12.C解析:
设三条弦长分别是a,2a,h,则a2+(2a)2+h2=25,即5a2+h2=25,三条弦长之和S=3a+h,将h=S-3a代入5a2+h2=25,得14a2-6aS+S2-25=0,由≥0得S2≤70.故选C.
二、填空题
13.(-∞,-2)∪(0,
].解析:
直线y+2=k(x+1)过定点(-1,-2),作图得k的取值范围是
(-∞,-2)∪(0,
].
14.
解析:
在36对可能的结果中,和为7的有6对:
(1,6),(2,5),(2,5),(3,4),(3,4),(4,3).∴得到两数之和为7的概率是
.
15.(1,-1)解析:
由题意可知b的终点在直线x=1上,可设b=(1,y),则
,
,17y2+48y+31=0,∴y=-1或y=-
(增解,舍去),∴b=(1,-1).
16.
解析:
∵{an}是等差数列,∴a=0,Sn=n2,∴a2=3,a3=5,a4=7.
设三角形最大角为,由余弦定理,得cos=-
,∴=120°.∴该三角形的面积
S=
×3×5×sin120°=
.
三、解答题
17.(Ⅰ)解:
a1=2,a2=2+k,a3=2+3k,由a22=a1a3得,(2+k)=2(2+3k),
∵k≠0,∴k=2.······················································································2分
由an+1=an+2n,得an-an-1=2(n-1),
∴an=a1+(a2-a1)+(a3-a2)+···+(an-an-1)=2+2[1+2+···+(n-1)]=n2-n+2.·························6分
(Ⅱ)解:
.·······························································8分
∴Tn=
,
,························································10分
两式相减得,
,
∴Tn=1-
.·······················································································12分
18.(Ⅰ)证明:
设O为AB的中点,连结A1O,
∵AF=
AB,O为AB的中点,∴F为AO的中点,
又E为AA1的中点,∴EF∥A1O.
又∵D为A1B1的中点,O为AB的中点,∴A1D=OB.
又A1D∥OB,∴四边形A1DBO为平行四边形.
∴A1O∥BD.又EF∥A1O,∴EF∥BD.
又EF平面DBC1,BD平面DBC1.
∴EF∥平面DBC1.…………………6分
(Ⅱ)解:
∵AB=BC=CA=AA1=2,
D、E分别为A1B1、AA1的中点,AF=
AB,
∴C1D⊥面ABB1A1.
而
,
=
.
∵C1D=
.
∴
.………………………………12分
19.(Ⅰ)解:
样本中体重在区间(45,50]上的女生有a×5×20=100a(人),·····················1分
样本中体重在区间(50,60]上的女生有(b+0.02)×5×20=100(b+0.02)(人),··············2分
依题意,有100a=
×100(b+0.02),即a=
×(b+0.02).①·································3分
根据频率分布直方图可知(0.02+b+0.06+a)×5=1,②··········································4分
解①②得:
a=0.08,b=0.04.······································································6分
(Ⅱ)解:
样本中体重在区间(50,55]上的女生有0.04×5×20=4人,分别记为
A1,A2,A3,A4,··················································································7分
体重在区间(55,60]上的女生有0.02×5×20=2人,分别记为B1,B2.··················8分
从这6名女生中随机抽取两人共有15种情况:
(A1,A2),(A1,A3),(A1,A4),(A1,B1),(A1,B2),(A2,A3),(A2,A4),(A2,B1),
(A2,B2),(A3,A4),(A3,B1),(A3,B2),(A4,B1),(A4,B2),(B1,B2).·······10分
其中体重在(55,60]上的女生至少有一人共有9种情况:
(A1,B1),(A1,B2),(A2,B1),(A2,B2),(A3,B1),(A3,B2),(A4,B1),(A4,B2),
(B1,B2).····························································································11分
记“从样本中体重在区间(50,60]上的女生随机抽取两人,体重在区间(55,60]上的女生
至少有一人被抽中”为事件M,则P(M)=
.··········································12分
20.(Ⅰ)解:
设圆心C(a,b),则
,解得
.·······················3分
则圆C的方程为x2+y2=r2,将点P的坐标代入得r2=2,
故圆C的方程为x2+y2=2.·····································································5分
(Ⅱ)解:
由题意知,直线PA和直线PB的斜率存在,且互为相反数,
故可设PA:
y-1=k(x-1),PB:
y-1=-k(x-1),且k≠0,······································6分
由
,得(1+k2)x2-2k(k-1)x+k2-2k-1=0,······································7分
∵点P的横坐标x=1一定是该方程的解,故可得xA=
.····················8分
同理,xB=
.···········································································9分
∴
=1=kOP.······················11分
∴直线AB和OP一定平行.·····································································12分
21.(Ⅰ)解:
f(x)的定义域是(0,+∞),f′(x)=2+
+
.
依题设,f
(1)=5,f′
(1)=-3,∴a=-3,b=-2.···················································4分
∴f′(x)=2-
,令f′(x)>0,又x>0,∴x>
.
∴函数的单调增区间为(
,+∞).······················································6分
(Ⅱ)g(x)=f(x)-
=2x-2lnx,g′(x)=2-
.
设过点(2,2)与曲线g(x)的切线的切点坐标为(x0,y0),
则y0-2=g′(x0)(x0-2),即2x0-2lnx0-2=(2-
)(x0-2),∴lnx0+
2.·····················8分
令h(x)=lnx+
-2,则h′(x)=
,∴x=2.
∴h(x)在(0,2)上单调递减,在(2,+∞)上单调递增.······································10分
∵h(
)=2-ln2>0,h
(2)=ln2-1<0,h(e2)=
>0.
∴h(x)与x轴有两个交点,∴过点(2,2)可作2条曲线y=g(x)的切线.···············12分
22.(Ⅰ)证明:
∵∠CPD=∠ABC,∠D=∠D,
∴△DPC~△DBA.
∴
.
又∵AB=AC,∴
.·····································································5分
(Ⅱ)解:
∵∠ACD=∠APC,∠CAP=∠CAD,∴△APC~△ACD.
∴
,∴AC2=AP·AD=4.·······························································10分
23.(Ⅰ)解:
设动点A的直角坐标为(x,y),则
∴动点A的轨迹方程为(x-2)2+(y+2)2=9,
其轨迹是以(2,-2)为圆心,半径为3的圆.·····················································5分
(Ⅱ)解:
直线l的极坐标方程cos(-
)=a化为直角坐标方程是x+y=
a.
由
=3,得a=3,或a=-3.··························································10分
24.(Ⅰ)解:
由题设可得b=
>0,∴a>0.∴a+b=a+
=
≥3,
当a=2,b=1时,a+b取得最小值3,∴m的最大值为3.·································5分
(Ⅱ)解:
要使2|x-1|+|x|≤a+b对任意的a,b恒成立,须且只须2|x-1|+|x|≤3.
用零点区分法求得实数x的取值范围是-
≤x≤
.········································10分