高考数学一轮复习 课时规范练22　利用导数证明不等式（湘教版）

课时规范练22利用导数证明不等式(分值:61分)1.(15分)(2025·山西太原一模)已知函数f(x)=x-alnx,a∈R.(1)讨论函数f(x)的单调性;(2)当x∈(0,+∞)时,若f(x)≥cos(x-1)恒成立,求a的值.2.(12分)(2024·全国甲,文20)已知函数f(x)=a(x-1)-lnx+1.(1)讨论f(x)的单调区间;(2)若a≤2,证明:当x>1时,f(x)<ex-1恒成立.3.(17分)(2026·湖南长沙模拟)已知函数f(x)=ex·cosx(e为自然对数的底数).(1)求函数f(x)的图象在点(0,f(0))处的切线方程;(2)记g(x)=-x3-3x·cosx+(a+3)x,若∀x∈[0,1],有f(x)≤eg(x),求实数a的取值范围;(3)记n∈N+,且n≥2,证明:cos1+cos23+cos12+cos25+…+cos2n4.(17分)已知函数f(x)=lnx-mx2在x=1处的切线方程为x+my=0.(1)求实数m的值;(2)已知a>0,函数g(x)=f(x)+1-x-ba+x2,若g(x)≤0,求证:ab

参考答案课时规范练22利用导数证明不等式1.解(1)由题意得f'(x)=1-ax,x>0,当a≤0时,则f'(x)≥0,∴f(x)在区间(0,+∞)上单调递增;当a>0时,令f'(x)<0,则0<x<a;令f'(x)>0,则x>a,∴f(x)在区间(0,a)上单调递减,在区间(a,+∞)上单调递增(2)令g(x)=f(x)-cos(x-1)=x-alnx-cos(x-1),x>0,则g'(x)=1-ax+sin(x-1),∵g(1)=0,g(x)≥0,∴g'(1)=1-a=0,∴a=当a=1时,g(x)=x-lnx-cos(x-1),x>0,令h(x)=x-1-lnx,x>0,则h'(x)=1-1x,令h'(x)<0,则0<x<1;令h'(x)>0,则x>1,∴h(x)在区间(0,1)上单调递减,在区间(1,+∞)上单调递增,∴h(x)≥h(1)=0,∴x-lnx≥1,∴g(x)=x-lnx-cos(x-1)≥1-cos(x-1)≥0.综上,a=12.(1)解f(x)=a(x-1)-lnx+1,x>0,则f'(x)=ax-1x,x>0.若a≤0,则f'(x)<0恒成立,则f(x)的单调递减区间为(0,+∞),无单调递增区间.若a>0,则当0<x<1a时,f'(x)<0;当x>1a时,f'(x)>0.所以f(x)的单调递减区间为(0,1a),单调递增区间为综上,当a≤0时,f(x)的单调递减区间为(0,+∞),无单调递增区间;当a>0时,f(x)的单调递减区间为(0,1a),单调递增区间为(1a,+∞(2)证明因为a≤2,所以当x>1时,ex-1-f(x)=ex-1-a(x-1)+lnx-1≥ex-1-2x+lnx+1.令g(x)=ex-1-2x+lnx+1,x≥1,则g'(x)=ex-1-2+1x令h(x)=g'(x),则h'(x)=ex-1-1x2在(1,+∞)上单调递增,则h'(x)≥h'(1)=0且等号不恒成立,所以h(x)=g'(x)在(1,+∞)上单调递增,g'(x)≥g'(1)=0且等号不恒成立,故g(x)在(1,+∞)上单调递增,g(x)≥g(1)=0,所以当x>1时,ex-1-f(x)≥ex-1-2x+lnx+1>0恒成立,即f(x)<ex-13.(1)解由f(x)=ex·cosx,则f(0)=1,而f'(x)=ex·cosx(cosx-xsinx),则f'(0)=1,所以函数f(x)的图象在点(0,f(0))处的切线方程为y-1=x,即y=x+1.(2)解若∀x∈[0,1],有f(x)≤eg(x)⇔excosx≤e-即为xcosx≤-x3-3xcosx+(a+3)x在区间[0,1]上恒成立,当x=0时,不等式恒成立;当0<x≤1时,不等式等价于cosx≤-x2-3cosx+(a+3),即为4cosx+x2-(a+3)≤0恒成立,设M(x)=4cosx+x2-(a+3),x∈(0,1],则M'(x)=-4sinx+2x,设T(x)=-4sinx+2x,x∈(0,1],则T'(x)=2-4cosx,因为0<x≤1<π3,所以12<cosx<1,T'(x)<0,所以M'(x)在区间(0,1]上单调递减,则M'(x)<-4sin0+2×0=0,所以M(x)在区间(0,1]上单调递减,即M(x)<4cos0+0-(a+3)=4-(所以a≥1,则实数a的取值范围为[1,+∞).(3)证明由(2)知,当x∈(0,1]时,x2+4cosx-4≤0,即cosx≤1-x24,令x=2n(n≥2),得cos2n因为1n所以cos2n≤1-1n2<1-1n+1n+1,令n=2,cos1<1-12+13,令n=3,cos23<1-13+14,…,cos2n-1<1-1n-1+1n,cos2n<1-1n+14.(1)解因为f(x)=lnx-mx2,所以f'(x)=1x-2mx,当m=0时,f(x)=lnx,显然x=0不是f(x)的切线,不符合题意;当m≠0时,由题意得f(1)(2)证明g(x)=lnx+1-x-ba=lnx-xa+1+ba,g'(因为a>0,所以g(x)在区间(0,a)上单调递增,在区间(a,+∞)上单调递减,所以g(x)max=g(a)=lna+ba.若g(x)≤0,当且仅当g(x)max≤0,即lna+ba≤0,即b≤-alna,所以ab≤-a2ln设h(x)=-x2lnx,x>0,则