2019-2020学年山东省寿光现代中学高三上学期开学考试 数学(理)
参考答案
1-5 BCDCB 6-10 CDACB 11-12 AC 13.1 14.1 15.2 16.(?,?2ln2] 17.(1)an?2n?3
11113n2?5n1111?)?2(2)bn?n?2n,?(? ),Tn?(1??22n?1n?2bn2nn?24n?12n?824e?5??18.(1)f(x)?sin(2x?),单调增区间[??k?,?k?](k?Z)
31212519.(1)k?1,2x?2?x?,解集{x|x?1或x??1}
222x?2?2x?64x?xt?2?2,t?2(2)m?,令,g(t)?t??4,当且仅当t?2取最小值 x?xt2?2m?4,即m的最大值4.
20.以AB,AC,AA1分别为x轴,y轴,z轴建立空间直角坐标系 (1)A1B?(2,0,?4),C1D?(1,?1,4),cos?A1B,C1D??310 10?????310 10所以A1B与C1D所成角的余弦值
?(2)面ADC1的一个法向量n?(2,?2,1),面ABA1的一个法向量AC?(0,2,0)
52 cos?n,AC???,所以正弦值33??21.如图,以AB为x轴,以AB的垂直平分线为y轴建立平面直角坐标系,
y2x2椭圆方程2?2?1,S?2(x?r)r2?x2,定义域(0,r)
4rr(2)令g(x)?(x?r)2(r2?x2),g?(x)?0,x?rrx?(0,),g?(x)?0,x?(,r),g?(x)?0
22r 2g(x)maxr27r433r2?g()?,Smax?
2162 - 5 -
22.(1)h?(x)?ex(x?2a)(x?2?a)
当a?23时,h?(x)?0,h(x)在R上递增; 当a?23时,在(??,a?2),(?2a,??)上增,(a?2,?2a)上减。
当a?23时,在(??,?2a),(a?2,??)上增,在(?2a,a?2)上减。
(2)当x?1时,eex?2?x
当x?1时,令h(x)?ex?2?lnx,h?(x)?ex?2?1,h??(x)?ex?2?1xx2?0, 所以h?(x)在(1,??)上递增,h?(1)?0,h?(e)?0,存在唯一x0?(1,e),h?(x0)?0,当x?(1,x0)时,h?(x)?0,当x?(x0,e)时,h?(x)?0
h(x)?h(x0)?ex0?2?lnx0?x0?1x?2?0 0所以eex?2?x
综上可得eex?2?x
(另解)ex?2?x?1?lnx
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ex0?2?1x0