2020(理科数学)泉州市4月质检解答题参

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．三、填空题：本大题共4小题，每小题5分，共20分．将答案填在答题卡的相应位置。13．214．4315．(,0)，3427416．(27,2]7四、解答题：共70分。解答应写出文字说明，证明过程或演算步骤。第17～21题为必考题，每个试题考生都必须作答。第22、23题为选考题，考生根据要求作答。（一）必考题：共60分。17．本小题主要考查解三角形、三角恒等变换等基础知识，考查推理论证能力和运算求解能力等，考查数形结合思想和化归与转化思想等，体现综合性与应用性，导向对发展直观想象、逻辑推理、数算及数学建模等核心素养的关注．满分12分．【解析】思路探求1：在Rt△ABC中，由已知条件求出相关的边与角，由倍角关系推导求出△ADC为等边三角形，再利用余弦定理即求出BD（1）解法一：在Rt△ABC中，由∠ACB=

7的长度.π

，AC2........................................................................................2分3ππ

又∠DAC2∠ACB=，∠ADC=，33所以△ADC为等边三角形，所以AD2.......................................................................4分AB1，∠BAC=

在△ABD中，由余弦定理得，BDABAD2ABADcos∠BAD，2

2

2

π

，BC3得62π

=7，解得BD7..............................................6分3思路探求2：在Rt△ABC中，由已知条件求出相关的边与角，由倍角关系推导求出△ADC为等边三即BD12212cos

2

2

2

角形，再由角的关系推导出△ABD是直角三角形，利用勾股定理即求出BD

市质检数学（理科）参考解答与评分标准第1页共12页7的长度.（1）解法二：在Rt△ABC中，由∠ACB=

π

，AC2.........................................................................................2分3ππ

又∠DAC2∠ACB=，∠ADC=，33π

所以△ADC为等边三角形，所以CD2，∠ACD=，.............................................4分3ππππ

因为∠ACB=，所以∠BCD=ACB+ACD=；6362AB1，∠BAC=

（3）+2=7，解得BD在△ABD中，由勾股定理得BDCBCD

2

2

2

2

2

π

，BC3得67.....6分思路探求1：由题目已知条件∠DAC2∠ACB，可将所要的角转化到△ACD中，再将AC用Rt△ABC中边角来表示，利用正弦定理及三角恒等变换求解即可得.（2）解法一：设∠ACB=，ABx，则∠DAC2，DC在Rt△ABC中，AC=

3x......................................................................7分ABx

，......................................................8分sinsinACDC

在△ACD中，根据正弦定理得，，sin∠DACsin∠ADCx

3x即sin，......................................................................10分sin2sinπ3xπ3xsin23x2sincos3sin，2sin..........................11分33

，即cos∠ACB=.......................................................................12分443xsin解得cos

思路探求2：由题目已知条件∠DAC2∠ACB，可将所要的角转化到△ACD中，利用正弦定理求市质检数学（理科）参考解答与评分标准第2页共12页出AC，再将AC用Rt△ABC中边角来表示，最后再由等量代换求解即可得.（2）解法二设∠ACB=，ABx，则∠DAC2，DC

3x.......................................................................7分ACDC

，sin∠DACsin∠ADC在△ACD中，根据正弦定理得，3x

3xAC3x

得AC2........................................9分即

sin2sinπsin2θ4sinθcosθ

3ABx

在Rt△ABC中，AC=.............................................................10分sinsinx3x

所以，.............................................................................11分4sinθcosθsinθ33

解得cosθ即cos∠ACB=.....................................................................12分44思路探求3：作辅助线，过点C作CEAD于点E，构造两个直角三角形，将AC用边角来表示，再将AC用Rt△ABC中边角来表示，最后再由等量代换求解即可得.（2）解法三过点C作CEAD于点E

设∠ACB=，ABx，则∠DAC2，DC

3x........................................................................7分3

x；.........................................8分32在Rt△CDE中，CEDCsinD3xsin

在Rt△ACE中，AC

CE3x3x

...................................9分sinCAE2sin2θ4sinθcosθxAB.........................................10分sinsin在Rt△ABC中，AC=所以3xx

，.........................................................11分4sinθcosθsinθ33

解得cosθ即cos∠ACB=.................................................12分44市质检数学（理科）参考解答与评分标准第3页共12页18．本小题考查线面垂直的判定与性质、二面角的求解及空间向量的坐标运算等基础知识，考查空间想象能力、推理论证及运算求解能力，考查化归与转化思想、数形结合思想等，体现基础性、综合性与应用性，导向对发展数学抽象、逻辑推理、直观想象等核心素养的关注．满分12分．解：解法一：（1）依题意知，因为CD^BE，所以PE^BE，·················································1分当平面PBE

^平面ABED时,平面PBE平面ABCD=BE，PEÌ平面PBE，所以PE^平面ABCD，···········································································2分因为AB平面ABCD,所以PE^AB，······················································3分由已知，DBCD是等边三角形，且E为CD的中点，所以BE^CD，AB//CD，所以AB^BE，·············································4分又PEBE=E，所以AB^平面PBE，·····················································5分又ABÌ平面PAB，所以平面PAB^平面PBE.······（2）以E为原点，分别以ED，EB,EP的方向为x轴，y

轴，z轴的正方向，建立空间直角坐标系Exyz,则E(0,0,0)，P(0,0,1)，B(0,3,0)，A(2,3,0)，EP(0,0,1)，EA(2,3,0)，BA(2,0,0)，PA(2,3,1)，·······················································································7分设平面PAB的一个法向量m(x1,y1,z1)，平面PAE的一个法向量n(x2,y2,z2)

BAm02x10

由得；令y11，解得z13，x10，2xyz3011PAm01所以m(0,1,3)，·····················································································8分z20EPn0

由得；令y22，解得x23，z20，y02x32EAn02市质检数学（理科）参考解答与评分标准第4页共12页

所以n(3,-2,0)，·················································································9分

0207mn2cosm,n.····11分

222222727mn01(3)(3)(2)07易得所求二面角为锐角，所以二面角的余弦值为7.·······································12分解法二：（1）同解法一（2）由(1)知PE平面ABCD,且PE平面PAE,所以平面PAE平面ABCD.在ABE中，作BFAE,垂足F.因为平面PAE平面ABCDAE，BF平面ABCD，所以BF平面PAE.·············································7分所以BFAP.因为BPBCAB，所以ABP是等腰三角形.取AP的中点G，并连接FG,BG，则BGAP.又BFBGB,所以AP平面BFG.所以BGF为二面角BPAE的平面角.在RTABE中，AB2,BE

·····································9分····························8分3,AE7所以BF

237

221.7·10分同理，在RTABP中，ABBP2,AP22，所以BG在RTBFG中，GF

2.···11分147FG

,所以cosBGF.77BG7.7·············································12分所以二面角BPAE的余弦值为19．本小题主要考查椭圆的几何性质、直线与椭圆的位置关系等基础知识，考查推理论证能力、运算求解能力等，考查化归与转化思想、数形结合思想、函数与方程思想等，体现基础性、综合性与创新性，导向对发展逻辑推理、直观想象、数算、数学建模等核心素养的关注．满分12分．解法一：（1）由己知得a2b21，······················································································1分因点P(1,)在椭圆上，所以2

2

3

291

1····························································2分

a24b2所以a4,b3···························································································3分x2y2

1······································································4分所以椭圆的方程为：34市质检数学（理科）参考解答与评分标准第5页共12页（2）设直线l的方程为y1xt，············································································5分21yxt2，消去y得x2txt230，··············································7分联立22xy1342

在时，x1x2t，x1x2t3······························································8分11

tx4t3xxx340xxyy由12，即2121120，

22所以2x1x2tx1x22t0（*）.·······························································9分2

将x1x2t，x1x2t3代入（*）式，解得t22，·····································10分2

由于圆心O到直线l的距离为d

2t5

225，···················································11分所以直线l被圆O截得的弦长为l24d2248415.···························12分55解法二：（1）2a

2

33(11)2(0)2(11)2(0)24,所以a2·································2分222

2

所以bac413··············································································3分x2y2

1······································································4分所以椭圆的方程为：34（2）同解法一.····································································································12分20．本小题主要考查等高条形图、性检验、分布列与期望等基础知识，考查数据处理能力、运算求解能力、应用意识等，考查统计与概率思想等，考查数学抽象、数学建模、数据分析等核心素养，体现基础性、综合性与应用性.解：（1）根据所给等高条形图，得列联表：A材料成功不成功合计45550A材料或B材料的数据对，给1分，全对2分B材料302050合计7525100市质检数学（理科）参考解答与评分标准第6页共12页································································································································2分K2的观测值k

100(4520530)

=12，·························································4分50507525最后回答对，无大小比较，不扣分2

列式1分，结果1分由于126.635，故有99%的把握认为试验成功与材料有关．································································5分（2）生产1吨的石墨烯发热膜，所需的修复费用为X万元．······································6分单位：元，0,1000,2000…也可易知X可取0，0.1，0.2，0.3，0.4，0.5．·······························································7分22240111，CXPPX0C2=，=0.12

121233221112201=，PX0.2C=，PX0.3C2

2312231222

2

2

2

2

22

概率值没化简不扣分对1-3个，1分全对，2分12111121，==，··························9分0.5CXPX0.4C2P222331212

则X的分布列为：XP0分布列没写，不扣分0.10.20.30.40.516131611216112修复费用的期望：EX0

111111

0.10.20.30.40.50.2.11分12612636列式1分，结果1分所以石墨烯发热膜的定价至少为0.2+1+1=2.2万元/吨，才能实现预期的利润目标．·······12分22000元/吨，也可21．本小题主要考查导数的综合应用，利用导数研究函数的单调性、最值和零点等问题，考查抽象概括、推理论证、运算求解能力，考查应用意识与创新意识，综合考查化归与转化思想、分类与整合思想、函数与方程思想、数形结合思想、有限与无限思想以及特殊与一般思想，考查数学抽象、逻辑推理、直观想象、数算、数学建模等核心素养．满分12分．（1）当a0时，fxexcosx2，········································································1分解：记gxfx，则gxesinx,x当x0时，e1,1sinx1，所以gxesinx0，gx在0，单调递增，···············································2分xx市质检数学（理科）参考解答与评分标准第7页共12页所以gxg00，因为fxgx0，所以fx在0，为增函数；···············································3分xx当x0时，e1,1cosx1，所以fxecosx-20，所以fx在0，为减函数.·················································································4分综上所述，fx的递增区间为0，，递减区间为0，.·······································5分x（2）由题意可得fxecosx2ax2，f00.记gxfx，则gxesinx-2a.x再令hxgx,则hxecosx.x下面证明hxecosx在

x

,0有零点：2



,0是增函数，所以x0.22

令xhx,则xesinx在

x又



,00，····················································································6分0，

2



,0,x10,且当x,x1,x0,xx1,0，x0，22

所以存在x1

所以x，即hx在



,x1为减函数，在x1,0为增函数，2

又h



0,h00,所以hx10，2



,x1,hx00················································7分2

根据零点存在性定理，存在x0所以当xx0,0,hx0,又x0,hxecosx0，x市质检数学（理科）参考解答与评分标准第8页共12页所以hx,即gxesinx-2a在x0,0单调递减，在0，单调递增，x所以gxg0esin0-2a12a.·····························································8分0①当12a0，a

1

，gx0恒成立，所以gx，即fx为增函数，21

满足题意.············································10分2又f00，所以当xx0,0,fx0,fx为减函数，x0，,fx0,fx为增函数，x0是fx的极小值点，所以a②当a

1

，g01-2a0,令u(x)exx1,x02因为x0，所以u(x)ex10，故u(x)在(0,)单调递增，故u(x)u(0)0，即有exx1故g2ae2asin2a2a2a1sin2a2a0，单调递增，又gxexsinx-2a在0，

,gm0,由零点存在性定理知，存在唯一实数m0，

当x0,m,gx0,gx单调递减，即fx递减，所以fxf00，此时fx在0,m为减函数，所以fxf00,不合题意，应舍去.综上所述，a的取值范围是a

11分1

.········································································12分2（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：坐标系与参数方程本小题主要考查圆的直角坐标方程与极坐标方程的互化，直线的参数方程及参数的几何意义、直线与圆的位置关系等基础知识，考查推理论证能力与运算求解能力，考查函数与方程思想、转化与化归思想、数形结合思想，体现基础性与综合性，导向对发展直观想象、逻辑推理、数算等核心素养的关注．满分10分．xt,

，所以l的普通方程为3xy40，···········································1分解：（1）因为

y43t又xcos,ysin,x2y22，l的极坐标方程为3cossin40，··························································3分市质检数学（理科）参考解答与评分标准第9页共12页C的方程即为x2y22y0，对应极坐标方程为2sin．·································5分（2）由己知设A(1,),B(2,)，则1

43cossin,22sin，·······················6分所以，OA|211|OB

2sin(3cossin)[3sin2cos21]·················7分

4OB|14|OA1[2sin(2)1]·······························································8分52，2，又361266当2|OB|1，即时，取得最小值；·················································9分|OA|6662|OB|3取得最大值．················································10分当2，即时，|OA|6234所以，|OB|13

的取值范围为[,]．········································································10分|OA|2423．选修45：不等式选讲本小题主要考查绝对值不等式的解法、不等式解集的概念、绝对值的意义等基础知识，考查抽象概括能力、运算求解能力，考查分类与整合的思想，转化与化归的思想，体现基础性与综合性，导向对发展逻辑运算、数算、直观想象等核心素养的关注．满分10分．11

3x,x，

22对1个给1分，全对2分311

解法一：（1）fxx,x，··········································································2分222

3x1,x1.

22

当x

11

时，fxf2，22

当

111

x，fxf1，222

对1个给1分，全对2分当x

1

时，fx21

f1，········································································4分2

所以mfminx1·························································································5分市质检数学（理科）参考解答与评分标准第10页共12页11

，xx3,

22对1个给1分，全对2分311

（1）fxx,x，补充：解法二：·································································2分222

3x1,x1.

22

如图··································································································4分当x

1

时，mfminx1·································································5分2（1）f(x)x解法三：111111

xx≥xxx································1分2222221

≥1························································································2分21x

11

xx≤0122即x=时，等号成立.·········································4分当且仅当

21x=0

2

(列式1分，x值1分，或直接给出x值，2分)1

时，mfminx1···········································································5分2111

解法一：（2）由题意可知，abbcca，·····························································6分cab9

，因为a0,b0,c0，所以要证明不等式abbcca

abc

111

abc9，·································································7分（）只需证明cab当x

两个基本不等式各1分因为（

1

c11

abc33133abc9成立，··········································9分）abcab所以原不等式成立.·························································································10分市质检数学（理科）参考解答与评分标准第11页共12页（2）因为a0,b0,c0,所以abbcca3abc0,······································6分解法二：3222abc33abc0，··················································································7分又因为abc1,所以abcabbcac33abc3abc9，·········································8分踩空回补3222abbcacabc≥9············································································9分9

，原不等式得证.·····················································10分abc111

（2）由题意可知，abbcca，····················································6分补充：解法三：cab9

因为a0,b0,c0，所以要证明不等式abbcca，abc所以abbcca

只需证明

111

abc≥9，······················································7分abc柯西2分2

111111

由柯西不等式得：abc≥a+b+c9成cbaabc立，·····································································································9分所以原不等式成立.················································································10分市质检数学（理科）参考解答与评分标准第12页共12页