1、2018 年中考第一次模擬調(diào)研九年級數(shù)學(xué)學(xué)科注意事項：1本試卷共 6 頁全卷滿分 120 分考試時間為 120 分鐘考生答題全部答在答題卡上，答在本試卷上無效2請認(rèn)真核對監(jiān)考教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合，再將自己的姓名、考試證號用 0.5 毫米黑色墨水簽字筆填寫在答題卡及本試卷上3答選擇題必須用 2B 鉛筆將答題卡上對應(yīng)的答案標(biāo)號涂黑如需改動，請用橡皮擦干凈后，再選涂其他答案答非選擇題必須用 0.5 毫米黑色墨水簽字筆寫在答題卡上的指定

2、位置，在其他位置答題一律無效4作圖必須用 2B 鉛筆作答，并請加黑加粗，描寫清楚一、選擇題（本大題共 6 小題，每小題 2 分，共 12 分在每小題所給出的四個選項中，恰有一項是符合題目要求的，請將正確選項前的字母代號填涂在答題卡相應(yīng)位置上）1下列計算結(jié)果為負(fù)數(shù)的是（）A(3)(4)B(3)(4)C(3) (4)4D(3)2計算 a6×(a2)3÷a4 的結(jié)果是（）Aa3Ba7Ca8Da93若銳角三角函數(shù) tan55°a，則 a

3、0;的范圍是（）A0a1B1a2C2a3D3a44下列各數(shù)中，相反數(shù)、絕對值、平方根、立方根都等于其本身的是（）A0B1C0 和 1D1 和1圖                    圖5把球放在長方體紙盒內(nèi)，球的一部分露出盒外，其截面如圖所示，已知 EF=CD=4 cm，則球的半徑長是（）A2 cmB2.5 cmC3 

4、;cmD4 cm6如圖，是一個每條棱長均相等的三棱錐，圖是它的主視圖、左視圖與俯視圖若邊 AB 的長度為 a，則在這三種視圖的所有線段中，長度為 a 的線段條數(shù)是（）A12 條B9 條C6 條D5 條EFDAAB主視圖左視圖OBC俯視圖（第 5 題）（第 6 題）二、填空題（本大題共 10 小題，每小題 2 分，共 20 分不需寫出解答過程，請把答案直接填寫在答題卡相應(yīng)位置上）7函數(shù) y

5、 1x中，自變量 x 的取值范圍是8分解因式 a3a 的結(jié)果是9若關(guān)于 x 的一元二次方程 x2kx20 有一個根是 1，則另一個根是10遼寧號是中國人民解放軍海軍第一艘可以搭載固定翼飛機的航空母艦，其滿載排水量為67 500 噸用科學(xué)記數(shù)法表示 67 500 是13如圖，四邊形 ABCD 是O 的內(nèi)接四邊形，若O 的半徑為 3 cm，A110°，則劣弧BD的長為11一組數(shù)據(jù)&

6、#160;1、2、3、4、5 的方差為 S12，另一組數(shù)據(jù) 6、7、8、9、10 的方差為 S22，那么 S12S22（填“”、“”或“”）k012在同一平面直角坐標(biāo)系中，反比例函數(shù)y1x（k 為常數(shù)，k）的圖像與一次函數(shù) y2 xa（a 為常數(shù)，a0）的圖像相交于 A、B 兩點若點 A 的坐標(biāo)為(m，n)，則點 B 的坐標(biāo)為cm14如圖，點 F、G 在正五邊形 ABCDE 的邊上，BF、CG&#

7、160;交于點 H，若 CFDG，則BHG°ADABEBOHGCCFD（第 13 題）（第 14 題）16如圖，以 AB 為直徑的半圓沿弦 BC 折疊后，AB 與 相交于點 D若CD  BD，則B      °1CB15如圖，正八邊形 ABCDEFGH 的邊長為 a，I、J、K、L 分別是各自所在邊的中點，且四邊形 I

8、JKL 是正方形，則正方形 IJKL 的邊長為（用含 a 的代數(shù)式表示）3AHILBGADD      E三、解答題（本大題共 11 小題，共 88 分請在答題卡指定區(qū)域內(nèi)作答，解答時應(yīng)寫出文字說明、證明過17（6 分）計算：æèa2 öø÷æèa öøïî21    

9、，CFOJKCB（第 15 題）（第 16 題）程或演算步驟）11aaìï2x0，18（7 分）解不等式組í5x12x1并把它的解集在數(shù)軸上表示出來3-3 -2 -10123（第 18 題）19（7 分）如圖，四邊形 ABCD 是平行四邊形，線段 EF 分別交 AD、AC、BC 于點 E、O、F，EFAC，AOCO（1）求證：四邊形 AFCE 是平行四邊形；（2）在本題三個已知

10、條件中，去掉一個條件， 1）的結(jié)論依然成立，這個條件是（直接寫出這個條件的序號）AEDOBFC（第 19 題）20（8 分）某天,一蔬菜經(jīng)營戶用 180 元錢從蔬菜批發(fā)市場批了西紅柿和豆角共 40 千克到菜市場去賣,西紅柿和豆角這天的批發(fā)價與零售價如下表所示：品名批發(fā)價(單位:元/千克)零售價(單位:元/千克)問：他當(dāng)天賣完這些西紅柿和豆角能賺多少錢?西紅柿3.65.4豆角4.67.521（8 分）超市水果貨架上有四個蘋果，重量分別是 100 g、110 g、120 

11、;g 和 125 g（1）小明媽媽從貨架上隨機取下一個蘋果恰是最重的蘋果的概率是；（2）小明媽媽從貨架上隨機取下兩個蘋果它們總重量超過 232 g 的概率是多少？22（8 分）河西中學(xué)九年級共有 9 個班，300 名學(xué)生，學(xué)校要對該年級學(xué)生數(shù)學(xué)學(xué)科學(xué)業(yè)水平測試成績進行抽樣分析，請按要求回答下列問題：收集數(shù)據(jù)（1）若從所有成績中抽取一個容量為 36 的樣本，以下抽樣方法中最合理的是 在九年級學(xué)生中隨機抽取 36 名學(xué)生的成績；按男、女各隨機抽取&#

12、160;18 名學(xué)生的成績；按班級在每個班各隨機抽取 4 名學(xué)生的成績整理數(shù)據(jù)（2）將抽取的 36 名學(xué)生的成績進行分組，繪制頻數(shù)分布表和成績分布扇形統(tǒng)計圖如下請根據(jù)圖表中數(shù)據(jù)填空：C 類和 D 類部分的圓心角度數(shù)分別為°、°；估計九年級 A、B 類學(xué)生一共有名成績（單位：分） 頻數(shù)頻率九年級學(xué)生數(shù)學(xué)成績分布扇形統(tǒng)計圖B 類A 類（80100）181225%B 類（6079）C 類（4059）961416A 類50%

13、D 類（039）3112數(shù)據(jù)來源：學(xué)業(yè)水平考試數(shù)學(xué)成績抽樣（第 22 題）分析數(shù)據(jù)（3）教育主管部門為了解學(xué)校教學(xué)情況，將河西、復(fù)興兩所中學(xué)的抽樣數(shù)據(jù)進行對比，得下表：學(xué)校平均數(shù)（分）  極差（分）  方差A(yù)、B 類的頻率和河西中學(xué)復(fù)興中學(xué)717152804324970.750.82你認(rèn)為哪所學(xué)校本次測試成績較好，請說明理由23（8 分）下圖是投影儀安裝截面圖教室高 EF3.5 m，投影儀 A 發(fā)出的光線夾角BAC30°，投影屏幕高 BC1.2&

14、#160;m固定投影儀的吊臂 AD0.5 m，且 ADDE，ADEF，ACB45°求屏幕下邊沿離地面的高度 CF（結(jié)果精確到 0.1 m）（參考數(shù)據(jù)：tan15°0.27，tan30°0.58）DA（第 23 題）EBCF24（9 分）一輛貨車從甲地出發(fā)以每小時 80 km 的速度勻速駛往乙地，一段時間后，一輛轎車從乙地出發(fā)沿同一條路勻速駛往甲地貨車行駛 2.5 h 后，在距乙地 160 km&#

15、160;處與轎車相遇圖中線段 AB 表示貨車離乙地的距離 y1 km 與貨車行駛時間 x h 的函數(shù)關(guān)系（1）求 y1 與 x 之間的函數(shù)表達式；（2）若兩車同時到達各自目的地，在同一坐標(biāo)系中畫出轎車離乙地的距離y2 與 x 的圖像，求該圖像與x 軸交點坐標(biāo)并解釋其實際意義ykmA160O2.5B xh（第 24 題）25（8 分）某超市欲購進一種今年新上市的產(chǎn)品，購進價為 20

16、0;元/件，該超市進行了試銷售，得知該產(chǎn)品每天的銷售量 t（件）與每件銷售價 x（元/件）之間有如下關(guān)系：t3x90（1）請寫出該超市銷售這種產(chǎn)品每天的銷售利潤 y（元）與 x 之間的函數(shù)表達式；（2）當(dāng) x 為多少元時，銷售利潤最大？最大利潤是多少？26（9 分）ABC 中，ACB90°，AC:BC4:3，O 是 BC 上一點，O 交 AB 于點 D，交 BC 延長線于點 E連接 ED

17、，交 AC 于點 G，且 AGAD（1）求證：AB 與O 相切；（2）設(shè)O 與 AC 的延長線交于點 F，連接 EF，若 EFAB，且 EF5，求 BD 的長EAGCODB（第 26 題）F（P27 10 分）圖是一張AOB45°的紙片折疊后的圖形， 、Q 分別是邊 OA、OB 上的點，且 OP2 cm將AOB 沿 P

18、Q 折疊，點 O 落在紙片所在平面內(nèi)的 C 處（1）當(dāng) PCQB 時，OQcm；在 OB 上找一點 Q，使 PCQB（尺規(guī)作圖，保留作圖痕跡）；（2）當(dāng)折疊后重疊部分為等腰三角形時，求 OQ 的長BBBQCOP        AO        P     &

19、#160; A備用圖 1O        P        A備用圖 2（第 27 題）2018年中考第一次模擬調(diào)研數(shù)學(xué)參考答案及評分標(biāo)準(zhǔn)說明：本評分標(biāo)準(zhǔn)每題給出了一種或幾種解法供參考，如果考生的解法與本解答不同，參照本評分標(biāo)準(zhǔn)的精神給分一、選擇題（每小題 2 分，共計 12 分）題號答案1A2C3B4A5B6D12（n，m）  

20、;  13               14108°        15    a         16. 18°a      &

21、#160;aa     a21a    (a1)(a1)a1二、填空題（每小題 2 分，共計 20 分）7x18a(a+1)(a-1)9-2106.75×1041172+ 232三、解答題（本大題共 10 小題，共計 88 分）17（本題 6 分）a22a1a21解：原式 ÷a22a1a·(a1) 2a·a1 ·

22、83;·················································

23、83;······································ 6 分18（本題 7 分）解：解不等式，得 x2 · &

24、#183;·················································&

25、#183;························· 2 分解不等式，得 x 1 ·················

26、83;·················································

27、83;····· 4 分所以，不等式組的解集是1x2········································

28、;················ 5 分畫圖-1    012    · ····················&#

29、183;·········································· 7 分19（本題 7 分）解：（1）四邊形&#

30、160;ABCD 是平行四邊形AECFDACBCA· ···········································

31、···························· 1 分在AOE 和COF 中ìïDAC=ACBí AO=COïîAOE=COFAOECOF（ASA） ····&

32、#183;·················································&

33、#183;············ 3 分AECF四邊形 AFCE 是平行四邊形· ····························

34、····························· 5 分（2）··················&

35、#183;·················································&

36、#183;························ 7 分   解：（1）···················

37、··················································

38、···························· 2 分    因此，總重量超過 232g 的概率是· ·········

39、········································· 8 分   tan  BAPtan1

40、5°0.27 ···············································&

41、#183;··················· 5 分20（本題 8 分）解：設(shè)批發(fā)了西紅柿 x 千克，豆角 y 千克ì x + y = 40由題意得： í·······

42、3;············································· 3 分î3.6 

43、;x + 4.6 y = 180ì x = 4解得： í6 分î y = 36(5.4  3.6)× 4(7.5  4.6)× 36  111.6（元） ··············

44、3;························· 7 分答：賣完這些西紅柿和豆角能賺 111.6 元 ················

45、···································· 8 分21（本題 8 分）14（2）共有 6 種等可能出現(xiàn)的結(jié)果，分別為···

46、;··················································

47、;· 3 分（100，110）；（100，120）；（100，125）；（110，120）；（110，125）；（120，125）····································

48、3;······················ 6 分總重量超過 232g 的結(jié)果有 2 種，即（110，125），（120，125） ··············&#

49、183;···· 7 分1322（本題 8 分）解：（1）······································

50、83;·················································

51、83;··· 2 分（2） 60°，30°········································

52、;··········································· 4 分 225···

53、··················································

54、······························ 6 分（3）兩所學(xué)校都可以選擇只要理由正確皆可得分··············

55、······························· 8 分選擇河西中學(xué)，理由是平均分相同，河西中學(xué)極差和方差較小，河西中學(xué)成績更穩(wěn)定選擇復(fù)興中學(xué)，理由是平均分相同，復(fù)興中學(xué) A，B 類頻率和高，復(fù)興中學(xué)高分人數(shù)更多23（本題 8

56、 分）解：過點 A 作 APEF，垂足為 PADDE，ADE90°ADEF，DEP90°APEF，APEAPC90°,ADEDEPAPE90°四邊形 ADEP 為矩形EPAD0.5m ························

57、3;·················································

58、3;·········· 2 分APC90°，ACB45°CAP45°ACB，BAPCAPCAB45°30°15°APCP·························

59、;··················································

60、;···················· 4 分在 APB 中BPAPBP0.27AP0.27CP,BCCPBPCP0.27CP0.73CP1.2mCP1.64m················&#

61、183;·················································&#

62、183;························ 7 分CFEFEPCP3.50.51.641.361.4m··················

63、3;··················· 8 分24（本題 9 分）解：（1）由條件可得 k1801 分設(shè) y180x+b1，過點（2.5，160），可得方程 16080×2.5+b1解得 b1360 ········

64、··················································

65、··························· 3 分y1 80x+360··················&

66、#183;·················································&

67、#183;··········· 4 分（2）當(dāng) y1 0 時，可得 x4.5轎車和貨車同時到達，終點坐標(biāo)為（4.5，360）設(shè) y2 k2 x+b2 ，過點（2.5，160）和（4.5，360）解得 k2 100，b2 90y2 100x90 圖像如下圖·······

68、················································· 

69、7 分與 x 軸交點坐標(biāo)為(0.9，0) ···········································&

70、#183;························ 8 分說明轎車比貨車晚出發(fā) 0.9h ··················

71、83;··············································· 9 分2

72、5（本題 8 分）解：（1）表達式為 y(3x+90)(x20)化簡為 y3x²+150x1800·····································

73、3;··························· 4 分（2）把表達式化為頂點式 y3(x25)² +75·············

74、3;···································· 6 分當(dāng) x25 時，y 有最大值 75答：當(dāng)售價為 25 元時，有最大利潤

75、60;75 元 · ·············································

76、··· 8 分26（本題 9 分）（1）證明：連結(jié) ODACB90°，OED+EGC90° ·································&

77、#183;································ 1 分O，ODOE，ODEOEDAGAD，ADGAGD ··········

78、;··················································

79、;···················· 3 分AGDEGCOED+EGCADG+ODEADO90°ODAB ····················

80、3;·················································

81、3;················· 4 分OD 為半徑AB 是O 的切線································