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

巢湖市2009屆高三第一次教學(xué)質(zhì)量檢測(cè)試題學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

一、DABAD   CCCBB   AD學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

二、13.  14.     15      16. 學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

三、17.(Ⅰ)∵,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

,         (2分)學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

.                     (4分)學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

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

, ∴.               (6分)學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

(Ⅱ)由,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

    整理得,∴.              (10分)學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

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

18.由題意知,Ea⊥平面ABC,DC⊥平面ABC,AE∥DC,ae=2,dc=4,ab⊥ac,且AB=AC=2.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

學(xué)科網(wǎng)(Zxxk.Com)(Ⅰ)∵Ea⊥平面ABC,∴ea⊥ab,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

又∵ab⊥ac,   ∴ab⊥平面acde,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

        ∴四棱錐b-acde的高h(yuǎn)=ab=2,梯形acde的面積S=6,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

,即所求幾何體的體積為4.  (4分)學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

(Ⅱ)證明:取bc中點(diǎn)G,連接em,mG,aG.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

學(xué)科網(wǎng)(Zxxk.Com)∵m為db的中點(diǎn),∴mG∥DC,且學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

      ∴mG  ae,∴四邊形aGme為平行四邊形,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

      ∴em∥aG.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

又∵AG平面ABC,∴EM∥平面ABC.           (8分)學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

(Ⅲ)解法1:由(Ⅱ)知,em∥aG.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

又∵平面BCD⊥底面ABC,aG⊥bc,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

∴AG⊥平面BCD,∴EM⊥平面BCD.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

又∵EM平面BDE,∴平面BDE⊥平面BCD.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

在平面BCD中,過M作MN⊥DB交DC于點(diǎn)N,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

∴MN⊥平面BDE  點(diǎn)n即為所求的點(diǎn).學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

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

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

∴邊DC上存在點(diǎn)N,當(dāng)DN=DC時(shí),NM⊥平面BDE.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

解法2:以A為原點(diǎn),建立如圖所示的空間直角坐標(biāo)系,則A(0,0,0),B(0,2,0),C(-2,0,0),D(-2,0,4),E(0,0,2),M(-1,1,2),(2,2,-4),(2,0,-2),(0,0,-4),(1,1,-2).學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

學(xué)科網(wǎng)(Zxxk.Com)    假設(shè)在DC邊上存在點(diǎn)N滿足題意.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

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

∴邊DC上存在點(diǎn)N,當(dāng)DN=DC時(shí),NM⊥平面BDE.        (12分)學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

19.(Ⅰ)由題意知,        (2分)學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

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

當(dāng)時(shí),不等式的解集為學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

當(dāng)時(shí),不等式的解集為.      (6分)學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

(Ⅱ)

,且

,

,即.                          (12分)

20. (Ⅰ),

,∴.                (4分)

.

.

0

極大值

極小值

故函數(shù)的單調(diào)增區(qū)間為,,單調(diào)減區(qū)間為.  (8分)

(Ⅱ)由(Ⅰ)知,上遞增,在上遞減,在上遞增,在時(shí),取極大值.

又∵,,

∴在上,.

又∵,

(當(dāng)且僅當(dāng)時(shí)取等號(hào)).

的最小值為.

        ∵,∴對(duì)于,.        (12分)

21.(Ⅰ)動(dòng)點(diǎn)的軌跡的方程為;                         (3分)

(Ⅱ)解法1

當(dāng)直線的斜率不存在時(shí),,不合題意;

當(dāng)直線的斜率存在時(shí),設(shè)過的直線,代入曲線方程得

.

設(shè),則,

,

解得 ,

∴所求的直線的方程為.                  (9分)

解法2

當(dāng)直線軸時(shí),, ,不合題意;

當(dāng)直線不為軸時(shí),設(shè)過的直線,代入曲線方程得

.

設(shè),則,

=,解得,

∴所求的直線的方程為.                  (9分)

(Ⅲ)設(shè),

處曲線的切線方程為,

;令.

.

(當(dāng),時(shí)取等號(hào)).

,∴面積的最小值為2.   (14分)

22.(Ⅰ)由,即.

,∴,∴.

,∴,

即數(shù)列的通項(xiàng)公式為.                    (5分)

(Ⅱ)由(Ⅰ)知,.

設(shè)     ①

  ②

①-②,得

          ,

,即數(shù)列的前項(xiàng)和為.   (10分)

(Ⅲ)假設(shè)存在實(shí)數(shù),使得對(duì)一切正整數(shù),總有成立,

總成立.

設(shè),

當(dāng) 時(shí),,且遞減;當(dāng)時(shí),,且遞減,

最大,∴,∴.

故存在,使得對(duì)一切正整數(shù),總有成立.       (14分)

命題人:廬江二中   孫大志

柘皋中學(xué)   孫  平

巢湖四中   胡善俊

                                      審題人:和縣一中   賈相偉

巢湖市教研室  張永超

 

 


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