(完整)2015年臨沂市中考數(shù)學試題及答案整理版,推薦文檔

試卷類型：A絕密★啟用前2015年臨沂市初中學生學業(yè)考試試題數(shù)學注意事項：1．本試卷分第Ⅰ卷（選擇題）和第Ⅱ卷（非選擇題），共8頁，滿分120分，考試時間120分鐘．答卷前，考生務必用0.5毫米黑色簽字筆將自己的姓名、準考證號、座號填寫在試卷和答題卡規(guī)定的位置．考試結(jié)束后，將本試卷和答題卡一并交回．2．答題注意事項見答題卡，答在本試卷上不得分．第Ⅰ卷（選擇題共42分）一、選擇題（本大題共14小題，每小題3分，共42分）在每小題所給出的四個選項中，只有一項是符合題目要求的．1．1的絕對值是2(A)1.2(B)12.(C)2.(D)2.2．如圖，直線a∥b，∠1=60°，∠2=40°，則∠3等于ab12(A)40°.(B)60°.(C)80°.(D)100°.3（第2題圖）3．下列計算正確的是(A)a2a22a4.(C)a2a3a6.(B)(a2b)3a6b3.(D)a8a2a4.4．某市6月份某周內(nèi)每天的最高氣溫數(shù)據(jù)如下（單位：℃）：24262926293229則這組數(shù)據(jù)的眾數(shù)和中位數(shù)分別是(A)29，29.(B)26，26.(C)26，29.(D)29，32.數(shù)學試題第1頁（共14頁）5．如圖所示，該幾何體的主視圖是(A)(B)(C)(D)（第5題圖）6．不等式組2x6,的解集，在數(shù)軸上表示正確的是x2≤0－3－2－10122－3－2－1012(A)(B)－3－2－101－3－2－1012(C)(D)7．一天晚上，小麗在清洗兩只顏色分別為粉色和白色的有蓋茶杯時，突然停電了，小麗只好把杯蓋和茶杯隨機地搭配在一起.則其顏色搭配一致的概率是(A)1.4(B)1.2(C).3(D)1.48．如圖A，B，C是eO上的三個點，若AOC100o，則ABC等于O(A)50°.(B)80°.CA(C)100°.(D)130°.B9．多項式mx2m與多項式x22x1的公因式是（第8題圖）(A)x1.(C)x21.(B)x1.(D)x1.210．已知甲、乙兩地相距20千米，汽車從甲地勻速行駛到乙地，則汽車行駛時間t（單位：小時）關(guān)于行駛速度v（單位：千米／小時）的函數(shù)關(guān)系式是20v(C)t20v(D)t10v.(A)t20v.(B)t..11．觀察下列關(guān)于x的單項式，探究其規(guī)律：x，3x2，5x3，7x4，9x5，11x6，….數(shù)學試題第2頁（共14頁）按照上述規(guī)律，第2015個單項式是(A)2015x2015.(B)4029x2014.(C)4029x2015.(D)4031x2015.E12．如圖，四邊形ABCD為平行四邊形，延長AD到E，使DE=AD，連接EB，EC，DB.添加一個條件，不能使四邊形DBCE．．DC成為矩形的是(A)AB=BE.(B)BE⊥DC.(D)CE⊥DE.AB(C)∠ADB=90°.（第12題圖）13．要將拋物線yx22x3平移后得到拋物線yx2，下列平移方法正確的是(A)向左平移1個單位，再向上平移2個單位.(B)向左平移1個單位，再向下平移2個單位.(C)向右平移1個單位，再向上平移2個單位.(D)向右平移1個單位，再向下平移2個單位.14．在平面直角坐標系中，直線y=－x＋2與反比例函數(shù)y1的圖象有唯一公共點.若直線yxbxy1與反比例函數(shù)的圖象有2個公共點，則b的取值范圍是yx(A)b﹥2.(B)－2﹤b﹤2.2(C)b﹥2或b﹤－2.(D)b﹤－2.O2x（第14題圖）數(shù)學試題第3頁（共14頁）第Ⅱ卷（非選擇題共78分）注意事項：1．第Ⅱ卷分填空題和解答題．2．第Ⅱ卷所有題目的答案，考生須用0.5毫米黑色簽字筆答在答題卡規(guī)定的區(qū)域內(nèi)，在試卷上答題不得分．二、填空題（本大題共5小題，每小題3分，共15分）15．比較大?。?_______3（填“”，“”，“”）.﹤＝﹥a42a16．計算：a2a2____________.3，則Y17．如圖，在ABCD中，連接BD，ADBD,AB4,sinAYABCD的面積是________.4ADCEDOABBC（第17題圖）（第18題圖）18．如圖，在△ABC中，BD，CE分別是邊AC，AB上的中線，BD與CE相交于點O，則OB_________.OD19．定義：給定關(guān)于x的函數(shù)y，對于該函數(shù)圖象上任意兩點（x1，y1），（x2，y2），當x﹤﹤x2時，都有yy2，稱該函數(shù)為增函數(shù).根據(jù)以上定義，可以判斷下面所給的函數(shù)中，是增函數(shù)的有11______________（填上所有正確答案的序號）.④y1.①y=2x；②y=x＋1；③y=x＞(x0)；2x三、解答題（本大題共7小題，共63分）20．（本小題滿分7分）計算：(321)(321).數(shù)學試題第4頁（共14頁）21．（本小題滿分7分）“保護環(huán)境，人人有責”，為了了解某市的空氣質(zhì)量情況，某校環(huán)保興趣小組，隨機抽取了2014年內(nèi)該市若干天的空氣質(zhì)量情況作為樣本進行統(tǒng)計，繪制了如圖所示的條形統(tǒng)計圖和扇形統(tǒng)計圖（部分信息未給出）.請你根據(jù)圖中提供的信息，解答下列問題：（1）補全條形統(tǒng)計圖；（2）估計該市這一年（365天）空氣質(zhì)量達到“優(yōu)”和“良”的總天數(shù)；（3）計算隨機選取這一年內(nèi)的某一天，空氣質(zhì)量是“優(yōu)”的概率.某市若干天空氣質(zhì)量情況條形統(tǒng)計圖某市若干天空氣質(zhì)量情況扇形統(tǒng)計圖天數(shù)3636輕微污染302418126輕度中度5%良12優(yōu)重度3210優(yōu)良輕微輕度中度重度空氣質(zhì)量類別（第21題圖）22．（本小題滿分7分）小強從自己家的陽臺上，看一棟樓頂部的仰角為30°，看這棟樓底部的俯角為60°，小強家與這棟樓的B水平距離為42m，這棟樓有多高？AαDβ（第22題圖）C數(shù)學試題第5頁（共14頁）23．（本小題滿分9分）如圖，點O為Rt△ABC斜邊AB上的一點，以OA為半徑的⊙O與BC切于點D，與AC交于點E，連C接AD.（1）求證：AD平分∠BAC；（2）若∠BAC=60°，OA=2，求陰影部分的面積（結(jié)果保留）.AEDBO（第23題圖）24．（本小題滿分9分）新農(nóng)村社區(qū)改造中，有一部分樓盤要對外銷售.某樓盤共23層，銷售價格如下：第八層樓房售價為4000元／米2，從第八層起每上升一層，每平方米的售價提高50元；反之，樓層每下降一層，每平方米的售價降低30元，已知該樓盤每套樓房面積均為120米2.若購買者一次性付清所有房款，開發(fā)商有兩種優(yōu)惠方案：方案一：降價8%，另外每套樓房贈送a元裝修基金；方案二：降價10%，沒有其他贈送.（1）請寫出售價y（元／米2）與樓層x（1≤x≤23，x取整數(shù)）之間的函數(shù)關(guān)系；式（2）老王要購買第十六層的一套樓房，若他一次性付清購房款，請幫他計算哪種優(yōu)惠方案更加合算.數(shù)學試題第6頁（共14頁）25．（本小題滿分11分）如圖1，在正方形ABCD的外側(cè)，作兩個等邊三角形ADE和DCF，連接AF，BE.（1）請判斷：AF與BE的數(shù)量關(guān)系是，位置關(guān)系是；（2）如圖2，若將條件“兩個等邊三角形ADE和DCF”變?yōu)椤皟蓚€等腰三角形ADE和DCF，且EA=ED=FD=FC”，第（1）問中的結(jié)論是否仍然成立?請作出判斷并給予證明；（3）若三角形ADE和DCF為一般三角形，且AE=DF，ED=FC，第（1）問中的結(jié)論都能成立嗎？請直接寫出你的判斷.AABBCBCAEECDDDFF圖1圖2（第25題圖）備用圖數(shù)學試題第7頁（共14頁）26.（本小題滿分13分）在平面直角坐標系中，O為原點，直線y=－2x－1與y軸交于點A，與直線y=－x交于點B,點B關(guān)于原點的對稱點為點C.（1）求過A，B，C三點的拋物線的解析式；y（2）P為拋物線上一點，它關(guān)于原點的對稱點為Q.①當四邊形PBQC為菱形時，求點P的坐標；＜＜②若點P的橫坐標為t（－1t1），當t為何B值時，四邊形PBQC面積最大，并說明理由.OxCAyxy2x1（第26題圖）數(shù)學試題第8頁（共14頁）參考答案及評分標準說明：解答題給出了部分解答方法，考生若有其它解法，應參照本評分標準給分.一、選擇題（每小題3分，共42分）題號答案1234567891011121314ACBADCBDABCBDC二、填空題（每小題3分，共15分）a2；17．3716．；18．2；19．①③.15．>；a三、解答題20．解：方法一：(321)(321)=[3(21)][3(21)]············································1分=(3)2(21)2······························································3分3(2221)······························································5分32221································································6分22.············································································7分方法二：(321)(321)(3)2323123(2)221131211·········3分363622321··················································5分22.····························································································7分21．解：（1）圖形補充正確.·······································································2分某市若干天空氣質(zhì)量情況條形統(tǒng)計圖天數(shù)36363024181212663210優(yōu)良輕微輕度中度重度空氣質(zhì)污染污染污染污染量類別（2）方法一：由（1）知樣本容量是60，∴該市2014年（365天）空氣質(zhì)量達到“優(yōu)”、“良”的總天數(shù)約為：123660365292（天）.··········································································5分方法二：由（1）知樣本容量是60，∴該市2014年（365天）空氣質(zhì)量達到“優(yōu)”的天數(shù)約為：126036573（天）.·················································································3分該市2014年（365天）空氣質(zhì)量達到“良”的天數(shù)約為：數(shù)學試題第9頁（共14頁）3660365219（天）.···············································································4分∴該市2014年（365天）空氣質(zhì)量達到“優(yōu)”、“良”的總天數(shù)約為：73+219=292（天）.···················································································5分（3）隨機選取2014年內(nèi)某一天，空氣質(zhì)量是“優(yōu)”的概率為：126015.··································································································7分B22．解：如圖，α=30°，β=60°，AD=42.ABDADCDADα，tan，D∵tanβ∴BD=AD·tanα=42×tan30°=42×3=143.···························3分3CD＝ADtanβ＝42×tan60°＝423.·········································6分C∴BC＝BD＋CD＝143＋423＝563(m).因此，這棟樓高為563m.········································································7分C23．（1）證明：連接OD.ED∵BC是⊙O的切線，D為切點，∴OD⊥BC.···································1分又∵AC⊥BC，ABO∴OD∥AC，·································2分∴∠ADO=∠CAD.··························3分又∵OD=OA，∴∠ADO=∠OAD，··················································································4分∴∠CAD=∠OAD，即AD平分∠BAC.··························································5分（2）方法一：連接OE，ED.C∵∠BAC=60°，OE=OA，∴△OAE為等邊三角形，DE∴∠AOE=60°，∴∠ADE=30°.ABO又∵1OADBAC30o，2數(shù)學試題第10頁（共14頁）∴∠ADE=∠OAD，∴ED∥AO，···································6分∴S△AED＝S△OED，3606042∴陰影部分的面積=S扇形ODE=3.············································9分方法二：同方法一，得ED∥AO，·······························································6分∴四邊形AODE為平行四邊形，12∴SVSV233.································································7分AEDOAD360又S扇形ODE－S△OED=6043323.···············································8分∴陰影部分的面積=(S扇形ODE－S△OED)+S△AED2323=3.················9分324．解：（1）當1≤x≤8時，y＝4000－30（8－x）＝4000－240＋30x＝30x＋3760；···············································2分＜當8x≤23時，y＝4000＋50（x－8）＝4000＋50x－400＝50x＋3600.30x3760（1≤x≤8，x為整數(shù)），∴所求函數(shù)關(guān)系式為y≤23，x為整數(shù)）.······················4分＜50x3600（8x（2）當x＝16時，方案一每套樓房總費用：w1＝120（50×16＋3600）×92%－a＝485760－a；····································5分方案二每套樓房總費用：w2＝120（50×16＋3600）×90%＝475200.···············································6分＜＜＞∴當w1w2時，即485760－a475200時，a10560；當w＝＝＝w2時，即485760－a475200時，a10560；1當w＞＞＜w2時，即485760－a475200時，a10560.1因此，當每套贈送裝修基金多于10560元時，選擇方案一合算；當每套贈送裝修基金等于10560元時，兩種方案一樣；當每套贈送裝修基金少于10560元時，選擇方案二合算.···································9分25．解：（1）AF=BE，AF⊥BE.·································································2分（2）結(jié)論成立.························································································3分證明：∵四邊形ABCD是正方形，ABE數(shù)學試題第11頁（共14頁）CDF∴BA=AD=DC，∠BAD=∠ADC=90°.在△EAD和△FDC中，EAFD,EDFC,ADDC,∴△EAD≌△FDC.∴∠EAD=∠FDC.∴∠EAD+∠DAB=∠FDC+∠CDA，即∠BAE=∠ADF.······································4分在△BAE和△ADF中，BAAD,BAEADF,AEDF,∴△BAE≌△ADF.∴BE=AF，∠ABE=∠DAF.········································································6分∵∠DAF+∠BAF=90°，∴∠ABE+∠BAF=90°，∴AF⊥BE.······························································································9分（3）結(jié)論都能成立.·················································································11分y2x1，x1，得26．解：（1）解方程組yx，y1.∴點B的坐標為（-1，1）.········································································1分∵點C和點B關(guān)于原點對稱，∴點C的坐標為（1，-1）.········································································2分又∵點A是直線y=-2x-1與y軸的交點，∴點A的坐標為（0，-1）.········································································3分設拋物線的解析式為y=ax2+bx+c，abc1，a1，∴abc1，解得b1，c1.c1.∴拋物線的解析式為y=x2-x-1.····································································5分（2）①如圖1，∵點P在拋物線上，∴可設點P的坐標為（m，m2-m-1）.當四邊形PBQC是菱形時，O為菱形的中心，∴PQ⊥BC，即點P，Q在直線y=x上，∴m=m2-m-1，······················································································7分數(shù)學試題第12頁（共14頁）

解得m=1±2.······················································································8分∴點P的坐標為（1+2，1+2）或（1-2，1-2）.································9分yyyx2x1yx2x1EBBQQDOAxxOPPFCCAyxyxy2x1圖1y2x1圖2②方法一：如圖2，設點P的坐標為（t，t2-t-1）.過點P作PD∥y軸，交直線y=-x于點D，則D（t，-t）.分別過點B，