2019级资阳市高三数学第一次诊断性考试 答案

2019级资阳市高三数学第一次诊断性考试

理科数学参和评分意见

评分说明：

1． 各阅卷组阅卷前组织阅卷教师细化评分细则。 2． 本解答只给出了一种解法供参考．如果考生的解法与本解答不同，可根据试题的主要

考查内容比照评分参考制定相应的评分细则。 3． 对计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内

容和难度，可视影响程度决定后继部分的给分，但不得超过该正确部分解答得分的一半；如果后继部分的解得有严重错误，就不再给分。 4． 只给整数分。选择题和填空题不给中间分。 一、选择题：本题共12小题，每小题5分，共60分。

1.C 2.A 3.C 4.B 5.A 7.B 8.D 9.B 10.C 11.D 二、填空题：本题共4小题，每小题5分，共20分。

413．1 14．5 15．511 16.

3三、解答题：共70分。解答应写出文字说明、证明过程或演算步骤。 （一）必考题：共60分。 17.（12分）

3113···································· 2分 sin2xcos2xcos2xsin2x ·

2222π·························································· 4分 3sin2xcos2x2sin(2x)． ·6π令f(x)0，即sin(2x)0，

6π1π则2xkπ，kZ，得xkπ，kZ， ······································ 6分

21265π11π由于x[0，；令k2，得x． ·························· 7分 ]，令k1，得x12125π11π所以，f(x)在[0，，． ················································ 8分 ]上的零点为

1212ππ2π（2）由x[，]，则2x[,]． ················································ 9分

446333π所以，······························································· 11分 ≤sin(2x)≤1， ·

262]． ·故f(x)在[，]上的取值范围是[3，············································· 12分

4418.（12分）

6.C 12.D

（1）f(x)（1）由Snan(n1)2， 得Sn1an1n2，

两式相减，得an1an1an2n1，

理科数学答案 第1页（共4页）

所以，an2n1． ················································································ 4分

1352n1（2）由题Tn23n，

2222111352n1两边同乘以，有Tn234n+1， ······································ 6分

222222两式相减，得

1122222n1TnTn234nn+1

222222211(1n1)12n132n32················································· 10分 24n+1n+1． ·

12222122n3所以，Tn3n． ··········································································· 12分

219．（12分）

π（1）由bsinAasin(B)，

3π根据正弦定理，有sinBsinAsinAsin(B)， ··········································· 2分

3π13即有sinBsin(B)sinB················································ 3分 cosB， ·

322则有tanB3，又0Bπ，

π所以，B． ····················································································· 4分

3π2π（2）由（1），B，则AC，又△ABC为锐角三角形，

33π2ππ所以，0A且0A，

2323ππ所以A，于是tanA． ···························································· 6分

3622π31sin(A)cosAsinAcsinC31322则······················· 8分 2． asinAsinAsinA2tanA2311又， 2tanA22c1所以，的取值范围是(,2)． ································································ 12分

a220．（12分）

12（1）由题可知a≠0，所以函数f(x)2ax2x1的对称轴为x，

2a由于yf(x1)是偶函数，

2所以f(x1)f(x1)，即f(x)2ax2x1关于x＝1对称， ····················· 2分

理科数学答案 第2页（共4页）

111，即a． 2a22所以f(x)x2x1． ·········································································· 4分 所以

mmexf(x)有三个不同实数根． x有三个不同的实数根，即方程ex2x令g(x)ef(x)，由（1）有g(x)(x2x1)e， （2）方程f(x)2x所以g(x)(x1)e，令g(x)0，则x1或x1．

当x1时，g(x)0；当1x1时，g(x)0；当x1时，g(x)0． ······ 6分

故当x1时，g(x)单调递增；当1x1时，g(x)单调递减；当x1时，g(x)单调递增．

4；当x1时，g(x)取得极小值g(1)0． e ·········································································································· 8分 所以，当x1时，g(x)取得极大值g(1)又由于g(x)≥0，且当x时，g(x)0；当x时，g(x)．

4x所以，方程mef(x)有三个不同实数根时，m的范围是(0,)． ·················· 12分

e21．（12分）

a（1）f(x)2(1a)xb，由题f(1)a2(1a)b0，

x······························································ 2分 解得ab2，由a＝1，得b=1.

1(x1)因为fx的定义域为(0,)，所以f(x)1，

xx故当x(0,1)时，f(x)0，fx为增函数，

············································ 4分 当x(1,)时，f(x)0，fx为减函数， ·（2）由（1）知b＝2－a，

a2(1a)x2(a2)xa2(1a)xa(x1). 所以f(x)2(1a)x(2a)xxx·············· 5分 （i）若a1，则由（1）知fxmaxf(1)0，即fx≤0恒成立． ·

aa2(1a)x(x1)0， ii（）若a1，则且2(1a)xa(x1)2(1a)2(1a)f(x)xx故当x(0,1)时，f(x)0，fx为增函数， 当x(1,)时，f(x)0，fx为减函数，

fxmaxf(1)0，即fx≤0恒成立． ··················································· 7分

a2(1a)x2(x1)且a1， 2(1a)xa(x1)（iii）若a1，则2(1a)2(1a)3f(x)xx故当x(0,1)时，f(x)0，fx为增函数，

理科数学答案 第3页（共4页）

a)时，f(x)0，fx为减函数，

2(1a)a,)时，f(x)0，fx为增函数， 当x(2(1a)当x(1,e22e1由题只需fe≤0即可，即a(1a)e(2a)e+1≤0，解得a≥2，

ee1e22e12(e2)21e22e12e20，且2120， 而由2ee133e3e3ee1ee12e2e1··········································································· 9分 ≤a1． ·得2ee122(x1)20，fx为增函数，且f10， （iv）若a，则f(x)33x····································· 10分 所以x(1,e)，f(x)f10，不合题意，舍去； ·

a21，f(x)在x(1,e)上都为增函数，且f10， （v）若a，则

2(1a)32所以x(1,e)，f(x)f10，不合题意，舍去；

e22e1·················································· 12分 ,)． 综上所述，a的取值范围是[2ee1（二）选考题：共10分。请考生在第22、23题中任选一题作答。如果多做，则按所做的第一题计分。

22．[选修4－4：坐标系与参数方程]（10分）

（1）直线l的普通方程为yx1． ························································· 2分

422222222sin4xyy4x2y4， 由2，得，则有，即

1sin2x2y2则曲线C的直角坐标方程为·················································· 5分 1． ·

423（2）将l的参数方程代入x22y24，得t222t20，设两根为t1，t2， ··· 7分

242则t1，t2为M，N对应的参数，且t1t2，

3tt22所以，线段MN的中点为Q对应的参数为12． ······························· 9分

2322所以，|OPOQ||PQ|． ······························································· 10分

323．[选修4－5：不等式选讲]（10分）

（1）(abc)2abc2ab2bc2ca

≤abc(ab)(bc)(ca)≤3(abc)3． 1当且仅当abc取“＝”．所以，abc的最大值为3． ············ 5分

3111abcabcabc（2）(1)(1)(1)(1)(1)(1)

abcabc理科数学答案 第4页（共4页）

2bc2ac2abbcacab≥8． abcabc1当且仅当abc取“＝”． ······························································ 10分

3理科数学答案 第5页（共4页）