阶段性检测试题
一、选择题(共9小题,每题4分)
1、已知全集U=R,集合A={x|lg x≤0},B={x|2x≤word/media/image1.gif},则A∪B=( D )
A.∅B.(0,word/media/image2.gif]C.[word/media/image2.gif,1] D.(-∞,1]
(1)由题意知,A=(0,1],B=(-∞,word/media/image2.gif],∴A∪B=(-∞,1].故选D.
2.已知等比数列{an}的公比为正数,且a3a9=2a52,a2=2,则a1=( C )
A. word/media/image3.gif B. word/media/image4.gif
C. word/media/image5.gif D.2
解析:选C.由等比数列的性质得 ,
∵q>0,
∴a6=word/media/image5.gifa5,q=word/media/image6.gif=word/media/image5.gif,a1=word/media/image7.gif=word/media/image5.gif,故选C.
3.已知f(x)=3sin x-πx,命题p:∀x∈word/media/image8.gif,f(x)<0,则( D )
A.p是假命题,word/media/image9.gif p:∀x∈word/media/image8.gif,f(x)≥0
B.p是假命题,word/media/image9.gif p:∃x0∈word/media/image8.gif,f(x0)≥0
C.p是真命题,word/media/image9.gif p:∀x∈word/media/image8.gif,f(x)>0
D.p是真命题,word/media/image9.gif p:∃x0∈word/media/image8.gif,f(x0)≥0
解析:选D.因为f′(x)=3cos x-π,所以当x∈word/media/image8.gif时,f′(x)<0,函数f(x)单调递减,所以∀x∈word/media/image8.gif,f(x)
4.已知向量a,b满足|a|=3,|b|=2word/media/image10.gif,且a⊥(a+b),则a与b的夹角为(D )
A. word/media/image11.gif B. word/media/image12.gif C. word/media/image13.gif D. word/media/image14.gif
解析:选D.a⊥(a+b)⇒a·(a+b)=a2+a·b=|a|2+|a||b|cos〈a,b〉=0,故cos〈a,b〉=-word/media/image15.gif,故所求夹角为word/media/image14.gif.
5.下列函数中,既是偶函数又在区间(-∞,0)上单调递增的是( A )
A.f(x)=word/media/image16.gif B.f(x)=x2+1
C.f(x)=x3 D.f(x)=2-x
解析:选A.A中f(x)=word/media/image17.gif是偶函数,且在(-∞,0)上是增函数,故A满足题意.B中f(x)=x2+1是偶函数,但在(-∞,0)上是减函数.C中f(x)=x3是奇函数.D中f(x)=2-x是非奇非偶函数.故B,C,D都不满足题意.
6.已知lg a+lg b=0,则函数f(x)=ax与函数g(x)=-logbx的图象可能是( B)
解析:选B.∵lg a+lg b=0,∴ab=1,
∵g(x)=-logbx的定义域是(0,+∞),故排除A.
若a>1,则0<b<1,
此时f(x)=ax是增函数,
g(x)=-logbx是增函数,
结合图象知选B.
7、已知数列{an}的前n项和为Sn,a1=1,Sn=2an+1,则Sn=( B )
A.2n-1 B. word/media/image19.gif word/media/image20.gif
C. word/media/image21.gif word/media/image20.gif D. word/media/image22.gif
[解析] (1)由已知Sn=2an+1,得Sn=2(Sn+1-Sn),即2Sn+1=3Sn,word/media/image23.gif=word/media/image24.gif,而S1=a1=1,所以Sn=word/media/image19.gifword/media/image20.gif.
[答案] B
8.设正实数x,y,z满足x2-3xy+4y2-z=0.则当word/media/image25.gif取得最大值时,word/media/image26.gif+word/media/image27.gif-word/media/image28.gif的最大值为( B )
A.0 B.1C. word/media/image29.gif D.3
解析:选B.z=x2-3xy+4y2(x>0,y>0,z>0),
∴word/media/image25.gif=word/media/image30.gif=word/media/image31.gif≤word/media/image32.gif=1.
当且仅当word/media/image33.gif=word/media/image34.gif,即x=2y时等号成立,此时z=x2-3xy+4y2=4y2-6y2+4y2=2y2,∴word/media/image26.gif+word/media/image27.gif-word/media/image28.gif=word/media/image35.gif+word/media/image27.gif-word/media/image36.gif=-word/media/image37.gif+word/media/image38.gif=-word/media/image39.gifword/media/image40.gif+1,∴当y=1时,word/media/image26.gif+word/media/image27.gif-word/media/image28.gif的最大值为1.
9.已知{an}为等差数列,a10=33,a2=1,Sn为数列{an}的前n项和,则S20-2S10等于( C )
A.40 B.200C.400 D.20
解析:选C.S20-2S10=word/media/image41.gif-2×word/media/image42.gif
=10(a20-a10)=100d.
又a10=a2+8d,
∴33=1+8d,
∴d=4.
∴S20-2S10=400.
二、填空题(共8小题,每题4分)
1、函数f(x)=word/media/image43.gif的定义域为( )
解析:要使函数有意义,
则x需满足word/media/image44.gif即word/media/image45.gif
解①得-1≤x≤10.
所以不等式组的解集为(1,2)∪(2,10].
2、函数y=word/media/image46.gif的单调减区间为________.
(3)由y=cosword/media/image47.gif=cosword/media/image48.gif,得
2kπ≤2x-word/media/image49.gif≤2kπ+π(k∈Z),
故kπ+word/media/image50.gif≤x≤kπ+word/media/image51.gif (k∈Z).
所以函数的单调减区间为word/media/image52.gif (k∈Z).
3、函数f(x)=word/media/image53.gif在[0,2]上的最小值是( )
A.-word/media/image54.gif B.-word/media/image55.gif
C.-4 D.-word/media/image56.gif
解析:选A.f′(x)=x2+2x-3,
令f′(x)=0,得x=1(x=-3舍去),
又f(0)=-4,f(1)=-word/media/image54.gif,f(2)=-word/media/image55.gif,
故f(x)在[0,2]上的最小值是f(1)=-word/media/image54.gif.
4、某三棱锥的三视图如图所示,则该三棱锥最长棱的棱长为________.
解析:根据三视图还原几何体,得如图所示的三棱锥PABC.由三视图的形状特征及数据,可推知PA⊥平面ABC,且PA=2.底面为等腰三角形,AB=BC,设D为AC中点,AC=2,则AD=DC=1,且BD=1,易得AB=BC=word/media/image5.gif,所以最长的棱为PC,PC=word/media/image58.gif=2word/media/image5.gif.
答案:2word/media/image5.gif
5、若数列{an}满足a1=15,且3an+1=3an-4,则an=________.
解析:由3an+1=3an-4,得an+1-an=-word/media/image59.gif,
所以{an}是等差数列,首项a1=15,公差d=-word/media/image59.gif,
所以an=15-word/media/image59.gif (n-1)=word/media/image60.gif.
答案:word/media/image61.gif
6、若命题“∃x0∈R,2xword/media/image62.gif-3ax0+9<0”为假命题,则实数a的取值范围是________.
因为“∃x0∈R,2xword/media/image62.gif-3ax0+9<0”为假命题,则“∀x∈R,2x2-3ax+9≥0”为真命题.因此Δ=9a2-4×2×9≤0,故-2word/media/image5.gif≤a≤2word/media/image5.gif.
7、若函数f(x)(x∈R)是周期为4的奇函数,且在[0,2]上的解析式为f(x)=word/media/image63.gif则 fword/media/image64.gif+fword/media/image65.gif=________.
∵f(x)是以4为周期的奇函数,∴fword/media/image64.gif=fword/media/image66.gif=fword/media/image67.gif,fword/media/image65.gif=fword/media/image68.gif=fword/media/image69.gif.
∵当0≤x≤1时,f(x)=x(1-x),∴fword/media/image70.gif=word/media/image71.gif×word/media/image72.gif=word/media/image73.gif.∵当1<x≤2时,f(x)=sin πx,
∴fword/media/image74.gif=sinword/media/image75.gif=-word/media/image3.gif.又∵f(x)是奇函数,
∴fword/media/image67.gif=-fword/media/image70.gif=-word/media/image73.gif,fword/media/image69.gif=-fword/media/image74.gif=word/media/image3.gif.
∴fword/media/image64.gif+fword/media/image65.gif=word/media/image3.gif-word/media/image73.gif=word/media/image76.gif.
8.设函数f(x)=ax3-3x+1(x∈R),若对于任意x∈[-1,1],都有f(x)≥0成立,则实数a的值为________.
解析:(构造法)若x=0,则不论a取何值,f(x)≥0显然成立;
当x>0时,即x∈(0,1]时,f(x)=ax3-3x+1≥0可化为a≥word/media/image77.gif-word/media/image78.gif.
设g(x)=word/media/image77.gif-word/media/image78.gif,
则g′(x)=word/media/image79.gif,
所以g(x)在区间word/media/image80.gif上单调递增,在区间word/media/image81.gif上单调递减,
因此g(x)max=gword/media/image82.gif=4,从而a≥4.
当x<0时,即x∈[-1,0)时,同理a≤word/media/image77.gif-word/media/image78.gif.
g(x)在区间[-1,0)上单调递增,
∴g(x)min=g(-1)=4,
从而a≤4,综上可知a=4.
答案:4
三.计算下列各题:(18分)
(1) word/media/image3.gif lgword/media/image83.gif-word/media/image59.giflgword/media/image84.gif+lgword/media/image85.gif;
解:(1) word/media/image3.gif lgword/media/image83.gif-word/media/image59.giflgword/media/image84.gif+lg word/media/image85.gif
=word/media/image3.gif×(5lg 2-2lg 7)-word/media/image59.gif×word/media/image24.giflg 2+word/media/image3.gif (lg 5+2lg 7)
=word/media/image86.giflg 2-lg 7-2lg 2+word/media/image3.giflg 5+lg 7
=word/media/image3.giflg 2+word/media/image3.giflg 5=word/media/image3.giflg(2×5)=word/media/image3.gif.
(2)在△ABC中,a,b,c分别为内角A,B,C的对边,且2asin A=(2b+c)sin B+(2c+b)sin C.求角A的大小;
[解] (1)由题意知,
根据正弦定理得2a2=(2b+c)b+(2c+b)c,
即a2=b2+c2+bc.①
由余弦定理得a2=b2+c2-2bccos A,
故cos A=-word/media/image3.gif,A=120°.
四、(12分)已知word/media/image87.gif,word/media/image88.gif,若word/media/image89.gif的必要不充分条件,求实数m的取值范围。
五、证明:(1)连接AD1,由ABCDA1B1C1D1是正方体,知AD1∥BC1,
因为F,P分别是AD,DD1的中点,所以FP∥AD1.
从而BC1∥FP.
而FP⊂平面EFPQ,且BC1⊄平面EFPQ,
故直线BC1∥平面EFPQ.
(2)如图,连接AC,BD,则AC⊥BD.word/media/image90.gif
由CC1⊥平面ABCD,BD⊂平面ABCD,可得CC1⊥BD.
又AC∩CC1=C,
所以BD⊥平面ACC1.
而AC1⊂平面ACC1,所以BD⊥AC1.
因为M,N分别是A1B1,A1D1的中点,
所以MN∥BD,从而MN⊥AC1.
同理可证PN⊥AC1.
又PN∩MN=N,所以直线AC1⊥平面PQMN.
(12分)六、已知函数word/media/image91.gif的最小正周期为word/media/image92.gif,将函数word/media/image93.gif的图像上各点的横坐标缩短到原来的word/media/image94.gif,纵坐标不变,得到函数word/media/image95.gif的图像,求函数word/media/image95.gif在区间word/media/image96.gif上的最小值。
(14分)七、已知数列word/media/image97.gif满足word/media/image98.gif,且word/media/image99.gif
(1)设word/media/image100.gif,证明数列word/media/image101.gif为等比数列;
(2)设word/media/image102.gif,求数列word/media/image103.gif的前n项和word/media/image104.gif。
(14分)八、已知函数f(x)=word/media/image105.gif
(1)求函数f(x)的单调区间;
(2)设g(x)=xf(x)-ax+1,若g(x)在(0,+∞)上存在极值点,求实数a的取值范围.
解:(1)f(x)=word/media/image106.gif,x∈(-∞,0)∪(0,+∞),
∴f′(x)=word/media/image107.gif.
当f′(x)=0时,x=1.
f′(x)与f(x)随x的变化情况如下表:
故f(x)的增区间为(1,+∞),减区间为(-∞,0)和(0,1).
(2)g(x)=ex-ax+1,x∈(0,+∞),∴g′(x)=ex-a,
①当a≤1时,g′(x)=ex-a>0,即g(x)在(0,+∞)上递增,此时g(x)在(0,+∞)上无极值点.
②当a>1时,令g′(x)=ex-a=0,得x=ln a;
令g′(x)=ex-a>0,得x∈(ln a,+∞);
令g′(x)=ex-a<0,得x∈(0,ln a).
故g(x)在(0,ln a)上递减,在(ln a,+∞)上递增,
∴g(x)在(0,+∞)有极小值无极大值,且极小值点为x=ln a.
故实数a的取值范围是a>1.
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