全品高考备战2027年数学一轮学生用书08第19讲导数与不等式第2课时利用导数证明不等式【答案】作业手册

第2课时利用导数证明不等式1.解:(1)∵f(x)=(2-x)ex,∴f'(x)=-ex+(2-x)ex=(1-x)ex.当-1≤x<1时,f'(x)>0,f(x)单调递增;当1<x≤2时,f'(x)<0,f(x)单调递减.又f(-1)=3e,f(1)=e,f∴函数f(x)在区间[-1,2]上的最大值为e,最小值为0.(2)证明:记g(x)=f(x)-12x2+x-2e=(2-x)ex-12x2+x-2e,x>0,则g'(x)=(1-x)(ex当0<x<1时,g'(x)>0,g(x)单调递增;当x>1时,g'(x)<0,g(x)单调递减.则g(x)的最大值为g(1)=12-e,因为12-e<0,所以g(x)<0,所以当x>0时,f(x)<12x2-2.解:(1)因为f(x)=a(ex+a)-x,所以定义域为R,f'(x)=aex-1.当a≤0时,aex≤0,故f'(x)=aex-1<0恒成立,此时f(x)在R上单调递减;当a>0时,令f'(x)=aex-1=0,解得x=-lna,当x<-lna时,f'(x)<0,则f(x)在(-∞,-lna)上单调递减,当x>-lna时,f'(x)>0,则f(x)在(-lna,+∞)上单调递增.综上,当a≤0时,f(x)在R上单调递减;当a>0时,f(x)在(-∞,-lna)上单调递减,在(-lna,+∞)上单调递增.(2)证明:方法一:由(1)得,f(x)min=f(-lna)=a(e-lna+a)+lna=1+a2+lna,要证f(x)>2lna+32,即证1+a2+lna>2lna+32,即证a2-12-ln令g(a)=a2-12-lna(a>0),则g'(a)=2a-1a=令g'(a)<0,可得0<a<22令g'(a)>0,可得a>22所以g(a)在0,22上单调递减,在22,+∞上单调递增,所以g(a)min=g22=222-所以当a>0时,f(x)>2lna+32方法二:令h(x)=ex-x-1,则h'(x)=ex-1,因为y=ex在R上单调递增,所以h'(x)=ex-1在R上单调递增,又h'(0)=e0-1=0,所以当x<0时,h'(x)<0,当x>0时,h'(x)>0,所以h(x)在(-∞,0)上单调递减,在(0,+∞)上单调递增,所以h(x)≥h(0)=0,即ex≥x+1,当且仅当x=0时,等号成立.因为f(x)=a(ex+a)-x=aex+a2-x=ex+lna+a2-x≥x+lna+1+a2-x,当且仅当x+lna=0,即x=-lna时,等号成立,所以要证f(x)>2lna+32,即证x+lna+1+a2-x>2lna+32,即证a2-12-ln令g(a)=a2-12-lna(a则g'(a)=2a-1a=2令g'(a)<0,可得0<a<22令g'(a)>0,可得a>22所以g(a)在0,22上单调递减,在22,+∞上单调递增,所以g(a)min=g22=222-所以当a>0时,f(x)>2lna+323.解:(1)函数f(x)的定义域为(0,+∞),由已知得f'(x)=ax-b则f(1(2)由题意得g(x)=x·f(x)=xlnx+2e(x>0),则g'(x)=lnx+1.当x∈0,1e时,g'(x)<0,g(x)单调递减,当x∈1e,+∞时,所以g(x)的单调递减区间为0,1e,单调递增区间为1e,+∞,所以g(x(3)证明:要证f(x)>1ex,即证lnx+2ex>1ex,只需证xln令h(x)=xex(x>0),则h'(x)=1-xex,当x∈(0,1)时,h'(当x∈(1,+∞)时,h'(x)<0,h(x)单调递减,所以h(x)≤h(1)=1e由(2)知,当x∈(0,+∞)时,g(x)≥g1e=1e,又g(x)的最小值点与h(所以xlnx+2e>xex,即lnx+2所以f(x)>1e4.解:(1)函数f(x)的定义域为(0,+∞),f'(x)=a(lnx+1),令f'(x)>0,得x>1e,令f'(x)<0,得0<x<1e,所以函数f(x)在1e,(2)①由(1)知f'(x)=a(lnx+1).设与直线2x-y-11e=0平行的直线与曲线y=f(x)相切于点P(x0,y0),由f'(x0)=2,得a(lnx0+1)=2,则lnx0=2a-1,所以x0=e又f(x0)=ax0lnx0=ae2a-1×2a-1=(2-a)e2a-1,所以切点的坐标为(e所以-2e2a-1设2a-1=t,则a=2t+1,所以2ett设H(t)=2ett+1,t>0,则H'(t)=2tet(t+1)2>0在(0,+∞)上恒成立.所以H(t)=2ett+1在(0,+∞)上单调递增.又因为H(1)=e,所以2a-1=1,解得a=1,所以方程a②证明:当0<x≤1时,xlnx≤0,令m(x)=ex+cosx-2,0<x≤1,因为m'(x)=ex-sinx>0,所以m(x)在(0,1]上单调递增,所以m(x)>m(0)=1+1-2=0,所以xlnx≤ex+cosx-2成立;当x>1时,要证xlnx<ex+cosx-2,只需证xlnx-ex-cosx+2<0,设h(x)=xlnx-ex-cosx+2,x>1,则h'(x)=lnx-ex+sinx+1,设φ(x)=lnx-ex+sinx+1,x>1,则φ'(x)=1x-ex+cosx,当x>1时,ex>e,0<1x<1,-1≤cosx≤1,所以φ'(x)=1x-ex所以φ(x)在(1,+∞)上单调递减,所以φ(x)<φ(1)=1-e+sin1<0,即h'(x)<0,所以h(x)在(1,+∞)上单调递减,所以h(x)<h(1)=2-e-cos1<0,即当x>1时,xlnx<ex+cosx-2成立.综上,当a=1时,f(x)<ex+cosx-2.5.解:(1)函数f(x)=lnx+ax+1(a∈R)的定义域为(0,+∞),且f'(x)=1x+a.当a≥0时,对任意x∈(0,+∞),f'(x)=1x+a>0恒成立,所以f(当a<0时,令f'(x)=1x+a=1+axx=0,解得x=-1a,当x∈0,-1a时,f'(x)>0,则f(x)在区间0,-1a上单调递增,当x∈-1a,+∞时,f'(x)<0,则f(x)在区间-1a,+∞(2)证明:当a≤2时,因为x>0,所以要证f(x)x≤e2x,只需证lnx+ax+1≤xe2x,只需证lnx+2x+1≤xe2x,只需证e2x+lnx≥lnx+2x令g(x)=ex-x-1,则g'(x)=ex-1,当x∈(-∞,0)时,g'(x)<0,g(x)单调递减,当x∈(0,+∞)时,g'(x)>0,g(x)单调递增,所以g(x