2013年科技大學專升本高數微分中值證明題集f(xexex,f(xexe0xx＞1f(xf(1)=0,即exf(x在[0，1]f(1)f(0)0F(x)x2f(x，證明：在（0，1）內至少存在一點F()0F(1f(10F(00,1(0,1),使得F(1F(x2xf(xx2f(xF(00

f(xg(x在(a,bg(x)0f(x)g(xf(x)g(x)0x(a,b。證明：存在常數k，使f(x)kg(x),x(a,b)證：要證f(x) f(x)k,即[f(x)]0 F(x)

f(x,則F'(x)

fx)g(xg'(xf(x)0F(x)k(常數2f(x在[0，1]上連續，在（0，1）f(01,f(10,（0，1）內至少有一點f'(f(1f(xex1x1cosx,f(0f'(x)ex1sinx,f'(0)0,f''(x)excosx0(xf'(x)單調增,故當x0時f'(xf'(0f(x)單調增,故f(xf(00x2、證：設(x)xf(x)，則(xC[0,1],(x且(0(1)0,由羅爾定理得:(0,即：f(f(0

pba(ab)與向量a a

證：pa[b ]aba(a

baab22 22 2sinnxcosnxdx2n2cosnxdx，n0/

t2

tt2

/證：左端0

(sin2x)ndx

0

sinn

/

cosn2n2cosnudu2n2cosnxdx=

exex 1、證明：x≠0時 1 2f(x在[a,b]f(x＞0,證明：在[a,b]內有唯一的一點 b使得f(x)dx

fa1F(x)

exe2

1

x,知F(0222F(x)

exex2x

,F(0)0；F

(x)

12

ex

0,xF(x（1）x0F(xF(00，從而F(xF(0)0F(x)F(0)(2)x0F(x)F(0)0F(xF(0)0F(x)F(0)x0時F(x)0，即得證 b2F(x)f(t)dtf(t,則F(x)在[a,b]上連續,在(a,b) b ∵F(a)f(t)0,F(b)f(x)dx b

(a,b),使F(0,即f(x)dx

f (a,b),使F()0,由羅爾定理知,在與之間存在一點使F(

f()

0

f1f(x

)3

(

f(xxc(c0f(0f(00，求最小的常數kx0f(xkx22f(xf(xf(xf(x的零點。證：1（1）∵f(xxcf(c)0又

1ecc

x

1exx

1x

＞ 時

1x

2

x21xf(x)1,

1ex 2

1

即當

fx

x0 且k=1/2為所求最小常數121、設(xf(x)exxxf(x的兩個零點，亦為(xf(x12故(x(x1x2或(x2x1，使得(0[f()f()]e0f()f()aca,b,aca,b,

1 (abc

4 a,bcF(x)f(xxF(x∵0<f(x) ∴F(0)f(0)0f(0)0,F(1)f(1)1∴由根的存在定理知，至少存在一點（0,1）使F(0f( ．

f(xC[a,),f(xD2 ，

f(x) ，F(x)

f(xf(a(xaF(x在(a,x2.設f(x)dxF(xC,f(xf(xf1(xf1(x)dxxf1(x)F[f1(x)]

()(

a)2

由日中值定理：(a,x)

f(x)f(a)x

∴F(x)

x

[f(x)f(f(x)0,知f(x在(a,f(xf(從而F(x) ∴F(x)在(a,)內單調增加證：令tf1(xxff1(x)dxxf1(x)xdf1(x)=xf1(x)f

=xf1(x)F(t)

=xf1(x)F(f1(x))設f(x) ,且f(x)0,又G(x)

xf(t)dtx(ab，試證G(x(a,b)f(x)arcsinxg(x)

xa1 ,x1，證明:f(x)g(1xG(x)f(x)x

f(t)dt(xa)f(x)

fx

(x

x(xxxF(x)(xa)f(xx

f(t)dtF(x)f(x)(xa)f(x)f(x)(xa)f(x) F(xF(x)F(a)0,(x1 G(x)0,即G(x)在(a,b1F(x)

f(x)g(x)arcsinx

,x∵F(x)

1x21x2x11x2

x1

1x21x21x211x21x21x2∴F(x)

(xx=0F(00c∴F(xf(xg(x故arcsinx

,x11

f(xg(x在[a,b]上連續，在(a,b)f(af(b=0g(x證明：至少存在一點(a,b)f()g()fx>0(x21lnxx證：1F(x)f(xg(x≠0f(xg(x在[a,b上連續，在(a,bF(x在[a,b上連續，在(a,b又F(a)

f(a)g(a)0,F(b)

f(b)g(b)∴由羅爾定理知，在(a,b內至少存在一點F(而F(x)f(x)g(xf(x)g(xg2F(0f()g(f證：2．令(xx21lnxx1)2則(1(x)2xlnxx12,(1)x(x)2lnx

(x)

2(x20<x<1(x0；當1x時，(x0x(0,)時,(x)在x=1處取最小值(120故(x)(1)0(x在(0,內單調增加(x)在x1取最小值(10，因此(x(1

∴當x>0時，(x21)lnx(xb1

arctanbarctana

ba,(0ab)1a2設f(xC[a,b],且acdb，試證：在[a,b]上必有一點，使得mf(c)nf(d)(mn)f證：1f(xarctanx在[a,b]arctanbarctana

(ba),(a又b1

b

b1a∴b

arctanbarctana

b

1 1a2證：2f(x

f(x在[a,b]上取最大值M和最小值(mn)mmf(cnf(d)(m∴mmf(c)nf(d)m由介值定理：在[a,b]上必存在一點mf(c)nf(d)f(m14.

mf(c)nf(d)(mn)fln(11)x

1

,0x

f(xC[a,b],f(xD(a,b),且f(x0,記F(x)1

xa在(a,b)內F(x)證：1．即證ln(1xlnx

11f(tlnt，在區間[x,1+x]ln(1x)