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7.把5名志愿者分配到三个不同的社区,每个社区至少有一个志愿者,其中甲社区恰有1名志愿者的分法有( ).
A.14种 B.35种 C.70种 D.100种
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A.(-∞,7) B.(-∞,7]
C.(-∞,8) D.(-∞,8]
10.某几何体的三视图如图1所示,图中两个点M为直观图中的同一个点M,两个点N也为直观图中的同一个点N,且分别为所在棱的中点,则在此几何体表面上,从点M到点N的路径中,最短路径的长度为( ).
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图1
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图1
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11.卡西尼卵形线是1675年卡西尼在研究土星及其卫星的运行规律时发现的.在数学史上,同一平面内到两个定点(叫做焦点)的距离之积为常数的点的轨迹称为卡西尼卵形线.已知卡西尼卵形线是中心对称图形且有唯一的对称中心.若某卡西尼卵形线C两焦点间的距离为2,且C上的点到两焦点的距离之积为1,则C上的点到其对称中心距离的最大值为( ).
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第Ⅱ卷
本卷包含必考题和选考题两部分,第13~21题为必考题,每个试题必须作答。第22~23为选考题,考生根据要求作答.
二、填空题(本小题共4小题,其中13~15题每小题5分,16题第一空2分,第二空3分.)
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14.在数列{an}中,若a1=1,a2=3,an+2=an+1-an(n∈N+),则该数列的前2021项之和为____.
15.不等式(x2-2)log0.5x<0的解集是____.
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三、解答题(解答题写出文字说明、证明过程或演算步骤,17~21每题12分,选做10分.)
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(1)求角A;
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18.如图2,在四棱锥P-ABCD中,侧棱PD⊥底面ABCD,底面ABCD为长方形,且PD=CD=1,E是PC的中点,作EF⊥PB交PB于点F.
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图2
(1)证明:PB⊥平面DEF;
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(1)求椭圆C的方程;
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20.已知各项均为正数的等差数列{an}与等比数列{bn}满足a2=b2=4,又a1,a3,a7+30成等比数列且b5=b1b4.
(1)求数列{an},{bn}的通项公式:
(2)若cn=anbn-nan-kbn(k∈R),数列{cn}是严格递增数列,求k的取值范围.
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(1)若a<0时恒有f(x)≥0,求实数a的取值范围;
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(1)求曲线C1,C2的极坐标方程;
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23.已知函数f(x)=|2x+3|-|x-a|(a∈R).
(1)当a=1时,解不等式f(x)≥2;
(2)若关于x的不等式f(x)≥|x-3|的解集包含[3,5],求a的取值范围.
参考答案
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2.B.若x≠1或y≠1,则x+y≠2的逆否命题为:若x+y=2,则x=1且y=1,①当x=4,y=-2时,满足x+y=2,所以充分性不成立,②当x=1且y=1时,x+y=2,所以必要性成立,所以x+y=2是x=1且y=1的必要不充分条件,即x≠1或y≠1是x+y≠2的必要不充分条件,
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因为l与OM的斜率的乘积等于2,
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所以2a2=c2-a2.
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图3
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10.B.该三视图对应的几何体是正三棱柱,如图4所示.仅考虑以下三种情况即可:
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图4
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(3)经侧面、底面到点N,将底面ABC沿AB,BC剪开,并展开至和ACC1A1在同一平面上,则
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11.B.设左、右焦点分别为F1,F2,以线段F1F2的中点为坐标原点,F1,F2所在的直线为x轴建立平面直角坐标系,则F1(-1,0),F2(1,0).设曲线上任意一点P(x,y),则
化简得该卡西尼卵形线的方程为
(x2+y2)2=2(x2-y2).
显然其对称中心为(0,0).
由(x2+y2)2=2(x2-y2),得
(x2+y2)2-2(x2+y2)=-4y2≤0,
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14.2.由an+2=an+1-an=(an-an-1)-an=-an-1=-(an-2-an-3)=-(an-3-an-4)+an-3=an-4,得{an}是周期为6的周期函数.又a3=a2-a1=3-1=2,a4=2-3=-1,a5=-1-2=-3,a6=-3+1=-2,因为2021=336×6+5,所以S2021=336×(1+3+2-1-3-2)+(1+3+2-1-3)=2.
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由不等式(x2-2)·log0.5x<0,得
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18.(1)因为PD⊥面ABCD,BC⊂面ABCD,所以PD⊥BC.因为底面ABCD为长方形,所以CD⊥BC.因为PD∩CD=D,所以BC⊥平面PCD.因为DE⊂平面PCD,所以DE⊥BC.因为PD=CD,E为PC的中点,所以DE⊥PC.
因为PC∩BC=C,所以DE⊥平面PBC,所以DE⊥PB.又EF⊥PB,DE∩EF=E,所以PB⊥平面DEF.
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图6
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20.(1)设数列{an}的公差为d,数列{bn}的公比为q,则由条件有
又q>0,d>0,解得a1=1,d=3,b1=2,q=2.
所以an=3n-2,bn=2n.
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又x2+y2=ρ2,y=ρsinθ,所以曲线C2的极坐标方程为ρ=4sinθ.
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23.(1)当a=1时,不等式f(x)≥2,即|2x+3|-|x-1|≥2,
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解得x≤-6或0≤x≤1或x>1.
所以不等式f(x)≥2的解集为(-∞,-6]∪[0,+∞).
(2)关于x的不等式f(x)≥|x-3|的解集包含[3,5],即|2x+3|-|x-3|≥|x-a|在x∈[3,5]时恒成立,即x+6≥|x-a|在x∈[3,5]时恒成立,即-6≤a≤2x+6在x∈[3,5]时恒成立,则-6≤a≤12,所以a的取值范围是[-6,12].