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一、选择题2021年哈三中高三学年第一次模拟考试理科数学答案BDCABCCBADAD二、填空题13.14.−1515.y2=2x16.(5.5,8)三、解答题17.（1）a(4a+6d+6)=(2a+d+1)2,a+d=8·······

·································3分1111整理得(a1−3)(a1−9)=0又a1a2,a1=3,d=5an=5n−2································

··········································6分（2）b=1=11−1n(5n−2)(5n+3)55n−25n+3T=11−1=1−1··································10分n535n+3155

(5n+3)1Tn15················································································12分18.（1）平面EDAF⊥平面ABCD，DE平面EDAF平面EDAF平面ABCD=AD，

DE⊥ADDE⊥平面ABCD·······························································································2分AC平面ABCDDE⊥AC四边形ABCD

是正方形AC⊥BD··································································3分DE、BD平面BDE，DEBD=D，AC⊥平面BDE·················

···········4分2EH,n=2a−232a2+46图2AC平面ACE平面AEC⊥平面BDE····························································5分（2）建系如图平面BEF的法向量n=(

1,1,1)···············································································7分E(0,0,2)设H(a,a,0

)，EH=(a,a,−2)17cos=，解得a=或a=（舍）············10分924H11，DH=2···········································12

分,,022219.（1）yˆ=0.5x2+0.8··········································································6分（2）模型二的残差平

方和为0.423.7，模型二的拟合效果更好········6分x2220.（1）M:+y4=1············································································3分（2）F2(3,0)，设直线l:x=m

y+3，A(x1,y1)，B(x2,y2)，x2+4y2−4=0x=my+，代入得：(m2+4)y2+23my−1=0由于=16(m2+1)0恒成立3m2+1313e−e则有y1+y2=m2+4，y1y2=m2+4··························

···················5分4(m2+1)AB=y−y==6分12点O到直线l的距离d=m2+4······································7分则S=1AOB2A

Bd=m2+4231·············10分3当且仅当：=3，即m=+m2+1时取等号，又由于OC=−OA，知S2ABC=5S2AOB5，此时l:x=22y+.···12分21.（1）a=1,f(x)的最

大值为1·······························································3分（2）方法一：由题意知，ex22ex+e−x恒成立只需x2ln2ex+e−x恒成立设

g(x)=x2−ln2+ln(ex+e−x)由于g(x)为偶函数，只需x0时g(x)0，而g(0)=0x−xg(x)=+2ex+e−xx,g(0)=0g(x)=4(ex+e−x)2+2−23m1+m21+m2y+y−4yy(12)21231+m223m2+1m2+1m2+12

=−2+1−1−(0,lng(x)=4(ex+e−x)2+2=4e2x+2+e−2x+21+2①当−1时，g(x)0，对x0,g(x)单调递减2x0,g(x)g(0)

=0，对x0,g(x)单调递减x0,g(x)0恒成立···················5分②当0时，g(x)0，对x0,g(x)单调递增x0,g(x)g(0)=0，对x0,g(x)

单调递增x0,g(x)0恒成立，不成立.·····················································6分③当−10时，当x1−2+1−1−)时

，g(x)0221−(0,ln212+1−1−)，g(x)g(0)=0，g(x)单调递增x(0,ln)，g(x)g(0)=0，不成立.······················8分2综上−12方法二：F(x

)=ex2+x+ex2−x2(=F(0))对x0恒成立F(x)=ex2−xe2x(2x+1)+2x−1=ex2−xh(x)h(x)=e2x(4x+4+2)+2，h(x)=4e2x(2x+2+1)讨论：①−1时；②2

+10即−10时；③0时.2202x方法三：ln2−ln(ex+e−x)=F(x)x21e−x−exx−x1则F(x)=3xxxe+e−x−2ln2+2ln(e+e)=x3h(x)则h(x)

=e2x−e−2x−4x(ex+e−x)2−,−12x−xx2（3）设h(x)=ln(pep+qeq)−8p2q2由对称性，只需x0时h(x)0x=epq−1−x=h(x)xp

epq+q4p2q2,h(0)0xepqxh(x)=pqx(pepq+q)2−1=−4p2q2(pepq−q)2x04p2q2(pepq+q)2所以x0,h(x)递减，h(x)h(0)=0x0,h(x)递减，h(x)h(0

)=0不等式得证.······················································································12分22.（1）曲线C的普

通方程为x4+y2=1，直线l直角坐标方程为2x+3y−8=0·········································5分（2）由参数方程设点P(2cos,sin)，12则点P到直线

l的距离为d=|4cos+3sin−8|=|8−5sin(+)|其中tan=431313所以d313，此时sin=3,cos=41355所以点P的坐标为P(8,3)···············································10分5

523.（1）①当x−1时，2−2xx+3x−1，所以无解3②当−1x3时，x1，所以1x3③当x3时，2x−2x+3x5，所以无解综上，不等式的解集为x|1x55分（2）因为f(x)=|x+1|+|x−3||x+1−x+3|=4，所

以m=4，所以a+b+c=m=4，1a+b+1b+c+1c+a=1[(a+b)+(b+c)+(c+a)](81a+b+1b+c+1c+a)98···························10分