考研数学三经济ch10各章复习题目

第十章函数方程与不等式证明11a一.证明不等式an1an(n1)2lna

11n1ann2.(a>1,n1)证明:令f(x)ax,在1n1,1n上使用拉格朗日定理f(1)f(1n)n111alnaanan1

f'()(11)1nn1n(n1)1ana即lna因此

1n1

an(n1)11aan1an(n1)2lna二.若a0,b0,0<p<1,证明

11n1ann2.(a>1,n1)(ab)papbp证明:令f(x)(xb)pxpbp明显f(0)=0.当x0时,由于0<p<1f'(x)p(xb)p1pxp10因此当x0时,f(x)单减,因此f(a)f(0)=0.因此(ab)即得(ab)

papbp0papbp三.设函数f(x)在[0,1]上有连续导数,知足0f'(x)1且f(0)0.求证1213(x)dxf(x)dx0f0x2x3(t)dt,明显F(0)证明:令F(x)=0.由于0f'(x)1且f(0)0,f(t)dtf00因此当x>0时f(x)>0.F'(x)xf(t)dtf3(x)2f(x)0=f(x)x2(x)(1)2f(t)dtf0xf(t)dtf2(x),明显(0)=0.令(x)20'(x)2f(x)2f(x)f'(x)2f(x)(1f'(x))0因此当x>0时,(x)>0.由(1)知F'(x)0(x>0).当x>0时F(x)F(0)=0.因此F(1)F(0)=0.立刻获得121f(x)dx0f3(x)dx0四.求证|a|p|b|p21p(|a||b|)p,(0<p<1).求证:先证当0x1,0<p<1时,有21pxp(1x)p1令F(x)xp(1x)pF'(x)pxp1p(1x)p1.F'(x)0得x1.F(1)21p，F(1)F(0)1.22因此F(1)21p为最大值，F(1)F(0)1为最小值.因此当0x1,0<p<1时,2有21pxp(1x)p12令x|a|,则1x|b|.代入上述结论,立刻获得|a||b||a||b|21p|a|p|b|p1(|a||b|)p(|a||b|)p即(|a||b|)p|a|p|b|p21p(|a||b|)p,(0<p<1).五.求证:若x+y+z=6,则x2y2z212,(x0,y0,z0).证明：方法1:2(x2y2z2)2xy2yz2xzx2y2z2(xyz)22xy2yz2xz362(x2y2z2)因此3(x2y2z2)36,x2y2z212方法2:解以下条件极值问题:s(x,y,z)x2y2z2条件：xyz6222－(x+y+z－6)令F(x,y,z,)=x+y+zFx'2x0,Fy'2y0,Fz'2z0解得x=y=z=2.只有一个驻点,当x=y=z=2时达到最小值12.因此x2y2z212,(x0,y0,z0)六.证明:1若f(x)在[a,b]上是增添的，且在其上f''(x)0，则(ba)f(a)b(ba)f(a)f(b)f(x)dx2a2若f(x)在[a,b]上是增添的，且在其上f''(x)0，则(ba)f(b)f(x)dx(ba)f(a)f(b)ba2证明：1方法1:由于f(x)是增添的,因此关于[a,b]中的全部x,有f(x)>f(a),因此bf(x)dxf(a)(ba)af(a)f(x)令F(x)f(t)dt(xa)xa2F'(x)f(x)f(a)f(x)(xa)f'(x)f(x)f(a)(xa)f'(x)2222f'( )(xa)f'(x)(xa)(ax)22=1(xa)[f'()f'(x)]0(由于f''(x)0)2因此F(x)单增.又由于F(a)=0,因此F(b)>F(a)=0.立刻可得bf(x)dx(ba)f(a)f(b)a2方法2:将f(x)台劳睁开t,x,因此f(a)

f(t)f(x)f'(x)(tx)f''( )(tx)22!f(a)f(x)f'(x)(ax)f''(1)(ax)22!f(b)f(x)f'(x)(bx)f''(2)(bx)22!f''(1)(ax)2f''(2)(bx)2f(b)2f(x)f'(x)(ab)2xf'(x)2!2!(f(b)f(a))(ba)2b(ab)b上式二边积分得f(x)dxf'(x)dxaa2xf'(x)dxbf''(1)(ax)2f''(2)(bx)2dxbaa2!2!因此bbb(f(b)f(a))(ba)f(x)dx(ab)(f(b)f(a))2xf(x)f(x)dx22aaabf(x)dxaf(b)af(a)bf(b)bf(a)2bf(b)2af(a)4abf(x)dx(ba)(f(a)f(b))4a2(ba)(f(b)f(a))bf(x)dx于是4aba(f(b)f(a))b即f(x)dx2a证法同1.注:不论方法1,2,右侧的不等式都不需要f(x)单增的条件.七.证明:1x1x2xnx12x22xn2nnx1x2xnnx1x2xn2nn2nn证明:1方法一:先证akbkak2bk2k1k1k1nnnn由(akxbk)2ak2x22akbkxbk20k1k1k1k1n2nn获得akbkak2bk2k1k1k1上述不等式中令akxk，bk1,获得nn2xkn1k1xk2nnk1n即x1x2xnx12x22xn2.nn方法二:令f(x)x2,p1p2pn1n由于f''(x)20因此f(p1x1pnxn)p1f(x1)pnf(xn)即(x1nxn)2x12nxn2即x1x2xnx12x22xn2nn2取f(x)=lnx,f''(x)10.令p=p==p=1/n.12nx2因此lnx1nxn1lnx11lnxnlnnx1xnnn立刻获得x1x2xnnx1x2xn.n八.设f''(x)c[a,b],且f(a)f(b)0,求证:b(ba)3max|f''(x)|f(x)dxa12axb证明:方法1=(x1=2

1b1b1:f''(x)(xa)(xb)dx2(xa)(xb)df'(x)2aab1a)(xb)f'(x)f'(x)(2xab)dx2a1bb1b(2xab)df(x)f(x)(2xab)2f(x)2dxa2aabf(x)dxab1ba)(x1b因此|f(x)dx|=f''(x)(xb)dx|f''(x)(xa)(xb)|dxa2a2a1max|f''(x)|(xa)(bx)dx=(ba)3b2axba12axb方法2：t,x,f(t)f(x)f'(x)(tx)f''()(tx)22!因此0=f(a)f(x)f'(x)(ax)f''()(ax)2f''( )(a2!f(x)f'(x)(xa)x)22!f''()bbba)2dxf(x)dxf'(x)(xa)dx2!(xaaab1=f(x)(xa)bbf''( )(xa)2dxf(x)dx2aaa因此于是(b=.

b1b)(xa)2dxf(x)dxf''(a4ab1b|f''()|(xa)2dx1ba)2dx|f(x)dx|max|f''(x)|(xa4a4axbaa)3max|f''(x)|12axb若f'(x)在[0,2]上连续,且f'(x)0,n(正整数)有22[f(2)f(0)]f(x)sinnxdx0n21证明:f(x)sinnxdx=0n

2f(x)dcosnx0=1)f(0))12f'(x)cosnxdx(f(2n0n因此2f(2)f(0)12f'(x)dx2[f(2)f(0)]f(x)sinnxdxnn0n0十.设在[a,b]上f''(x)0,a<x1<x2<b,0<<1,试证:f(x1)(1)f(x2)f[x1(1)x2]证明:f(x1(1)x2)f(x2)(1)(x2x1)f'(1)(1)f(x2)f(x1(1)x2)(x2x1)f'(2)(2)×－(2)×