學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

1.C   2.D   3.D   4.C   5.D   6.D   7.A   8.A   學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

9.D提示:由得, , 即 ,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

所以 是等差數(shù)列.故.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

10.A    11.   12.   13.①②③   14.16   15.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

16.解:學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

 ⑴ .學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

⑵ 函數(shù)上單調(diào)遞增,在上單調(diào)遞減.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

所以,當(dāng)時(shí),;當(dāng)時(shí),.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

的值域?yàn)?sub>.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

17.解:⑴若,則,的圖象與軸的交點(diǎn)為,滿足題意.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

,則依題意得:,即.   故. 學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

⑵顯然.若,則由可知, 學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

方程有一正一負(fù)兩根,此時(shí)滿足題意.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

,則時(shí),,不滿足題意. 時(shí),方程有兩負(fù)根,也不滿足題意.故.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

18.解:由題意可知圓的方程為,于是.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

時(shí),設(shè),則由得,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

,. 所以的中點(diǎn)坐標(biāo)為.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

又由,且,可知直線與直線垂直,即直線的斜率為.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

此時(shí)直線的方程為,即.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

時(shí),同理可得直線的方程為.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

故直線的方程為.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

19.證明:⑴由函數(shù)的圖象關(guān)于直線對(duì)稱,有,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

即有. 又函數(shù)是定義在R上的奇函數(shù),有.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

. 從而. 即是周期為的周期函數(shù).學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

⑵由函數(shù)是定義在R上的奇函數(shù),有.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

時(shí),,.故時(shí),.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

時(shí),.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

從而,時(shí),函數(shù)的解析式為.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

20.解:⑴設(shè)第年新城區(qū)的住房建設(shè)面積為,則學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

當(dāng)時(shí),;當(dāng)時(shí),.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

所以, 當(dāng)時(shí),學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

當(dāng)時(shí),.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

時(shí),,,顯然有.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

   時(shí),,,此時(shí).學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

   時(shí),,,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

. 所以,時(shí),時(shí),.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

  時(shí),顯然.  故當(dāng)時(shí),;當(dāng) 時(shí),. 學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

21.解:⑴方法一  設(shè)動(dòng)點(diǎn)的坐標(biāo)為,則,.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

,得,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

化簡(jiǎn)得(當(dāng)時(shí)也滿足).學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

顯然,動(dòng)點(diǎn)在線段的中垂線的左側(cè),且,故軌跡的方程為 .學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

方法二  作的平分線交,則有,且 學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

,得 .學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

設(shè)動(dòng)點(diǎn)的坐標(biāo)為,則,即有,且.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

,故軌跡的方程為.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

⑵設(shè),,中點(diǎn).學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

學(xué)科網(wǎng)(Zxxk.Com)由點(diǎn)差法有 ;即.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

,所以,.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

①由, 得學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

②設(shè)直線的方程為,代入.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

所以,.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

四點(diǎn)共圓,則,由到角公式可得  ,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

,即,即.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

又由得,;所以,即.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

此外時(shí),存在,關(guān)于直線對(duì)稱,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

且滿足四點(diǎn)共圓.  故可能有四點(diǎn)共圓,此時(shí).學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

 


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