【文档说明】浙江省温州新力量联盟2020-2021学年高二下学期期中联考数学答案.pdf,共(4)页,195.848 KB,由小赞的店铺上传
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2020学年第二学期温州新力量联盟期中联考高二年级数学学科参考答案一、本大题共10题,每小题4分,共40分,在每个小题给出的四个选项中,只有一项是符合题目要求的。题号12345678910答案BACCCDCABC二、填空题:本大题共7小题,多空题每题6分,单空题每题4分,共36分
,51.17152.16365.152290.145.123.13.12212.11exyei二、解答题:本大题共5小题,共74分.解答应写出文字说明、证明过程或演算步骤.分,,的单调递增
区间为函数分得由分分解析:7................2125125....212512,2232223...............................................
.............2332sin2.................................................22cos132sin21sin3cossin1.182ZkkkxfkxkZkkxkxxxxx
xxf分或的值为综上所述,分时,当分不合题意时,当分时当分上的最大值为在则,上的最大值为,在14.
..........................................................................23113......23343323,332323,0011..........................................
......................0)(010.......13332332,3323,008...........................................233,032sin33,02332sin)(2
aaaaaxxaxgaaaaaxxaaxaxaxfxg
19(1)证明:在该组合体中,平面PAD底面ABCD,且平面PAD底面ABCDAD,CDADCD底面PAD故有CDPD……………………………………..4分同理可证ADPD………………………………………..5分又AD,CD是底面ABCD的两条相交直线,
PD底面ABCD……………………………………….7分(2)取AD的中点G,连GF,GB则GF底面ABCD,故EF在底面的射影是GB………………….9分当点E与B重合时,所成的线面角最大……………………………..11分122GFDP172GB在RtBGF中,GFGB,故045G
BF…………………….13分故不可能在BE上存在点,满足条件。……………………...15分(说明:其它解法,按相应给分)分解得分分由题可知,解析:6..............................
......................................1ln1,03...................................................................111111...............
.......................................11ln11.20''xxfnmmfmxxfnmf分恒成立对定义域内的
时,当分上单调递减,上单调递增,在单调递减,在又由分且分的最大值不大于即求分恒成立对任意恒成立对定义域内15.........................................114..............
.................................................................1111,001ln111..................................
................................011ln19..........................................................1ln8.......01ln21max'1'111xaexfagxgaxgxxygexxxg
aexxgxexaxaexfxxxxx分分的方程为设直线抛物线方程为分故由题知,解析:8....................1441111115................
.....................................................4,404441,,,,1.42.....................................................20,11.21222121212
121212222112yyxxxxxxBFAFBFAFDFBFFCAFBDACyymyymyyxymyxyxByxAmyxlxypF分即的方程为分故分,,可求得知由分15....
...............................042414213................................422142522212211............................212119..............
......................................044222121yxyxlmmyymxxCDBDACABBDACBD
ACmBDACSSBODAOC分都成立,可知猜想对任何和根据分时猜想也成立当那么分猜想成立,即假设分满足通项公式时,当数学归纳法证明:分猜想计算解法解析:7......................................216.............
.......................................................11111212124.....................................123........................
....................2111112.................145,34,2311.22*1*1432NnknkkkkkkaakkaNkknannnaaaakkkn
分求得分的等差数列公差为是以首项为分:解法7.............................................................11115.....
........................11114............................................11111111122111nanaaaaaaaaaaannnnnnnnnnnn
分分则知,由15...........................................................................................333213332112171
51513112....................................................................................32112112321232123212322123232112321,1212321
nnnccccSnnnnnnnnnnnnnnncnbnnnn