21.已知A.M.N是圓O:x2+y2=1上的兩個(gè)動(dòng)點(diǎn).且M.N關(guān)于x軸對(duì)稱.直線AM與BN交于P點(diǎn).⑴求P點(diǎn)的軌跡C的方程, 查看更多

 

題目列表(包括答案和解析)

已知二次函數(shù)f(x)=ax2+bx(a、b為常數(shù),且a≠0)滿足條件:f(x-1)=f(3-x),且方程f(x)=2x有等根.
(1)求f(x)的解析式;
(2)是否存在實(shí)數(shù)m、n(m<n),使f(x)定義域和值域分別為[m,n]和[4m,4n]?如果存在,求出m、n的值;如果不存在,說(shuō)明理由.

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已知圓M的方程為:x2+y2-2x-2y-6=0,以坐標(biāo)原點(diǎn)為圓心的圓N與圓M相切.
(1)求圓N的方程;
(2)圓N與x軸交于E、F兩點(diǎn),圓內(nèi)的動(dòng)點(diǎn)D使得|DE|、|DO|、|DF|成等比數(shù)列,求·的取值范圍;
(3)過(guò)點(diǎn)M作兩條直線分別與圓N相交于A、B兩點(diǎn),且直線MA和直線MB的傾斜角互補(bǔ),試判斷直線MN和AB是否平行?請(qǐng)說(shuō)明理由

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已知二次函數(shù)f(x)ax2bx(ab為常數(shù),a≠0)滿足條件:f(x1)f(3x)且方程f(x)2x有等根.

(1)f(x)的解析式;

(2)是否存在實(shí)數(shù)mn(mn),使f(x)定義域和值域分別為[m,n][4m4n]?如果存在,求出m、n的值;如果不存在,說(shuō)明理由.

 

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已知二次函數(shù)f(x)=ax2+bx(a、b為常數(shù),且a≠0)滿足條件:f(x-1)=f(3-x),且方程f(x)=2x有等根.
(1)求f(x)的解析式;
(2)是否存在實(shí)數(shù)m、n(m<n),使f(x)定義域和值域分別為[m,n]和[4m,4n]?如果存在,求出m、n的值;如果不存在,說(shuō)明理由.

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已知圓M的方程為:x2+y2-2x-2y-6=0,以坐標(biāo)原點(diǎn)為圓心的圓N與圓M相切.
(1)求圓N的方程;
(2)圓N與x軸交于E、F兩點(diǎn),圓內(nèi)的動(dòng)點(diǎn)D使得|DE|、|DO|、|DF|成等比數(shù)列,求·的取值范圍;
(3)過(guò)點(diǎn)M作兩條直線分別與圓N相交于A、B兩點(diǎn),且直線MA和直線MB的傾斜角互補(bǔ),試判斷直線MN和AB是否平行?請(qǐng)說(shuō)明理由

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一、選擇題

1.D  2.A  3.C  4.D  5.B  6.C  7.D  8.B  9.A  10.A

二、填空題

11.148  12.-4  13.  14.-6  15.①②③④

三、解答題

16.解:⑴

                                                                                                                 3分

=1+1+2cos2x

=2+2cos2x

=4cos2x

∵x∈[0,]  ∴cosx≥0

=2cosx                                                                                                    6分

⑵ f (x)=cos2x-?2cosx?sinx

      =cos2x-sin2x

      =2cos(2x+)                                                                                           8分

∵0≤x≤  ∴

  ∴

,當(dāng)x=時(shí)取得該最小值

 ,當(dāng)x=0時(shí)取得該最大值                                                                  12分

17.由題意知,在甲盒中放一球概率為,在乙盒放一球的概率為                    3分

①當(dāng)n=3時(shí),x=3,y=0的概率為                                              6分

②|x-y|=2時(shí),有x=3,y=1或x=1,y=3

它的概率為                                                                12分

18.解:⑴證明:在正方形ABCD中,AB⊥BC

又∵PB⊥BC  ∴BC⊥面PAB  ∴BC⊥PA

同理CD⊥PA  ∴PA⊥面ABCD    4分

⑵在AD上取一點(diǎn)O使AO=AD,連接E,O,

則EO∥PA,∴EO⊥面ABCD 過(guò)點(diǎn)O做

OH⊥AC交AC于H點(diǎn),連接EH,則EH⊥AC,

從而∠EHO為二面角E-AC-D的平面角                                                             6分

在△PAD中,EO=AP=在△AHO中∠HAO=45°,

∴HO=AOsin45°=,∴tan∠EHO=,

∴二面角E-AC-D等于arctan                                                                   8分

⑶當(dāng)F為BC中點(diǎn)時(shí),PF∥面EAC,理由如下:

∵AD∥2FC,∴,又由已知有,∴PF∥ES

∵PF面EAC,EC面EAC  ∴PF∥面EAC,

即當(dāng)F為BC中點(diǎn)時(shí),PF∥面EAC                                                                         12分

19.⑴f '(x)=3x2+2bx+c,由題知f '(1)=03+2b+c=0,

f (1)=-11+b+c+2=-1

∴b=1,c=-5                                                                                                    3分

f (x)=x3+x2-5x+2,f '(x)=3x2+2x-5

f (x)在[-,1]為減函數(shù),f (x)在(1,+∞)為增函數(shù)

∴b=1,c=-5符合題意                                                                                      5分

⑵即方程:恰有三個(gè)不同的實(shí)解:

x3+x2-5x+2=k(x≠0)

即當(dāng)x≠0時(shí),f (x)的圖象與直線y=k恰有三個(gè)不同的交點(diǎn),

由⑴知f (x)在為增函數(shù),

f (x)在為減函數(shù),f (x)在(1,+∞)為增函數(shù),

,f (1)=-1,f (2)=2

且k≠2                                                                                               12分

20.⑴∵

                                                                                         3分

∴{an-3n}是以首項(xiàng)為a1-3=2,公比為-2的等比數(shù)列

∴an-3n=2?(-2)n1

∴an=3n+2?(-2)n1=3n-(-2)n                                                                        6分

⑵由3nbn=n?(3n-an)=n?[3n-3n+(-2)n]=n?(-2)n

∴bn=n?(-)n                                                                                                    8分

<6

∴m≥6                                                                                                                   13分

21.⑴設(shè)M(x0,y0),則N(x0,-y0),P(x,y)

AM:y=  、

BN:y=  、

聯(lián)立①②  ∴                                                                                      4分

∵點(diǎn)M(xo,yo)在圓⊙O上,代入圓的方程:

整理:y2=-2(x+1)  (x<-1)                                                                             6分

⑵由

設(shè)S(x1、y1),T(x2、y2),ST的中點(diǎn)坐標(biāo)(x0、y0)

則x1+x2=-(3+)

x1x2                                                                                                          8分

中點(diǎn)到直線的距離

故圓與x=-總相切.                                                                                        14分

⑵另解:∵y2=-2(x+1)知焦點(diǎn)坐標(biāo)為(-,0)                                                  2分

頂點(diǎn)(-1,0),故準(zhǔn)線x=-                                                                              4分

設(shè)S、T到準(zhǔn)線的距離為d1,d2,ST的中點(diǎn)O',O'到x=-的距離為

又由拋物線定義:d1+d2=|ST|,∴

故以ST為直徑的圓與x=-總相切                                                                      8分

 


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