2020年江蘇南京市聯(lián)合體中考數(shù)學(xué)一模試卷

．．．．．．．．．．．．．．知識(shí)像燭光，能照亮一個(gè)人，也能照亮無(wú)數(shù)的人。培根．．．．．．．．．．．．．．?dāng)?shù)學(xué)注意事項(xiàng)：．本試卷共6頁(yè)．全卷分120分．考試時(shí)間為120分．考生答題全部答在答題卡上，答在本試卷上無(wú)效．．請(qǐng)認(rèn)真核對(duì)監(jiān)考師在答題卡上所粘貼條形碼的姓名、考試證號(hào)是否與本人相符，再將自己的姓名、準(zhǔn)考證號(hào)用0.5毫黑色墨水簽字筆填寫(xiě)在答題卡及本試卷上．．答選擇題必須用2B鉛筆將答題卡上對(duì)應(yīng)的答案標(biāo)號(hào)涂黑．如需改動(dòng)，請(qǐng)用橡皮擦干后，再選涂其他答案答非選擇題必用毫米黑色墨水簽字筆寫(xiě)在答題卡上的定位置其他位置答題一律無(wú)效．．作圖必須用2B鉛筆作答，并請(qǐng)加黑加粗，描寫(xiě)清楚．一、選擇題（大題共6小題，每小題，共12分在每小題所給出的四個(gè)選項(xiàng)中，恰有項(xiàng)是符合題目要求的，請(qǐng)將正確項(xiàng)前的字母代號(hào)填涂在答題相應(yīng)位置上）4的算術(shù)平方根是A2B－2CD±2．2019年江省糧食總產(chǎn)達(dá)噸居全國(guó)第四位．用科學(xué)記數(shù)法表示000是的結(jié)果是A4054×104．計(jì)算－a2Aa5

B．4B．－a

C．7C．a(chǎn)6

D．D．．已eq\o\ac(△,)ABC∽△DEFeq\o\ac(△,)eq\o\ac(△,)DEF面之比為．若BC，則的是A2B．C．4．．下列整數(shù)中，與7最接近是A1B．已一次函數(shù)y＝＋圖像如圖所示，則y＝－2kx－b的圖像可能是

OOxOx

OA

B．

C．

D．

（6題）二、填空題（本大題共0小題，每小2分，共20分．不需寫(xiě)出解答程，請(qǐng)把答案直接填寫(xiě)在答題卡相應(yīng)位置上）．使式子＋x－1意義的x取值范圍是▲．．計(jì)算27－3

的結(jié)果是．/

⌒x．．．．．．．知識(shí)像燭光，能照亮一個(gè)人，也能照亮無(wú)數(shù)的人。培根⌒x．．．．．．．．分解因式(－－＋1的結(jié)果是．．已知關(guān)于x的方程＋mx－3一個(gè)根，則另一個(gè)根為，＝▲．11.若一組據(jù)，3，，，x的差比另一組數(shù)據(jù)，67，，9的方差小，可為舉個(gè)滿足條件的值）．如圖，四邊形ABCD是的接四邊形，若⊙O半為4且C＝2∠，則BD長(zhǎng)為▲．．如圖，將正六邊形ABCDEF點(diǎn)D逆針旋轉(zhuǎn)得六邊形′C′F′，則1＝

▲．

F′

F

O

′

1

′

′

D

′

DD

′

O

（第12題）

（第13題）

（第15）

（第16題）k14.反例數(shù)＝的像過(guò)點(diǎn)（2，ba＝－6則ab▲．15.如圖，在eq\o\ac(△,Rt)ACB中，C＝°，BC4，＝5BD平∠AC于D＝．．如圖，在平面直角坐標(biāo)系中，點(diǎn)A的坐標(biāo)(2，點(diǎn)的標(biāo)是(，0)．作點(diǎn)關(guān)的對(duì)稱點(diǎn)B，點(diǎn)B的標(biāo)是（▲，三解答本大題共題共88分請(qǐng)?jiān)诖痤}卡指定區(qū)域內(nèi)作答解時(shí)應(yīng)寫(xiě)出文字說(shuō)明、證明過(guò)程或演算步驟）x－1x＋6＋9分計(jì)(2．x＋1x－＋3≥x＋，分解不等式組2＞－，

并解集在數(shù)軸上表示出來(lái)．－4－3－－013/

知識(shí)像燭光，能照亮一個(gè)人，也能照亮無(wú)數(shù)的人。培根分課外興趣小組為了某段路上機(jī)動(dòng)車的車速查了一段時(shí)間內(nèi)若干輛車的車車速取整數(shù)，單位：千/時(shí)）并制成如圖示的頻數(shù)分布直方圖．已知車速在41千米時(shí)到50千米時(shí)的車輛數(shù)占車輛總數(shù)的．（1在這段時(shí)間內(nèi)他們抽查的車有▲輛；（2被抽查車輛的車速的中位數(shù)所在速度段（單位：千時(shí)）是（▲）AB．．50.5~60.5D．60.5~70.5（3補(bǔ)全頻數(shù)分布直方圖；（4如果全天超速（車速大于千米時(shí)）的車有輛，則當(dāng)天的車流量約為多少輛？車輛數(shù)20161284

128533030.5（第19）

車速(千米/時(shí))．（）甲、乙、丙醫(yī)生志愿報(bào)名參加新冠肺炎救治工.（1隨機(jī)抽取，則恰是甲的概率是▲；（2隨機(jī)抽取，求甲在其中的概率．現(xiàn)有120臺(tái)小兩種型號(hào)的挖掘機(jī)同時(shí)工作型挖掘機(jī)每小時(shí)可挖掘土方360立米小型挖掘機(jī)每小時(shí)可挖掘土立方米小共挖掘土方704立米求小型號(hào)的挖掘機(jī)各多少臺(tái)？/

112知識(shí)像燭光，能照亮一個(gè)人，也能照亮無(wú)數(shù)的人。培根11222.（8分）一輛貨車從地發(fā)以每小時(shí)的度勻速駛往地一段時(shí)間后，一輛轎車從B地發(fā)沿同一條路勻速駛往A地車駛小后距B地與轎車相遇圖中線段表示貨車離B地距離y與車行駛的時(shí)間的系．（1）兩之間的距離▲

；（2求y與之的函數(shù)關(guān)系式；（3若兩車同時(shí)到達(dá)各自目的地，在同一坐系中畫(huà)出轎車離B地距離與車行駛時(shí)間x的數(shù)圖像，用字說(shuō)明該圖像與x軸點(diǎn)所表示的實(shí)際意義．y∕160O3（第22題）

x∕h23.（8分）圖①，在四邊形中∠＝C＝90°，AB＝，求證：四邊形是矩形；（2如圖②，若四邊形滿∠A＝∠C＞90°＝，求證：四邊形是行四邊形DD

B

（圖①）

（圖②）/

知識(shí)像燭光，能照亮一個(gè)人，也能照亮無(wú)數(shù)的人。培根24.（8分如圖，位于南偏西37°方向港口C位于A偏東35°方向，B位于西方向輪船甲從A出發(fā)沿正南向行駛40海里到點(diǎn)D，此時(shí)輪乙從出發(fā)沿正東向行駛海里至處，E位于D南偏西45°方向這時(shí)處距離港口C有多？（參考據(jù)：tan37°，）北37°D45

東E

（第24題）25.（）如圖①，在矩形ABCD中，AB，BC，點(diǎn)E是BC邊一點(diǎn)連接、，eq\o\ac(△,)的外接⊙O，交AD于，交于點(diǎn)，接FG.（1求eq\o\ac(△,)AFG∽△；（2當(dāng)?shù)拈L(zhǎng)為時(shí)eq\o\ac(△,)為腰三角形；（3如圖②，若＝，求證ABO相F

D

F

DGG

O

O

E

（圖①）

（圖②）26.（分）已知二次函數(shù)y＝x22＋m＋m－（是常數(shù)（1求證：不論為值，該函數(shù)的圖像的頂點(diǎn)都在函數(shù)＝－1的像．（2若該函數(shù)的圖像與函數(shù)y＝x＋的像有兩個(gè)交點(diǎn)，則b的值范圍為（▲）（

AbB．b＞

C．＞－

D．＞－2（3該函數(shù)圖像與坐標(biāo)軸交點(diǎn)的個(gè)數(shù)隨的值變化而變化，直接寫(xiě)出交點(diǎn)個(gè)數(shù)及對(duì)應(yīng)的m的值范圍．/

．．知識(shí)像燭光，能照亮一個(gè)人，也能照亮無(wú)數(shù)的人。培根．．27.（分）【概念認(rèn)識(shí)】在同一個(gè)圓中兩條互相垂直且相等的弦定義等垂”兩弦所在直線交點(diǎn)為等垂弦的分割點(diǎn)．如圖①AB是O的AB＝CD，⊥，垂足為，AB是等垂弦為等垂弦AB、的割．

O

OD（圖①）（圖②）

D【數(shù)學(xué)理解】（1如圖②，AB是⊙的，⊥OAOD⊥OB，分別交O點(diǎn)、，連接．求證：AB、CD是的垂弦．（在⊙中⊙O的半徑為，E為垂弦AB、的割點(diǎn)，

BE1＝．AB的長(zhǎng)度．AE3【題決（3）、⊙O的條弦，＝AB，且⊥AB，垂足為F．①在圖③中，利用直尺和圓規(guī)作弦CD保留作圖痕跡，不寫(xiě)作法②若⊙O的徑為r，＝（m為數(shù)足與O的置關(guān)系隨m的變化而化，直接寫(xiě)出點(diǎn)F與⊙O的置關(guān)系及對(duì)應(yīng)點(diǎn)的值范圍．BO（圖③）/

知識(shí)像燭光，能照亮一個(gè)人，也能照亮無(wú)數(shù)的人。培根南市2020初畢生模試數(shù)試參答及分準(zhǔn)說(shuō)明本分標(biāo)準(zhǔn)每題給出了一或幾種解法供參考果考生的解法與本解答不同參照本評(píng)分標(biāo)準(zhǔn)的精神給分．（錯(cuò)扣！該的了出要分明筆出1次面回扣；筆一直改來(lái)扣；面錯(cuò)，面法確算沒(méi)的后面低一的）一、選擇題（本大題共6小，每小題2分共12分題號(hào)答案

35ACDBC二、填空題（本大題共10題，每小題2分共分．，．

11．x≥．

．3．

．a(chǎn)－．（有，1）（案唯，＜＜6即．π．．

．．

．．

．．

三、解答題（本大題共題，共88分）題6）x＋2x－＋1)(－1)解：原式＝（－·························································x＋1x＋1(x＋3)2（分1分除乘1分分、母式解各1分x＋3(＋1)(x－1)＝·······························································5（加分x＋1(＋2＝

x－1x＋3

············································································6分（約分題6）解：解不等式①，得x≤1········································································解不等式②，得x＞－．·····································································4分∴原不等式組的解集為－2≤1····························································5分－－－2－014··············································································································分題8）解）40················································································（2）································································································4分（3圖（方1分，數(shù)分）························································（4200÷＝（過(guò)、果分）·················································題8）解）．··························································································分（2）所有可能出現(xiàn)的結(jié)有乙）共，它們出現(xiàn)的可能性相同．所有的結(jié)果中，滿足“選中甲事件）結(jié)/

1225知識(shí)像燭光，能照亮一個(gè)人，也能照亮無(wú)數(shù)的人。培根1225果有種，所以(A)＝＝．······································································分（解所可出的果（甲乙甲，（乙，）共3種它出的能相．有結(jié)中滿“中（為件）的果種，以P(＝（舉列或狀過(guò)正3；子分正各1分；果1分無(wú)過(guò)僅正確果得分結(jié)沒(méi)約不分結(jié)正但沒(méi)列所結(jié)或有明可性1分；樹(shù)圖不結(jié)扣1分；表用勾扣1分列法舉不但中有確果只有個(gè)的只1）題7）（設(shè)沒(méi)方不分設(shè)且面思才設(shè)1分列程解問(wèn)，個(gè)程2分解：設(shè)大型挖掘機(jī)臺(tái)則小型挖掘(120－x)．·········································根據(jù)題意得［360＋200(120)］＝704000·····································5分解得x＝，120＝答：大型挖掘機(jī)臺(tái)，小型挖掘機(jī)50臺(tái)·············································722.（題8分解）；………（2y＝400x＋400；………………分（過(guò)程，結(jié)果分結(jié)果的2個(gè)子都對(duì)）（3如圖，線段y即所求的像線失扣1分…………………6分貨車行駛的時(shí)間為÷80＝5h可求出y的數(shù)表達(dá)式：y＝120－200該圖像與x軸點(diǎn)標(biāo)為（0分13

y∕400160

它表示的實(shí)際意義：貨車從A地發(fā)時(shí)后，轎車從B地出發(fā)．……………………8

O

353（第22題）

x∕h題8）（1證明：如圖①，連接，∵∠A＝∠＝90°在eq\o\ac(△,Rt)和eq\o\ac(△,Rt)中＝，＝，∴eq\o\ac(△,Rt)≌eq\o\ac(△,Rt)（···································································2分∴，∴四邊形是行四邊形，··································································3分∵∠A＝90°，∴四邊形是形．············································································分D

A

D

（圖①）

B

（圖②）

F/

知識(shí)像燭光，能照亮一個(gè)人，也能照亮無(wú)數(shù)的人。培根（2如圖②，分別過(guò)點(diǎn)B、作⊥點(diǎn)E，⊥BC于，····················分∵∠BAD∠BCD，∴∠＝∠，在△ABE△CDF中∠AEB＝∠＝，∠＝∠，AB＝CD∴△≌△AAS······································································6分∴BEDF，＝CF，由（）可得四邊形EBFD是形，····························································7分∴＝BF，∴＝BC∵ABCD，＝BC∴四邊形是行四邊形·································································題8）解：如圖，延長(zhǎng)交于點(diǎn)，⊥.設(shè)EF＝x里．在eq\o\ac(△,Rt)DEF中∠＝，∵∠EDF＝，………分DFx∴tan45＝，DF，…………DF在eq\o\ac(△,Rt)中，∠DFE＝90°，∵∠＝，………分∴＝AFtan37°…………4分∴20≈0.75(40＋)，…

37°40D45°x

北

東∴x＝，………分∴AF＝ADDF＝80．

20E

x

F在eq\o\ac(△,Rt)中∠AFC＝，CF∵tan∠CAF＝，

（第24題）CF∵＝，························································································∴＝≈800.70∴＝EF＋＝40＋＝96······································································答：E處離港口約海里．題9）（1證明：∵四邊形FGED是O的接四邊形，∴∠FGE∠ADE＝180．·······································································1分∵∠AGF∠FGE＝180，∴∠AGF∠ADE．··················································································2分又∠GAFDAE，∴AFG∽△；···················································································3分（25、、9分·····························································6分（3證明：圖②，過(guò)O作OHAB于H，反向延長(zhǎng)OH交CD于點(diǎn)I∴∠AHI＝90°，/

F

DGH

O

I

O知識(shí)像燭光，能照亮一個(gè)人，也能照亮無(wú)數(shù)的人。培根O在矩形ABCD中∠BAD∠＝，∴∠AHI＝∠BAD＝∠ADC＝，∴四邊形為形，∴HI＝＝，OID90°，即OICD∴DI＝CD3∵BE1，＝，∴＝，∵∠＝90，∴DE為直徑OD為徑，在eq\o\ac(△,Rt)DEC中，由勾股定理得DE＝10∴OD5在eq\o\ac(△,Rt)DIO中由勾股定理得∴IOOD2

－DI

＝4…………分∴OHHI－OI＝－4，……………8分∴是⊙O的徑，又OH，∴AB與O相．···················································································題10分（1證明：∵y＝x

－2＋m＋m＝(x－)2＋m1··································································1分∴該函數(shù)的圖像的頂點(diǎn)坐標(biāo)為，m－························································2分將x＝代入＝－1得＝－1··································································∴不論為何值，該函數(shù)的圖像的頂點(diǎn)都在數(shù)＝－1的像上．························（2）．································································································（3①當(dāng)＞時(shí)，該函數(shù)圖像與坐標(biāo)軸交點(diǎn)的個(gè)數(shù)為；······························－15－1－②當(dāng)m，＝，m時(shí)，該函數(shù)圖像與坐標(biāo)軸交點(diǎn)的個(gè)數(shù)為2；8分－15－－5－15－＋5③當(dāng)m，＜m＜，＜＜1時(shí)該函數(shù)圖像與標(biāo)軸交點(diǎn)222的個(gè)數(shù)為3······················································································10題10分（1如圖①，連接BC∵⊥O、⊥，∴∠＝∠BOD90°，

∴∠AOB，∴ABCD············································································