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课时跟踪检测(四) 演绎推理

课时跟踪检测(四)  演绎推理
课时跟踪检测(四)  演绎推理

课时跟踪检测(四)演绎推理

一、选择题

1.给出下面一段演绎推理:

有理数是真分数,……………………………大前提

整数是有理数,……………………………小前提

整数是真分数.……………………………结论

结论显然是错误的,是因为()

A.大前提错误B.小前提错误

C.推理形式错误D.非以上错误

解析:选A推理形式没有错误,小前提也没有错误,大前提错误.举反例,如2是有理数,但不是真分数.

2.“所有金属都能导电,铁是金属,所以铁能导电”这种推理方法属于()

A.演绎推理B.类比推理

C.合情推理D.归纳推理

解析:选A是由一般到特殊的推理,故是演绎推理.

3.下面几种推理过程是演绎推理的是()

A.两条直线平行,同旁内角互补,如果∠A与∠B是两条平行直线的同旁内角,则∠A+∠B=180°

B.某校高三(1)班有55人,(2)班有54人,(3)班有52人,由此得高三所有班人数超过50人

C.由三角形的性质,推测四面体的性质

D.在数列{a n}中,a1=1,a n=1

2?

?

?

?

a n-1+

1

a n-1(n≥2),由此归纳出a n的通项公式

解析:选A B项是归纳推理,C项是类比推理,D项是归纳推理.

4.“∵四边形ABCD是矩形,∴四边形ABCD的对角线相等.”补充以上推理的大前提()

A.正方形都是对角线相等的四边形

B.矩形都是对角线相等的四边形

C.等腰梯形都是对角线相等的四边形

D.矩形都是对边平行且相等的四边形

解析:选B推理的大前提应该是矩形的对角线相等,表达此含义的选项为B.

5.有一段演绎推理是这样的:直线平行于平面,则直线平行于平面内所有直线;已知

直线b ?平面α,直线a ?平面α,直线b ∥平面α,则直线b ∥直线a .结论显然是错误的,这是因为( )

A .大前提错误

B .小前提错误

C .推理形式错误

D .非以上错误

解析:选A 大前提是错误的,直线平行于平面,则不一定平行于平面内所有直线,还有异面直线的情况.

二、填空题

6.若有一段演绎推理:“大前提:整数是自然数.小前提:-3是整数.结论:-3是自然数.”这个推理显然错误,则推理错误的是________(填“大前提”“小前提”或“结论”).

解析:整数不全是自然数,还有零与负整数,故大前提错误.

答案:大前提

7.已知推理:“因为△ABC 的三边长依次为3,4,5,所以△ABC 是直角三角形.”若将其恢复成完整的三段论,则大前提是____________________.

解析:大前提:一条边的平方等于其他两条边的平方和的三角形是直角三角形. 小前提:△ABC 的三边长依次为3,4,5,满足32+42=52.结论:△ABC 是直角三角形. 答案:一条边的平方等于其他两条边的平方和的三角形是直角三角形

8.若不等式ax 2+2ax +2<0的解集为空集,则实数a 的取值范围为________.

解析:①a =0时,有2<0,显然此不等式解集为?.

②a ≠0时需有??? a >0,Δ≤0?????? a >0,4a 2-8a ≤0??

????

a >0,0≤a ≤2, 所以0<a ≤2.

综上可知,实数a 的取值范围是[0,2].

答案:[0,2]

三、解答题

9.如图,在直四棱柱ABCD -A 1B 1C 1D 1中,底面是正方形,E ,F ,G 分别是棱B 1B ,D 1D ,DA 的中点.求证:

(1)平面AD1E∥平面BGF;

(2)D1E⊥AC.

证明:(1)∵E,F分别是B1B和D1D的中点,∴D1F綊BE,

∴四边形BED1F是平行四边形,

∴D1E∥BF.

又∵D1E?平面BGF,BF?平面BGF,

∴D1E∥平面BGF.

∵F,G分别是D1D和DA的中点,

∴FG是△DAD1的中位线,

∴FG∥AD1.

又∵AD1?平面BGF,FG?平面BGF,

∴AD1∥平面BGF.

又∵AD1∩D1E=D1,

∴平面AD1E∥平面BGF.

(2)连接BD,B1D1,

∵底面ABCD是正方形,

∴AC⊥BD.

∵D1D⊥AC,BD∩D1D=D,

∴AC⊥平面BDD1B1.

∵D1E?平面BDD1B1,

∴D1E⊥AC.

10.在数列{}a n 中,a 1=2,a n +1=4a n -3n +1,n ∈N *.

(1)证明数列{}a n -n 是等比数列.

(2)求数列{}a n 的前n 项和S n .

(3)证明不等式S n +1≤4S n ,对任意n ∈N *皆成立. 解:(1)证明:因为a n +1=4a n -3n +1, 所以a n +1-(n +1)=4(a n -n ),n ∈N *. 又a 1-1=1,

所以数列{}a n -n 是首项为1,且公比为4的等比数列.

(2)由(1)可知a n -n =4n -1,

于是数列{}a n 的通项公式为a n =4n -1+n .

所以数列{}a n 的前n 项和S n =4n -13+n (n +1)2

. (3)证明:对任意的n ∈N *,

S n +1-4S n =4n +1-13+(n +1)(n +2)2-44n -13+n (n +1)2=-12

(3n 2+n -4)≤0. 所以不等式S n +1≤4S n ,对任意n ∈N *皆成立.

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