第19練導(dǎo)數(shù)與函數(shù)的單調(diào)性(分值：40分)一、單項(xiàng)選擇題(每小題5分,共20分)1.(2024·浙江Z20名校聯(lián)盟聯(lián)考)函數(shù)f(x)=ln(2x-1)-x2+x的單調(diào)遞增區(qū)間是()A.(0,1) B.1C.1-22,答案D解析函數(shù)f(x)=ln(2x-1)-x2+x的定義域?yàn)?且f'(x)=22x-1-2x+1=令f'(x)>0,解得12<x<所以f(x)的單調(diào)遞增區(qū)間為122.若函數(shù)f(x)=lnx+ax2-2在區(qū)間12,2上存在單調(diào)遞增區(qū)間,則實(shí)數(shù)a的取值范圍是(A.(-∞,-2] B.-C.-2,-18 D.(-2,+答案D解析∵f(x)=lnx+ax2-2,∴f'(x)=1x+2ax若f(x)在區(qū)間12,2上存在單調(diào)遞增區(qū)間,則f'(x)>0在1即a>-12x2在∴a>-由x∈12,2,得-故a>-2,即實(shí)數(shù)a的取值范圍是(-2,+∞).3.(2025·滄州模擬)已知f(x)=3-x2-2cosx,設(shè)a=2-0.5,b=33,c=log213,則f(a),f(b),f(c)A.f(c)>f(a)>f(b) B.f(b)>f(a)>f(c)C.f(b)>f(c)>f(a) D.f(c)>f(b)>f(a)答案B解析由題意知,函數(shù)f(x)的定義域?yàn)镽,且f(-x)=3-(-x)2-2cos(-x)=f(x),所以f(x)為偶函數(shù).又f'(x)=2(sinx-x),當(dāng)x≥0時(shí),令g(x)=2(sinx-x),則g'(x)=2(cosx-1)≤0,所以g(x)在[0,+∞)上單調(diào)遞減,所以g(x)≤g(0)=0,即f'(x)≤0,所以f(x)在[0,+∞)上單調(diào)遞減.a=2-0.5=22>33=|c|=log213故|c|>a>b,從而f(b)>f(a)>f(|c|),即f(b)>f(a)>f(c).4.已知函數(shù)f(x)=exx-ax,當(dāng)0<x1<x2時(shí),不等式f(x1)x2-f(A.(-∞,e) B.(-∞,e]C.-∞,e答案D解析因?yàn)楫?dāng)0<x1<x2時(shí),不等式f(x1)x2-f(即x1f(x1)<x2f(x2),令g(x)=xf(x)=ex-ax2,則g(x1)<g(x2),又因?yàn)?<x1<x2,所以g(x)在(0,+∞)上單調(diào)遞增,所以g'(x)=ex-2ax≥0在(0,+∞)上恒成立,分離參數(shù)得2a≤exx在(0,+∞)令h(x)=exx(x>0則只需2a≤h(x)min,而h'(x)=ex·x令h'(x)>0,得x>1,令h'(x)<0,得0<x<1,所以h(x)在(0,1)上單調(diào)遞減,在(1,+∞)上單調(diào)遞增,所以h(x)≥h(1)=e,故2a≤e,即a≤e2二、填空題(共5分)5.(2024·洛陽(yáng)模擬)已知f(x)在R上是可導(dǎo)函數(shù),f(x)的圖象如圖所示,則不等式(x2-2x-3)f'(x)>f(-2)的解集為.

答案(-∞,-1)∪(-1,1)∪(3,+∞)解析由題圖可知f(-2)=0,且當(dāng)x∈(-∞,-1)和(1,+∞)時(shí),f'(x)>0,當(dāng)x∈(-1,1)時(shí),f'(x)<0,則原不等式等價(jià)于(x2-2x-3)f'(x)>0,等價(jià)于x2-2等價(jià)于x>3或解得x<-1或x>3或-1<x<1.即原不等式的解集為(-∞,-1)∪(-1,1)∪(3,+∞).三、解答題(共15分)6.(15分)已知函數(shù)f(x)=ax2+(2-a)x-lnx(a∈R).(1)當(dāng)a=2時(shí),求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程；(5分)(2)求函數(shù)y=f(x)的單調(diào)區(qū)間.(10分)解(1)由a=2,得f(x)=2x2-lnx,f(1)=2,切點(diǎn)為(1,2),f'(x)=4x-1x,f'(1)則切線方程為y-2=3(x-1),即3x-y-1=0.(2)f(x)=ax2+(2-a)x-lnx,定義域?yàn)?0,+∞),則f'(x)=2ax+(2-a)-1x=2a①當(dāng)a≥0時(shí),由f'(x)<0,得0<x<12由f'(x)>0,得x>1則f(x)的單調(diào)遞減區(qū)間為0,12,②當(dāng)a<0時(shí),令f'(x)=0,解得x=12或x=-當(dāng)-2<a<0時(shí),-1a>由f'(x)>0,得12<x<-1由f'(x)<0,得0<x<12或x>-則f(x)的單調(diào)遞增區(qū)間為12,-1當(dāng)a=-2時(shí),f'(x)=-(2x-1)2f(x)的單調(diào)遞減區(qū)間為(0,+∞),無(wú)單調(diào)遞增區(qū)間.當(dāng)a<-2時(shí),-1a<由f'(x)>0,得-1a<x<1由f'(x)<0,得0<x<-1a或x>則f(x)的單調(diào)遞增區(qū)間為-1a,綜上,當(dāng)a≥0時(shí),f(