【定稿】2017福州市普通高中毕业班质量检测参考答案及评分标准(理科数学)

2017年福州市高中毕业班质量检测

理科数学试题答案及评分参考

评分说明：

1．本解答给出了一种或几种解法供参考，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标准制定相应的评分细则．

2．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，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分．

3．解答右端所注分数，表示考生正确做到这一步应得的累加分数．

4．只给整数分数．选择题和填空题不给中间分．

一、选择题：本大题考查基础知识和基本运算．每小题5分，满分60分．

（1）D （2）B （3）C （4）C （5）A （6）C

（7）A （8）D （9）D （10）A （11）C （12）D

二、填空题：本大题考查基础知识和基本运算．每小题5分，满分20分．

（13）3 （14）36 （15）80m （16）2

三、解答题：本大题共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) 本小题主要考查正弦定理、余弦定理、三角形面积公式、三角恒等变换及基本不等式等基础知识，考查运算求解能力，考查化归与转化思想、函数与方程思想等．满分12分．

【解析】（Ⅰ）∵()tan 3cos cos c c a B b A =+，

∴()sin tan 3sin cos sin cos C C A B B A =+， ················································ 2分 ∴()sin tan 3sin 3sin C C A B C =+=， ···················································· 4分 ∵0πC <<，∴sin 0C ≠，

∴tan 3C =，∴60C =? ． ···································································· 6分 （Ⅱ）∵23c =，60C =?，

由余弦定理2222cos c a b ab C =+-得：22122a b ab

ab ab =+-- ···················· 9分 ∴12ab ，∴1=sin 332ABC S ab C ? ······················································ 11分 当且仅当23a b ==时，ABC ?面积取最大值33． ·································· 12分

(18) 本小题主要考查空间直线与直线、直线与平面的位置关系及二面角等基础知识，考查空间想象能力、推理论证能力、运算求解能力，考查化归与转化思想等．满分12分．

【解析】（Ⅰ）连接BD 交AC 于N ，连接MN ，

依题意知//AB CD ，∴ABN

CDN ??， ·················· 1分 ∴

2BN BA ND CD ==，∵12PM MB =， ∴2BN BM ND MP

==，∴在BPD ?中，//MN PD ， ·········· 3分 又∵PD MAC ?平面，MN MAC ?平面.

∴//PD MAC 平面． ·············································· 5分

（Ⅱ）∵平面PAD ⊥平面ABCD ，PAD ABCD AD =平面平面，PA AD ⊥，PA PAD ?平面，∴PA ABCD ⊥平面，又AD AB ⊥，从而,,PA AD AB 两两垂直， 以A 为原点，分别以,,AD AB AP 的方向为,,x y z 轴的正方向建立空间直角坐标系，如图所示．依题意12AP AD AB ===，，又12

PM MB =， ∴()()()()220,0,00,2,00,0,101,1,033A B P M C ?? ???

，，，，，，， ····························· 7分 ∴()()22001,0=1,1,033AP AM AC ??== ???

，，，，，， ∵PA ABCD ⊥平面，∴()10,0,1n AP ==为平面BAC 的一个法向量． ··············· 8分 设()2,,n x y z =为平面MAC 的法向量， 则220,0,n AM n AC ??=???=?? ∴220,330,

y z x y ?+=???+=? 令1,1,1x y z ==-=则，∴()21,1,1n =-为平面MAC 的一个法向量， (10)

分 ∴121212cos ,1n n n n

n n ?===??， ∴二面角M AC B --． ···················································· 12分 (19) 本小题主要考查随机抽样方法、随机变量分布列和数学期望等基础知识，考查运算求解

能力、数据处理能力、应用意识，考查必然与或然思想、化归与转化思想．满分12分．

【解析】（Ⅰ）抽出的前7人的后三位考号分别为:310,503,315,571,210,142,188.

·········································································································· 4分 （Ⅱ）ξ的可能取值为0,1,2,3. ································································· 5分

()0P ξ=3437435C C ==,()1P ξ=2143371835

C C C ==, ()2P ξ=1243371235C C C ==, ()3P ξ=3337135

C C ==. ··············································· 9分 故随机变量ξ的概率分布列为：

········································································································ 10分

所以418121012335353535E ξ=?+?+?+?=97

. ············································ 12分 (20) 本小题主要考查直线与圆、直线与抛物线位置关系等基础知识，考查运算求解能力，考

查化归与转化思想．满分12分．

【解析】 (Ⅰ)设(),P x y 为曲线C 上任意一点,

因为曲线C 上的点(),P x y 到点()0,1F 的距离比它到直线3y =-的距离小2，

所以点P 到点F 的距离等于它到直线1y =-的距离， ···································· 2分 所以曲线C 是以F 为焦点,直线1y =-为准线的抛物线，其方程为24x y =．······· 4分 (Ⅱ)（ⅰ）依题意直线l 的方程为1y kx =+，代入24x y =得2440x kx --=，2(4)160k ?=-+>

设()11,A x y ，()22,B x y ，则12124, 4.x x k x x +=?=- ········································ 5分

因为BF BA λ=,即122121221(1)(,),1x x y x x y y x λλ

--=--=-， 2212121221()16121241x x x x k x x x x λλλ+==++=-++--，即21142111k λλ

+=-+-； ········· 7分 因为12[]23λ∈，,所以111[1]2λ-∈，,又函数1()f x x x =+在1[,1]2

单调递减, 所以2542[2]2k +∈，

,44

k ≤≤ ··························································· 8分 （ⅱ）因为2

4x y =，所以2

4x y =，'2x y = 则切线PA 方程为2111()24

x x y x x =-+……① PB 方程为2222()24

x x y x x =-+……② ②－①得221212244

x x x x x -=-， 121()22

x x x k =+=……③， 将③代入①得1y =-，所以()2,1P k - ······················································· 10分 P 到直线AB

的距离2d = 1||2AMP S AM d ?=?，1||2

BNP S BN d ?=?， 21||||4AMP BNP S S AM BN d ???=

?? 因为12||||1,||||1,AM AF y BN BF y =-==-= 所以221212||||116

x x AM BN y y ??=== 2214

AMP BNP d S S k ???==+， 当且仅当0k =时, AMP BNP S S ???取最小值1． ·············································· 12分

(21) 本小题主要考查导数及其应用、不等式等基础知识，考查推理论证能力、运算求解能力、

创新意识等，考查函数与方程思想、化归与转化思想、分类与整合思想、数形结合思想等．满分12分．

【解析】（Ⅰ）()x f x e a '=-， ································································· 1分 ∵()f x 在0x =处的极小值为2，

∴()()00,0 2.f f '?=??=??

即10,1 2.a b -=??+=? ······································································ 3分 ∴1,1.a b =??=?

···························································································· 4分 （Ⅱ）()()()()+ln 1ln 1x g x f x x e ax b x =+=-+++ ∴1()1x g x e a x '=

+-+， ········································································· 5分 设()11x h x e a x =

+-+，则()()211x h x e x '=-+， 当0x 时，()

211,

11x e x +， ······························································· 6分 ∴()()2101x h x e x '=-

+，∴()11x h x e a x =+-+在[)0,+∞上为增函数． ·········· 7分 ∴()()02h x h a =-，即1()21x g x e a a x '=+--+ ···································· 8分 ∴当2a

时，()0g x '，∴()()ln 1x g x e ax b x =-+++在[)0,+∞上为增函数， ∴当0x 时，()()01g x g b =+，符合题意； ·

·········································· 9分 当2a >时，由()11

x h x e a x =+-+在[)0,+∞上为增函数， 且()020h a =-<，()ln 11ln e 0ln 1ln 1a h a a a a =

+-=>++， 则存在0(0,ln )x a ∈，使得0()0h x =， ······················································· 10分 于是()g x 在0(0,)x 上单调递减，在0(,)x +∞上单调递增，则有0()(0)1g x g b <=+， 此时()1g x b +不恒成立，不符合题意． ·

················································ 11分 综上可得实数a 的取值范围为(,2]-∞． ···················································· 12分

(22) 本小题考查极坐标方程和参数方程等基础知识，考查运算求解能力，考查数形结合思想、

化归与转化思想等．满分10分．

【解析】（Ⅰ）将曲线C 的极坐标方程2222cos 3sin 12ρθρθ+=化为普通方程为2

21124

x y +=

，则其左焦点为()-，则m =- ·································

· 2分 将直线

l 的参数方程,x m y ?=??????与曲线C 的方程221124x y +=联立， 化简可得2220t t --=， ········································································· 4分

则12||||||2FA FB t t ?==. ············································································ 5分

（Ⅱ）由曲线C 的方程22

1124

x y +=，可设曲线C 上的任意一点P 的坐标

为()

,2sin θθ， ··············································································· 7分 则以P 为顶点的内接矩形周长为

()

42sin 16sin 032θθθθππ?????+=+<< ???????， ······································· 9分 因此该内接矩形周长的最大值为16． ······················································· 10分

(23) 本小题考查绝对值不等式的解法与性质、不等式的证明等基础知识，考查运算求解能力、

推理论证能力，考查分类与整合思想、化归与转化思想等． 满分10分．

【解析】解：（Ⅰ）令()1,1,1223,12,1,2,x f x x x x x x -??=---=-<

························ 2分

则()11f x -， ·

················································································ 3分 因为0x ?∈R 使不等式|1||2|

x x t ---成立，所以1t ，即{}|1T t t =． ······· 5分 （Ⅱ）由（Ⅰ）知，33log log 1m n ?， ······················································ 6分 根据基本不等式3333log log 2log log 2m n

m n +， ···································· 8分 所以9mn

，当且仅当3m n ==时取等号， 所以mn 的最小值为9． ········································································ 10分