=0,函數(shù)f(x)=12xm在[2,1]上是減函數(shù),所以f(x1)min=f(1)=2m,所以2m≥0,所以m5.函數(shù)f(x)=log12(1+x2)11-2|x|,A.(∞,1] B.13,C.13,1 D.【解析】選B.由題意知函數(shù)的定義域?yàn)?∞,0)∪(0,+∞),當(dāng)x>0時(shí),f(x)=log12(1+x2)11因?yàn)閒(x)=log12[1+(x)2]11-2|-x所以f(x)是偶函數(shù),所以f(x)在(∞,0)上單調(diào)遞增.因?yàn)閒(x)≤f(2x1),所以|x|≥|2x1|≠0,兩邊平方后化簡(jiǎn)得3x24x+1≤0,x≠0且x≠12,解得13≤x<12或12<x≤1,故使不等式成立的x的取值范圍是6.已知函數(shù)f(x)=ex+x3,若f(x2)<f(3x2),則實(shí)數(shù)x的取值范圍是________.

【解析】因?yàn)閒′(x)=ex+3x2>0,所以函數(shù)f(x)為增函數(shù)