新版高中數(shù)學(xué)人教A版選修1-1習(xí)題第三章導(dǎo)數(shù)及其應(yīng)用3.3.1

3.3導(dǎo)數(shù)在研究函數(shù)中的應(yīng)用3.3.1函數(shù)的單調(diào)性與導(dǎo)數(shù)課時過關(guān)·能力提升基礎(chǔ)鞏固1.函數(shù)y=f(x)的圖象如圖,則導(dǎo)函數(shù)y=f'(x)的圖象可能是()解析:由函數(shù)y=f(x)的圖象可知,在區(qū)間(∞,0)和(0,+∞)內(nèi),函數(shù)f(x)均為減函數(shù),故在區(qū)間(∞,0)和(0,+∞)內(nèi),f'(x)均小于0,故選D.答案:D2.函數(shù)f(x)=x3+x在(1,+∞)內(nèi)為()A.減函數(shù) B.增函數(shù)C.常數(shù)函數(shù) D.不能確定解析:當(dāng)x∈(1,+∞)時,f'(x)=3x2+1<0,故f(x)在(1,+∞)內(nèi)為減函數(shù).答案:A3.已知函數(shù)f(x)=ax3x在R上為減函數(shù),則()A.a≤0 B.a<1 C.a<2 D.a≤1解析:f'(x)=3ax21.∵f(x)在R上為減函數(shù),∴f'(x)≤0在R上恒成立.∴a≤0.經(jīng)檢驗a=0符合題意.故選A.答案:A4.函數(shù)f(x)=A.(∞,1) B.(1,+∞)C.(1,1) D.(∞,1)∪(1,+∞)解析:函數(shù)f(x)=x(1-x)2f'(x)==令f'(x)=0,則1+x(1-x當(dāng)x∈(∞,1)時,f'(x)=當(dāng)x∈(1,1)時,f'(x)=當(dāng)x∈(1,+∞)時,f'(x)=故函數(shù)f(x)=x(1答案:C5.若函數(shù)y=f(x)在R上可導(dǎo),且滿足不等式xf'(x)>f(x)恒成立,常數(shù)a,b滿足a<b,則下列不等式一定成立的是()A.af(b)>bf(a) B.af(a)>bf(b)C.af(a)<bf(b) D.af(b)<bf(a)答案:C6.函數(shù)f(x)=x315x233x+6的單調(diào)遞減區(qū)間是.

解析:f'(x)=3x230x33=3(x11)(x+1),由(x11)(x+1)<0,得單調(diào)遞減區(qū)間為(1,11).答案:(1,11)7.函數(shù)y=(3x2)ex的單調(diào)遞增區(qū)間是.

解析:y'=2xex+(3x2)ex=(x22x+3)ex,令(x22x+3)ex>0,由于ex>0,則x22x+3>0,解得3<x<1,所以函數(shù)的單調(diào)遞增區(qū)間是(3,1).答案:(3,1)8.已知函數(shù)f(x)=2axsinx在其定義域上是增函數(shù),則實數(shù)a的取值范圍是.

解析:f'(x)=2acosx.∵f(x)在其定義域上是增函數(shù),∴f'(x)≥0恒成立,即2acosx≥0,∴a≥cos答案:19.已知函數(shù)f(x)=ax-解:f'(x)=a+∵f(x)在其定義域(0,+∞)內(nèi)為增函數(shù),∴當(dāng)x∈(0,+∞)時,f'(x)≥0恒成立.∴a+ax2-2∵x>0,∴2x10.已知函數(shù)y=f(x)=x3+bx2+cx+d的圖象經(jīng)過點P(0,2),且在點M(1,f(1))處的切線方程為6xy+7=0.(1)求函數(shù)y=f(x)的解析式;(2)求函數(shù)y=f(x)的單調(diào)區(qū)間.解:(1)由y=f(x)的圖象經(jīng)過點P(0,2),知d=2,所以f(x)=x3+bx2+cx+2,f'(x)=3x2+2bx+c.由在點M(1,f(1))處的切線方程為6xy+7=0,知6f(1)+7=0,即f(1)=1,f'(1)=6.所以即2b-故所求的解析式是y=f(x)=x33x23x+2.(2)f'(x)=3x26x3.令f'(x)>0,得x<1-2或x>令f'(x)<0,得1-故f(x)=x33x23x+2的單調(diào)遞增區(qū)間為(∞,1-2)和(1能力提升1.若函數(shù)y=x3ax2+4在(0,2)內(nèi)單調(diào)遞減,則實數(shù)a的取值范圍為()A.a≥3 B.a=3C.a≤3 D.0<a<3解析:y'=3x22ax≤0在(0,2)內(nèi)恒成立,即3x2≤2ax,a≥3x22答案:A2.已知f(x)滿足f(4)=f(2)=1,f'(x)為其導(dǎo)函數(shù),且導(dǎo)函數(shù)y=f'(x)的圖象如圖所示,則f(x)<1的解集是()A.(2,0)B.(2,4)C.(0,4)D.(∞,2)∪(0,4)答案:B3.函數(shù)f(x)的定義域為R,f(1)=2,對任意x∈R,f'(x)>2,則f(x)>2x+4的解集為()A.(1,1) B.(1,+∞)C.(∞,1) D.(∞,+∞)解析:由f(x)>2x+4,得f(x)2x>4,令g(x)=f(x)2x,則g'(x)=f'(x)2>0,∴g(x)在R上為增函數(shù),又g(1)=f(1)+2=4,即g(x)>g(1),∴x>1,故選B.答案:B4.若函數(shù)y=-解析:若函數(shù)y=-43x3+bx有三個單調(diào)區(qū)間,則其導(dǎo)數(shù)y'=4x2+b=0有兩個不相等的實根,所以Δ=16答案:(0,+∞)5.對于R上可導(dǎo)的任意函數(shù),若滿足(x1)f'(x)≥0,則f(0)+f(2)與2f(1)的大小關(guān)系是.

解析:∵(x1)f'(x)≥0,∴當(dāng)x>1時,f'(x)≥0,∴f(2)≥f(1);當(dāng)x<1時,f'(x)≤0,∴f(0)≥f(1).∴f(0)+f(2)≥2f(1).答案:f(0)+f(2)≥2f(1)★6.若函數(shù)f(x)=mx2+lnx2x在定義域內(nèi)是增函數(shù),則實數(shù)m的取值范圍是.

解析:f'(x)=2mx+1x-2,因為函數(shù)f(x)在其定義域(0,+∞)內(nèi)為增函數(shù),所以2mx+1x-2≥0在(0,+∞)內(nèi)恒成立,即2m≥-1x2+2x在(0,+∞)內(nèi)恒成立,由于函數(shù)φ答案:17.已知函數(shù)f(x)=xcosxsinx,x∈(0,π).(1)判斷函數(shù)f(x)的單調(diào)性;(2)判斷函數(shù)g(x)=sinxx,x∈解:(1)f'(x)=cosx+(xsinx)cosx=xsinx.∵x∈(0,π),∴sinx>0,∴f'(x)<0,∴f(x)在(0,π)內(nèi)是減函數(shù).(2)由(1)知,當(dāng)x∈(0,π)時,f(x)<f(0)=0.∴g'(x)=∴g(x)在(0,π)內(nèi)是減函數(shù).★8.設(shè)f(x)=x(ex1)ax2.(1)若a=(2)若當(dāng)x≥0時,f(x)≥0,求a的取值范圍.解:(1)當(dāng)a=12時,f(x)=x(e當(dāng)x∈(∞,1)時,f'(x)>0;當(dāng)x∈(1,0)時,f'(x)<0;當(dāng)x∈(0,+∞)時,f'(x)>0.故f(x)在(∞,1),(0,+∞)內(nèi)單調(diào)遞增,在(1,0)內(nèi)單調(diào)遞減.(2)f(x)=x(ex1ax),令g(x)=ex1ax,則g'(x)=exa.若a≤1,則當(dāng)x∈(0,+∞)時,g'(x)>0