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】听课手册

第2课时利用导数证明不等式●课堂考点探究例1[思路点拨](1)对f(x)求导得f'(x)=(2x+1)(ax+1)x,分a≥0和a<0两种情况讨论f(x)的单调性即可;(2)要证f(x)≤-2a-2+a,只需证f(x)max≤-2a-2+a,结合(1)的结论得,即证ln-1a+1a+1≤0恒成立,令t=-1a>0,g解:(1)f(x)=lnx+ax2+(a+2)x+a,其定义域为(0,+∞),则f'(x)=1x+2ax+a+2=2ax2+(①当a≥0时,f'(x)>0,f(x)在(0,+∞)上单调递增.②当a<0时,若x∈0,-1a,则f'(x)>0,f(x)在0,-1a上单调递增;若x∈-1a,+∞,则综上,当a≥0时,f(x)在(0,+∞)上单调递增;当a<0时,f(x)在0,-1a上单调递增,在(2)证明:当a<0时,要证f(x)≤-2a-2+a,只需证f(x)max≤-2a-2+a,由(1)得,f(x)max=f-1a=ln-1a+1a-a+2a即证ln-1a+1a+1≤0恒成立.令t=-1a>0,g(t)=lnt-t+1(t>0),则g'(t)=1t-1=1-tt,当t∈(0,1)时,当t∈(1,+∞)时,g'(t)<0,g(t)单调递减,∴g(t)的最大值为g(1)=0,即g(t)≤0,∴ln-1a+1变式题解:(1)函数f(x)=cosx+xsinx,x∈(-π,π),求导得f'(x)=-sinx+sinx+xcosx=xcosx.当-π<x<-π2时,f'(x)>0,f(x)单调递增;当-π2<x<0时,f'(x)<0,f(x)单调递减;当0<x<π2时,f'(x)>0,f(x)单调递增;当π2<x<π时,f'(x)<0,f(x)单调递减.所以f(x)的单调递增区间为-π,-π2,0,π2(2)证明:当x∈[0,π)时,令F(x)=ex+e-x-2(cosx+xsinx),求导得F'(x)=ex-e-x-2xcosx≥ex-e-x-2x,令φ(x)=ex-e-x-2x,x∈[0,π),求导得φ'(x)=ex+e-x-2≥2ex·e所以函数φ(x)在[0,π)上单调递增,则φ(x)≥φ(0)=0,所以F'(x)≥0,所以F(x)在[0,π)上单调递增,所以F(x)≥F(0)=0,所以2f(x)≤ex+e-x.例2[思路点拨](1)求导,分类讨论a与0的关系,判断导函数的符号,从而得出原函数的单调区间;(2)利用主元思想,根据题设条件a≤2将问题转化成证明当x>1时,ex-1-2x+1+lnx>0恒成立即可.解:(1)由题可得,f(x)的定义域为(0,+∞),f'(x)=a-1x=ax当a≤0时,f'(x)=ax-1x<0恒成立,故f(x)在(0,+∞)上单调递减;当a>0时,若x∈1a,+∞,则f'(x)>0,f(x)单调递增,若x∈0,1a综上所述,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)的单调递增区间为1a,+(2)证明:当a≤2,且x>1时,ex-1-f(x)=ex-1-a(x-1)+lnx-1≥ex-1-2x+1+lnx,令g(x)=ex-1-2x+1+lnx(x>1),要证ex-1>f(x)恒成立,只需证g(x)>0恒成立.易知g'(x)=ex-1-2+1x(x>1),令h(x)=g'(x则h'(x)=ex-1-1x2(x显然h'(x)在(1,+∞)上单调递增,则h'(x)>h'(1)=e0-1=0,即g'(x)=h(x)在(1,+∞)上单调递增,故g'(x)>g'(1)=e0-2+1=0,即g(x)在(1,+∞)上单调递增,故g(x)>g(1)=e0-2+1+ln1=0,得证.变式题解:(1)由题可得f'(x)=(ax2+4ax+2a+1)ex,当a=0时,f'(x)=ex>0恒成立,则f(x)的单调递增区间为(-∞,+∞).当a≠0时,关于x的方程ax2+4ax+2a+1=0的判别式Δ=4a(2a-1),当Δ>0时,方程的两根x1=-2a-2a2-aa,x2=-2a+2a2当0<a≤12时,f(x)的单调递增区间为(-∞,+∞)当a<0时,f(x)的单调递增区间为-2a+2a(2)证明:f(x)=(ax2+2ax+1)ex-2=[(x2+2x)ex]a+ex-2.当x≥0时,令g(a)=[(x2+2x)ex]a+ex-2,a≤-17,则g(a)在-∞,-17上单调递增,故g(a)≤g-17=-17(x2+2x)ex+ex-2.当x≥0时,要证f(x)<0,只需证-17(x2即证(x2+2x-7)ex+14>0.设h(x)=(x2+2x-7)ex+14,x≥0,则h'(x)=(x2+4x-5)ex=(x-1)(x+5)ex.当x∈[0,1)时,h'(x)<0,故h(x)在[0,1)上单调递减;当x∈(1,+∞)时,h'(x)>0,故h(x)在(1,+∞)上单调递增.所以当x≥0时,h(x)≥h(1)=14-4e>0,即当x≥0时,f(x)<0.故原不等式得证.例3[思路点拨](1)根据导数的运算和导数的几何意义即可求得结果;(2)由(1)可得f(x)的解析式,对要证的不等式进行变形,构造两个函数,分别利用导函数求出两函数在定义域内的最值,进而得证.解:(1)函数f(x)的定义域为(0,+∞),f'(x)=aexlnx+axex-bx2ex-1+bx由题意可得f(1)=b=2,f'(1)=ae-b+b=e,故a=1,b=2.(2)证明:由(1)知,f(x)=exlnx+2xex-1,又x>0,故f(x)>1等价于xlnx>xe-x-2设函数g(x)=xlnx,则g'(x)=1+lnx.当x∈0,1e时,g'(x)<0,当x∈1e,故g(x)在0,1e上单调递减,在1e,+∞上单调递增,故g(设函数h(x)=xe-x-2e,x>0,则h'(x)=e-x(1-x)当x∈(0,1)时,h'(x)>0,当x∈(1,+∞)时,h'(x)<0,故h(x)在(0,1)上单调递增,在(1,+∞)上单调递减,故h(x)的最大值为h(1)=-1e综上,当x>0时,g(x)>h(x),即f(x)>1.变式题解:(1)由函数f(x)=eln(x-1)-a(x-1),可得其定义域为(1,+∞),且f'(x)=ex-1-a当a≤0时,f'(x)>0,此时f(x)在区间(1,+∞)上单调递增;当a>0时,由f'(x)>0,可得1<x<1+ea,由f'(x)<0,可得x>1+e此时f(x)在区间1,1+ea(2)证明:当x>1时,要证f(x)-ex-1x-1+2e≤0,只需证f(x)≤ex-1x-1-2e,由(1)知,当a=e时,f(x)