数学参考答案
一、单选题：本题共8小题，每小题5分，共40分． 在每小题给出的四个选项中，只有一项是符合题目要求的． 1．C 2．D 3．A 4．B 5．C（教材P140．3） 6．B 7．A B10） （教材P222．例6） 8．（教材P58．
二、多选题：本题共4小题，每小题5分，共20分． 在每小题给出的四个选项中，有多个选项符合题目要求，全部选对的得5分，选对但不全的得3分，有选错的得0分． 9．ABD 10．CD 11．BC 12．BC
三、填空题：本题共4小题，每小题5分，共20分． 13．2 14．π 15．32；22（教师用书P52．12） 16． （教材P231问题、教材P245例2） 解析：以枢轮中心为原点建立坐标系，则
16．13
P点纵坐标：y1=1.7sinπππ+x=1.7cosx； 水面纵坐标：y2=−1.19−0.017x，21515P点进入水中，则1.7cosx−1.19−0.017x，即cosx−0.7−0.01x，
1515y=cos作出x和y=−0.7−0.01x的图象，在10,1515存在一个交点，令h(x)=cosx+0.7+0.01x， 15因为h(12)0，h(13)0，所以点P至少经过13分钟进入水中．
四、解答题：本题共6小题，共70分． 解答应写出文字说明、证明过程或演算步骤．
17．本题考查函数的基本性质、二次不等式、韦达定理等基础知识；考查推理论证、运算求解等能力；考查函数与方程、化归与转化、数形结合等思想．满分10分．
···················································· 1分 解：由g(x)=f(x)+2x=x+(b+2)x+c
2因为g(x)为偶函数，所以g(x)对称轴x=−2b+2=0，得b=−2． 2·············································································· 4分 所以f(x)=x−2x+c ·方案一：选条件①．
······················································· 5分 因为f(x)的对称轴为x=1，且开口向上 ·
···························································· 7分 所以当x=−2时，f(x)取得最大值5·
······················································· 9分 所以f(−2)=4+4+c=5，解得c=−3 ·
高一数学试题答案 第1页（共6页）
·············································································· 10分 所以f(x)=x2−2x−3 ·方案二：选条件②．
因为f(x)≤0的解集为1，且函数f(x)图象开口向上，
·································································· ７分 所以f(x)有且仅有一个零点为1 ·
············································································· ８分 所以f(1)=1−2+c=0 ·
································································································ 9分 所以c=1 ·
················································································ 10分 所以f(x)=x2−2x+1·方案三：选条件③．
因为x1,x2为方程x2−2x+c=0的两根． 所以=4−4c0，即c1．
·············································································· 7分 且x1+x2=2，x1x2=c ·
··············································· 8分 所以x1+x2=(x1+x2)−2x1x2=4−2c=10 ·
······························································································ 9分 解得c=−3 ·
·············································································· 10分 所以f(x)=x−2x−3 ·
18．本题考查三角函数的图象和性质等知识；考查推理论证能力和运算求解能力；考查数形
结合，化归与转化等数学思想． 本题满分12分． 解：
（1）由图可知，
2222Tπππ=−−= ····························································· 1分 4362····························································································· 2分 解得T=2π ·因为T=2π············································································· 3分 ，所以=1
所以f(x)=sin(x+)．
ππ−,0sinfx··································· 4分 因为()的图象过点，所以−+=0 ·66所以−ππ+=kπ，kZ得=kπ+， 66因为ππ············································································ 5分 ，所以= ·
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所以f(x)=sinx+π············································································· 6分 ·6π················································· 8分 ·6（2）解法一：由题意，g(x)=sin2x+令t=2x+ππ13π ············································· 9分 ，因为0xπ，所以t66611π5π13π，得sint=，得t=，，． 22666由g(x)=即2x+ππ5π13ππ=或或，解得x=0，，π． 666631π················································ 12分 在0,π的解为0，，π ·
231πππ5π，得2x+=2k1π+，k1Z或2x+=2k2π+，k2Z 26666所以方程g(x)=解法二：令g(x)= ··············································································································· 8分 解得x=k1π，k1Z或x=k2π+π················································· 10分 ，k2Z ·
3因为x0,π，所以x=0，
π，π 3所以方程g(x)=1π················································ 12分 在0,π的解为0，，π ·
2319． （教材P161．12）本题考查函数单调性的证明及其应用，对数函数的图象与性质，对
数不等式的求解等知识，考查分类讨论、化归与转化等思想．
··············································································· 1分 解：（1）f(x)是减函数 证明如下：x1,x2R，且x1x2 则f(x1)−f(x2)=11− ·································································· ２分 x1x21+21+22x2−2x1 ································································· 4分 =(1+2x1)(1+2x2)·························· 5分 因为x1x2，所以2x2−2x10，又因为1+2x10，1+2x20 ·
高一数学试题答案 第3页（共6页）
所以f(x1)−f(x2)0，即f(x1)f(x2)．
····················································································· 6分 所以f(x)是减函数 （2）由题意得f(loga2)1························ 7分 =f(1)，由（1）知f(x)是减函数 ·
3·························································································· 8分 所以loga21 ·
··································· 10分 当a1时，由loga21=logaa，得a2，所以a2 当0a1时，由loga21=logaa，得a2，所以0a1． 综上所述：a的取值范围为(0,1)····················································· 12分 (2,+) ·
20． （教材P255．22）本题考查三角函数图象与性质，诱导公式． 考查运算求解，推理论
证能力． 考查化归与转化，数形结合等数学思想． 本题满分12分． 解：（1）f(x)=311··············································· 2分 sin2x+cos2x++m ·
2221=sin2x+++m ·························································· 3分
62当2x+πππ=−+2kπ，kZ，即x=−+kπ，kZ时，
36215=−3，得m=− ······················································ 4分 22f(x)的最小值为m−fx=sin2x+因为()−2，令z=2x+，
66函数y=sinz−2的单调递减区间是2kπ+π3π,2kπ+，kZ ······················· 5分 22且由2kπ+ππ3ππ2π2x+2kπ+，得kπ+xkπ+ 26263π2π,kπ+，kZ ······························· 6分 63所以函数f(x)的单调递减区间是kπ+（2）由题意得：asinx+sin2x+π−20在(0,π)上恒成立 2················································· 7分 所以asinx+cos2x−20在(0,π)上恒成立 ·
高一数学试题答案 第4页（共6页）
················································ 8分 所以asinx−1−2sin2x0在(0,π)上恒成立 ·
································································ 9分 因为x(0,π)，所以sinx(0,1 ·
2sin2x+11所以a在(0,π)上恒成立 =2sinx+sinxsinx又因为2sinx+11π3π22，当且仅当2sinx=，即x=或时，等号成立．
sinxsinx44··································································· 12分 所以a的取值范围为−,22 ·
21．（教材P156．11）
本题考查指数函数模型应用，对数运算等知识；考查运算求解和推理论证等能力、应用意识与创新意识；考查化归与转化、函数与方程等数学思想．本题满分12分． 解：（1）记2017−2019年全球年产生数据量的年增长率分别为p1，p2，p3 依题意得p1=()263341··············· 3分 −10.44，p2=−10.27，p3=−10.24 ·
182633所以p=1·························································· 4分 (0.44+0.27+0.24)0.32 ·
3··························································································· 5分 又因为a=18 ·
··············································· 6分 所以f(t)=18(1+0.32)=181.32(tN) ·
tt（2）设从2020年起，经过n年我国的数据量将达到全球数据总量的30%，
············································ 7分 由（1）知2020年全球年产生数据量为181.324 ·依题意得0.2181.32(4)(1+0.5)n181.32n+40.3 ···································· 9分
1.5所以1.32即nlogn3 233lg3−lg20.477−0.3010.1762====3.2
1.52lglg3−lg2−lg1.320.477−0.301−0.1210.0551.32lg1.51.32 ··············································································································· 11分
······················ 12分 答：估计到2024年，我国的数据量将达到全球数据总量的30%． ·
22．本题考查函数单调性与最值、零点与基本不等式等基本知识；考查推理论证能力、运算求解能力；考查化归与转化、函数与方程、分类讨论等数学思想方法． 解：(1) 因为y=a(a1)，y=−x1在(0,+)上单调递增.............................................. 1分 x高一数学试题答案 第5页（共6页）
1(a1)在(0,+)上单调递增 ................................................................. 2分 x12所以f(x)在[1,2]的最大值为f(2)=a− ....................................................................... 3分
2172所以a−=，所以a=2 ................................................................................................... 4分
221x（2）证明：因为x(−,0)，所以f(x)=a−0
x1x所以f(x)=a−在(−,0)不存在零点 ............................................................................. 5分
x1x由(1)得f(x)=a−在(0,+)上单调递增，
x所以f(x)=a−x11又因为f()=aa−a0，f(1)=a−10，
a所以f(x)在(0,+)上有唯一零点x0，且x0(,1) ......................................................... 7分
0方法一：因为a−1ax1=0，所以x0ax0=1，logax0+x0=0 .......................................... 8分 x011x+2， 因为x0(,1)，所以0xa0所以2−x0由x0ax011=−logax0=x0 .............................................. 10分 ，loga(2−x0)logax0x0=1，loga(2−x0)x0
2x0所以loga(2−x0)+x0−2x0ax0+x02−2 ................................................................... 11分
2因为0x01，所以x0+x02，得证． ........................................................................ 12分
x0方法二：因为a−1=0，有x0ax0=1 x02x02=loga(2−x0)+x0−2 ..................................................... 8分
所以loga(2−x0)+x0−2x0a因为g(x)=loga(2−x)在(,1)单调递减， 所以loga(2−x0)loga(2−)， 当a1时，a+1a1a112，所以2−a aa有loga(2−)logaa=1，即loga(2−x0)1 ................................................................ 10分
1a高一数学试题答案 第6页（共6页）
2因为h(x)=x2−2在(,1)单调递增，所以x0−2−1 .................................................. 11分
1a所以loga(2−x0)+x0−20，得证 .................................................................................. 12分
2高一数学试题答案 第7页（共6页）
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