．．．．．．．．．．．．．．知識像燭光，能照亮一個人，也能照亮無數的人。培根．．．．．．．．．．．．．．數學注意事項：．本試卷共6頁．全卷分120分．考試時間為120分．考生答題全部答在答題卡上，答在本試卷上無效．．請認真核對監考師在答題卡上所粘貼條形碼的姓名、考試證號是否與本人相符，再將自己的姓名、準考證號用0.5毫黑色墨水簽字筆填寫在答題卡及本試卷上．．答選擇題必須用2B鉛筆將答題卡上對應的答案標號涂黑．如需改動，請用橡皮擦干后，再選涂其他答案答非選擇題必用毫米黑色墨水簽字筆寫在答題卡上的定位置其他位置答題一律無效．．作圖必須用2B鉛筆作答，并請加黑加粗，描寫清楚．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選項中，恰有項是符合題目要求的，請將正確項前的字母代號填涂在答題相應位置上）4的算術平方根是A2B－2CD±2．2019年江省糧食總產達噸居全國第四位．用科學記數法表示000是的結果是A4054×104．計算－a2Aa5

B．4B．－a

C．7C．a6

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．．下列整數中，與7最接近是A1B．已一次函數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．不需寫出解答程，請把答案直接填寫在答題卡相應位置上）．使式子＋x－1意義的x取值范圍是▲．．計算27－3

的結果是．/

⌒x．．．．．．．知識像燭光，能照亮一個人，也能照亮無數的人。培根⌒x．．．．．．．．分解因式(－－＋1的結果是．．已知關于x的方程＋mx－3一個根，則另一個根為，＝▲．11.若一組據，3，，，x的差比另一組數據，67，，9的方差小，可為舉個滿足條件的值）．如圖，四邊形ABCD是的接四邊形，若⊙O半為4且C＝2∠，則BD長為▲．．如圖，將正六邊形ABCDEF點D逆針旋轉得六邊形′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＝的像過點（2，ba＝－6則ab▲．15.如圖，在eq\o\ac(△,Rt)ACB中，C＝°，BC4，＝5BD平∠AC于D＝．．如圖，在平面直角坐標系中，點A的坐標(2，點的標是(，0)．作點關的對稱點B，點B的標是（▲，三解答本大題共題共88分請在答題卡指定區域內作答解時應寫出文字說明、證明過程或演算步驟）x－1x＋6＋9分計(2．x＋1x－＋3≥x＋，分解不等式組2＞－，

并解集在數軸上表示出來．－4－3－－013/

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

128533030.5（第19）

車速(千米/時)．（）甲、乙、丙醫生志愿報名參加新冠肺炎救治工.（1隨機抽取，則恰是甲的概率是▲；（2隨機抽取，求甲在其中的概率．現有120臺小兩種型號的挖掘機同時工作型挖掘機每小時可挖掘土方360立米小型挖掘機每小時可挖掘土立方米小共挖掘土方704立米求小型號的挖掘機各多少臺？/

112知識像燭光，能照亮一個人，也能照亮無數的人。培根11222.（8分）一輛貨車從地發以每小時的度勻速駛往地一段時間后，一輛轎車從B地發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與車行駛的時間的系．（1）兩之間的距離▲

；（2求y與之的函數關系式；（3若兩車同時到達各自目的地，在同一坐系中畫出轎車離B地距離與車行駛時間x的數圖像，用字說明該圖像與x軸點所表示的實際意義．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）/

知識像燭光，能照亮一個人，也能照亮無數的人。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出發沿正南向行駛40海里到點D，此時輪乙從出發沿正東向行駛海里至處，E位于D南偏西45°方向這時處距離港口C有多？（參考據：tan37°，）北37°D45

東E

（第24題）25.（）如圖①，在矩形ABCD中，AB，BC，點E是BC邊一點連接、，eq\o\ac(△,)的外接⊙O，交AD于，交于點，接FG.（1求eq\o\ac(△,)AFG∽△；（2當的長為時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y＝x22＋m＋m－（是常數（1求證：不論為值，該函數的圖像的頂點都在函數＝－1的像．（2若該函數的圖像與函數y＝x＋的像有兩個交點，則b的值范圍為（▲）（

AbB．b＞

C．＞－

D．＞－2（3該函數圖像與坐標軸交點的個數隨的值變化而變化，直接寫出交點個數及對應的m的值范圍．/

．．知識像燭光，能照亮一個人，也能照亮無數的人。培根．．27.（分）【概念認識】在同一個圓中兩條互相垂直且相等的弦定義等垂”兩弦所在直線交點為等垂弦的分割點．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【數學理解】（1如圖②，AB是⊙的，⊥OAOD⊥OB，分別交O點、，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，

BE1＝．AB的長度．AE3【題決（3）、⊙O的條弦，＝AB，且⊥AB，垂足為F．①在圖③中，利用直尺和圓規作弦CD保留作圖痕跡，不寫作法②若⊙O的徑為r，＝（m為數足與O的置關系隨m的變化而化，直接寫出點F與⊙O的置關系及對應點的值范圍．BO（圖③）/

知識像燭光，能照亮一個人，也能照亮無數的人。培根南市2020初畢生模試數試參答及分準說明本分標準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分標準的精神給分．（錯扣！該的了出要分明筆出1次面回扣；筆一直改來扣；面錯，面法確算沒的后面低一的）一、選擇題（本大題共6小，每小題2分共12分題號答案

35ACDBC二、填空題（本大題共10題，每小題2分共分．，．

11．x≥．

．3．

．a－．（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，數分）························································（4200÷＝（過、果分）·················································題8）解）．··························································································分（2）所有可能出現的結有乙）共，它們出現的可能性相同．所有的結果中，滿足“選中甲事件）結/

1225知識像燭光，能照亮一個人，也能照亮無數的人。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．有結中滿“中（為件）的果種，以P(＝（舉列或狀過正3；子分正各1分；果1分無過僅正確果得分結沒約不分結正但沒列所結或有明可性1分；樹圖不結扣1分；表用勾扣1分列法舉不但中有確果只有個的只1）題7）（設沒方不分設且面思才設1分列程解問，個程2分解：設大型挖掘機臺則小型挖掘(120－x)．·········································根據題意得［360＋200(120)］＝704000·····································5分解得x＝，120＝答：大型挖掘機臺，小型挖掘機50臺·············································722.（題8分解）；………（2y＝400x＋400；………………分（過程，結果分結果的2個子都對）（3如圖，線段y即所求的像線失扣1分…………………6分貨車行駛的時間為÷80＝5h可求出y的數表達式：y＝120－200該圖像與x軸點標為（0分13

y∕400160

它表示的實際意義：貨車從A地發時后，轎車從B地出發．……………………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/

知識像燭光，能照亮一個人，也能照亮無數的人。培根（2如圖②，分別過點B、作⊥點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，延長交于點，⊥.設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O作OHAB于H，反向延長OH交CD于點I∴∠AHI＝90°，/

F

DGH

O

I

O知識像燭光，能照亮一個人，也能照亮無數的人。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∴該函數的圖像的頂點坐標為，m－························································2分將x＝代入＝－1得＝－1··································································∴不論為何值，該函數的圖像的頂點都在數＝－1的像上．························（2）．································································································（3①當＞時，該函數圖像與坐標軸交點的個數為；······························－15－1－②當m，＝，m時，該函數圖像與坐標軸交點的個數為2；8分－15－－5－15－＋5③當m，＜m＜，＜＜1時該函數圖像與標軸交點222的個數為3······················································································10題10分（1如圖①，連接BC∵⊥O、⊥，∴∠＝∠BOD90°，

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