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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、CABDA   DCCCD   BA學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

二、13.4   14.  15.     16. 學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

三、17.(Ⅰ)∵,∴,         (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)

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)取bc的中點(diǎn)n,連接em,mn,an.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

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

∴mn∥ae,且mn=ae,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

∴四邊形aNme為平行四邊形,∴aN∥em,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

∴em與ac所成的角即為aN與ac所成的角,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

∵在中,∠CAN=,學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

∴em與ac所成的角為.                  (8分)學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

(Ⅲ)由(Ⅱ)知,em∥aN.學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

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

∴AN⊥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.    (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)學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

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

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

,即.                          (12分)學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

20.(Ⅰ)動(dòng)點(diǎn)的軌跡的方程為;             (3分)學(xué)科網(wǎng)(Zxxk.Com)學(xué)科網(wǎng)

(Ⅱ)解法1:(1)當(dāng)直線的斜率不存在時(shí),;   (6分)

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

.

設(shè),則,

                               (9分)

,

又∵當(dāng) 時(shí),取最小值

.

根據(jù)(1)、(2)得的取值范圍為.     (12分)

解法2:當(dāng)直線軸時(shí),,.     (6分)

當(dāng)直線不為軸時(shí),設(shè)過的直線的方程為,代入曲線方程得

設(shè),則,

                                 (9分)

=,

,即的取值范圍為.  (12 分)

21.(Ⅰ).

  ∴              (4分)

.

;由

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

(Ⅱ)函數(shù)的圖象有唯一的交點(diǎn),等價(jià)于方程

,即有唯一解.

由(Ⅰ)知,上遞減,在上遞增,

∴當(dāng)時(shí),取極小值(最小值).                     (11分)

從而方程有唯一解的充要條件是,∴函數(shù)的圖象有唯一交點(diǎn)時(shí),.    (14分)

22.(Ⅰ)由得,,∴,

.

,∴

即數(shù)列的通項(xiàng)公式為.                    (4分)

(Ⅱ).

設(shè)     ①

  ②

①-②,得

          ,

,即數(shù)列的前項(xiàng)和為.   (9分)

(Ⅲ)解法1:.

不等式恒成立,即對(duì)于一切恒成立.

設(shè).

當(dāng)時(shí),由于對(duì)稱軸,且,而函數(shù)是增函數(shù),

∴不等式恒成立,即當(dāng)時(shí),不等式對(duì)于一切恒成立.                                                 (14分)

解法2:.

不等式恒成立,即對(duì)于一切恒成立.

.

,∴,而,∴恒成立.

∴當(dāng)時(shí),不等式對(duì)于一切恒成立.     (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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