高考数学一轮复习 课后习题 课时规范练21　利用导数证明不等式（黑龙江版）

课时规范练21利用导数证明不等式(分值:62分)1.(15分)已知函数f(x)=x-lnx-exx.(1)求f(x)的最大值;(2)证明:1+x3+exx≥3x-f(x2.(15分)已知函数f(x)=aex-12x2-x(1)若f(x)在R上单调递增,求实数a的取值范围;(2)当a=1时,证明:∀x∈(-2,+∞),f(x)>sinx.3.(15分)已知函数f(x)=12ax2-xlnx(1)若函数f(x)在(0,+∞)上单调递增,求实数a的取值范围;(2)若a=2e,证明:f(x)<xex+1.4.(17分)(2025·江苏南通模拟)已知函数f(x)=exsinx,g(x)=2ln(x+1)-x.(1)求曲线f(x)在点(0,f(0))处的切线方程;(2)求证:当x∈(0,π2)时,f(x)>exg(x(3)求证:12ln(n+1)<sin12+sin14+…+sin12n<12×(1+12+13+参考答案1.(1)解f(x)=x-lnx-exx,定义域为(0,+∞),则f'(x)=1令g(x)=ex-x,则g'(x)=ex-1.当x∈(0,+∞)时,g'(x)>0恒成立,即g(x)在(0,+∞)内单调递增,所以g(x)>g(0)=1>0,即当x>0时,ex>x.令f'(x)>0,得0<x<1,令f'(x)<0,得x>1,所以f(x)在(0,1)内单调递增,在(1,+∞)内单调递减,所以f(x)max=f(1)=1-e.(2)证明要证1+x3+exx≥3x-f(x),即证x3-2x+令h(x)=x-1-lnx,h'(x)=1-1令h'(x)>0,得x>1,令h'(x)<0,得0<x<1,所以h(x)在(0,1)内单调递减,在(1,+∞)内单调递增,h(x)≥h(1)=0,即x-1≥lnx,即欲证x3-2x+1≥lnx,只需证x3-2x+1≥x-1.也就是证明x3-3x+2≥0.(*)设φ(x)=x3-3x+2(x>0),则φ'(x)=3x2-3,令φ'(x)=0,得x=1.当x∈(0,1)时,φ'(x)<0,φ(x)单调递减;当x∈(1,+∞)时,φ'(x)>0,φ(x)单调递增.所以当x=1时,φ(x)取得最小值φ(1)=0.故(*)式成立,从而1+x3+exx≥3x-f(x2.(1)解因为f(x)=aex-12x2-x,所以f'(x)=aex-x-1,由f(x)在R上单调递增,得f'(x)≥0在R上恒成立,即aex-x-1≥0在R上恒成立,所以a≥x+1ex在R上恒成立,令h(x)=x+1ex,可得h'(x)=-xex,当x>0时,h'(x)<0,h(x)单调递减;当x<0时,h'(x)>0,h(x)单调递增,所以当x=0时,函数h(x)取得极大值且为最大值,最大值h(0)=(2)证明当a=1时,f(x)=ex-12x2-x,所以f'(x)=ex-x-1,且f(0)=1要证f(x)>sinx,只需证f(x)>1.①当x∈(0,+∞)时,令g(x)=f'(x)=ex-x-1,可得g'(x)=ex-1>0,所以g(x)单调递增,所以g(x)>g(0)=0,因此f(x)单调递增,所以f(x)>f(0)=1;②当x=0时,可得f(0)=1且sin0=0,所以f(0)>sin0,满足f(x)>sinx;③当-2<x<0时,可得sinx<0,因为ex>0,且-12x2-x=-12(x+1)2+12>0,所以f(x)>0,因此f(x)>综上,∀x∈(-2,+∞),都有f(x)>sinx.3.(1)解由已知得f'(x)=ax-lnx-1.因为f(x)在(0,+∞)内单调递增,所以当x>0时,f'(x)≥0,即a≥lnx+1x恒成立.令h(x)=lnx+1x(x>0),则h'(x)=-lnxx2,所以当0<x<1时,h'(x)>0,当x>1时,h'(x)<0,即h(x)在(0,1)内单调递增,在(1,+∞)内单调递减,因此h(x)max=h(1)=1,(2)证明若a=2e,要证f(x)<xex+1,只需证ex-lnx<ex+1x,即ex-ex<lnx+1x.令t(x)=lnx+1x(x>0),则t'(x)=x-1x2,所以当0<x<1时,t'(x)<0,当x>1时,t'(x)>0,所以t(x)在(0,1)内单调递减,在(1,+∞)内单调递增,则t(x)min=t(1)=1,所以lnx+1x≥1.令φ(x)=ex-ex(x>0),则φ'(x)=e-ex,所以当0<x<1时,φ'(x)>0,当x>1时,φ'(x)<0,所以φ(x)在(0,1)内单调递增,在(1,+∞)内单调递减,则φ(x)max=φ(1)=0,所以ex-ex≤0.所以ex-ex<lnx+1x,4.(1)解因为f(x)=exsinx,所以f(0)=e0sin0=0,f'(x)=exsinx+excosx,所以切线斜率为f'(0)=e0sin0+e0cos0=1,所以曲线f(x)在点(0,f(0))处的切线方程为y-0=1×(x-0),即x-y=0.(2)证明因为ex>0,所以f(x)>exg(x)等价于sinx>2ln(x+1)-x.设t(x)=sinx+x-2ln(x+1),所以t'(x)=cosx+1-2x+1,且t(0)=0,t'(0)=当x∈(0,π2)时,设q(x)=t'(x)=cosx+1-2x+1,则q'(x)=-sinx+2(x+1)2.因为函数y=-sinx,则q'(x)在(0,π2)内单调递减,又因为q'(0)=2>0,q'(π2)=-1+2(π2+1)

2<0,所以∃x0∈(0,π2)且当0<x<x0时,q'(x)>0,当x0<x<π2时,q'(x)<0此时q(x)在(0,x0)内单调递增,在(x0,π2)内单调递减又q(0)=0,q(π2)=1-2π所以对任意的x∈(0,π2),t'(x)=q(x)>0,则t(x)在(0,π2)所以t(x)>t(0)=0.综上,当x∈(0,π2)时,t(x)>即得sinx>2ln(x+1)-x,所以f(x)>exg(x).(3)证明因为n∈N*,所以0<1接下来证明sinx>ln(x+1),其中x∈(0,π2)设m(x)=sinx-ln(x+1),则m'(x)=cosx-1x设n(x)=m'(x),则n'(x)=-sinx+1因为函数y=-sinx,y=1(x+1)2在(则n'(x)=-sinx+1(x+1)2在区间(0,π2)内单调递减,因为n'(0)=1>0,n'(π2所以∃x1∈(0,π2),使得n'(x1)=当0<x<x1时,n'(x)>0;当x1<x<π2时,n'(x)<所以m'(x)在区间(0,x1)内单调递增,在区间(x1,π2)内单调递减又因为m'(0)=0,m'(x1)>0,m'(π2)=-1π所以∃x2∈(x1,π2),使得m'(x2)=0当0<x<x2时,m'(x)>0;当x2<x<π2时,m'(x)<所以m(x)在区间(0,x2)内单调递增,在区间(x2,π2)内单调递减因为m(0)=0,m(π2)=1-ln(π2+1)>0,所以sinx>ln(x+1)在区间(0,π2)内恒成立.令所以sin12n>ln(12n+1)所以sin12>ln32,sin14>ln54,sin16>ln76,…所以sin12+sin14+…+sin12n>ln32+ln5对∀n∈N*,2n+12n-2n+22所以2(sin12+sin14+…+sin12n)>2(ln32+ln54+…+ln2n+12n)>ln32+ln43+ln54+ln65+…+ln2n+12n+ln2n+22n+1=ln3-ln2+ln4-ln3+…+ln(2n+2)设p(x)=sin