2025屆高三一輪復(fù)習(xí)數(shù)學(xué)試題(人教版新高考新教材)考點規(guī)范練15　利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性

考點規(guī)范練15利用導(dǎo)數(shù)研究函數(shù)的單調(diào)性一、基礎(chǔ)鞏固1.函數(shù)f(x)=(x-3)ex的單調(diào)遞增區(qū)間是()A.(-∞,2) B.(0,3)C.(1,4) D.(2,+∞)答案:D解析:函數(shù)f(x)=(x-3)ex的導(dǎo)數(shù)為f'(x)=[(x-3)ex]'=ex+(x-3)ex=(x-2)ex.由函數(shù)導(dǎo)數(shù)與函數(shù)單調(diào)性的關(guān)系,得當(dāng)f'(x)>0時,函數(shù)f(x)單調(diào)遞增,此時由不等式f'(x)=(x-2)ex>0,解得x>2.2.已知函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),且函數(shù)f(x)的圖象如圖所示,則函數(shù)y=xf'(x)的圖象可能是()答案:C解析:由題圖可知函數(shù)f(x)在區(qū)間(-∞,-1)上單調(diào)遞減,在區(qū)間(-1,+∞)上單調(diào)遞增,則當(dāng)x∈(-∞,-1)時,f'(x)<0,當(dāng)x∈(-1,+∞)時,f'(x)>0,且f'(-1)=0.對于函數(shù)y=xf'(x),當(dāng)x∈(-∞,-1)時,xf'(x)>0,當(dāng)x∈(-1,0)時,xf'(x)<0,當(dāng)x∈(0,+∞)時,xf'(x)>0,且當(dāng)x=-1時,xf'(x)=0,當(dāng)x=0時,xf'(x)=0,顯然選項C符合.3.(2023新高考Ⅱ,6)已知函數(shù)f(x)=aex-lnx在區(qū)間(1,2)上單調(diào)遞增,則a的最小值為()A.e2 B.e C.e-1 D.e-2答案:C解析:由題意可知f'(x)=aex-1x≥0在區(qū)間(1,2)內(nèi)恒成立,即a≥1x設(shè)g(x)=xex,則g'(x)=(x+1)ex>0在區(qū)間(1,2)內(nèi)恒成立,所以函數(shù)g(x)=xex在區(qū)間(1,2)內(nèi)單調(diào)遞增,所以g(x)>g(1)=e,則0<1g(x)<1e,即a≥4.(多選)下列函數(shù)中,在R上為增函數(shù)的有()A.f(x)=x4 B.f(x)=x-sinxC.f(x)=xex D.f(x)=ex-e-x-2x答案:BD解析:A選項,由f(x)=x4,得f'(x)=4x3,當(dāng)x>0時,f'(x)=4x3>0,f(x)單調(diào)遞增;當(dāng)x<0時,f'(x)=4x3<0,f(x)單調(diào)遞減,故排除A;B選項,由f(x)=x-sinx,得f'(x)=1-cosx,因為f'(x)≥0恒成立,且不恒為零,所以f(x)=x-sinx在R上為增函數(shù),故B滿足題意;C選項,由f(x)=xex,得f'(x)=(1+x)ex,當(dāng)x>-1時,f'(x)>0,f(x)單調(diào)遞增;當(dāng)x<-1時,f'(x)<0,f(x)單調(diào)遞減,故排除C;D選項,由f(x)=ex-e-x-2x,得f'(x)=ex+e-x-2,因為f'(x)≥2ex·e-x-2=0恒成立,且不恒為零,所以f(x)=ex-e-x-2x在R上為增函數(shù)5.已知函數(shù)f(x)=2x-log2x,則不等式f(x)>0的解集是(A.(0,1) B.(-∞,2) C.(2,+∞) D.(0,2)答案:D解析:f(x)=2x-log2x的定義域為(0,+∞),因為f'(x)=-2x所以f(x)=2x-log2x在區(qū)間(0,+∞)上單調(diào)遞減,又f(2)=22-log22所以不等式f(x)>0的解集是(0,2).6.已知函數(shù)f(x)=kx3+3(k-1)x2-k2+1(k>0).(1)若f(x)的單調(diào)遞減區(qū)間是(0,4),則實數(shù)k的值為;

(2)若f(x)在區(qū)間(0,4)內(nèi)單調(diào)遞減,則實數(shù)k的取值范圍是.

答案:(1)13(2)解析:(1)f'(x)=3kx2+6(k-1)x(k>0),由題意知f'(4)=0,解得k=13(2)f'(x)=3kx2+6(k-1)x(k>0),由題意知f'(4)≤0,解得k≤13又k>0,故0<k≤137.已知函數(shù)f(x)=12x2-(a+1)x+alnx(a≥1)(1)當(dāng)a=1時,求函數(shù)f(x)的圖象在點(1,f(1))處的切線方程;(2)討論函數(shù)f(x)的單調(diào)性.解:(1)當(dāng)a=1時,f(x)=12x2-2x+lnx由f'(x)=x-2+1x,得f'(1)=0,又f(1)=-3故f(x)的圖象在點(1,f(1))處的切線方程為2y+3=0.(2)函數(shù)f(x)的定義域為(0,+∞),f'(x)=x-(a+1)+ax若a=1,則f'(x)=(x-1)2x≥0恒成立,函數(shù)f(x)在區(qū)間若a>1,則當(dāng)x∈(1,a)時,f'(x)<0,函數(shù)f(x)單調(diào)遞減;當(dāng)x∈(a,+∞)∪(0,1)時,f'(x)>0,函數(shù)f(x)單調(diào)遞增;綜上可知,當(dāng)a=1時,函數(shù)f(x)在區(qū)間(0,+∞)內(nèi)單調(diào)遞增,當(dāng)a>1時,函數(shù)f(x)在區(qū)間(1,a)內(nèi)單調(diào)遞減,在區(qū)間(a,+∞),(0,1)內(nèi)單調(diào)遞增.8.設(shè)函數(shù)f(x)=3x2+axex(1)若f(x)在x=0處取得極值,確定a的值,并求此時曲線y=f(x)在點(1,f(1))處的切線方程;(2)若f(x)在區(qū)間[3,+∞)內(nèi)單調(diào)遞減,求a的取值范圍.解:(1)對f(x)求導(dǎo)得f'(x)=(6x+因為f(x)在x=0處取得極值,所以f'(0)=0,即a=0.當(dāng)a=0時,f(x)=3x2ex,f'(x故f(1)=3e,f'(1)=3從而f(x)圖象在點(1,f(1))處的切線方程為y-3e=3化簡得3x-ey=0.(2)由(1)知f'(x)=-3令g(x)=-3x2+(6-a)x+a,由g(x)=0,解得x1=6-x2=6-當(dāng)x>x2時,g(x)<0,即f'(x)<0,故f(x)單調(diào)遞減.由f(x)在區(qū)間[3,+∞)內(nèi)單調(diào)遞減,知x2=6-a+a2+366≤3,解得a≥-二、綜合應(yīng)用9.已知函數(shù)f(x)=m3x3+2x2-x在區(qū)間13,+∞A.[0,+∞) B.(-4,+∞)C.[-3,+∞) D.-答案:B解析:f'(x)=mx2+4x-1,由題意可知mx2+4x-1≥0在區(qū)間13,當(dāng)m>0時,二次函數(shù)的圖象開口向上,當(dāng)m=0時,函數(shù)為y=4x-1,在區(qū)間13,即當(dāng)m≥0時,mx2+4x-1≥0在區(qū)間13,當(dāng)m<0時,由Δ>0,即16+4m>0,得-4<m<0,又二次函數(shù)圖象的對稱軸為x=-2m∈12,+∞,綜上所述,m>-4.10.(多選)已知定義在R上的函數(shù)f(x)的導(dǎo)函數(shù)為f'(x),且f(x)+xf'(x)<xf(x)對x∈R恒成立,則下列選項不正確的是()A.2ef(2)>f(1) B.2ef(2)C.f(1)>0 D.f(-1)>0答案:ACD解析:構(gòu)造函數(shù)F(x)=xf(因為F'(x)=ex[所以函數(shù)F(x)=xf(x)e因為2>1,所以F(2)<F(1),即2f(2)e2<f(1)因為F(1)<F(0),即f(1)e<0,所以f(1)<0,因為F(-1)>F(0),即-f(-1)e-1>0,所以f(-11.已知函數(shù)f(x)=-12x2+4x-3lnx在區(qū)間[t,t+1]上不單調(diào),則實數(shù)t的取值范圍是.答案:(0,1)∪(2,3)解析:由題意知f'(x)=-x+4-3x=-由f'(x)=0,得x1=1,x2=3,可知1,3是函數(shù)f(x)的兩個極值點.則只要這兩個極值點有一個在區(qū)間(t,t+1)內(nèi),函數(shù)f(x)在區(qū)間[t,t+1]上就不單調(diào),由t<1<t+1或t<3<t+1,得0<t<1或2<t<3.12.已知函數(shù)f(x)=xex+xex,且f(1+a)+f(-a2+a+2)>0,則實數(shù)a的取值范圍是答案:(-1,3)解析:因為x∈R,f(-x)=-xe-x+-xe-x=-xex+xex=-ff'(x)=ex+xex+1-xex=e2x(1+x)+1-xex,x∈R則g'(x)=e2x(3+2x)-1,令h(x)=e2x(3+2x)-1,則h'(x)=e2x(8+4x).當(dāng)x≥0時,h'(x)>0,所以h(x)在區(qū)間[0,+∞)上單調(diào)遞增,h(x)≥h(0)=2>0,即g'(x)>0,所以當(dāng)x≥0時,g(x)單調(diào)遞增,g(x)≥g(0)=2>0,所以f'(x)>0,f(x)在區(qū)間[0,+∞)上單調(diào)遞增.因為f(x)是奇函數(shù),所以f(x)在x∈R上是增函數(shù).由f(1+a)+f(-a2+a+2)>0,得f(1+a)>-f(-a2+a+2)=f(a2-a-2),所以1+a>a2-a-2,解得-1<a<3.13.若函數(shù)g(x)=lnx+ax2+bx,且g(x)的圖象在點(1,g(1))處的切線與x軸平行.(1)確定a與b的關(guān)系;(2)若a≥0,試討論函數(shù)g(x)的單調(diào)性.解:(1)因為g(x)=lnx+ax2+bx,所以g'(x)=1x+2ax+b由題意,得g'(1)=1+2a+b=0,所以2a+b=-1.(2)由(1)知g'(x)=(2ax-1當(dāng)a=0時,g'(x)=-x-由g'(x)>0,解得0<x<1,由g'(x)<0,解得x>1,即函數(shù)g(x)在區(qū)間(0,1)內(nèi)單調(diào)遞增,在區(qū)間(1,+∞)內(nèi)單調(diào)遞減;當(dāng)a>0時,令g'(x)=0,得x=1或x=12a,若12a<1,即a>12,則由g'(x)>0,解得x>1由g'(x)<0,解得12a即函數(shù)g(x)在區(qū)間0,12a,(1,+∞)內(nèi)單調(diào)遞增,若12a>1,即0<a<12,則由g'(x)>0,解得x>12由g'(x)<0,解得1<x<12即函數(shù)g(x)在區(qū)間(0,1),12a,+∞內(nèi)單調(diào)遞增若12a=1,即a=12,則在區(qū)間(0,+∞)內(nèi)恒有g(shù)'(x)≥0,即函數(shù)g(x)在區(qū)間(0,+∞綜上可得,當(dāng)a=0時,函數(shù)g(x)在區(qū)間(0,1)內(nèi)單調(diào)遞增,在區(qū)間(1,+∞)內(nèi)單調(diào)遞減;當(dāng)0<a<12時,函數(shù)g(x)在區(qū)間(0,1)內(nèi)單調(diào)遞增,在區(qū)間1,12a內(nèi)單調(diào)遞減當(dāng)a=12時,函數(shù)g(x)在區(qū)間(0,+∞)內(nèi)單調(diào)遞增當(dāng)a>12時,函數(shù)g(x)在區(qū)間0,12a內(nèi)單調(diào)遞增,在區(qū)間12a,1三、探究創(chuàng)新14.(多選)若函數(shù)f(x)在定義域D內(nèi)的某個區(qū)間I上單調(diào)遞增,且F(x)=f(x)x在區(qū)間I上也單調(diào)遞增,則稱y=f(x)在區(qū)間I上“一致單調(diào)遞增”.已知f(x)=x+exx,若函數(shù)f(x)在區(qū)間I上“一致單調(diào)遞增A.(-∞,-2) B.(-∞,0)C.(0,+∞) D.(2,+∞)答案:AD解析:f(x)=x+exx,則f'(x)=x2+ex(x-1)x2;F(x)=f當(dāng)x∈(-∞,-2)時,f'(x)=x2+ex(x-1)x2>x2+(x-1)x2>0,函數(shù)f(xf'-12=14-F'(1)=-e<0,故C不滿足;當(dāng)x∈(2,+∞)時,f'(x)=x2+ex(x-1)x2>0,F'15.定義在區(qū)間(0,+∞)內(nèi)的函數(shù)f(x)滿足f(x)>0,且當(dāng)x∈(0,+∞)時,2f(x)<xf'(x)<3f(x)恒成立,其中f'(x)為f(x)的導(dǎo)函數(shù),則()A.116<f(C.14<f(答案