(Ⅱ)若.試問數(shù)列中是否存在整數(shù).使得對(duì)任意的正整數(shù)都有成立?并證明你的結(jié)論. 河西區(qū)2008―2009學(xué)年度第二學(xué)期高三年級(jí)總復(fù)習(xí)質(zhì)量調(diào)查(二) 查看更多

 

題目列表(包括答案和解析)

已知數(shù)列中,,且點(diǎn)P在直線x-y+1=0上。
(1)求數(shù)列的通項(xiàng)公式;
(2)設(shè),求數(shù)列的前n項(xiàng)和Tn;
(3)設(shè)表示數(shù)列的前n項(xiàng)和。試問:是否存在關(guān)于n的整式,使得對(duì)于一切不小于2的自然數(shù)n恒成立? 若存在,寫出的解析式,并加以證明;若不存在,試說明理由。

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已知數(shù)列{an}是以d為公差的等差數(shù)列,{bn}數(shù)列是以q為公比的等比數(shù)列.
(Ⅰ)若數(shù)列的前n項(xiàng)和為Sn,且a1=b1=d=2,S3<a1003+5b2-2010,求整數(shù)q的值;
(Ⅱ)在(Ⅰ)的條件下,試問數(shù)列中是否存在一項(xiàng)bk,使得bk恰好可以表示為該數(shù)列中連續(xù)p(p∈N,p≥2)項(xiàng)的和?請(qǐng)說明理由;
(Ⅲ)若b1=ar,b2=as≠ar,b3=at(其中t>s>r,且(s-r)是(t-r)的約數(shù)),求證:數(shù)列{bn}中每一項(xiàng)都是數(shù)列{an}中的項(xiàng).

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(1)若對(duì)于任意的n∈N*,總有
n+2
n(n+1)
=
A
n
+
B
n+1
成立,求常數(shù)A,B的值;
(2)在數(shù)列{an}中,a1=
1
2
,an=2an-1+
n+2
n(n+1)
(n≥2,n∈N*),求通項(xiàng)an;
(3)在(2)題的條件下,設(shè)bn=
n+1
2(n+1)an+2
,從數(shù)列{bn}中依次取出第k1項(xiàng),第k2項(xiàng),…第kn項(xiàng),按原來的順序組成新的數(shù)列{cn},其中cn=bkn,其中k1=m,kn+1-kn=r∈N*.試問是否存在正整數(shù)m,r使
lim
n→+∞
(c1+c2+…+cn)=S
4
61
<S<
1
13
成立?若存在,求正整數(shù)m,r的值;不存在,說明理由.

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已知數(shù)列{an}的前n項(xiàng)和為Sn,且滿足2Sn=pan-2n,n∈N*,其中常數(shù)p>2.
(1)證明:數(shù)列{an+1}為等比數(shù)列;
(2)若a2=3,求數(shù)列{an}的通項(xiàng)公式;
(3)對(duì)于(2)中數(shù)列{an},若數(shù)列{bn}滿足bn=log2(an+1)(n∈N*),在bk與bk+1之間插入2k-1(k∈N*)個(gè)2,得到一個(gè)新的數(shù)列{cn},試問:是否存在正整數(shù)m,使得數(shù)列{cn}的前m項(xiàng)的和Tm=2011?如果存在,求出m的值;如果不存在,說明理由.

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已知數(shù)列{an}是以d為公差的等差數(shù)列,數(shù)列{bn}是以q為公比的等比數(shù)列.
(1)若數(shù)列{bn}的前n項(xiàng)的和為Sn,且a1=b1=d=2,S3<a1003+5b2,求整數(shù)q的值;
(2)在(1)的條件下,試問數(shù)列{bn}中是否存在一項(xiàng)bk,使得bk恰好可以表示為該數(shù)列中連續(xù)p(p∈N,p≥2)項(xiàng)的和?請(qǐng)說明理由;
(3)若b1=a1,b2=as≠arb3=at,(其中t>s>r,且(s-r)是(t-r)的約數(shù)),求證:數(shù)列{bn}中每一項(xiàng)都是數(shù)列{an}中的項(xiàng).

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一、選擇題:(每小題5分,共50分)

題號(hào)

1

2

3

4

5

6

7

8

9

10

答案

B

D

B

A

C

C

D

A

A

B

二、填空題:(每小題4分,共24分)

11.6ec8aac122bd4f6e;  12.6ec8aac122bd4f6e;   13.6ec8aac122bd4f6e;    14.6ec8aac122bd4f6e;   15.4    16.120

三、解答題:(共76分,以下各題為累計(jì)得分,其他解法請(qǐng)相應(yīng)給分)

17.解:(I)6ec8aac122bd4f6e

        6ec8aac122bd4f6e  6ec8aac122bd4f6e

        由6ec8aac122bd4f6e,得6ec8aac122bd4f6e。

        又當(dāng)6ec8aac122bd4f6e時(shí)6ec8aac122bd4f6e,得6ec8aac122bd4f6e

        6ec8aac122bd4f6e

       (Ⅱ)當(dāng)6ec8aac122bd4f6e

        即6ec8aac122bd4f6e時(shí)函數(shù)遞增。

        故6ec8aac122bd4f6e的單調(diào)增區(qū)間為6ec8aac122bd4f6e6ec8aac122bd4f6e

        又由6ec8aac122bd4f6e,得6ec8aac122bd4f6e

        由6ec8aac122bd4f6e

        解得6ec8aac122bd4f6e

        故使6ec8aac122bd4f6e成立的6ec8aac122bd4f6e的集合是6ec8aac122bd4f6e

18.解:(I)設(shè)袋中有白球6ec8aac122bd4f6e個(gè),由題意得6ec8aac122bd4f6e

        即6ec8aac122bd4f6e

        解得6ec8aac122bd4f6e6ec8aac122bd4f6e(舍),故有白球6個(gè)

       (法二,設(shè)黑球有6ec8aac122bd4f6e個(gè),則全是黑球的概率為6ec8aac122bd4f6e   由6ec8aac122bd4f6e

        即6ec8aac122bd4f6e,解得6ec8aac122bd4f6e6ec8aac122bd4f6e(舍),故有黑球4個(gè),白球6個(gè)

       (Ⅱ)6ec8aac122bd4f6e,6ec8aac122bd4f6e

6ec8aac122bd4f6e

0

1

2

3

P

6ec8aac122bd4f6e

6ec8aac122bd4f6e

6ec8aac122bd4f6e

6ec8aac122bd4f6e

       故分布列為

 

 

 

 

 

 

       

       數(shù)學(xué)期望6ec8aac122bd4f6e

 

19.解:(I)取AB的中點(diǎn)O,連接OP,OC         6ec8aac122bd4f6ePA=PB   6ec8aac122bd4f6ePO6ec8aac122bd4f6eAB

       又在6ec8aac122bd4f6e中,6ec8aac122bd4f6e,6ec8aac122bd4f6e

       在6ec8aac122bd4f6e中,6ec8aac122bd4f6e,又6ec8aac122bd4f6e,故有6ec8aac122bd4f6e

       6ec8aac122bd4f6e   又6ec8aac122bd4f6e6ec8aac122bd4f6e面ABC

       又PO6ec8aac122bd4f6e面PAB,6ec8aac122bd4f6e面PAB6ec8aac122bd4f6e面ABC

      (Ⅱ)以O(shè)為坐標(biāo)原點(diǎn), 分別以O(shè)B,OC,OP為6ec8aac122bd4f6e軸,6ec8aac122bd4f6e軸,高考資源網(wǎng)(www.ks5u.com),中國(guó)最大的高考網(wǎng)站,您身邊的高考專家。軸建立坐標(biāo)系,

如圖,則A6ec8aac122bd4f6e

6ec8aac122bd4f6e       6ec8aac122bd4f6e

       6ec8aac122bd4f6e

       設(shè)平面PAC的一個(gè)法向量為6ec8aac122bd4f6e。

       6ec8aac122bd4f6e     得6ec8aac122bd4f6e

       令6ec8aac122bd4f6e,則6ec8aac122bd4f6e

       6ec8aac122bd4f6e

       設(shè)直線PB與平面PAC所成角為6ec8aac122bd4f6e

       于是6ec8aac122bd4f6e

20.解:(I)由題意設(shè)C的方程為6ec8aac122bd4f6e6ec8aac122bd4f6e,得6ec8aac122bd4f6e。

       6ec8aac122bd4f6e

       設(shè)直線6ec8aac122bd4f6e的方程為6ec8aac122bd4f6e,由6ec8aac122bd4f6e

       ②代入①化簡(jiǎn)整理得   6ec8aac122bd4f6e

       因直線6ec8aac122bd4f6e與拋物線C相交于不同的兩點(diǎn),

       故6ec8aac122bd4f6e

       即6ec8aac122bd4f6e,解得6ec8aac122bd4f6e6ec8aac122bd4f6e時(shí)僅交一點(diǎn),6ec8aac122bd4f6e

      (Ⅱ)設(shè)6ec8aac122bd4f6e,由由(I)知

       6ec8aac122bd4f6e

       6ec8aac122bd4f6e

       6ec8aac122bd4f6e

21.解:(I)當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e

       設(shè)曲線6ec8aac122bd4f6e6ec8aac122bd4f6e在公共點(diǎn)(6ec8aac122bd4f6e)處的切線相同,則有6ec8aac122bd4f6e

       即6ec8aac122bd4f6e    解得6ec8aac122bd4f6e6ec8aac122bd4f6e(舍)

       又6ec8aac122bd4f6e故得6ec8aac122bd4f6e公共點(diǎn)為6ec8aac122bd4f6e

       6ec8aac122bd4f6e切線方程為   6ec8aac122bd4f6e,即6ec8aac122bd4f6e

      (Ⅱ)6ec8aac122bd4f6e,設(shè)在(6ec8aac122bd4f6e)處切線相同,

       故有6ec8aac122bd4f6e

       即6ec8aac122bd4f6e

       由①6ec8aac122bd4f6e,得6ec8aac122bd4f6e(舍)

       于是6ec8aac122bd4f6e

       令6ec8aac122bd4f6e,則6ec8aac122bd4f6e

       于是當(dāng)6ec8aac122bd4f6e6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e,故6ec8aac122bd4f6e6ec8aac122bd4f6e上遞增。

       當(dāng)6ec8aac122bd4f6e,即6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e,故6ec8aac122bd4f6e6ec8aac122bd4f6e上遞減

       6ec8aac122bd4f6e6ec8aac122bd4f6e處取最大值。

       6ec8aac122bd4f6e當(dāng)6ec8aac122bd4f6e時(shí),b取得最大值6ec8aac122bd4f6e

22.解:(I)6ec8aac122bd4f6e的對(duì)稱軸為6ec8aac122bd4f6e,又當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e

       故6ec8aac122bd4f6e在[0,1]上是增函數(shù)

       6ec8aac122bd4f6e6ec8aac122bd4f6e

     (Ⅱ)6ec8aac122bd4f6e

       由6ec8aac122bd4f6e

       得6ec8aac122bd4f6e

       ①―②得6ec8aac122bd4f6e   即6ec8aac122bd4f6e

       當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e,當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e

       6ec8aac122bd4f6e

       于是6ec8aac122bd4f6e

       設(shè)存在正整數(shù)6ec8aac122bd4f6e,使對(duì)6ec8aac122bd4f6e,6ec8aac122bd4f6e恒成立。

       當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e,即6ec8aac122bd4f6e

       當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e

       6ec8aac122bd4f6e。

       6ec8aac122bd4f6e當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e,當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e,當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e

       6ec8aac122bd4f6e存在正整數(shù)6ec8aac122bd4f6e或8,對(duì)于任意正整數(shù)6ec8aac122bd4f6e都有6ec8aac122bd4f6e成立。


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