2025高考數(shù)學(xué)一輪復(fù)習(xí)-3.2-導(dǎo)數(shù)與函數(shù)的單調(diào)性-專項(xiàng)訓(xùn)練【含答案】

2025高考數(shù)學(xué)一輪復(fù)習(xí)-3.2-導(dǎo)數(shù)與函數(shù)的單調(diào)性-專項(xiàng)訓(xùn)練【A級(jí)基礎(chǔ)鞏固】1.函數(shù)f(x)=2x-sinx在(-∞,+∞)上是()A.增函數(shù) B.減函數(shù)C.先增后減 D.不確定2.函數(shù)f(x)=xlnx+1的單調(diào)遞減區(qū)間是()A.(-∞,1e) B.(1C.(0,1e) 3.函數(shù)y=f(x)的導(dǎo)函數(shù)y=f′(x)的圖象如圖所示,則函數(shù)y=f(x)的圖象可能是()4.已知函數(shù)f(x)=3x+2cosx.若a=f(32),b=f(2),c=f(log27),則a,b,c的大小關(guān)系是()A.a<b<c B.c<b<aC.b<a<c D.b<c<a5.已知函數(shù)f(x)=aex-lnx在區(qū)間(1,2)單調(diào)遞增,則a的最小值為()A.e2 B.e C.e-1 D.e-26.已知函數(shù)f(x)=13x3+mx2+nx+1的單調(diào)遞減區(qū)間是(-3,1),則m+n的值為7.已知函數(shù)f(x)=ex-e-x-2x,則不等式f(x2-2x)+f(x-2)<0的解集為.

8.若f(x)=-12x2+aln(x+2)在[-1,+∞)上是減函數(shù),則實(shí)數(shù)a的取值范圍是9.討論函數(shù)f(x)=x3-x2+ax+1的單調(diào)性.INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【B級(jí)能力提升】10.已知函數(shù)f(x)是偶函數(shù),其導(dǎo)函數(shù)f′(x)的圖象如圖所示,且f(x+2)=f(2-x)對(duì)x∈R恒成立,則下列說法正確的是()A.f(-1)<f(12)<f(5B.f(52)<f(1C.f(-1)<f(52)<f(1D.f(12)<f(-1)<f(511.寫出一個(gè)同時(shí)具有下列性質(zhì)①②③的函數(shù)f(x):.

①f(x1x2)=f(x1)f(x2);②當(dāng)x∈(0,+∞)時(shí),f′(x)>0;③f′(x)是奇函數(shù).12.已知函數(shù)f(x)=12ax2單調(diào)性.13.已知g(x)=2x+lnx-ax(1)若函數(shù)g(x)在區(qū)間[1,2]內(nèi)單調(diào)遞增,求實(shí)數(shù)a的取值范圍;(2)若g(x)在區(qū)間[1,2]上存在單調(diào)遞增區(qū)間,求實(shí)數(shù)a的取值范圍.INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【C級(jí)應(yīng)用創(chuàng)新練】14.已知函數(shù)f(x)是定義在R上的偶函數(shù),其導(dǎo)函數(shù)為f′(x).若x>0時(shí),f′(x)<2x,則不等式f(2x)-f(x-1)>3x2+2x-1的解集是.參考答案【A級(jí)基礎(chǔ)鞏固】1.解析:因?yàn)閒(x)=2x-sinx,所以f′(x)=2-cosx>0在(-∞,+∞)上恒成立,所以f(x)在(-∞,+∞)上是增函數(shù).故選A.2.解析:f(x)的定義域?yàn)?0,+∞),f′(x)=1+lnx,令f′(x)<0,得0<x<1e,所以f(x)的單調(diào)遞減區(qū)間為(0,13.解析:f′(x)>0的解集對(duì)應(yīng)y=f(x)的單調(diào)遞增區(qū)間,f′(x)<0的解集對(duì)應(yīng)y=f(x)的單調(diào)遞減區(qū)間,驗(yàn)證只有D符合.故選D.4.解析:由題意,得f′(x)=3-2sinx.因?yàn)?1≤sinx≤1,所以f′(x)>0恒成立,所以函數(shù)f(x)是增函數(shù).因?yàn)?>1,所以32>3.又log24<log27<log28,即2<log27<3,所以2<log27<32,所以f(2)<f(log27)<f(32),即b<c<a.故選D.5.解析:依題意可知,f′(x)=aex-1x≥0在(1,2)上恒成立,顯然a>0,所以xex≥1設(shè)g(x)=xex,x∈(1,2),所以g′(x)=(x+1)ex>0,所以g(x)在(1,2)上單調(diào)遞增,g(x)>g(1)=e,故e≥1a,即a≥1e=e-1,即a的最小值為e故選C.6.解析:f′(x)=x2+2mx+n,由f(x)的單調(diào)遞減區(qū)間是(-3,1),得f′(x)<0的解集為(-3,1),則-3,1是f′(x)=0的解,所以-2m=-3+1=-2,n=1×(-3)=-3,可得m=1,n=-3,故m+n=-2.答案:-27.解析:f(x)=ex-e-x-2x,定義域?yàn)镽,且為奇函數(shù),f′(x)=ex+e-x-2≥2ex·e-x所以x2-2x<2-x,解得-1<x<2,所以原不等式的解集為(-1,2).答案:(-1,2)8.解析:f(x)=-12x2+aln(x+2)?f′(x)=-x+ax+2,因?yàn)閒(x)=-12x2+aln(x+2)在[-1,+∞)上是減函數(shù),所以f′(x)=-x+ax當(dāng)x∈[-1,+∞)時(shí),(x+1)2-1的最小值為-1,所以a≤-1.答案:(-∞,-1]9.解:由題意知f(x)的定義域?yàn)镽,f′(x)=3x2-2x+a,對(duì)于f′(x)=0,Δ=(-2)2-4×3a=4(1-3a).①當(dāng)a≥13②當(dāng)a<13即3x2-2x+a=0,解得x1=1-1-3a令f′(x)>0,則x<x1或x>x2;令f′(x)<0,則x1<x<x2,所以f(x)在(-∞,x1)上單調(diào)遞增,在(x1,x2)上單調(diào)遞減,在(x2,+∞)上單調(diào)遞增.綜上,當(dāng)a≥13當(dāng)a<13時(shí),f(x)在(-∞,1-1在(1-1-INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【B級(jí)能力提升】10.解析:因?yàn)閒(-x)=f(x),所以f(-1)=f(1),又f(x+2)=f(2-x),所以f(52)=f(32).由導(dǎo)函數(shù)的圖象得,當(dāng)x∈(0,2)時(shí),f′(x)>0,f(x)單調(diào)遞增,所以f(12)<f(1)<f(32),所以f(故選D.11.解析:取f(x)=x4,則f(x1x2)=(x1x2)4=x14x24f′(x)=4x3,當(dāng)x>0時(shí),有f′(x)>0,滿足②,f′(x)=4x3的定義域?yàn)镽,又f′(-x)=-4x3=-f′(x),故f′(x)是奇函數(shù),滿足③.答案:f(x)=x4(答案不唯一,f(x)=x2n(n∈N*)均滿足)12.解:函數(shù)f(x)的定義域?yàn)?0,+∞),f′(x)=ax-(a+1)+1x=ax2令f′(x)=0,得x=1a①當(dāng)0<a<1時(shí),1a所以x∈(0,1)或(1a,+∞)時(shí),f′(x)>0;x∈(1,1所以函數(shù)f(x)在(0,1)和(1a,+∞)上單調(diào)遞增,在(1,1②當(dāng)a=1時(shí),1a所以f′(x)≥0在(0,+∞)上恒成立,所以函數(shù)f(x)在(0,+∞)上單調(diào)遞增;③當(dāng)a>1時(shí),0<1a所以x∈(0,1a)或(1,+∞)時(shí),f′(x)>0;x∈(1所以函數(shù)f(x)在(0,1a)和(1,+∞)上單調(diào)遞增,在(1綜上,當(dāng)0<a<1時(shí),函數(shù)f(x)在(0,1)和(1a,+∞)上單調(diào)遞增,在(1,1當(dāng)a=1時(shí),函數(shù)f(x)在(0,+∞)上單調(diào)遞增;當(dāng)a>1時(shí),函數(shù)f(x)在(0,1a)和(1,+∞)上單調(diào)遞增,在(1遞減.13.解:(1)g′(x)=2+1x+a所以g′(x)≥0在[1,2]上恒成立,即2+1x+a所以a≥-2x2-x在[1,2]上恒成立,所以a≥(-2x2-x)max,x∈[1,2],(-2x2-x)max=-3,所以a≥-3.所以實(shí)數(shù)a的取值范圍是[-3,+∞).(2)g(x)在[1,2]上存在單調(diào)遞增區(qū)間,則g′(x)>0在[1,2]上有解,即a>-2x2-x在[1,2]上有解,所以a>(-2x2-x)min,又(-2x2-x)min=-10,所以a>-10,所以實(shí)數(shù)a的取值范圍是(-10,+∞).INCLUDEPICTURE"B組.TIF"INCLUDEPICTURE"E:\\大樣\\人教數(shù)學(xué)\\B組.TIF"INET【C級(jí)應(yīng)用創(chuàng)新練】14.解析:令g(x)=f(x)-x2,則g(x)是