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数学试题参考答案及评分标准一、选择题:本大题共12个小题,每小题5分,共60分。题目123456789101112题号ADCDBACCDBAB二、填空题:本大题共5个小题,每小题4分,共20分。13.a≥514.x(x+3)(x—3)15.(1,3)16.—217.(—2023,202
2)三、解答题:本大题共7个小题,共70分。18.(本题满分8分)解:整理方程组得232313xyxy−=+=①②,·········································2分①×2—②得—7y=7
,y=1,····································································4分把y=1代入①得x—2=3,解得x=5,··························
······································6分∴方程组的解为51xy==.··················································8分19.(本题满分8分)证明
:∵△ABC是等腰三角形,∴∠EBC=∠DCB,·······················································2分在△EBC与△DCB中,∴BE=CD,BC=CB∴△EBC≌△DCB(SAS),···········
·······································6分∴BD=CE.·························································
·····8分20.(本题满分10分)解:(1)将点A(1,2)代入y=mx,得m=2,∴双曲线的表达式为:y=2x,···············································1分把A(1,2)和B(4,0)代入y=kx+b得:y=240kbkb+=+=,
解得:2383kb=−=,·········································3分∴直线的表达式为:y=23−x+83;·········································4分(2)联立22833yxyx==
−+,解得12xy==,或323xy==,·························5分∵点A的坐标为(1,2),∴点B的坐标为(3,23),······················
·························6分∵S△AOB=S△AOB—S△AOB=12OC·Ay—12OC·By=12×4×2—12×4×23=83,∴△AOB的面积为83;······························
·····················8分(3)1<m<3.····························································10分21.(本题满分10分)解:(1)120
99························································4分(2)如图:··············8分(3)把“礼仪”“陶艺”“园艺”“厨艺”及“编程”等五门校本课程分别记为A、B、C、D、E,画树状图如下:共有
25种等可能的结果,其中小刚和小强两人恰好选到同一门课程的结果有5种,∴小刚和小强两人恰好选到同—门课程的概率P=51255=.············································10分22.(本题满分10分)解:小明能运用以上数据,得到综合楼的高度,
理由如下:作EG⊥AB,垂足为G,作AH⊥CD,垂足为H,如图:····················2分由题意知,EG=BF=40米,EF=BG=12.88米,∠HAE=16°=∠AEG=16°,∠CAH=9°,在Rt△AEG中,tan∠AEG=AGEG,∴t
an16°=40AG,即0.287≈40AG,············································4分∴AG=40×0.287=11.48(米),∴AB=AG+BG=11.48+12.88=24.36(米),····················
················6分∴HD=AB=24.36米,在Rt△ACH中,AH=BD=BF+FD=80米,tan∠CAH=CHAH,∴tan9°=80CH,即0.158≈80CH,············
································8分∴CH=80×0.158=12.64(米),∴CD=CH+HD=12.64+24.36=37.00(米),则综合楼的高度约是37.00米.·····································
········10分注:结果精确到0.01米,须保留两位小数,未保留最后一步不得分!23.(本题满分12分)解:(1)证明:由题意得,AI、BI分别平分∠BAC、∠ABC,∴∠BAD=∠CAD,∠ABI=∠C
BI,············································2分又∵∠CAD=∠CBD,∴∠BAD=∠CBD,∵∠BID=∠BAD+∠ABI,∠IBD=∠CBI+∠CBD,∴∠BID=∠IBD,∴BD=DI;······················
·········································4分22题答案图24题(2)答案图(2)证明:如图,连接OD,∵∠CAD=∠BAD,∴BDCD=,∴OD⊥BC,·······································5分∵DE∥B
C,∴OD⊥DE,······························································6分∴DE是⊙O的切线;·······················
································7分(3)证明:如图,连接BH,CH,∵GH是⊙O的切线,∴∠CHG=∠HBG,·····························8分∵∠CGH=∠BGH,∴△HCG∽△BHG,∴GH2=BG•CG,·······
····················································9分∵AD∥GF,∴∠AFG=∠CAD,∵∠CAD=∠FBG,∴∠FBG=∠AFG,························
·············10分∵∠CGF=∠BGF,∴△CGF∽△FGB,∴FG2=BG•CG,··························································11分∴FG=HG.·······
·······················································12分24.(本题满分12分)解:(1)∵抛物线的顶点D(1,4)∴根据顶点式,抛物线的解析式为y=—(x—1)2+4=—x2+2x+3;············2
分(2)如图,设直线l交x轴于点T,连接PT,BD,BD交PM于点J.设P(m,—m2+2m+3).···························3分点D(1,4)在直线l:y=43x+t上,∴4=43x+t,∴t=83,∴直线DT的解析式为y=4
3x+83,···········································4分23题(2)答案图23题(3)答案图24题(3)答案图令y=0,得到x=—2,∴T(—2,0),∴O
T=2,∵B(3,0),∴OB=3,∴BT=5,························································5分∵DT=2234+=5,∴TD=TB,∵
PM⊥BT,PN⊥DT,∴S四边形DTBP=S△PDT+S△PBT=12×DT×PN+12×TB×PM=(PM+PN),∴四边形DTBP的面积最大时,PM+PN的值最大,······························
··········································6分∵D(1,4),B(3,0),∴直线BD的解析式为y=—2x+6,································
···········7分∴J(m,—2m+6),∴PJ=—m2+4m—3,∵S四边形DTBP=S△DTB+S△BDP=12×5×4+12×(—m2+4m—3)×2=—m2+4m+7=—(m—2)2+11∵—1<0,∴m=2时
,四边形DTBP的面积最大,最大值为11,∴PM+PN的最大值=25×11=225;··········································8分(3)四边形AFBG的面积不变.理由:如图,设P(m,—m2+2m+3),················9分∵
A(—1,0),B(3,0),∴直线AP的解析式为y=—(m—3)x—m+3,··········10分∴E(1,—2m+6),·∵E,G关于x轴对称,∴G(1,2m—6),∴直线PB的解析式y=—(m+1)x+3(m+1),·······························
·11分∴F(1,2m+2),∴GF=2m+2—(2m—6)=8,∴四边形AFBG的面积=12×AB×FG=12×4×8=16.∴四边形AFBG的面积是定值.·····················
·······················12分数学试题参考解析一、选择题1.【答案】A【解析】∵实数a的相反数是—1,∴a=1,∴a+1=2.2.【答案】D【解析】A选项不是中心对称图形,也不是轴对称图形,故此选项不合题意;B选项不是中心对称图形,是轴对称图形,故此选项不合题意;C选项不
是中心对称图形,是轴对称图形,故此选项不合题意;D选项既是轴对称图形,又是中心对称图形,故此选项符合题意.3.【答案】C【解析】因为图中两个空白面不是相对面,所以图中的四个字不能恰好环绕组成一个四字成语,故A不符合题意;因为图中两个空白面不是相对面,
所以图中的四个字不能恰好环绕组成一个四字成语,故B不符合题意;因为金与题是相对面,榜与名是相对面,所以正方体侧面上的字恰好环绕组成一个四字成语金榜题名,故C符合题意;因为图中两个空白面不是相对面,所以图中的四个字不能恰好环绕组成一个四字成语,故D不符合题意.4.【答案】D【解析】中位数为
第10个和第11个的平均数15152+=15,众数为15.5.【答案】B【解析】∵AB∥CD,∴∠DFE=∠BAE=50°,∵CF=EF,∴∠C=∠E,∵∠DFE=∠C+∠E,∴∠C=12∠DFE=12×50°=25°.6.【答案】
A【解析】A选项≈3.1416,B选项≈3.1408,C选项≈3.14,D选项≈3.1428,π≈3.14159≈3.1416,A选项符合题意.7.【答案】C【解析】连接AD,如图,∵AB=AC,∠A=120°,∴∠B=∠C=30°,由作法得DE垂
直平分AC,∴DA=DC=3,∴∠DAC=∠C=30°,∴∠BAD=120°—30°=90°,在Rt△ABD中,∵∠B=30°,∴BD=2AD=6.8.【答案】C【解析】原式=4a6b2—3a6b2=a6b2,9.【答案】D【分析】设第二次采购单价为x元,则第一次采购单
价为(x+10)元,根据单价=总价÷数量,结合总费用降低了15%,采购数量与第一次相同,即可得出:2000020000(115%)10xx−=+10.【答案】【解析】连接AC交BD于O,如图,∵四边形ABCD为菱形,∴AD∥BC,CB=CD=AD=4,AC⊥AB,BO=OD,O
C=AO,∵E为AD边的中点,∴DE=2,∵∠DEF=∠DFE,∴DF=DE=2,∵DE∥BC,∴∠DEF=∠BCF,∵∠DFE=∠BFC,∴∠BCF=∠BFC,∴BF=BC=4,∴BD=BF+DF=4+2=6,∴OB=OD=3,在Rt△
BOC中,OC=2243−=7∴AC=2OC=27∴菱形ABCD的面积=12AC·BD=12×2×6=6711.【答案】A【解析】∵二次函数y=ax2+2的图象经过P(1,3),∴3=a+2,∴a=1,∴y=x2+2,∵Q(m,n)在y=x2+2上,∴n=m2+2,
∴n2—4m2—4n+9=(m2+2)2—4m2—4(m2+2)+9=m4—4m2+5=(m2—2)2+1,∵(m2—2)≥0,∴n2—4m2—4n+9的最小值为1.12.【答案】B【解析】如图,连接AI,BI,CI,DI,过点I作IT⊥AC于点T.∵I是△ABD的内心,∴∠BAI=∠CA
I,∵AB=AC,AI=AI,∴△BAI≌△CAI(SAS),∴IB=IC,∵∠ITD=∠IED=90°,∠IDT=∠IDE,DI=DI,∴△IDT≌△IDE(AAS),∴DE=DT,IT=IE,∵∠BEI=∠CTI=90°,∴Rt△BEI≌Rt△CTI(HL),∴BE=CT,设BE=CT=x
,∵DE=DT,∴10—x=x—4,∴x=7,∴BE=7.二、填空题13.【答案】a≥5【解析】∵a-5≥0,∴a≥5.14.【答案】x(x+3)(x—3)【解析】原式=x(x2-9)=x(x+3)(x—3).15.【答案】(1,3)【解
析】∵点A(—3,4)的对应点是A1(2,5),∴点B(—4,2)的对应点B1的坐标是(1,3).16.【答案】—2【解析】原式=22222(1)1111xxxxxxx−−−−==−−−−=—2.17.【答案】(—2023,2022)【解析】∵将顶点D(1,0)绕点A(0,1)
逆时针旋转90°得点D1,∴D1(1,2),∵再将D1绕点B逆时针旋转90°得点D2,再将D2绕点C逆时针旋转90°得点D3,再将D3绕点D逆时针旋转90°得点D4,再将D4绕点A逆时针旋转90°得点D5……∴D2(-3,2),D
3(-3,-4),D4(5,-4),D5(5,6),D6(-7,6),……,观察发现:每四个点一个循环,D4n+2(-4n-3,4n+2),∵2022=4×505+2,∴D2022(-2023,202
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