(1)證明:當(dāng)時(shí).對(duì)所有實(shí)數(shù)都有意義,反之.若對(duì)所有實(shí)數(shù)都有意義.則, 查看更多

 

題目列表(包括答案和解析)

設(shè)m是實(shí)數(shù),記M={m|m>1},f(x)=log3(x2-4mx+4m2+m+) 

(1)證明: 當(dāng)mM時(shí),f(x)對(duì)所有實(shí)數(shù)都有意義;反之,若f(x)對(duì)所有實(shí)數(shù)x都有意義,則mM。 

(2)當(dāng)mM時(shí),求函數(shù)f(x)的最小值。

(3)求證: 對(duì)每個(gè)mM,函數(shù)f(x)的最小值都不小于1。

查看答案和解析>>

設(shè)m是實(shí)數(shù),記M={m|m>1},f(x)=log3(x2-4mx+4m2+m+) 
(1)證明: 當(dāng)mM時(shí),f(x)對(duì)所有實(shí)數(shù)都有意義;反之,若f(x)對(duì)所有實(shí)數(shù)x都有意義,則mM。 
(2)當(dāng)mM時(shí),求函數(shù)f(x)的最小值。
(3)求證: 對(duì)每個(gè)mM,函數(shù)f(x)的最小值都不小于1。

查看答案和解析>>

已知定義在實(shí)數(shù)集上的函數(shù)y=f(x)滿足條件:對(duì)于任意的x、y∈R,都有f(x+y)=f(x)+f(y).
(1)求證:f(0)=0;
(2)若f(x)是奇函數(shù),試舉出兩個(gè)這樣的函數(shù);
(3)若當(dāng)x≥0時(shí),f(x)<0,
1)試判斷函數(shù)f(x)在R上的單調(diào)性,并證明之;
2)判斷函數(shù)|f(x)|=a.所有可能的解的個(gè)數(shù),并求出對(duì)應(yīng)的a的范圍;

查看答案和解析>>

已知定義在實(shí)數(shù)集上的函數(shù)y=f(x)滿足條件:對(duì)于任意的x、y∈R,都有f(x+y)=f(x)+f(y).
(1)求證:f(0)=0;
(2)試證明f(x)是奇函數(shù),試舉出兩個(gè)這樣的函數(shù);
(3)若當(dāng)x≥0時(shí),f(x)<0,
1)試判斷函數(shù)f(x)在R上的單調(diào)性,并證明之;
2)判斷函數(shù)|f(x)|=a.所有可能的解的個(gè)數(shù),并求出對(duì)應(yīng)的a的范圍.

查看答案和解析>>

設(shè)m是常數(shù),集合
(1)證明:當(dāng)m∈M時(shí),f(x)對(duì)所有的實(shí)數(shù)x都有意義;
(2)當(dāng)m∈M時(shí),求函數(shù)f(x)的最小值;
(3)求證:對(duì)每個(gè)m∈M,函數(shù)f(x)的最小值都不于1.

查看答案和解析>>

一、選擇題(每題5分共50分)

1.D            2.A            3.B           4.C            5.C           

6.C       7.B        8.C    9.C    10.D

二、填空題(每題5分共20分)

       11.6ec8aac122bd4f6e          12.6ec8aac122bd4f6e                 13.6ec8aac122bd4f6e                  

14.(0,2),6ec8aac122bd4f6e               15.3

三、解答題(共80分)

16.解:(Ⅰ)由已知得:6ec8aac122bd4f6e,  

6ec8aac122bd4f6e是△ABC的內(nèi)角,所以6ec8aac122bd4f6e.    

(2)由正弦定理:6ec8aac122bd4f6e,6ec8aac122bd4f6e

又因?yàn)?sub>6ec8aac122bd4f6e,6ec8aac122bd4f6e,又6ec8aac122bd4f6e是△ABC的內(nèi)角,所以6ec8aac122bd4f6e

 

17.證明:連結(jié)AB,A1D,在正方形中,A1B=A1D,O是BD中點(diǎn),

∴A1O⊥BD;                 

連結(jié)OM,A1M,A1C1,設(shè)AB=a,則AA1=a,MC=6ec8aac122bd4f6ea=MC1

OA=OC=6ec8aac122bd4f6ea,AC=6ec8aac122bd4f6ea,

∴A1O2=A1A2+AO2=a2+6ec8aac122bd4f6ea2=6ec8aac122bd4f6ea2,OM2=OC2+MC2=6ec8aac122bd4f6ea2,A1M2=A1C12+MC12=2a2+6ec8aac122bd4f6ea2=6ec8aac122bd4f6ea2,∴A1M2=A1O2+OM2,

∴A1O⊥OM,  

∴AO1⊥平面MBD

18解:(Ⅰ)6ec8aac122bd4f6e,

因?yàn)楹瘮?shù)6ec8aac122bd4f6e6ec8aac122bd4f6e6ec8aac122bd4f6e取得極值,則有6ec8aac122bd4f6e,6ec8aac122bd4f6e

6ec8aac122bd4f6e

解得6ec8aac122bd4f6e6ec8aac122bd4f6e

(Ⅱ)由(Ⅰ)可知,6ec8aac122bd4f6e,

6ec8aac122bd4f6e

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

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

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

所以,當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e取得極大值6ec8aac122bd4f6e,又6ec8aac122bd4f6e6ec8aac122bd4f6e

則當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e的最大值為6ec8aac122bd4f6e

因?yàn)閷?duì)于任意的6ec8aac122bd4f6e,有6ec8aac122bd4f6e恒成立,

所以 6ec8aac122bd4f6e

解得 6ec8aac122bd4f6e6ec8aac122bd4f6e,

因此6ec8aac122bd4f6e的取值范圍為6ec8aac122bd4f6e

19.解(Ⅰ)由題意知6ec8aac122bd4f6e,6ec8aac122bd4f6e   6ec8aac122bd4f6e  

當(dāng)n≥2時(shí),6ec8aac122bd4f6e,6ec8aac122bd4f6e,

兩式相減得 6ec8aac122bd4f6e

整理得:6ec8aac122bd4f6e    

∴數(shù)列{6ec8aac122bd4f6e}是以2為首項(xiàng),2為公比的等比數(shù)列。

6ec8aac122bd4f6e   

(Ⅱ)由(Ⅰ)知6ec8aac122bd4f6e,∴bn=n6ec8aac122bd4f6e  

6ec8aac122bd4f6e, …………①

6ec8aac122bd4f6e, …………②

①-②得

6ec8aac122bd4f6e,   

6ec8aac122bd4f6e,    

6ec8aac122bd4f6e,   

20.解:設(shè)這臺(tái)機(jī)器最佳使用年限是n年,則n年的保養(yǎng)、維修、更換易損零件的總費(fèi)用為:

6ec8aac122bd4f6e

6ec8aac122bd4f6e

6ec8aac122bd4f6e

6ec8aac122bd4f6e

等號(hào)當(dāng)且僅當(dāng)6ec8aac122bd4f6e6ec8aac122bd4f6e

答:這臺(tái)機(jī)器最佳使用年限是12年,年平均費(fèi)用的最小值為1.55萬元.

21.⑴c=2, a=3 雙曲線的方程為

⑵ 得 (1?3k2)x2?6kx?9=0

  x1+x2= , x1x2=

由△>0 得 k2<1

  由= x1x2+y1y2=(1+k2) x1x2+k(x1+x2)+2>2得 <k2<3

  所以,<k2<1

即k∈(?1, )∪( , 1 )

附加題

(1)證明:先將6ec8aac122bd4f6e變形:6ec8aac122bd4f6e,

當(dāng)6ec8aac122bd4f6e,即6ec8aac122bd4f6e時(shí),∴6ec8aac122bd4f6e恒成立,

6ec8aac122bd4f6e的定義域?yàn)?sub>6ec8aac122bd4f6e。                                     

反之,若6ec8aac122bd4f6e對(duì)所有實(shí)數(shù)6ec8aac122bd4f6e都有意義,則只須6ec8aac122bd4f6e

6ec8aac122bd4f6e,即6ec8aac122bd4f6e,解得6ec8aac122bd4f6e,故6ec8aac122bd4f6e。  

(2)解析:設(shè)6ec8aac122bd4f6e,

6ec8aac122bd4f6e是增函數(shù),

∴當(dāng)6ec8aac122bd4f6e最小時(shí),6ec8aac122bd4f6e最小。

6ec8aac122bd4f6e,                               

 顯然,當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e取最小值為6ec8aac122bd4f6e,

此時(shí)6ec8aac122bd4f6e為最小值。                      

(3)證明:當(dāng)6ec8aac122bd4f6e時(shí),6ec8aac122bd4f6e,

當(dāng)且僅當(dāng)m=2時(shí)等號(hào)成立。                                  

6ec8aac122bd4f6e。                               

 

 

 


同步練習(xí)冊(cè)答案