2018屆寧德市(二檢)文科答案

2018屆寧德市普通高中畢業(yè)班第二次質(zhì)量檢查數(shù)學(xué)（文科）試題參考答案及評(píng)分標(biāo)準(zhǔn)評(píng)分說(shuō)明：1、本解答指出了每題要考查的主要知識(shí)和能力，并給出了一種或幾種解法供參考，如果考生的解法與本解法不同， 可根據(jù)試題的主要考查內(nèi)容比照評(píng)分標(biāo)準(zhǔn)指定相應(yīng)的評(píng)分細(xì)則．2、對(duì)計(jì)算題,當(dāng)考生的解答在某一步出現(xiàn)錯(cuò)誤時(shí)，如果后繼部分的解答未改變?cè)擃}的內(nèi)容和難度，可視影響的程度決定后繼部分的給分， 但不得超過(guò)該部分正確解答應(yīng)給分?jǐn)?shù)的一半；如果后繼部分的解答有較嚴(yán)重的錯(cuò)誤，就不再給分．3、解答右端所注分?jǐn)?shù)，表示考生正確做到這一步應(yīng)得的累加分?jǐn)?shù)．4、只給整數(shù)分?jǐn)?shù)，選擇題和填空題不給中間分．一、選擇題：本題考查基礎(chǔ)知識(shí)和基本運(yùn)算，每小題5分,滿分60分．1.C2.A3.B4.D5.B6.C7.B8.A9.C10.B11.D12.C二、填空題：本題考查基礎(chǔ)知識(shí)和基本運(yùn)算，每小題5分,滿分20分．13.2π14.4+2215.416.0a13三、解答題：本大題共 6小題，滿分 70分,解答須寫出文字說(shuō)明、證明過(guò)程和演算步驟．17.本小題主要考查正弦定理、余弦定理及兩角和與差的三角函數(shù)公式等基礎(chǔ)知識(shí)，考查運(yùn)算求解能力，考查化歸與轉(zhuǎn)化思想、函數(shù)與方程思想等．滿分 12分．解：（1）由正弦定理有：sinBsinA(sinCcosC)sinBsin(AC)sinAcosCcosAsinC·················2分cosAsinCsinAsinC·································3分0csinC0··································4分cosAsinA·······································5分tanA10A···········································6分4（2）a13,b22,A4由余弦定理有：a2b2c22bccosA································8分c24c50·····································9分c5或c1（舍去）································10分BDcsinA252分5····························122218.本小題主要考查了概率、頻率、平均數(shù)等概率、統(tǒng)計(jì)基礎(chǔ)知識(shí)，考查數(shù)據(jù)處理能力、抽象概括能力、運(yùn)算求解能力以及應(yīng)用意識(shí)，考查或然與必然的思想、化歸與轉(zhuǎn)化思想等．滿分12分.解：（1）設(shè)“陳先生一次租用新能源租賃汽車的時(shí)間不低于30分鐘”的事件為A則所求的概率為P(A)1P(A)11219·················2分502530分鐘的概率為19所以陳先生一次租用新能源租賃汽車的時(shí)間不低于.··4分25文科數(shù)學(xué)答案與評(píng)分細(xì)則 第1頁(yè)共6頁(yè)（2）每次開車所用的平均時(shí)間為2512352845855235·······6分50505050每次租用新能源租賃汽車的平均費(fèi)用為1120.1235=16.2············8分每個(gè)月的費(fèi)用為16.2222=712.8，712.8<800·················10分因此公車補(bǔ)貼夠上下班租用新能源分時(shí)租賃汽車.················12分19.本小題主要考查空間中直線與直線、直線與平面的位置關(guān)系、平面與平面位置關(guān)系，幾何體的體積等基礎(chǔ)知識(shí)， 考查空間想象能力、 推理論證能力、運(yùn)算求解能力，考查化歸與轉(zhuǎn)化思想等．滿分12分．證明：(1)取BD中點(diǎn)O,連接AO,PO····························1分ABAD,O為BD中點(diǎn)PAOBD又PBPD,O為BD中點(diǎn)FPOBDE又AOPOOADBD面PAO····················3分O又PA面PAOBPABD······················4分C(2)(i)取PD中點(diǎn)F,連接CF,EF,則CF//BE，CF即為所作直線l··········5分理由如下:在PAD中E、F分別為PA、PD中點(diǎn)EF//AD,且EF11AD2又AD//BC,BC11AD2EF//BC且EF=BC四邊形BCFE為平行四邊形.·······························6分CF//BE·············································7分PAAB,PABD,ABBDBPA 面ABD···········································8分又在 ABD中,AB AD 2,BD 2 2,AB2 AD2 BD2AB AD又PAAB,PAADAAB面PAD···········································9分方法一：VP11223ACD32332VCAEFD11(133322)2····························10分22VPECF2333··································11分32=6VPECF316·······································12分VC=AEFD332方法二：在PAD中，EF為中位線文科數(shù)學(xué)答案與評(píng)分細(xì)則 第2頁(yè)共6頁(yè)SPEF1?????????????????????????????分..10SPAD4VCPEF1SPEFAB13..分11??????????????????????VCPAD14SPADAB3VPECF=1....................................................................................................................12分VCAEFD3方法三：EF1AD2VFPEC1SPECEF13?????????????????????..分11VDPAC14SPACAD3VPECF=1????????????????????????????分.12VCAEFD320．本題主要考查橢圓的標(biāo)準(zhǔn)方程及其簡(jiǎn)單的幾何性質(zhì)，直線與圓錐曲線的位置關(guān)系等基礎(chǔ)知識(shí)，考查運(yùn)算求解能力、推理論證能力，考查數(shù)形結(jié)合思想、特殊與一般思想、函數(shù)與方程思想、化歸與轉(zhuǎn)化思想，考查考生分析問(wèn)題和解決問(wèn)題的能力，滿分12分．解：（1）x2y21,S412ab22,ab2①·········1分a2b2ab2又e=2②,聯(lián)立①②得a2,b1．·················3分2x2橢圓方程為y21．····························4分2（2）由（1）得橢圓方程為x2y21,2依題意,設(shè)直線l的方程為y2xt,B(x1,y1),C(x2,y2),點(diǎn)A（1，2）l:y2xt的距離為d,到直線y，2xt聯(lián)立x22可得9x28tx(2t22)0,y21x1x28t9顯然0,2t2·································6分2x1x29BD1k2(x1x2)24x1x25(8t)24(2t22)2109t2,··7分99922tt·····································8分d55SABD1BDd12109t2t,·························9分22959t20d29t2t2(9t2)t22(9t2)t22······10分99922文科數(shù)學(xué)答案與評(píng)分細(xì)則 第3頁(yè)共6頁(yè)當(dāng)且僅當(dāng)t29時(shí),即t32時(shí)取等號(hào),(SABD)max2,············11分222此時(shí)直線l的方程為4x2y320或4x2y320．············12分21.本小題主要考查導(dǎo)數(shù)及其應(yīng)用、不等式等基礎(chǔ)知識(shí)，考查推理論證能力、運(yùn)算求解能力、創(chuàng)新意識(shí)等，考查函數(shù)與方程思想、化歸與轉(zhuǎn)化思想、分類與整合思想、數(shù)形結(jié)合思想等．滿分 12分．解：(1)f(x) 3x2 6ax=3x(x 2a）????????????????????1 分令f(x)0,則x=0或x=2a,當(dāng)a0時(shí),f'(x)0,f(x)在R上是增函數(shù)；??????????????..2分當(dāng)a0時(shí),令f'(x)0,得x0,x2a,所以f(x)在(,0),(2a,)上是增函數(shù)；令f'(x)0,得0x2a,所以f(x)在(0,2a)上是減函數(shù)???????...??3分當(dāng)a0時(shí),令f'(x)0,得x2a,x0,所以f(x)在(,2a),(0,)上是增函數(shù)；令f'(x)0,得2ax0,所以f(x)在(2a,0)上是減函數(shù)???????..??4分綜上所述：當(dāng)a0時(shí),f(x)在R上是增函數(shù)；當(dāng)a0時(shí),f(x)在(,0),(2a,)上是增函數(shù),在(0,2a)上是減函數(shù).當(dāng)a0時(shí),f(x)在(,2a),(0,)上是增函數(shù),在(2a,0)上是減函數(shù).??????????????.???5分（2）由(Ⅰ)可知：當(dāng)a0時(shí),f(x)在R上是增函數(shù)，函數(shù)f(x)不可能有三個(gè)零點(diǎn)；當(dāng)a0時(shí),f(x)在(,2a),(0,)上是增函數(shù),在(2a,0)上是減函數(shù).f(x)的極小值為f(0)=40，函數(shù)f(x)不可能有三個(gè)零點(diǎn)當(dāng)a0時(shí),f(x)minf(2a)44a3,要滿足f(x)有三個(gè)零點(diǎn),則需44a30,即a1?????????..???6分當(dāng)x0時(shí),要證明：f(x)6(aa2)ea等價(jià)于要證明f(x)min6(aa2)ea即要證：44a36(aa2)ea?????????????????????7分由于a1,故等價(jià)于證明：1aa23aea,證明如下：2法1：構(gòu)造函數(shù)g(a)3aea22a2a2(a(1,))????.???????8分g(a)(33a)ea24a????????????????????9分令h(a) (3 3a)ea 2 4ah(a) (6 3a)ea 4 0, 函數(shù)h(a)在(1, )單調(diào)遞增文科數(shù)學(xué)答案與評(píng)分細(xì)則 第4頁(yè)共6頁(yè)h(a)minh(1)6e60,函數(shù)g(a)在(1,)單調(diào)遞增????.?10分g(a)ming(1)3e60,??????????????????分111a23aaae2∴f(x)6(aa2)ea．????????????????????12分法2：構(gòu)造函數(shù)g(a)ea1a(a(1,))????????????????8分∵g(a)ea10,函數(shù)g(a)在(1,)單調(diào)遞增?????????9分∴ea1a?????????????????????????10分∴3aea(1aa2)3a(a1)(1aa2)???????????11分223aea(1aa2)1(a1)(a2)022∴f(x)6(aa2)ea．????????????????????12分22．選修44；坐標(biāo)系與參數(shù)方程本小題考查直線和圓的極坐標(biāo)方程、參數(shù)方程等基礎(chǔ)知識(shí)，考查運(yùn)算求解能力，考查數(shù)形結(jié)合思想、化歸與轉(zhuǎn)化思想等．滿分10分．解法一：（1）由C1：(4cos)r24，得24cos4r2，即x2y24x4r20，?????????????????????2分曲線C2化為一般方程為：(x4)2y23r2，即x2y28x163r2，???4分化為極坐標(biāo)方程為：28cos163r20．?????????.???5分（2）由x2y24x4r2及x2y28x163r2，消去r2，得曲線C3的直角坐標(biāo)方程為x2y22x2．?????????????7分x1t,設(shè)直線l的參數(shù)方程為2（t為參數(shù)），????????????.?8分3t2文科數(shù)學(xué)答案與評(píng)分細(xì)則 第5頁(yè)共6頁(yè)與x2y22x2聯(lián)立得1t23t2t2，44即t2t20，????????????????.????????9.分故t1t21，t1t220，∴OAOBt1t21．??????????.?????????..分?10（或由t2t20得，(t2)(t1)0,得t12,t21，∴OAOB211．????????????????????10分）解法二：（1）同解法一；（2）由24cos4r2及28cos163r20，消去r2，得曲線C3的極坐標(biāo)方程為22cos20(R)．???..?.?????7分將代入曲線C3的極坐標(biāo)方程，可得220，??????????8分故121，1220，???????????..?????????9分故OAOB121．?????????????..????????分10（或由220得(2)(1)0得12,21，???????9分故OAOB211???????????????????10分）23．選修45：不等式選講本小題考查絕對(duì)值不等式的解法與性質(zhì)等基礎(chǔ)知識(shí)，考查運(yùn)算求解能力、推理論證能力，考查分類與整合思想、化歸與轉(zhuǎn)化思想等．滿分10分．解法一：（1）xy1,|x2||x1|5，??????????????????????分..1當(dāng)x2時(shí)，原不等