8540

=-(3+ii13iq:2x2x30x<-1x1.5an2n-1,b11,b22,b34,b43

4的數(shù)列

f

=

kq44m0m1fx2lnx1ln(1x為奇函數(shù)fx的圖象關于(2,0對稱f

=2,f-

=0,f(

既不是奇函數(shù)也不是偶函數(shù)f

+f(-

2sinxfx(0,1對稱f(πxcosx1f(πxcosx1f(π=fπ+

f

x

對稱

ab02ab0,a2b0m2ab0,na2b0m2n4a3b=1,故 +2 =(1+2(m+2n=5+2(n+m ≥9,當且僅當m=n=1取2a- a+ 號a1,b1取等號 3618

=2cos2x-

=3sin2xcos2x,AD正確f(xx22axa在區(qū)間(-∞,1上有最小值函數(shù)圖象的對稱軸應當位于區(qū)間-∞,1內a1,g(xx+

2a(x1),1x1x2,g(x1g(x2x1

-x2-axx+a(x2-x1)xx)x1x2-a,a1,1xxxx0,xx10,xxa

1 1g(x1g(x20g(x1g(x2),g(xx+

2a在區(qū)間[1上單調遞增g(11aan+22an+1an+12an2an+1=-(an+1-2anan+12an(-1n,a102a9=-an+2an+12anSn+2Sn+1Sn+1Sn2(Sn-Sn-1(nn2時,Sn+22Sn+1Sn2Sn-1S22S10,S32S2a1所以Sn+1-2Sn01.S122S110a12S11.3515

=-

a,b

α

∈(π,

所以sin(α

22即sinαcosα4,1

,sin2α=7

3,(e-2,11,e2F(xf(x11ax(a0G(xg(xa1F(x1G(xx22a1F(xG(xaxx2HH(x在(-∞,0H(010,H(-1

-1<H(x在(-∞,01F(xG(x在(-∞,0=h(x2lnx(x0hr(x2(1-ln x0,e)時,hr(x0,h(x)xe時,hr(x0,h(x)h(xh(e20<lna

2個公共點,a的范圍是(1,ee 0a13ae-2- 2綜上所述3個公共點,aee,11,ee【解析】(1)由已知f(x=2cos2x+ 2=1+cos2x+sin2x=2sin2x+

4所以函數(shù)f(x的最小正周期為T=π 6(2由(1f

=

2kπ,k 8x

kπ,k 9f

x(xxπkπ,k

此時,f(xmax=2+ 13(1f(-a2e2ae.f

2ex 4(2x0時,f

ex.A

(

,fr(

6所以fr(x0=ex0.故切線OA的方程為y-ex0=ex0(x- 8因為過原點O(0,0,故-ex0=ex0(-x0,解得x0=1.所以切線OA的方程為y= 10fxex為偶函數(shù)y軸對稱故切線OB的方程為y=-ex.所以切線OA,OB的方程為y=ex,y=- 12②由①可知,A(1,eB(-1,e, 15 -1 =1 -1 =an-1 =1???4分an+1-

an-

1-

an-

an-

an- =1,所以數(shù)列1 是首項為1公差為1的等差數(shù)列 6a1- an-(2)由(1)知： =n,所以an=n+1.

an-(n+1

bn=an+1=n(n+

=1n(n+

=1+2(n-n+

9b1b2+?+bnn1(11(11(11+?+(12 n+n31(11 12 n31(11n2n29n160 nN1n9時n29n160不滿足條件當n≥10時,n2-9n-16單調遞增,所以n≥11.故n的最小值為 15在ΔAOB中AB2OA2OB22OAOBcosθ在ΔCOD中,CD2=OC2+OD2-2OC?ODcosθ 2在ΔAOD中,AD2OA2OD22OAODcosθ在ΔBOC中,BC2=OB2+OC2+2OB?OCcosθ 4故AD2+BC2-AB2-CD2=2AC? 7(2①由(1AD2BC2AB2CD22ACAB2,BCCD23,AD27,θ 9ABCD ②由若ΔABD與ΔBCD面積相等BD為公共底邊故兩個三角形BD上的高相等,即OAsin60°=OCsin60°,所以OA= 13OAOCm,OB在ΔAOB4m2n22mncos604m2n2在ΔBOC12m2n2mn.m2n28,2mn8所以(m-n2=m2+n2-2mn=0,故m= 15mn2AC2m4AD27,CD23AD2AC2CD2由勾股定理得：AC⊥ 17(1f(xaxlnx(0fr(xa因為x=1時f(x)取得最值,所以fr(1)=0,即:a+1=0,解得:a=- 4(2F(x

mf(x(1f(xxln因此:F(x)=

+m(-x+lnx)=

-mx+mlnx(x>Fr(x)

ex?x-ex-m+

m=ex(x-1)+m

=(x-1)(ex-Fr(x0x1exmx0（m=exF(xx1,x2,x3（x1x2m=

需有兩個不等于1的正 8mh(x

（x>0）,hr(x)=ex?x-

=ex(x-0x1時,hr(x0,h(x)x1時,hr(x0,h(x)單調遞增因此,h(xx1h(1em=

1的正根m故實數(shù)m的取值范圍為m> 12F(x10.05mFr(x0的根可知:x1x,xmex的兩個根（0x1x 1

=m=

(0<x<1<x F(x

+m(lnx-x)=

+m?lnx1=m-mlnm=F(x

所以F(x1)+0.05m<F(2x3)等價于F(2x3)-F(x3)>0.05m.(x3> 15