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1绵阳南山中学2023年高一新生入学考试数学试卷参考答案及评分标准一、选择题（每个小题3分，共36分）题号123456789101112选项ACBABDADCBDC二、填空题（每个小题4分，共24分）13.42614.21615.-116.1817.

5或218.2022.5三、解答：（共90分）19．解（1）原式43335143·····················································8分（2）原式4411xxxx

·············································································4分把52x代入得原式44(51)5151(51)(51)······························

··8分20．解（1）∵总人数302000.15（人）∴2000.4590m，600.3200n.……………………………………………3分（2）由（1）的计算知70至80分段的人数为90人，………………………4分90至100分段的人数

20030906020人，补全条形图如下图所示：……………………………………6分（3）比赛成绩的中位数落在：70～80分…………………………………9分（4）恰好抽中获奖选手的概率为：602022005.…………………………12分21．解（1）由

bkbk321，解得3534bk，所以3534xy································3分5(0)4C，，5(0)3D，．在Rt△OCD中，35OD，45OC，···············

····················5分∴OCDtan34OCOD．·········································································6分频数1209060300分数（分）90100806070

202（2）取点A关于原点的对称点(21)E，，则问题转化为求证45BOE．由勾股定理可得，5OE，5BE，10OB，∵222BEOEOB，∴△EOB是等腰直角三角形．∴45BOE．∴13

5AOB°…………………12分22．解（1）设改造一所A类学校和一所B类学校所需的改造资金分别为a万元和b万元．依题意得：22302205abab··········································

···························2分解之得6085ab···················································································3分答：改造一所A类学校和一所B类学校

所需的改造资金分别为60万元和85万元········································································································4分（2）设该县有A、B两

类学校分别为m所和n所．则60851575mn···········································································

····5分173151212mn················································································6分∵A类学校不超过5所∴1731551215n≤∴15n≥即：B类学校至少有

15所．·································································8分（3）设今年改造A类学校x所，则改造B类学校为6x所，依题意得：

507064001015670xxxx≤≥·······································································10分解之得14x≤≤·······

···········································································11分∵x取整数∴1234x，，，即：共有4种方案．································

···········12分23.证明（1）：∵AB是直径∴90ACB，90CABABC……………2分∵MACABC∴90MACCAB，即MAAB∴MN是半圆的切线.…………………………4分（2）如图∵DAC是弧的中点，

∴DBCDBA……………………………5分∵AB是直径∴90ACB，故90DBCCGB∵DEAB，∴90DBAFDG，∴FDGCGBFGD∴FDFG…………………………………………………………………8分（3）连结AD则90AD

B，∵DEAB，DAC是弧的中点∴ADFDBADAF∴AFDFFG………………………………………………………………9分∴29ADGDFGSS……………………………………………………………10分又∵90ADGBCGDGACGB，∴ADGBC

G∽………………………………………………………………11分∴22416()()39BCGADGSCGSDG∴169169BCGS.……………………………………………………………12分(其它解法，请酌情评分)24．解（1）∵点(

1,0)B,3OCOB，(0,3)C…………………………………1分MNAEDCGBF342-5xyAONMGFE把1,0B、(0,3)C代入23yaxaxc得：330caac解得：334ac，∴所求抛物线的解析式为239344yxx

……………………………………2分（2）过点D作DM∥y轴分别交线段AC和x轴于点MN、．∵对称轴3322axa，1,0B，∴点(4,0)A∴ABCACDABCDSSS四边形＝111553()2222DMANONDM

＝易得直线AC的解析式为334yx…………………4分令239(3)44Dxxx，，3(3)4Mxx，，其中40x，2233933(3)(2)34444DMxxxx………6分当2x时，DM有最大值3此时四边形AB

CD面积有最大值272．…………………7分（3）如图，有如下情况：①过点C作1CP∥x轴交抛物线于点1P，过点1P作11PE∥AC交x轴于点1E，此时四边形11ACPE为平行四边形，∵(03)C，，令2393344xx得：1203xx

，∴点1(33)P，……………………………………9分②平移直线AC交x轴于点E，交x轴上方的抛物线于点P，当ACPE时，四边形ACEP为平行四边形，∵(03)C，，∴由对称关系令(3)Px，，由2393344xx化简得：2380xx，解得3412x

或3412x，此时存在点2341(3)2P，和3341(3)2P－，…………………………11分综上，存在3个点符合题意，坐标分别是1(33)P，，2341(3)2P，，3341(3)2P－，.…………………………12分25.(1)△A

MN是直角三角形.……1分依题意得OA＝2，OM＝4，ON＝1，∴MN＝OM＋ON＝4＋1＝5在Rt△AOM中，AM＝22OAOM＝2224＝25在Rt△AON中，AN＝22OAON＝2221＝5ABCDxyNMOABCxy2

P1P3P1E2E3EO442-55FGEPlAxyONMB42-2-4-6-55GFEQ2Q1PlAxyONM∴MN2＝AM2+AN2∴△AMN是直角三角形(解法不惟一)………………………3分(2)答：(1)中的结论还成立.……

……………………………4分依题意得OA＝2，OM＝－m，ON＝n∴MN＝OM＋ON＝n－m∴MN2＝(n－m)2=n2－2mn＋m2∵mn＝－4∴MN2=n2－2×(－4)＋m2＝n2＋m2+8又∵在Rt△AOM中，AM＝22OAOM＝222()m＝24m在Rt△AON中，AN＝

22OAON＝222n＝24n∴AM2＋AN2=4＋m2＋4＋n2＝n2＋m2＋8∴MN2=AM2＋AN2∴△AMN是直角三角形.(解法不惟一)………………6分(3)∵mn＝－4，n＝4，∴1m.方法一：设抛物线的函数关系式为y＝ax2＋bx＋c.∵抛物线经过点M(

－1，0)、N(4，0)和A(0，2)2a-bc04402abcc12322abc∴所求抛物线的函数关系式为y＝－12x2＋32x＋2.………………8分方法二：设抛物线的函数关系式为

y＝a(x＋1)(x－4)．∵抛物线经过点A(0，2)∴－4a＝2解得a＝－12∴所求抛物线的函数关系式为y＝－12(x＋1)(x－4)即y＝－12x2＋32x＋2.……………8分(4)抛物线的对称轴与x轴的交点Q1符合

条件，∵l⊥MN，∠ANM＝∠PNQ1，∴Rt△PNQ1∽Rt△ANM∵抛物线的对称轴为x＝32，∴Q1(32，0)………………10分∴NQ1＝4－32＝52.………………11分过点N作NQ2⊥AN，交抛物线的对称轴于点Q2.∴Rt△PQ2N、Rt△NQ2Q1、Rt△PNQ1和Rt

△ANM两两相似∴AMNQANQQ121即Q1Q2＝221522455QNANAM…………12分∵点Q2位于第四象限，∴Q2(32，5)…………13分因此，符合条件的点有两个，分别是Q1(32，0)

，Q2(32，5).…………14分(解法不惟一)获得更多资源请扫码加入享学资源网微信公众号www.xiangxue100.com